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Given,
Now using the mirror formula,
we have,
∴ Image is formed at 6.7 cm at the back of the mirror ( because v is positive)
Now,
Magnification,
Hence, the image formed is erect, and of course virtual.
As, needle is moved farther away from the mirror then as a result, image moves away from the mirror (upto F) and keeps on decreasing in size.
(a)
Given,
Using Snell's law, we have
Refractive index of glass with respect to air,
(b)
Using Snell's law,
Refractive index of water wr.to air,
(c)
Angle of incidence in water,
Refractive index of glass wr.to water,
Given,
After the prism is placed in water,
Refractive index of water = 1.33
Using the formula,
i.e.,
which is the reqiured angle of minimum deviation.
Here,
Magnifying power of telescope,
Seperation between the objective and the eyepiece,
Given,
Diameter of moon = 3.48 x 106 m
Radius of lunar orbit, r = 3.8 x 108 m
Focal length of objective, fo = 15 cm
If d is diameter of the image, then
Angle subtended by diameter of moon
and,
Angle subtended by image
Therefore,
Diameter of the image of the moon formed by objective lens,
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(a) Given, a cross-section of a light pipe.
Critical angle is given by
Total internal reflection will occur if angle of incidence is greater than the critical angle, i’ > ic’
i.e., if i’ > 59° or,
when r < rmax, where r = 90°– 59° = 31°, which is the angle of reflection.
Now using Snell's law
Therefore, in the pipe the rays having incident angles in the range 0 < i < 60° with the axis of the pipe will suffer total internal reflection.
For the finite length of the pipe, the lower limit on i is determined by the ratio of the diameter to the length of the pipe. The lower limit of angle of incidence is not 0o.
(b) If there is no outer covering of the pipe then,
Hence,
Now, for angle of incidence, i = 36.5°
we have r = 36.5° and,
i’ = 90° – 36.5° = 53.5°, which is greater than the critical angle, ic.
Thus all rays which are incident at angle in the range 0 < i < 90° will suffer total internal reflection.
Here,
Distance between object and screen, D = 90 cm
Distance between two locations of convex lens, d = 20 cm
Since,
Focal length of the lens is,
∴ The parallel beam appears to diverge from a point 420 – 4 = 416 cm, on the left of the centre of the two lens system.
We finally conclude that the answer depends on the side of the lens system where the parallel beam is incident. Therefore, the notion of effective focal length does not seem to be meaningful here.
The image formed by the convex lens becomes object for concave lens at a distance of (120 – 8) = 112 cm on the other side.
For concave lens,
Focal length, f = – 20 cm
Object distance, u = + 112 cm (on the other side)
Image distance, v = ?
Using lens formula, we get
Now,
Using the formula of magnification
Net magnification,
and as,
The refracted ray in the prism is incident on the second face at critical angle ic.
Given, angle of prism, A = 60o
Refracting index of the material = 1.524
Now, 60° + 90° – r + 90° – ic = 180° [ sum of all angles of is 1800 ]
and,
Now,
Using Snell's law,
For a normal eye, near point of distinct vision = 25 cm.
i.e., u = –25 cm
Converging power = +40 D
To see objects at infinity, the eye uses its least converging power = 40 + 20 = 60 dioptres. This gives the rough idea of the distance between the retina and cornea eye-lens.
to focus an object at near point,
Now, using the formula,
i.e., corresponding to a converging power given by,
the power of the eye lens = 64 - 60 = 24 dioptre.
Hence, we can say that the range of accommodation of the eye-lens is roughly 20 to 24 dioptre.
No, a person may have normal ability of accommodation yet, he may be myopic or hypermetropic.
In fact, myopia arises when length of eye ball (from front to back) gets elongated and hypermetropia arises when length of eye ball gets shortened.
However, when eye ball has normal length, but the eye-lens losses partially its power of accommodation, this defect is called presbiopia.
Power of spectacles, P = -1D
Focal length, f = -100 cm
That is, the far point of the person is at 100 cm.
Near point of the eye might have been normal (i.e., 25 cm).
The objects at infinity produce virtual images at 100 cm (using spectacles).
To see objects between 25 cm to 100 cm, the person uses the ability of accommodation of his eye-lens. This ability is partially lost in old age.
The near point of the eye may recede to 50 cm. He has, therefore, to use glasses of suitable power for reading.
Here,
Object distance, u = –25 cm
Image distance, v = –50 cm
Since,
i.e., focal length,
Power,
The defect is called Astigmatism. When the curvature of the cornea plus eye-lens refracting system is not the same in different planes, this defect arises. As vertical lines are seen distinctly, the curvature in the vertical plane is enough, but the curvature of the lens system is insufficient in the horizontal plane and, the lines cannot be seen distinctly.
This defect is removed by using a cylindrical lens with its axis along the vertical plane.
For the closest distance,
Image distance, v = -25 cm
Focal length of thin convex lens, f = 5 cm
Using the formula,
i.e., 4.2 cm is the closest distance at which the man can read the book.
For the farthest image,
Using the formula,
which is the object distance.
This is the farthest distance at which the man can read the book.
Normal near point of human eye = 25 cm
Focal length , f = 5 cm
Maximum angular magnification is
Minimum angular magnification is
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|
Lenses |
Power (P) |
Aperture (A) |
|
L1 |
3D |
8 cm |
|
L2 |
6D |
1 cm |
|
L3 |
10D |
1 cm |
Laws of reflection:
(i) The incident ray, the reflected ray and the normal all three lie in the same plane.
(ii) The angle of incidence (i) is equal to the angle of reflection (r).
i.e., ∠i = ∠r.
Presbyopia is a progressive form of far-sightedness that affects most people with ageing. The power of accomodation of the eye decreases.
Presbyopia arises due to the gradual weakening of the ciliary muscles and diminishing flexibility of the crystalline lens.
Simple reading eyeglasses with convex lenses correct most cases of presbyopia.
Given, θ = 0°
Number of images,
What is the magnification produced by a single convex lens used as a simple microscope in normal use?
Which of the two main parts of an optical fibre has a higher value of refractive index?
A lens which is immersed in a transparent liquid becomes invisible when, refractive index of liquid is equal to or greater than the refractive index of glass.
Write the relation for the the refractive index μ of the prism in terms of the angle of minimum deviation δm and the angle A of prism.
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Ratio of the speed of light in air (vacuum) to the speed of light in the medium is called the refractive index of the medium.
Graph: The graph of angle of deviation (δ) versus angle of incidence (i) for a triangular prism is given as follow:
Given, a convex lens.
Refractive index of lens, = 1.5
Focal length of lens in air, fa = 18
Refractive index of water,
For the lens in air,
When the lens is immersed in water
Thus,
Hence, focal length changes from 18 to 32.
The figure below shows the ray diagram of an astronomical telescope in the near point adjustment. 
Magnifying power is given as,
We have,
Distance of object from the mirror, u = –30 cm;
Distance of image, v = ?
Focal length, f = + 20 cm
We know, using the mirror formula,
The image is virtual and erect as, it is formed at a distance of 12 cm behind the mirror.
Now, size if the object, O = 0.5 cm
sixe of the image = I
Magnification,
Hence, the height of the image = + 0.2 cm.
The positive sign indicates that the image is erect is smaller in size
We have,
Using formula,
The image is formed at a distance of 7.5 cm in front of mirror.
Now,
Magnification,
where,
I is the image size and,
O is the object size.
The negative sign indicates that the image is inverted and is diminished in size.
Here we are given an astronomical telescope.
Distance between objective and eyepiece, l = 101 cm
Magnification,
Since,
Then,
To prove: Least possible distance between an object and real image = 4f
Suppose, I is the real image of an object O.
Let, d be the distance between the image and the object.
If the image distance from the lens is x, the object distance from the lens will be (d – x).
Thus, u = – (d – x) and v = + x.
Sustituting in the lens formula
we have,
For a real image, the value of x must be real,
i.e., the roots of the above equation must be real. This is possible if,
d2 ≥ 4fd
i.e., d ≥ 4f
Hence, 4f is the minimum distance between the object and its real image formed by a convex lens.
The convex lens produces converging rays trying to meet at I1, at a distance of 20 cm from the convex lens, i.e., 15 cm behind the concave lens.
I1 will serve as a virtual object for the concave lens.
For refraction at the concave lens, we have
Object distance,
Focal length,
As per sign convention
Now using lens formula,
i.e., this image is in the side of object at a distance of 5 cm to the right of concave lens and at a distance of 10 cm(5 + 5) from convex lens.
Now,
u = + 15 cm; v =?, f = –10cm
Using lens formula, we have
Hence, the final image is virtual and is located at 30 cm to the left of the concave lens.
Refractive index of water = 4/3
Radius of spherical air bubble, r = 2 mm
(a) For refraction at the first surface, using the formula,
where
Thus,
which gives, v1 = –6 mm.
Negative sign indicates that the image I1 is virtual and is on the same side as the object at a distance of 6 mm from the first surface.
For refraction at the second surface, the image I1 serves as the virtual object which is at a distance of 6 mm + 4 mm = 10 mm from the second surface.
For refraction to take place, we use
where,
Thus, substituting the values in the formula, we get
which given v2 = –5 mm.
The final image I2 is virtual and is formed at a distance of 5 mm from the second surface to the left of the second surface, i.e., the final image is formed at a distance of 1 mm from the first surface.
(b) Figure given below shows the ray diagram. 
Here given an astronomical telescope.
From the formula,
i.e.,
which is the required distance from the objective.
Now, again using the formula,
i.e.,
Length of the telescope is given by,
Given, a slide projector.
Magnification,
which is the object distance.
Now, using the lens formula,
is the required focal length of lens in the projector.
Given, three lenses.
Power of first lens, P1 = 6 D
Power of second lens, P2 = 3D
Power of third lens, P3 = 12 D
Magnifying power of a microscope is given by,
Since power and magnification have direct dependence, lens of greater power is required for magnification.
Hence, we use P1 = 6D and P3 = 12D for constructing a microscope.
Therefore, P3 = 12D should serve as objective lens and P1 = 6D should serve as eye-lens.
Thus, convex lens will form the image I1, at a distance of 60 cm behind the lens.
As the mirror is at a distance of 10 cm from the lens, image I1 will be at a distance of (60 – 10) = 50 cm from the mirror, i.e., MI1 = 50 cm.
Now, as the final image I2 is formed at the object itself, the rays after reflection from the mirror retraces its path, i.e., the rays on the mirror are incident normally, i.e., I1 is the centre of the mirror so that
and
which is the focal length of the mirror.
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In the Fig. the direct image formed by the lens (f = 10 cm) of an object placed at O, and that formed after reflection from the spherical mirror are formed at same point O. What is the radius of curvature of the mirror?
A ray of light, incident on a equilateral glass prism moves parallel to the base of the prism inside it. What is the angle of incidence for this way?
Given, an equilateral glass prism.
Angle of prism = 60o
Refractive inex of the medium, =
When light ray moves parallel to the base of the prism inside it, then
r1 = r2
so that ,the minimum deviation takes place.
For minimum deviation r = A/2 =
Now, using Snell's law,
is the required angle of incidence so that, the ray travels parallel to the base ofthe prism.
An object is placed at a distance of 15 cm from a convex lens of focal length 10 cm. On the other side of the lens, a convex mirror is placed such that its distance, from the lens, equals the focal length of the lens. The image formed by this combination is observed to coincide with the object itself. Find the focal length of the convex mirror.
In the case of convex lens,
Object distance, u = -15 cm
Focal length, f = 10 cm
Now, using the lens formula,
=
i.e., the image is formed at a distance of 30 cm.
For convex mirror,
Object distance,
Image distance,
Now, using mirror formula,
i.e.,
Therefore, focal length of the convex mirror is 100 cm.
A 5 cm long needle is placed 10 cm from a convex mirror of focal length 40 cm. Find the position, nature and size of image of the needle. What happens to the size of the image when the needle is moved further away from the mirror?
A.
short sightedness or myopiaB.
total internal reflectionC.
ScatteringTotal internal reflection
Scattering
A.
Total internal reflection
Angle of prism, A = 60o
Refractive index of prism = 1.5
Given, angle of incidence , i1=i2 = = = 45o
As, A + = i1 + i2
60o + = 45o + 45o
Then, angle of minimum deviation, = 90o - 60o = 30o
We know that, velocity of light in vacuum = 3108 m/s
Now, using the formula,
That is,
is the required speed of light in the prism.
For minimum deviation to take place,
angle of incidence = angle of emergence.
The ray diagram is as shown below:
i.) Derive the lens formula, 1/ f = 1/v – 1/u for a concave lens, using the necessary ray diagram.
ii.) How does the angle of minimum deviation of a glass prism of refractive index 1.15 change, if it is immersed in a liquid of refractive index 1.3?
Fringe width is directly proportional to wavelength and inversely proportional to refractive index.
The formula is given by,
In air, Fringe width, x =
In medium,
where,
d is the distance between the slits and,
D is the distance between the screen and the slit.
Therefore, fringe width decreases by a factor of when placed in the liquid.
Wavefront is a locus of all particles of the medium, vibrating in the same phase.
Huygen's wave theory:
Consider two media namely, 1 and 2 and, let XY be the surface seperating two media.
The speed of the waves in these media will be v1 and v2. 
Suppose, a plane wavefront AB in first medium is incident obliquely on the boundary surface XY, and its end A touches the second surface at a point A' at time t = 0 while, the other end B reaches the second surface at point B' after time-interval t.
Here, BB' = t
As per Huygen’s principle, secondary spherical wavelets emanates from points A and B , which travel with speed v1 in the first medium and speed v2 in the second medium.
Distance traversed by secondary wavelets in medium 2 in time t , AA' = v2t
Distance traversed by point of wavefront in medium 1 in time t = BB' = v1t
Now, considering A as the centre we will draw an arc and a tangent( B'A') is drawn from B' to this point. So, as the incident wavefront advances, secondary wavelets start from points in-between A and B' and will reach A'B' simultaneously.
According to Huygen’s principle A'B' is the new position of wavefront AB in the second medium and, A'B' is the refracted wavefront.
Let the incident wavefront AB and refracted wavefront A'B' make angles i and r respectively with refracting surface XY.
In right ,
... (i)
Similarly, in right AA'B'
...(ii)
Dividing equations (i) by (ii) , we have
which is the required Snell's law.
Hence proved.
Given, a ray of light travelling from denser medium of refractive index n1 to a rarer medium of refractive index n2.
Snell's law is given by,
Now, critical angle is the angle of incidence in denser medium for which angle of refraction in rarer medium is 90o .
i.e., i = ic when r = 90o .
Thus, putting these values in snell's law , we have
i.e.,
Given, two convex lenses.
Aperture of first lens = 5 cm
Aperture of second lens = 10 cm
i) Resolving power is directly proportional to the diameter.
Therefore, the ratio of resolving power is,
ii) Intensity is directly proportional to the area of objective lens.
The refractive indices of glass and water with respect to air are 3/2 and 4/3 respectively. What will be the refractive index of glass with respect to water?
Refractive index of water,
Given,
Refractive index of glass w.r.t. air,
Refractive inex of water w.r.t air,
refractive index of glass w.r.t. water, = =
Given,
Wavelength of monochromatic light,
(a) For reflected light,
Wavelength is going to remain the same for incident and reflected light.
Wavelength,
Speed of lightwill remain invariant in two media.
Therefore,
(b) For refracted light,
Refractive index of water w.r.t. air,
i.e.,
So, wavelength of light
As frequency remains unaffected on entering another medium, therefore,
Speed,
i.e., speed of the light has decresed when it entered the second medium.
Given, refractive index of glass, = 1.5
Refractive index is given by,
Speed of light in glass
In a Young's double-slit experiment, the slits are separated by 0.28 mm and the screen is placed 1.4 m away. The distance between the central bright fringe and the fourth bright fringe is measured to be 1.2 m. Determine the wavelength of light used in the experiment.
Given,
Distance between the screen and slits, D = 1.4 m, Distance between slits, d = 0.28 mm =
Fringe width,
Using formula, we have
Fringe width,
which is the required wavelength.
Here,
Wavelength of the beam of light,
Wavelength of beam of light,
Suppose,
and,
D is the distance between the screen and slit.
For third bright fringe, n = 3
Distance of third bright fringe from the central maximum,
Here, a beam of light consists of two wavelengths.
Wavelength,
Wavelength,
Suppose,
Let n fringes of wavelength 650 nm coincide with (n + 1) fringes of wavelength 520 nm.
Hence,
Distance from the central maximum,
Given,
Refractive index of glass, μ = 1.5
Using formula,
tan ip = μ
ip = tan-1(μ)
= tan-1(1.5)
Thus, ip = 56.3°,
is the required polarizing angle or brewster angle.
Given,
Wavelength of incident light, = 5000
The wavelength and frequency of the reflected light are the same as that of the incident light.
∴ Wavelength of reflected light = 5000 Å
Frequency of reflected light, = c/λ
According to the law of reflection, i = r
When, the reflected ray is normal to the incident ray, we have
i + r = 90°
i + i = 90°
2i = 90°
i.e., i = 45°.
Here,
Aperture, a = 4 mm = 4 × 10-3 m
Wavelength, λ = 400 nm = 400 × 10-9 m = 4 × 10-7 m
Ray optics is good approximation upto distances equal to Fresnel's distance (ZF).
Fresnel's distance is given by,
According to Newton's Corpuscular theory of light, when corpuscles of light strike the interface XY, separating a denser medium from a rarer medium, the component of their velocity along XY remains the same.
If v1 is velocity of light in rarer medium (air),
v2 is velocity of light in denser medium (water),
i is angle of incidence, and
r is angle of refraction then,
Component of v1 along XY = v1 sin i
Component of v2 along XY = v2 sin r
As,
Since,
i.e., light should travel faster in water than in air.
This prediction of Newton s theory is opposite to the experimental result.
Huygens wave theory predicts that v2 < v1, which is consistent with experimental conclusion.
The reflected wavefront AB'C, appears to start from I. Hence, I becomes virtual image for O as real point object.
Draw AN normal to XY, hence parallel to OP.
Now, OA is the incident ray (being normal to incident wavefront ABC) and AD is reflected ray (being normal to reflected wavefront AB'C).
Thus, ∠OAN = ∠DAN = ө [i = r]
But, ∠OAN = alternate ∠AOP
and ∠DAN = corresponding ∠AIP
∠AOP = ∠AIP
Now, in ∆AIP and ∆AOP
∠AIP = ∠AOP (each ө)
∠APl = ∠APO = 90° (each 90°)
AP is common to both
∆s become congruent
Hence, PI = PQ
i.e., normal distance of image from the mirror = normal distance of object from the mirror.
Thus, virtual image is formed as much behind the mirror as the object in front of it
Let us list some of the factors which could possibly influence the speed of wave propagation:
(i) nature of source.
(ii) direction of propagation.
(iii) motion of source and/or observer.
(iv) wave length.
(v) intensity of the wave.
On which of these factors, if any, does
(a) the speed of light in vacuum,
(b) the speed of light in a medium (say, the glass or water) depend?
(a) Speed of light in vacuum is an absolute constant, according to Einstein's theory of relativity. It does not depend upon any of the factors listed above or any other factor.
(b) The speed of light in a medium like water or glass:
(i) does not depend upon the nature of the source.
(ii) does not depend upon the direction of propagation, when the medium is isotropic.
(iii) does not depend upon motion of the source w.r.t. the medium, but depends on motion of the observer relative to the medium.
(iv) depends on wavelength of light, being lesser for shorter wavelength and vice-versa.
(v) does not depend upon intensity of light.
For sound waves, the Doppler formula for frequency shift differs slightly between the two situations:
(i) source at rest; observer moving, and
(ii) source moving; observer at rest.
The exact Doppler formulas for the case of light wave in vacuum, are however, strictly identical for these situations. Explain why this should be so. Would you expect the formulas to be strictly identical for the two situations in case of light travelling in a medium?
Wavelength of light, = 600 nm = 600 10-9 m
Angular fringe width,
where, d is the distance between the slits.
Now,
is the required spacing between the two slits.
Distance between the two towers = 40 km
Size of aperture,
Distance of aperture from tower,
Fresnel distance is the half of the distance between the towers.
i.e., zf
Therefore, now using the formula,
Fresnel distance,
we have,
This is the required longest wavelength of radio waves, which can be sent in between the towers without considerable diffraction effects.
Given,
Distance of the screen from the slit,
Distance of the first minimum when the centre of screen,
n=1 is the first minimum.
Using formula,
we have,
is the required width of the slit.
Angular seperation of interference fringes is given by,
D is the distance between screen and slit and,
d is the distance between slits.
Now, when the distance between the slits and the screen is halved we have,
That is, angular seperation becomes half of the initial value.
Polarizing angle, = 60o
Therefore, refractive index of the medium is,
Given,
Angle between pass axis of polarizer and analyser,
Using formula,
i.e.,
is the required ratio of intensities of original light and transmitted lightafter passing through the analyzer.
How does the resolving power of microscope change on (i) decreasing wavelength of light, (ii) decreasing diameter of objective lens.
Here,
Distance between the slits, d = 0.3 mm = 0.3 × 10-3 m
Fringe width, β = 1.5 mm = 1.5 × 10-3 m
distance between screen and the slit, D = 7.5 cm = 0.75 m
Wavelength λ is given by
i.e.,
(a) Fringe width is given by, .
when D is doubled, fringe width is also doubled.
i.e.,
(b) When d is doubled, is reduced to half.
i.e.,
Here, we are given young's double slit experiment.
Wavelength of monochromatic light, λ1 = 600 nm = 600 × 10-9 m
Fringe width, β1 = 10 mm = 10 × 10-3 m
Fringe width, β2 = 8 mm = 8 × 10-3 m
Let d be the slit width and D the distance between slit and screen, then we have
i.e.,
is the required wavelength of light from the second source.
If the monochromatic source is replaced by white light, then we will not be able to see the interference fringes because white light is not a coherent source of light. The condition for interference to take place is, the availabilty of coherent sources of light.
In a Young's double-slit experiment, the two slits are kept 2 mm apart and the screen is positioned 140 cm away from the plane of the slits. The slits are illuminated with light of wavelength 600 nm. Find the distance of the third bright fringe, from the central maximum, in the interference pattern obtained on the screen.
If the wavelength of the incident light were changed to 480 nm, find out the shift in the position of third bright fringe from the central maximum.
If the wavelength of the incident light were changed to 480 nm, find out the shift in the position of third bright fringe from the central maximum.
We have Young's double slit experiment.
Distance between the slits, d = 2 mm = 2 × 10-3 m
Distance between the slit and the screen, D = 140 cm = 1.40 m
Wavelength of the monochromatic source of light, λ = 600 nm = 600 × 10-9m = 6 × 10-7
m
Position of bright fringes is given by,
Here, we are considering the third fringe.
∴ Distance of the third bright fringe is,
If the wavelength of incident light is changed to 480 nm, then
Distance of the third bright fringe is
Shift in the position of third bright fringe when there is a change in wavelngth,
What are coherent sources? How does the width of interference fringes in Young’s double-slit experiment change when
(a) the distance between the slits and screen is decreased?
(b) frequency of the source is increased?
Justify your answer in each case.
In a single slit diffraction experiment, a slit of width 'd' is illuminated by red light of wavelength 650 nm. For what value of 'd' will
(i) the first minimum fall at an angle of diffraction of 30°, and
(ii) the first maximum fall at an angle of diffraction of 30°?
(a) A ray of light is incident on a glass surface at an angle of 60°. If the reflected and refracted rays are perpendicular to each other, find the refractive index of glass.
(b) What is the value of the refractive index of a medium of polarizing angle 60°.
Given,
Distance between the slits, d = 3 mm = 3 × 10-3 m
Wavelength of light, λ = 600 nm = 6 × 10-7 m
Distance between the slit and the screen, D = 60 cm = 0.60 m
Distance of first order minima from central maximum is,
Therefore, distance between first order minimum on both sides of the central maximum is,
A wavefront is an imaginary surface over which an optical wave has a constant phase. The shape of a wavefront is usually determined by the geometry of the source.
Huygen's principle:
(i) Every point on a given wavefront acts as a fresh source of secondary wavelets which travel in all directions with the speed of light.
(ii) The forward envelope of these secondary wavelets gives the new wavefront at any instant.
Laws of reflection by Huygen's principle:
Let, PQ be reflecting surface and a plane wavefront AB is moving through the medium (air) towards the surface PQ to meet at the point B. 
Let, c be the velocity of light and t be the time taken by the wave to reach A' from A.
Then, AA' = ct.
Using Huygen's principle, secondary wavelets starts from B and cover a distance ct in time t and reaches B'.
To obtain new wavefront, draw a circle with point B as centre and ct (AA' = BB') as radius. Draw a tangent A'B' from the point A'.
Then, A'B' represents the reflected wavelets travelling at right angle. Therefore, incident wavefront AB, reflected wavefront A'B' and normal lies in the same plane.
Consider ∆ABA' and B'BA'
AA' = BB' = ct [∵ AA' = BB' = BD = radii of same circle]
BA' = BA' [common]
∠BAA' = ∆BB'A' [each 90°]
∴ ∆ABA' ≅ ∠DBA' [by R.H.S]
∠ABA' = ∠B'A'B [corresponding parts of congruent triangles]
∴ incident angle i = reflected angle r
i.e., ∠i = ∠r
Find the maximum intensity in case of interference of n identical waves each of intensity Io if the interference is
(i) coherent (ii) incoherent
Given,
Maximum intensity = Io
Distance between the slits, d = 5
Distance of screen from the slit, D = 10d
Using the formula,
Path difference,
Here,
So,
Corresponding phase difference will be
i.e.,
Intensity of light,
State two conditions to obtain sustained interference of light.
In Young's double-slit experiment, using light of wavelength 400 nm, interference fringes of width 'X' are obtained. The wavelength of light is increased to 600 nm and the separation between the slits is halved. If one wants the observed fringe width on the screen to be the same in the two cases, find the ratio of the distance between the screen and the plane of the interfering sources in the two arrangements.
Condition for sustained interference are:
(i) The two sources of light should emit light continuously.
(ii) The two waves must be in same phase or bear a constant phase difference.
Wavelength of light, = 400 nm.
Width of interference fringes = x
Fringe width in first case is,
i.e.,
In the second case,
Wavelength of light = 600 nm
Distance between the slits = d/2
Since, fringe width in second case is given to be the same, we have
Therefore, the ratio of the distance between the screen and the plane of interfering sources n the two arrangements is given as,
i.e.,
Why is interference pattern not detected, when two coherent sources are far apart?
In Young’s experiment, the width of the fringes obtained with light of wavelength 6000 Å is 2.0 mm. Calculate the fringe width if the entire apparatus is immersed in a liquid medium of refractive index 1.33.
The fringe width will be very small (almost negligible) and fringes will not be separately visible if the separation between the two coherent sources is large.
Given,
Fringe width, = 2 mm
Wavelength of light used, = 6000
Formula for fringe width in air,
Fringe width in liquid is,
Refractive index of the medium, (given)
Therefore,
i.e.,
Polaroid is a thin and large sheet made of crystalline polarizing material which is used to produces plane polarized beam of light.
(i) Polaroids can be used to demonstrate the transverse nature of light waves.
(ii) If an unpolarized light wave is incident on a polaroid then, the the light wave will get linearly polarised with the electric vector oscillating along a direction perpendicular to the aligned molecules.
Brewster's angle:
The partially reflected light gets partially plane polarized when ordinary light, is allowed to undergo refraction. However, there is an angle of incidence at which, an ordinary light undergoes refraction as well as reflection (partial) and then the partially reflected ray is richly plane polarized. Such an angle is known as polarizing angle or Brewster's angle. It is denoted by ip.
Light can be polarized by reflecting it from a transparent medium. The extent of polarization depends on the angle of incidence. At a particular angle of incidence, called Brewster's angle, the reflected light is completely polarized as shown in the diagram below: 
When reflected and refracted rays are perpendicular to each other, angle of incidence becomes equal to the polarizing angle.
where, is the refractive index of the medium.
(a) What are coherent sources of light? Two slits in Young's double-slit experiment are illuminated by two different sodium lamps emitting light of the same wavelength.
Why is no interference pattern observed?
(b) Obtain the condition for getting dark and bright fringes in Young's experiment. Hence write the expression for the fringe width.
(c) If s is the size of the source and its distance from the plane of the two slits, what should be the criteria for the interference fringes to be seen?
(a) Coherent sources:
Sources emitting waves of same frequency or wavelength having either zero or a constant phase difference are said to be coherent sources of light.
Two independent sources of light do not fulfil the requirement of constant phase difference. Since, the sodium lamps considered here are independent sources of light and are not coherent sources we cannot observe interference pattern.
(b) For bright fringes (maxima),
where, n = 0, 1, 2, 3... so on.
For dark fringes (minima),
Path difference,
Where n = 0, 1, 2, 3 ...so on.
The separation between the centre of two consecutive bright fringes is the width of a dark fringe.
Fringe width,
i.e.,
(c) The condition for interference fringes is to be:
.
where,
s is the source and,
d is the distance between the two slits.
A ray of light incident normally on one of the faces of a right-angled isosceles prism is found to be totally reflected as shown.
(a) What is the minimum value of the refractive index of the material of the prism?
(b) When the prism is immersed in water trace the path of the emergent ray for the same incident ray indicating the values of all the angles. (μ of water = 4/3).
(a)
ABC is the section of the prism right angled at B.
A and C are equal angles i.e., A = C = 45°. 
The ray PQ is normally incident on the face AB. Hence it is normally refracted and the ray QR strikes the face AC at an angle of incidence 45°.
It is given that the ray does not undergo refraction but is totally reflected at the face AC. This gives a maximum value for the critical angle as 45°.
Therefore,
Since,
i.e.,
Thus, the minimum value of refractive index
(b) When the prism is immersed in water the critical angle for the glass-water interface is given by
The angle of incidence at R continues to be 45° and obviously, 45° < 70.53°.
Hence, there is refraction taking place now and the refracted ray is RS.
The angle of refraction r is given by,
μg sin i = μw sin r 
Refractive index of water w.r.t glass is,
i.e.,
The angle of refraction in water = 48°36'.
What is interference of light? Write two essential condition for sustained interference pattern to be produced on the screen.
Draw a graph showing the variation of intensity versus the position on the screen in Young's experiment when (a) both the slits are opened and (b) one of the slit is closed.
What is the effect on the interference pattern in Young's double-slit experiment when:
(i) screen is moved closer to the plane of slits?
(ii) separation between two slits is increased. Explain your answer in each case.
Interference of light:
The phenomenon of redistribution of light energy in a medium on account of superposition of light waves from two coherent sources is called interference of light.
Conditions for sustained interference:
The two essential conditions of sustained interference are as follows:
(i) The two sources of light should emit light continuously.
(ii) The light waves should be of same wavelength. (Monochromatic).
When both the slits are open, we get interference pattern on the screen. Then the following intensity distribution curve is obtained.
When one of the slits is closed, diffraction pattern is obtained on the screen. The following intensity curve is obtained.
Fringe width,
where,
D is the distance between the screen and the slit and,
d is the distance between the slits.
(i) When screen is moved closer to the plane of slits, D decreases and hence, fringe with of the interference pattern also decreases.
(ii) Fringe width β decreases if, separation d between two slits is increased.
(a) What is plane polarised light? Two polaroids are placed at 90° to each other and the transmitted intensity is zero. What happens when one more polaroid is placed between these two, bisecting the angle between them? How will the intensity of transmitted light vary on further rotating the third polaroid?
(b) If a light beam shows no intensity variation when transmitted through a polaroid which is rotated, does it mean that the light is unpolarized? Explain briefly.
(b) Yes, if a light beam shows no variation when transmitted through a polaroid which is rotated, this means that light is unpolarized. If an unpolarized light from an ordinary source passes through a polaroid sheet P1, it is observed that its intensity is reduced by half. Rotating P1 has no effect on the transmitted beam and transmitted intensity remains constant. Now, let an identical piece of polaroid P2 be placed before P1. The light from the lamp is reduced in intensity on passing through P2 alone. In one position, the intensity transmitted by P2 is nearly zero. When turned by 90° from this position, P1 transmits nearly the full intensity emerging from P2.
What is diffraction of light? Draw a graph showing the variation of intensity with angle in a single slit diffraction experiment. Write one feature which distinguishes the observed pattern from the double-slit interference pattern.
How would the diffraction pattern of a single slit be affected when:
(i) the width of the slit is decreased?
(ii) the monochromatic source of light is replaced by a source of white light?
Diffraction of light:
Phenomenon of bending of light around the comers of an obstacle or aperture is called diffraction.
Variation of intensity with angle in a single slit diffraction experiment is as shown below :
In interference, all the bright fringes are of the same intensity whereas, in diffraction phenomenon, the intensity of bright fringes decreases as the distance from the central bright fringe increases.
(i) The diffraction pattern of a single slit becomes narrower if the width of the slit is decreased.
(ii) A coloured diffraction pattern is obtained if monochromatic source is replaced by white light source. The red fringe will be wider than the violet fringe because, the central band is white.
Two convex lens of same focal length but their aperture and focal lengths 5 cm and 10 cm are used as object lens in two astronomical telescope.
(a) What will be the ratio of their
(i) resolving power
(ii) magnifying power
(b) Compare the intensity of images formed in these cases.
Given, Young's double slit experiment with two coherent sources.
Intensity of first source, I1 = I
Intensity of second source, I2 = I + δI
Now, using the formula for resultant intensity,
For maxima,
As therefore,
For minima,
Hence proved.
A beam of light consisting of two wavelengths 6500 A and 5200 A, is used to obtain interference fringes in a Young's double-slit experiment.
(i) Find the distance of the third bright fringe on the screen from the central maximum for wavelength 6500 A.
(ii) What is the least distance from the central maximum where the bright fringes due to both the wavelengths coincide?
The distance between the slits is 2 mm and the distance between the plane of the slits and the screen is 120 cm.
In young's double slit experiment,
Angular separation is given by,
Angular seperation does not depend on D i.e., the distance of separation between slits and screen. Therefore, remains unaffected.
Given,
Change in fringe width, = 3 × 10-5 m
Change in distance between the screen and slit, = 5 × 10-2 m
Distance between the slits, d = 10-3 m
Fringe width,
Therefore,
i.e., is the wavelength of the light used in the experiment.
Intensity at the central maxima = Io
Distance OP is one-third of the fringe width, (given)
Fringe width is given by
Path difference, and,
Phase difference,
Intensity at point P,
Hence proved.
(a) A plane wavefront approaches a plane surface separating two media. If medium one is (optically) denser and medium two is optically rarer, construct the refracted wavefront using Huygen’s principle.
(b) Draw the shape of refracted wavefront when a plane wavefront is incident on (i) prism and (ii) convex mirror. Give a brief explanation for the construction.
The pattern shows a broader diffraction peak in which, there appear several fringes of smaller width, due to the double slit interference phenomenon.
The number of interference fringes depends upon the ratio of the distance between the two slits to the width of a slit.
Newton gave the corpuscular theory on the basis of
B.
Rectilinear motionBand spectrum is also called
A.
Molecular spectrumIn visible spectrum, red light has the greatest wavelength. And, fringe width is directly proportional to the wavelength of light. Therefore, the fringe width will maximum for red light.
In a single slit experiment,
As, size of central max
Therefore,
i) When slit width is halved, size of the central maxima is doubled. Hence, intensity reduces to one-fourth of it's initial value.
ii) When longer wavelength is used, size of central maximum is increased. Hence, intensity reduces.
Given,
Wavelength of light, = 600 nm = 600 10-9 m
Slit width, d = 3 mm = 3 10-3 m
Distance of the screen from the slit, D = 3 m
Now,
Linear width of central maximum =
=
= 1.2 mm
Given,
Polarising angle for a medium = 60o
Therefore, refractive index of the medium is given by Brewster's law.
Thus ,
A ray of light goes from medium 1 to medium 2. Velocity of light in the two media are c1 and c2 respectively. For an angle of incidence 0 in medium 1, the corresponding angle of refraction in medium 2 is ө/2.
(i) Which of the two media is optically denser and why?
(ii) Establish the relationship between ө, c1 and c2.
Given,
i) Angle of incidence in medium 1 =
Angle of refraction in medium 2 =
Angle of refraction in medium 2 is less than angle of incidence in medium 1. That is, the ray is bending towards the normal. Therefore, medium 2 is optically denser than medium 1.
ii) Using Snell's law,
=
= 2 cos
Also, refractive index
This implies,
i.e.,
which is the required relationship between ө, c1 and c2 .
Refracted spherical wavefront emerges out which is converging towards the focus of a lens.
The reflected and refracted ray is mutually perpendicular to each other at the polarising angle.
Therefore, according to Brewster's law , we have
where, is the refractive index of the medium.
A ray is a line extending outward from the source and representing the direction of propagation of the wave at any point along it. Rays are perpendicular to wave fronts.
A wave front is a line representing all parts of a wave that are in phase. The shape of the wave front depends upon the nature of the source; a point source will emit waves having circular or spherical wave fronts, while a large, extended source will emit waves whose wave fronts are effectively flat, or plane.
The objective and the eyepiece of a compound microscope must have short focal length so as to have larger angular magnification and magnifying power.
Energy flux of sunlight = Total energy per square metre per second = 1.388 x 103 Wm–2
Energy of each photon is given by,
Therefore,
Number of photons incident on earth's surface per square metre per second is,
In an experiment on photoelectric effect, the slope of the cut-off voltage versus frequency of incident light is found to be 4.12 x 10–15 Vs.
Calculate the value of Planck's constant.
Given,
Power, P (power) = 100 W
Wavelength of sodium light, λ = 589 x 10–9 m
(a) Energy of each photon assosciated with the sodium light,
(b) Number of photons delivered to sphere per second,
As, P= nE
Given,
Frequency of light, = 7.21 x 1014 Hz
Maximum speed of electrons, vmax = 6.0 x 105 ms–1
Mass of the electron, m = 9 x 10–31 kg
Applying Einstein's photoelectric equation,
Given,
Wavelength of light, λ = 488 nm = 488 x 10–9 m
Stopping potential, V0 = 0.38V
As,
Calculate the
(a) momentum, and
(b) de-Broglie wavelength of the electrons accelerated through a potential difference of 56 V.
What is the
(a) momentum,
(b) speed, and
(c) de-Broglie wavelength of an electron with kinetic energy of 120 eV.
given,
Kinetic energy of the electron, K.E = 120 eV
(a) Momentum of the electron is given as,
(b) Now, since
we have,
(c) De- broglie wavelength of electron is given by,
The wavelength of light from the spectral emission line of sodium is 589 nm. Find the kinetic energy at which
(a) an electron, and
(b) a neutron, would have the same de-Broglie wavelength.
Given,
Wavelength of sodium light, = 589 nm
We know that,
(a) For electron,
Energy,
(b) For neutron,
Energy,
What is the de-Broglie wavelength of
(a) a bullet of mass 0.040 kg travelling at the speed of 1.0 km/s.
(b) a ball of mass 0.060 kg moving at a speed of 1.0 m/s and
(c) a dust particle of mass 1.0 x 10–9 kg drifting with a speed of 2.2 m/s?
(a) Given,
De-broglie wavelength is,
(b) Mass of the ball,
Velocity with which the ball is moving,
Momentum of the particle,
Therefore,
De-broglie wavelength of the particle,
(c) Mass of the dust particle,
Here,
Potential, V = 10 MV =
Specific charge ratio,
Now,
i.e.,
The obtained speed is not possible because, it is greater than the speed of light. Now, since, therefore, as v approaches c, mass of the electron becomes infinite.
Hence, the formula needs to be modified.
A monoenergetic electron beam with electron speed of 5.20 x 106 ms–3 is subject to a magnetic field of 1.30 x 10–4 T normal to the beam velocity. What is the radius of the circle traced by the beam, given e/m for electron equals 1.76 x 1011C kg–1.
Given,
Velocity of electron, v = 5.20 x 106 m/s
Magnetic field, B = 1.30 x 10–4 T
specific charge ratio, e/m = 1.76 x 1011 C kg–1
Centripetal force is provided by the force exerted by magnetic field on electron,
Here,
Potential at collector, V = 100 V;
Magnetic field, B = 2.83 x 10–4 T
Radius of the circular orbit, r = 12.0 cm = 12.0 x 10–2 m
When electrons are accelerated through apotential of V volt, the gain in kinetic energy K.E. of the electron is given by,
...(i)
Since the electron moves in circular orbit under magnetic field, therefore, force on the electron due to magnetic field provides the centripetal force to the electron.
....(ii)
From equations (i) and (ii), we get
(a) An X-ray tube produces a continuous spectrum of radiation with its short wavelength end at 0.45 Å. What is the maximum energy of a photon in the radiation?
(b) From your answer to (a), guess what order of accelerating voltage (for electrons) is required in such a tube?
We are given an x-ray tube.
Wavelength of the radiation, λ = 0.45 Å = 0.45 x 10–10 m
Planck's constant, h = 6.62 x 10–34 Js
speed of light, c = 3 x 108 ms–1
(a) The maximum energy of photon is given by,
i.e.,
(b) To produce electrons of 27.5 keV, accelerating potential of 27.5 kV to 30 KV is required.
Given,
Wavelength of radio waves,
Energy of photon,
Now, using the formula for energy of a photon,
Number of photons emitted per second,
We can see, that the energy of a radiophoton is exceedingly small, and the number of photons emitted per second from a radio beam is enormously large. Thus, there is negligible error involved in ignoring the existence of a minimum quantum of energy of photon and treating the total energy of a radiowave as continuous.
Here,
Sopping potential,
Now, using Einstein's photoelectric equation,
where,
Taking,
We have,
i.e.,
Threshold wavelength is given by,
i.e.,
As calculated above, wavelength of red light is greater than threshold wavelength λ0. Therefore, photocell will not respond when red light of wavelength 6328 A produced by He-Ne laser is incident on the photocell.
The work function for the following metals is given:
Na: 2.75 eV; K: 2.30 eV; Mo: 4.17 eV; Ni: 5.15 eV.
Which of these metals will not give photoelectric emission for a radiation of wavelength 3300 A from a He-Cd laser placed 1m away from the photocell? What happens if the laser is brought nearer and placed 50 cm away?
Wavelength of the probe = 1
Mass of electron, me = 9.11 x 10–31 kg
Energy of photon,
i.e.,
For the case of electron,
i.e.,
From the above calculations, we can see that, for the same given wavelength, kinetic energy of a photon is much greater than that of electron.
We are given,
Room tempertaure, T = 27 + 273 = 300 K
Boltzmann's constant, k = 1.38 x 10–23 J mol–1 K–1
We know, average kinetic energy of neutron at absolute temperature T is given by,
where k is the Boltzmann's constant.
Now, wavelength,
Since this wavelength is comparable to interatomic spacing (~ 1 Å) in a crystal, therefore, thermal neutrons are suitable for diffraction experiments. So, a high energy neutron beam should be first thermalised before using it for diffraction.
Here,
Mass of electron,
Using formula,
Interelectronic separation,
Hence, . That is, interelectronic spacing is less than the wavelength.
We find that wave-packets in metals strongly overlap with one another whereas, this is not the case in gas atoms.
Given, de-broglie wavelength assosciated with electron and an alpha particle is same.
Now,
Kinetic energy,
... (1)
and,
where, pe and is the momentum of electron and alpha particle respectively.
So,
From equation (1)
is the required relation.
A mercury lamp is convenient source for studying frequency dependence of photoelectric emission, since it gives a number of spectral lines ranging from the UV to the red end of the visible spectrum. In our experiment with rubidium photocell, the following lines from a mercury source were used:
λ1 = 3650 Å, λ2 = 4047 Å, λ3 = 4358 Å, λ4 = 5461 Å, λ5 = 6907 Å
The stopping voltages, respectively were measured to be:
V01 = 1.28 V,V02 = 0.95 V,V03 = 0.74 V, V04 = 0.16 V,V05 = 0V
(a) Determine the value of Planck's constant h.
(b) Estimate the threshold frequency and work function for the material.
Using the given data, we first determine frequency in each case and then plot a graph between stopping potential V0 and frequency v.
Vo versus v plot is shown below: 
The first four points lie nearly on a straight line which intercepts the frequency axis at threshold frequency v0 = 5.0 x 1014 Hz.
The fifth point v (= 4.3 x 1014 Hz) corresponds to v < v0, so there is no photoelectric emission and not stopping voltage is required to stop the current.
Slope of V0 versus v graph is,
Now, from Einstein's photoelectric equation,
Kinetic energy of photon,
Hence,
Planck's constant,
(b) Threshold frequency,
i.e., is the required work function of the metal.
Given, work function of metal A = 2 eV
Work function of metal B = 5 eV
As,
i.e.
Thus, metal B has lower threshold wavelength of radiation since its work function is greater.
Maximum K.E of particle = 3 eV
Since,
Stopping potential,
Given, the frequency of the incident radiation is greater than the threshold frequency.
The formula gives us,
Therefore, the value of stopping potential (V0) increases with increase in frequency (v) of the incident radiation and K.E. increaseses.
Given, the kinetic energy of the emitted electrons is same in both cases.
Kinetic energy is given by,
is the frequency of radiation and is the work function of materials.
Therefore, as per given,
and,
Thus,
Given,
De-broglie wavelength, λ = 16.5 nm = 16.5 x 10–9 m
Mass of proton m = 1.6 x 10–27 kg
Using de-Broglie equation,
i.e.,
is the calculated kinetic energy of proton.
The Davisson and Germer’s experiment established famous de-Broglie hypothesis of wave-particle duality by confirming the wave nature of moving particle.
Given, an alpha particle and a proton are accelerated from rest by the same potential.
The energy acquired by alpha-particle, Eα = 2eV
The energy acquired by proton, Ep = eV
Hence, the ratio of wavelengths,
i.e.,
is the obtained ratio of the de-broglie wavelength asosciated with the particles.
Radiations of frequency 1015 Hz are incident on two photosensitive surfaces A and B. Following observations are recorded:
Surface A: No photoemission takes place.
Surface B: Photoemission takes place but photoelectrons have zero energy.
Explain the above observations on the basis of Einstein's photoelectric equation.
How will the observation with surface B change when the wavelength of incident radiations is decreased?
Sketch the graphs showing the variation of stopping potential with frequency of incident radiations for two photosensitive materials A and B having threshold frequencies v0 > v0respectively.
(i) Which of the two metals, A or B has higher work function?
(ii) What information do you get from the slope of the graphs?
(iii) What does the value of the intercept of graph ‘A’ on the potential axis represent ?
Given, a molecule of mass M.
It splits into two fragments of masses m/3 and 2m/3.
Let,
But,
Following the law of conservation of linear momentum,
Also,De-broglie wavelength is,
Hence,
Thus, the ratio of wavelength is 1.
The work function of lithium is 2.3 eV.
This means that to remove the outermost electron from the ground shell of a lithium atom, an energy of 2.3 eV is required.
The relation of work function (W) and wavelength (λ) is given as,
i.e.,
Red light, however bright it is, cannot produce the emission of electrons from a clean zinc surface. But even weak ultraviolet radiation can do so. Why?
X-ray of wavelength ‘λ’ fall on a photosensitive surface, emitting electrons. Assuming that the work function of the surface can be neglected, prove that the de-Broglie wavelength of the electrons emitted will be
For any material to emit photoelectrons, the threshold frequency has to be greater than or comparable to the frequency of incident radiation. Therefore, the frequency of incident light affects the emission of photoelectrons whereas, intensity does not.
Here, the threshold frequency of zinc surface is not greater than that of the frequency of red light. Hence, no emission of electrons take place.
The kinetic energy of photoelectron is given by,
But de-Broglie wavelength is given by,
Hence, proved.
The wavelength of light in the visible region is about 390 nm for violet colour, about 550 nm (average wavelength) for yellow-green colour and about 760 nm for red colour.
What are the energies of photons in (eV) at the (i) violet end, (ii) average wavelength, yellow-green colour, and (iii) red end of the visible spectrum? (Take h = 6.63 x 10–34 Js and 1 eV = 1.6 x 10–19J).
(i) Explain which metal has smaller threshold wavelengths.
(ii) Explain, giving reason, which metal emits photoelectrons having smaller kinetic energy.
(iii) If the distance between the light source and metal P is doubled, how will the stopping potential change?
Light of wavelength 2000 A falls on an aluminium surface (work function of aluminium 4.2 eV). Calculate
(a) the kinetic energy of the fastest and slowest emitted photoelectrons
(b) stopping potential
(c) cut-off wavelength for aluminium.
Energy of the fastest emitted photoelectron,
= h (v – v0) (Where v0 is the work function)
= (6.2 – 4.2) eV
= 2.0 eV
Since, the emitted electrons from a metal surface have an energy distribution, the minimum energy in this distribution being zero, the energy of slowest photoelectrons is also zero.
(b) Since,
where, is the maximum energy of the emitted photo electrons and V0 is the stopping potential, the stopping potential is 2V.
(c) The threshold frequency is related to the work function by the relation,
, is the required cut-off wavelength for aluminum.
Define the terms: (i) work function, (ii) threshold frequency and (iii) stopping potential, with reference to photoelectric effect.
Calculate the maximum kinetic energy of electrons emitted from a photosensitive surface of work function 3.2 eV, for the incident radiation of wavelength 300 nm.
Work function is the minimum amount of energy required to remove an electron to infinity, from the surface of a given metal.
Threshold frequency is the minimum frequency of the incident light which can cause the ejection of electrons without giving them additional energy.
The amount of potential that is required to stop the electron having the maximum kinetic energy from moving is known as stopping potential.
Numerical.
The maximum kinetic energy of emitted photoelectron is given by
Given,
Voltage, V = 54 V
Using the formula,
From Bragg's law, [For first order diffraction, n = 1]
For Ni crystal,
Inter-atomic spacing,
Substituting this in the formula,
We have,
But,
Hence,
for
Here,
Temperature, T = 27°C = 27 + 273 = 300 K.
Mass of neutron, m = 1.66 x 10–23 kg.
Energy of neutron at temperature Tk is,
Now,
i.e., , is the de-broglie wavelength of the neutron.
The two main observations are:
(i) The maximum kinetic energy of emitted photoelectron is independent of intensity of light.
(ii) For each photoelectron, there must be a threshold frequency of incident light below which no emission takes place.
For a metal of work function ɸ, the kinetic energy of photoelectron emitted due to falling of photon of frequency v is,
where is the threshold frequency.
For photoelectric emission,
or
gives us the relationship between threshold frequency and work function of the metal.
What reasoning led de-Broglie to put forward the concept of matter wave? The wavelength λ, of a photon, and de-Broglie wavelength associated with a particle of mass m has the same value, say λ. Show that the energy of photon is times the kinetic energy of the particle.
A.
equal to the circumference of the first orbit.B.
Particle nature of lightWavelength of the photon, = 5000
Rate at which the photons are emitted = 10-8 J/s
So, number of photons emitted, n =
i) In Davison and germer experiment,
The appearance of peak in a particular direction is due to constructive interference of electrons scattered from different layers of regularly spaced atoms of the crystal, i.e., the diffraction of electrons take place. This established the wave nature of electron.
ii) According to Davison- Germer experiment, the wavelength of the wave is found out to be 1.65 .
According to De- Broglie hypothesis, wavelength is given by 1.66 .
There is a close agreement with the estimated value of both experiments.
Hence, the De-broglie relation is established.
Consider an electron of mass 'm' and charge 'e'. Let v be the velocity acquired by electron when accelerated from rest through a potential difference of V volt.
Gain in K.E of electron =
Work done = eV
If is the De- Broglie wavelegth assosciated with the electron, then
.
In order to establish the wave nature of electrons an experiment was performed by Davison and Germer.
i) A fine beam of accelerated electrons obtained from electron gun is made to fall normally on the surface of nickel crystal.
ii) Incident electrons are scattered in different directions by the atoms of the crystals.
iii) Intensity of the elctron beam, is scattered in a different direction.
iv) By rotating the electron detector on circular scale at different positions, intensity of the scattered beam is measured for different values of scattering angle .
v) The experiment was performed by varying the accelerating voltage 44 V to 68 V.
vi) In an accelerating voltage at 54 V, a graph with axis of variation of intensity (I) and scattering angle () was plotted.
vi) The appearance of peak in a particular direction is due to constructive interference of electrons scattered from different layers of regularly spaced atoms of the crystal, i.e., the diffraction of electrons takes place.
This established the wave nature of electrons. 
where,
ɸ0 is the work function,
X wavelength of incident light and
V0 is the stopping potential.
For the first source,
Putting these values in equation 1,
...(I)
Let λ2 be the wavelength of the second source.
Stopping potential, V0’ = 1.1 V (given)
Therefore,
When the ejected photoelectrons are subjected to the action of a magnetic field no change in retarding potential will be observed because, a magnetic field does not alter the kinetic energy of the photoelectrons. The magnetic field only changes the direction of motion.
A classical atom based on __________ is doomed to collapse. (Thomson's model/Rutherford's model)
Given,
As oper the formula of radius of orbit,
(i) When,
Radius, r =
(ii) When
We have,
Radius, r =
(a) From the relation,
Therefore, the speed of electron in hydrogen atom in n=1, 2 and 3 is given by,
(b) Orbital period in each of the level is given by,
'
As,
Also as,
T1, T2 and T3 are respectively the orbital time period for n1, n2 and n3 .
In Thomson's model, a single collision causes very little deflection. Multiple scattering has to be taken into account inorder to explain the average angle of scattering. So it is wrong to ignore multiple scattering in Thomson's model.
On the contrary, in Rutherford's model, most of the scattering comes through a single collision and multiple scattering effects can be ignored as a first approximation.
The total energy of an electron in the first excited state of the hydrogen atom is about – 3.4 eV.
(a) What is the kinetic energy of the electron in this state?
(b) What is the potential energy of the electron in this state?
(c) Which of the answers above would change if the choice of the zero of potential energy is changed?
In Bohr's model, as per the quantisation of angular momentum we have,
mvr =
which gives
Kinetic energy,
Here, we have
Kinetic energy, K.E =
These relations have nothing to do with choice of the zero of potential energy.
Now, choosing the zero of potential energy at infinity, we have
Ep is the potential energy.
Thus, total energy,
(a) The quoted value of E = – 3.4 eV is based on the customary choice of zero of potential energy at infinity.
From the above result of E = – Ek, the kinetic energy of electron in this state is + 3.4 eV.
(b) Using Ep = – 2 Ek, potential energy of the electron is – 2 × 3.4 eV = – 6.8 eV.
(c) If the zero of potential energy is chosen differently, kinetic energy does not change. Its value is + 3.4 eV. This is independent of the choice of the zero of potential energy.
The potential energy, and the total energy of the state, however, would alter if a different zero of the potential energy is chosen.
Which of the following given transitions in a hydrogen atom emits the photon of lowest frequency?
(i) n = 2 to n = 1
(ii) n = 4 to n = 3.
Energy for the hydrogen atom is given by,
Putting
(i) Hence, from the above expressions it can be observed that the electron in the hydrogen atom cannot have an energy of – 2V.
(ii) As n increases, energies of the excited states come closer and closer together. Therefore, as n increases, En becomes less negative until at n = ∞, we get En = 0.
The ground state energy of hydrogen atom is - 13.6 eV.
(i) What is the potential energy of an electron in the 3rd excited state?
(ii) If the electron jumps to the ground state from the 3rd excited state, calculate the wavelength of the photon emitted.
The energy of an electron in nth orbit is given by,
(i) For 3rd excited state, n = 4
Energy at the third excited level is given by,
(ii) Required energy to jump electron to the ground state from the 3rd excited state is,
∴ Wavelength of the photon emitted is,
is the wavelength of the photon emitted.
Initially, hydrogen atom is in the ground state.
After absorption of photon, atom is excited to n=4th level.
USing the Rydberg's formula,
we have,
is the wavelength of the photon.
Frequency of the photon, is given by,
A doubly ionised lithium atom is hydrogen-like with atomic number 3:
(i) Find the wavelength of the radiation required to excite the electron in Li++ from the first to the third Bohr orbit. (Ionisation energy of the hydrogen atom equals 13.6 eV.)
(ii) How many spectral lines are observed in the emission spectrum of the above excited system?
(ii) The following three spectral lines are observed due to the following transitions:
3rd to 1st orbit
3rd to 2nd orbit
2nd to 1st orbit
A single electron, orbits around a stationary nucleus of charge ze, where z is a constant and e is the electronic charge. It requires 47.2 eV to excite the electron from the second Bohr orbit to 3rd Bohr orbit. Find,
(i) The value of z.
(ii) The energy required to excite the electron from the third to the fourth Bohr orbit.
(iii) The wavelength of electromagnetic radiation required to remove the electron from the first Bohr orbit to infinity.
(iv) The kinetic energy, potential energy and angular momentum of the electron in the first Bohr orbit.
(v) The radius of the first Bohr orbit.
(Ionisation energy of hydrogen atom = 13.6 eV. Bohr radius = 5.3 × 10–11 m, velocity of light = 3 × 108 m/s and Planck's constant = 6.6 × 10–34 Js)
Given,
Wavelength of the first member of lyman series = 1216 Å
Now, the rydberg's formula gives us,
For first member of Lyman series,
...(i)
For second member of Balmer series,
Therefore,
...(ii)
Dividing equation (ii) by (i), we get
is the wavelength of the second member of balmer series.
Radius of any orbit is given by the formula,
We have, and,
Therefore, putting the values in the formula,
we get,
, is the radius of the first orbit of hydrogen atom.
Now,
Velocity of electron in the first orbit is,
ie.,
Hence proved.
The ground state energy of hydrogen atoms is – 13.6 eV.
(i) Which are the potential and kinetic energy of an electron in the third excited state?
(ii) If the electron jumps to the ground state from the third excited state. Calculate the frequency of photon emitted.
B.
138 neutrons and 88 protons13.6 eV
D.
13.6 × (11)2 eVD.
Z protons and (A – Z) neutronsLimitation of Rutherford's alpha-particle scattering experiment:
In the model, it was proposed that the electron is revolving aroung the nucleus and is continuously experiencing a centripetal force. Thus, the electron has an accelerated motion. And, as per classical theory, the electron must radiate energy in the form of electromagnetic waves. So, the revolving electron is constantly losing energy and, it must spiral inwards eventually, falling into the nucleus.
But, since the matter is stable, the atom is not expected to collapse. This was a major drawback of rutherford's model.
No, Bohr's atomic theory is applicable for all single elctron atoms like He+, Li2+ etc.
In Bohr's theory we assume that centripetal force is provided by the electrostatic force of attraction by the nucleus.
In multiple elctron atoms, there will also be repulsion due to other electrons present in the nucleus.
The different spectral lines in hydrogen atom are:
n=1, Lyman series
n=2, Balmer series
n=3, Paschen series
n=4, Brackett series
n=5, Pfund series
One of the shortcomings of Bohr's Theory was that it could not explain the variation of intensity of the spectral lines of an element on the basis of frequency of photons emitted.
The intensity can be explained only by quantum mechanics.
In the third orbit, n =3
For hydrogen atom, energy of an electron is given by,
En =
The negative total energy of electron in it's excited state signify that the elctron is bound to the nucleus and is not free to leave it.
Given,
Mass of isotope 1 = 6.01512 u
Mass of isotope 2 = 7.01600 u
Abundance of isoptope 1 = 7.5 %
Abundance of isotope 2 = 92.5 %
Atomic mass of Li = weighted average of the isotopes
=
=
Boron has two stable isotopes, Their respective masses are 10.01294 u and 11.00931 u, and the atomic mass of boron is 10.81u. Find the abundances of
Let, the relative abundance of be x%
Relative abundance of
Atomic weight = weighted average of the isotopes,
i.e.,
Given, mass of the coin = 3.0 g
Mass of atom = 62.92960 u
Each atom of the copper contains 29 protons and 34 neutrons.
Mass of 29 electrons = 29 x 0.000548 u
= 0.015892 u
Mass of nucleus = (62.92960 - 0.015892) u
= 62.913708 u
Mass of 29 protons = 29 x 1.007825 u
= 29.226925 u
Mass of 34 neutrons = 34 x 1.008665 u
= 34.29461 u
Total mass of protons and neutrons = (29.226925 + 34.29461) u
= 63.521535 u
Binding energy = (63.521535 - 62.913708) x 931.5 MeV
= 0.607827 x 931.5 MeV
Required energy =
Given, the sample has a half life of T years.
(a) The fraction of the original sample left is ,
=
Hence, there are 5 half lives of T years spent. Thus, the time taken is 5T years.
(b) The fraction of the original sample left =
i.e., = 100
n log 2 = log 100
Hence, n = = 6.64
From, t= nT we have, t = 6.64 T.
Hence, there are 6.64 half lives of T years spent. Thus, the time taken is 6.64T years.
Given,
Half- life of = 28 years.
Using the formula,
90 g of Sr contains 6.023 x 1023 atoms.
15 mg of Sr contains,
Disintegration rate,
Using the relation between the radius of nucleus and atomic mass,
Atomic mass of gold, A1 = 197
Atomic mass of silver, A2 = 107
Now, taking log on both sides,
= 1.23, which is the required ratio of the nucleii.
For the given reaction, mass defect is,
Now, Q-value is ,
which, is the maximum energy of the positron.
We have,
The daughter nucleus is too heavy compared to e+ and v. So, it carries negligible energy (Ed ≈ 0). If the kinetic energy (Ev) carried by the neutrino is minimum (i.e., zero), the positron carries maximum energy, and this is practically all energy Q.
Hence, maximum Ee≈ Q.
The nucleus decays by β- emission. Write down the β-decay equation and determine the maximum kinetic energy of the electrons emitted from the following data:
The of may be represented as:
Ignoring the rest mass of antineutrino and electron , we get
Mass defect,
This energy of 4.3792 MeV, is shared by pair because, is very massive.
The maximum K.E. of when energy carried by is zero.
(i) Considering the first reaction,
Q-value is given by,
Since, Q-value is negative, this reaction is endothermic.
(ii) The second reaction is,
Q-value is given by,
Since, the Q-value is positive , the reaction is exothermic.
Given,
Average amount of energy released per fission,
Quantity of fissionable material = 1 kg
In 239 gm Pu, number of fissionable atom or nuclei
In 1 g of Pu, number of fissionable atom or nuclei
In 1000 gm of Pu, number of fissionable atom or nuclei,
Therefore,
Total energy released in fission of 25.2 x 1023 Pu nucleus or in fission of 1 kg pure Pu is,
= 180 x 25.2 x 1023
= 4536 x 1023 MeV
= 4.5 x 1026 MeV.
Given,
Power of reactor = 1000 MW = 103 MW
= 109W
= 109 Js-1
Energy generated by reactor in 5 Years = 5 x 365 x 24 x 60 x 60 x 109 J
Average energy generated = 200 MeV
= 200 x 1.6 x 10-13 J
Number of fission taking place or number of U235 nuclei required,
Mass of 6.023 x 1023 nuclei of U = 235 gm = 235 x 10-3 kg
Mass of 8.2125 x 1026 nuclei of U,
Power of the electric lamp = 100 W
When two nuclei of deuterium fuse together, energy released = 3.2 MeV
Number of deuterium atoms in 2 kg is,
Energy released when nuclei of deuterium fuse together,
If the lamp glows for time t, then the electrical energy consumed by the lamp is 100 t,
100 t =
which is the life span of an electric lamp.
We know that rate of disintegration is,
So, clearly the initial ratio of the phosphorous radio nuclides, is 1:9.
We have to find the time at which the ratio is 9:1.
Initially, if the amount of is x, then the amount of is 9x.
Finally, if the amount of is 9y, the amount of is y.
Now, from the formula,
Therefore,
Using,
On dividing,
i.e.,
(i) For the first decay process,
Mass defect,
Therefore,
Energy released,
(ii) Now, for second decay process we have,
Mass defect, = mass of - (mass of Rn219 + mass of He4)
Energy released,
Both the above given decay processes are energetically possible because the Q-value i.e., energy released in the reaction is positive.
(a) The reaction process is,
x 931 MeV
(b) Repulsive potential energy of two nuclei when they almost touch each other is given by,
Classically, this amount of K.E. is at least required to overcome Coulomb repulsion.
Now, using the relation,
though the temperature required for triggering the reaction is somewhat less practically.
To obtain the maximum kinetic energy of - particles = ?
Radiation frequency,
Frequency,
The emission process of decay may be represented as:
where, ( as per the fig.)
Now,
where,
are masses of the nuclei.
= (197.968233 - 197.966760) x 931.5 - 1.088
The emission process of decay may be represented as:
As in case of decay, it can be deduced in a similar manner.
(i)
(ii)
(iii)
(iv)
(v)
(vi)
(vii)
Given,
Half life , T = 5.3 years
Strength of radioactive source,
= 8.0 mCi = 8.0 x 10-3 Ci
= 8.0 x 10-3 x 3.7 x 10-10 disintegrations s-1
= 29.6 x 107 disintegrations s-1
Since the strength of the source decreases with time,
But
Number of atoms in 60 g of cobalt = 6.023 x 1023
Therefore,
Mass of 1 atom of cobalt =
Mass of is,
(a) The reaction invoved is,
The difference in mass between the original nucleus and the decay products = 226.02540 u - (222.01750 u + 4.00260 u)
= + 0.0053 u
Energy equivalent or Q-value = 0.0053 x 931.5 MeV
= 4.93695 MeV
= 4.94 MeV
The decay products would emerge with total kinetic energy 4.94 MeV.
Momentum is conserved. If the parent nucleus is at rest, the daughter and the -particle have momenta of equal magnitude p but, in the opposite direction.
Kinetic energy,
Since, p is the same for the two particles therefore the kinetic energy divides inversely as their masses.
The -particle gets of the total i.e.,
(b) The difference in mass between the original nucleus and the decay products = 220.01137 u - (216.00189 u + 4.00260 u)
= 0.00688 u
Q-value or Energy equivalent = 0.00688 x 931.5 MeV
= 6.41 MeV
Energy of the alpha particle,
Given,
Rato of mass numbers ,
Radius of nucleus is,
Therefore,
Thus, Ratio of radii is 3:5.
Heavy water is used an absorbant to control the reaction rate of neutrons in a nuclear reactor.
(i) Mass number of D = 172.
(ii) Atomic number of D = 70.
What is the nuclear radius of if that of Al27 is 3.6 fermi?
Therefore, Mass number of product nucleus = m - 4 and,
Atomic number of product nucleus = n - 1.
1. The ratio of neutron to proton in a nucleus increases after the emission of alpha-particle.
2. The ratio of neutron to proton in a nucleus decreases after the emission of beta- particle.
The number of radioactive disintegrations taking place per second in a given sample is called the activity of a sample.
Its SI unit is becquerel.
The graph between the activity of a radioactive species and time is given below: 
It is found from an experiment that the radioactive substance emits one beta particle for each decay process. Also an average of 8.4 beta particles are emitted each second by 2.5 milligram of substance. The atomic weight of substance is 230. What is the half-life?
The activity = 8.4 sec-1.
Atomic weight of the substance = 230 kg
Number of atoms in kilomole (i.e., 230 kg) = 6.02 x 1026
Therefore,
Number of atoms,
Half-life,
We have,
Mass + mass = 2.014102 + 3.016049
= 5.030151 u
and
As seen from above, since, mass of the reactants is larger than mass of the products therefore, energy is released in this reaction.
Now,
So, energy required =
Given,
Half life of tritium, T = 12.5 years
t = 37.5 years
Now, using the formula for decay rate,
we have,
Thus, one-eighth of the sample will remain undecayed after 13.5 years.
i) Nuclear reaction is given as,
ii) Mass defect ,
Energy released = 0.0111870 x 931
= 10.415 MeV.
The undertaking fission process can be wriiten as :
Total mass number of and neutron = 240
Mass number of each fragment =
Atomic number of each fragment =
Radius of each nucleus formed by the fission of is,
Distance between the centre of the two fragments
Electrostatic force between them is given by Coulomb's law,
Therefore,
F
The -decay of Uranium is given by,
Neutron to proton ratio before
Neutron to proton ratio after
Since,
Thus, the neutron to proton increases is an .
Classification of nuclides is as follows:
Isotopes: ; has the same atomic number but, different mass number.
Isobars ; has the same mass number but, different atomic number.
Isotones: ; has same neutron number but, different proton number.
Now,
Radius of nucleus, ...(i)
where, is the range of nuclear force (or Nuclear Unit Radius)
Density is given by, ...(ii)
But, ...(iii)
From (i) and (iii), we get
where,
...(iv)
From (ii) and (iv), we have
Thus, we can see that nuclear density is clearly independent of mass number A.
Nuclear reactions are given by:
(i) -decay :
(ii) decay :
(iii) decay :
(a) For, a radioactive sample, rate of disintegration is given by,
...(i)
According to radioactive decay law,
...(ii)
[By equation (i) and (ii)]
(b) Activity of a sample is,
We have,
Rate of increase of element =
That is,
The solution to this is the sum of the homogeneous solution,
where c is a constant, and
a particular solution,
Therefore, the required solution is,
The constant c is obtained from the requirement that the initial number of nuclei be zero,
so that,
The equilibrium value is
Setting N equal to 1/2 of this value gives,
The result is independent of R.
Half life of isotope X = 3 seconds.
Number of atoms at t=0 is 8000 atoms.
(a) Decay constant,
(b) Atoms remaining in the sample = 1000
Time t1 at which these many sample is remaining = ?
Now, using the formula of the law of radioactive decay we have,
(c) Number of decays per second in the sample is,
Why heavy stable nucleus must contain more neutrons than protons? What is the effect on neutron to proton ratio in a nucleus when
(i) an electron is emitted
(ii) a positron is emitted?
In case of ,
Neutron to proton ratio is,
In case of , neutron to proton ratio is given by,
For stability, this ratio has to be close to one.
Therefore, nucleus is more stable than the nucleus
The half-life period is,
T1/2 = 0.693 Tm
where, Tm is the mean half-life.
To decay by 50% of the sample one half-life is required.
Therefore, time taken is 0.693 Tm to decay by 50%.
Radioactivity is the physical quantity whose SI unit is Bq.
(i) Relation between radioactivity and disintegration constant is as follows:
R = R0e-λt or R = λN
where, λ is disintegration constant.
(ii) Relation between radioactivity and half life is,
where T1/2 is the half-life period.
(iii) Relation between radioactivity and mean-life of the element is,
where, Tm is the mean-life.
D.
Nuclear reactor-electronThe activity of a radioactive substance at any time 't' is given by,
... (1)
where,
A is the radioactivity material remaining after time 't,
Ao is the initial activity of the material, and
Given that,
t = 30 years
Now, putting these values in equation (1), we get
And, half life of a radioactive substance is given by,
,
is the required half-period of the radioactive substance.
Given, the ratio of mass number is 1:2.
Nuclear density is inversely proportional to mass number.
Therefore,
Potential energy of a pair of nucleons as a function of their seperation can be shown as below: 
The negative potential energy in the graph exhibits that the system is bounded.
Given,
Half life for X is the same as mean life time of another radio active element Y.
i.e.,
Now, as half life of Y is shorter than half life of X, therefore Y will decay faster than X.
Neutron to proton ratio of an atomic nucleus is the ratio of the number of neutrons to it's number of protons. This ratio generally increases with increase in atomic number, in a stable nuclei.
Beta decay decreases the ratio of neutron to proton.
e.g.,
It is clearly visible that the neutron-proton decreases.
How many disintegrations per second will occur in one gram of if half life against alpha decay 8is 1.42 × 1017 s?
Given, half life period (T) =
Therefore,
Avogadro's number =
n is the number of atoms present in 1 g of =
Number of disintegrations is given by,
In the series of radioactive disintegration of first one α-particle and then one β-particle is emitted. What is the atomic number and mass number of the new nucleus formed by these successive disintegrations?
When an alpha particle is emitted, mass number decreases by 2 and atomic number decreases by 2 and when a beta particle is emitted, mass number remains the same and atomic number increases by 1.
So, mass number becomes (A-4) and atomic number becomes (Z-1).
Given, contains 20 protons and (40-20) = 20 neutrons.
Mass of 20 protons = 20 1.007825 = 20.1565 a.m.u
Mass of 20 neutrons = 20 1.008665 = 20.1733 a.m.u
Therefore, total mass of 40 nucleons = 40.3298 a.m.u
Given that mass of = 39.962589 a.m.u
Therefore, mass defect,
Total binding energy per nucleon is given by,
Given,
Half life of RaB = 26.8 min
Activity of
If, is the disintegration constant of
Let N be the number of atoms of having an activity of 1 curie. Then ,
we have
Further we know that,
1g of atom = 6.02 1023 atoms.
Mass of 1 curie of atom = 214g =
Therefore, the mass of having an activity of 1 curie is,
(a) The alpha decay of is given by,
The energy released in this process is given by
... (1)
Substituting the atomic masses as given in the data in equation (1), we find
(b) The kinetic energy of the particle is,
(c) If spontaneously emits a proton, the decay process would be,
The Q for this process to happen is,
Since the Q-value for this process is negative, it cannot proceed spontaneously and will require 7.68 MeV energy.
In an n-type silicon, which of the following statement is true:
(a) Electrons are majority carriers and trivalent atoms are the dopants.
(b) Electrons are minority carriers and pentavalent atoms are the dopants.
(c) Holes are minority carriers and pentavalent atoms are the dopants.
(d) Holes are majority carriers and trivalent atoms are the dopants.
(c) is the correct statement.
Out of the given three elements, energy band gap is maximum for carbon, less for silicon and least for germanium.
A.
Raises the potential barrier.B.
Reduces the majority carrier current to zero.C.
Lowers the potential barrier.D.
None of the above.State True or False:
For transistor action, which of the following statements are correct:
A.
Base, emitter and collector regions should have similar size and doping concentrations.B.
The base region must be very thin and lightly doped.C.
The emitter junction is forward biased and collector junction is reverse biased.D.
Both the emitter junction as well as the collector junction are forward biased.C.
is low at high and low frequencies and constant at mid frequencies.For a transistor amplifier, the voltage gain is low at high and low frequencies and constant at mid frequencies.
Given, a common-emitter transistor configuration.
As, voltage gain,
Input signal voltage,
Base current, .
Given, two amplifiers are cascaded in series.
Voltage gain of first amplifier = 10
Voltage gain of second amplifier = 20
Input signal = 0.01 V
Therefore,
Total voltage gain,
is the output a.c. signal.
Given,
Energy band gap, Eg = 2.8 eV
Wavelength of the photon, = 600 nm
Using the formula for energy of a photon,
we have, E
As, E < Eg, so p-n junction cannot detect the radiation of given wavelength.
|
A |
B |
Y’ |
|
0 |
0 |
1 |
|
0 |
1 |
0 |
|
1 |
0 |
0 |
|
1 |
1 |
0 |
Now Y’ is feeded as input into NOT gate.
So output Y is represented in form of truth table as shown.
|
Input Y’ |
Output Y = Y’ |
|
1 |
0 |
|
0 |
1 |
|
0 |
1 |
|
0 |
1 |
which is similar to the truth table of OR gate.
The boolean expression for the circuit is as follows:
which is the expression for OR gate.
(b) Here in figure (b), input A and B are given to two NOT gates and these inverted input is provided to NOR gate.
Its truth table can be represented as:
|
Output of NOT gats |
|||
|
A |
B |
Ā |
|
|
0 |
0 |
1 |
1 |
|
0 |
1 |
1 |
0 |
|
1 |
0 |
0 |
1 |
|
1 |
1 |
0 |
0 |
Now output of NOT gate is fed as input to NOR gate. So its truth table can be represented as:
|
Input |
Output |
|
|
Ā |
Y |
|
|
1 |
1 |
0 |
|
1 |
0 |
0 |
|
0 |
1 |
0 |
|
0 |
0 |
1 |
which represents AND operation.
The boolean expression for circuit (b) is:
which the expression for AND gate.
|
A |
Y |
|
0 |
1 |
|
1 |
0 |
The truth table for first circuit:
(a)
|
A |
B |
Y’ |
Y |
|
0 |
0 |
1 |
0 |
|
0 |
1 |
1 |
0 |
|
1 |
0 |
1 |
0 |
|
1 |
1 |
0 |
1 |
If we remove the intermediate output Y’, the overall result stands for the AND gate.
|
A |
B |
Y |
|
0 |
0 |
0 |
|
0 |
1 |
0 |
|
1 |
0 |
0 |
|
1 |
1 |
1 |
Hence, the combinational circuit is equivalent to AND gate.
The truth table for second circuit:
(b)
|
A |
B |
Ā |
Y |
|
|
0 |
0 |
1 |
1 |
0 |
|
0 |
1 |
1 |
0 |
1 |
|
1 |
0 |
0 |
1 |
1 |
|
1 |
1 |
0 |
0 |
1 |
If we remove the intermediate output Ā and , the overall result stands for the OR gate.
Truth table for the OR gate is:
|
A |
B |
Y |
|
0 |
0 |
0 |
|
0 |
1 |
1 |
|
1 |
0 |
1 |
|
1 |
1 |
1 |
This can be represented as Y = A + B
Hence, the circuit is an equivalent OR gate.
Figure (a) represents NOT gate.
Here, when A = 1, Y = 0
and
when A = 0, Y = 1
Figure (b) represents two NOT gates whose outputs are given to NOR gate.
Truth table for this circuit is:
|
A |
B |
Ā |
Y |
|
|
0 |
0 |
1 |
1 |
0 |
|
0 |
1 |
1 |
0 |
0 |
|
1 |
0 |
0 |
1 |
0 |
|
1 |
1 |
0 |
0 |
1 |
The logic operation performed by circuit in figure (b) is AND operation.
Given,
Reverse saturation current,
Absolute temperature, T = 300 K
(a) Forward voltage, V = 0.6 V, then
and
(b) If the forward voltage is increased to V = 0.7 V, then
Therefore,
Thus, increase in current is,
(c) As,
Therefore, '
Dynamic resistance =
(d) For both the voltages, the current I will be almost equal to I0. This implies that the circuit offers infinite dynamic resistance in the reverse bias.
(i) Wavelength of photons emitted depends upon the band gap. Thus, energy is a factor controlling the wavelength of light.
(ii) The forward current increases as the intensity of light increases and reaches a maximum value. Further increase in the forward current results in decrease of light intensity.
LEDs are biased such that the light emitting efficiency is maximum.
Given,
Current gain,
Collector resistance, Rc = 5 = 5000
Input resistance, RB = 1 = 1000
Input voltage,
Using the formula for voltage gain, we have
Voltage gain,
Output voltage,
Given,
Current gain, = 120
Change in base current, = 100 A
A.c current gain,
is the change in the collector current.
In n-p-n transistor circuit the collector current is 10 mA. If 90% of the electrons emitted reach to the collector, find the base current and emitter current.
|
A |
B |
|
|
0 |
0 |
1 |
|
0 |
1 |
1 |
|
1 |
0 |
1 |
|
1 |
1 |
0 |
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Y’ = A + B |
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Two semiconductor materials X and Y shown in the given figure, are made by doping germanium crystal with indium and arsenic respectively. The two are joined end to end and connected to a battery as shown.
(i) Will the junction be forward biased or reverse biased?
(ii) Sketch a V - I graph for this arrangement. 
Semiconductors in which the number of electrons ne is equal to the number of holes nh are intrinsic semiconductors.
When a small amount of suitable impurity is added to the intrinsic semiconductor we can convert it into an extrinsic semiconductor of either p-type or n-type. Doping changes the concentration of charge carriers in the element.
i) P- type: When Si or Ge is doped with any trivalent impurity like Al, B etc we get a p-type semiconductor. The dopant has one valence electron less. Thus, this atom form covalent bond with the neighbouring three atoms and, is less of one electron to offer to the fourth silicon atom and as a result there is a vacancy. Hence, an electron in the outer orbit of the neighbouring atom may jump to fill this vacancy leaving a vacancy or hole at it's own site. This hole is the conducting charge carrier. Therefore, doping with a trivalent impurity gives us p-type semiconductor.
Energy band diagram for a p-type semiconductor is as shown below: 
ii) n - type: When we dope Si or Ge with pentavalent impurities like As, P then four electrons of this atom will form covalent bonds with the neighbouring Si atom in the lattice. Whereas, the fifth electron will remain loosely bound to it's parent atom. Hence, the ionisation energy which is required to make this electron free is very less and it will move around even in room temperature. Thus, the pentavalent dopant will donate one extra electron for conduction and this acts as an n-type semiconductor.
Energy band diagram for n-type semiconductor is as shown below: 
The gate shown in the figure is OR gate. 
(i) When A = 0, and B = 0, both diodes are reverse biased, hence Y = 0.
(ii) When A = 0, and B = 1, diode D2 is forward biased, hence Y = 1.
(iii) When A = 1, and B = 0, diode D1 is forward biased, hence Y = 1.
(iv) When A = 1, and B = 1, diode D1 and D2 both are forward biased, hence Y = 1.
When applied voltage is such that n-side of the diode is connected to the negative terminal of the battery and p-side is connected to positive terminal of the battery, the applied voltage is opposite to the barrier potential. The direction of applied potential is opposite to that of the built-in potential.
Hence, the effective barrier potential becomes VB - V, and the energy barrier across the junction decreases. Thus, the junction width decreases. 
In the figure given,
(i) p-n junction is forward biased.
(ii) p-n junction is reverse biased.
(i) In this case, the p-side of the diode is at -10 V, whereas the n-side is at 0 V.
Here, Vp < VN, therefore, the diode is reverse biased.
(ii) In this case, the p-side of the diode is at 0 V, whereas the n-side is at -10 V.
Thus, Vp > VN, hence, the diode is forward biased.
When applied voltage is such that n-side is connected to the negative terminal of the battery and p-side is connected to positive terminal, the applied voltage is opposite to the barrier potential. Hence, the effective barrier potential becomes VB-V, and the energy barrier across the junction decreases. Thus, the junction width decreases. 
|
Time interval |
Input A
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Input B |
Output, Y = A + B |
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t < t1 |
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t1 < t < t2 |
1
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t2 < t < t3 |
1
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t3 < t < t4 |
0
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t4 < t < t5 |
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t5 < t < t6 |
1
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t > t 6 |
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Draw a circuit diagram for use of n-p-n transistor as an amplifier in common emitter configuration. The input resistance of a transistor is 1000 Ω. On changing its base current by 10 μA, the collector current increases by 2 mA. If a load resistance of 5 kΩ is used in the circuit, calculate:
(i) the current gain,
(ii) voltage gain of the amplifier.
Advantages of semiconductor devices:
(i) Low cost
(ii) No noise
Disadvantages of semiconductor:
(i) Cannot operate at high voltage
(ii) Cannot operate at high current.
Draw the energy band diagrams of p-type and n-type semiconductors.
A semiconductor has equal electron and hole concentration 6 x 108 m–3. On doping with a certain impurity, electron concentration increases to 8 x 1012 m–3. Identify the type of semiconductor after doping.
Given,
As, change in emitter current,
This implies,
Change in base current,
and
i.e.,
Given,
Voltage gain = 2800
Resistance gain = 3000
Therefore,
Current gain,
Given, a common emitter configuration.
Collector resistance, Rc = 4 k
Internal resistance, Ri = 1 k
Current across collector, Ic = 1 mA
Current across base, Ib = 20
Using the formula for current gain,
Therefore,
Voltage gain is,
(i) The given logic gate is NOT gate.
(ii) Truth table of NOT gate.
|
Input A |
Output |
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0 |
1 |
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1 |
0 |
The variation of current on the input side with input voltage (IB versus VBE) is known as input characteristics, while the variation in the output current with output voltage (Icversus VCE) is known as output characteristics.
Current amplification factor of the transistor () - The ratio of change in collector current to the change in base current when, the collector-emitter voltage (VCE) is constant. The transistor is in active state.
Therefore,
As seen in the graph, we can say that, Ic is increasing almost linearly with IB.
Insulators: In insulators, the valence band is completely filled, the conduction band is empty and energy gap is quite large (Eg> 3 eV). As a result of huge energy gap, no electron will jump from the valence band to the conduction band even if electric field is applied. The energy band diagram of insulator is shown in figure (ii). 
Thus, electrical conduction is not possible and hence these materials behave as insulators. eg. diamond ( Eg = 6 eV).
Semiconductors: In semiconductors, the valence band is completely filled and conduction band is empty but, energy gap between them is less than 3 eV. The energy band diagram of a semiconductor is shown in figure (c.) 
At 0 K, conduction band remains totally empty because the electrons in the valence band are not able to get excited to the higher level. However, at room temperature, some electrons in the valence band acquire thermal energy greater than energy band gap and, jump over to the conduction band where they are free to move under the influence of even a small electric field. As a result of it, the semiconductor acquires small conductivity at room temperature.
The resistance of semiconductor would not be as high as that of insulator.
Given,
Wavelength, λ = 6000 nm = 6 x 10–6 m
Energy band gap, Eg = 2.8 eV
Using the formula for energy of a photon,
we have,
As, the energy of the photon is less than energy band gap (Eg= 2.8 eV) of the semiconductor, so a wavelength of 6000 nm cannot be detected.
Working: The negative terminal of the battery is gounded and corresponds to the 0 state and the positive (i.e., voltage 5 V in the present case) to the 1 state.
Input and output waveform: 
- When both A and B are connected to 0, no current passes through the diode and therefore no voltage develops across R and the output is zero.
- When input A is connected to zero and B to 1, the diode D2 is forward biased and the current through it is limited by a current limiting resistance. This current causes a 5 V drop across the resistance assuming the diode to be ideal and this gives an output of 5 V or 1.
- Now, interchanging A and B to 1 and 0 will still give a 5 V drop across the resistance as D1 will conduct.
- When the terminals A and B are connected to 1, then both the diodes D1 and D2 conduct. However, the voltage drop across R cannot exceed 5 V and the output is 1.
For an n-p-n transistor in the common emitter configuration, draw a labelled circuit diagram of an arrangement for measuring the collector current as a function of collector emitter voltage for at least two different values of base current. Draw the shape of the curves obtained. Define the terms:
(i) ‘output resistance’ and (ii) ‘current amplification factor’ .
Given, a common emitter configuration.
Current gain, = 59
Emitter current, IE = 6 mA
Now, using the relation,
we have,
But,
and,
Base current,
| Intrinsic | Extrinsic |
| 1. It is a pure semiconductor and no impurities are added to it. | 1. It is prepared by doping a small quantity of impurity atoms to the pure semiconducting material. |
| 2. Number of free electrons in conduction band is equal to the number of holes in valence band. | 2. Number of free electrons and holes vary in n-type and p-type semiconductor. There is excess of electrons in n-type and excess of holes in p-type. |
| 3. It's electrical conductivity is low. | 3. Electrical conductivity is high. |
| 4. Electrical conductivity is solely a function of temperature. | 4. Electrical conductivity depends upon temperature as well as on the quantity of dopants. |
Verfication:
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Y = Ā + B |
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The given circuit diagram shows a transistor configuration along with its output characteristics. Identify
(i) the type of transistor used and
(ii) the transistor configuration employed.
Using the curve, find the gain at VCE = 3 V 
(i) The transistor used is n-p-n transistor.
(ii) The transistor configuration employed is common emitter (CE) configuration.
Numerical:
Current amplification factor is given by,
From the characteristic curve,
Here,
Breakdown electric field of the Zener diode, E = 106Vm–1.
Width of the depletion region, d =2.4 = 2.4 x 10–6m
∴ Reverse biased potential, Vbreakdown = E x d
= 106 x 2.4 x 10–6
= 2.4 V.
Given,
VCE = 2 V
Current amplification factor,
Consider characteristics for any two values of IB (say, 10 and 60 μA).
Then for VCE = 2 V from the graph,
we have,
Therefore,
For a CE-transistor amplifier, the audio signal voltage across the collector resistance of 2 k Ω is 2 V. If the current amplification factor of the transistor is 100, calculate (i) input signal voltage
(ii) base current. Given that the value of the base resistance is 1 kΩ.
(i) Given,
Collector resistance, Rc = 2 k
Audio signal voltage, Vo = 2 V
Current amplification factor, = 100
Now, using the formula,
Output voltage, Vo =
is the input signal voltage.
(ii) Now, current amplification factor,
Base current,
Assume that the silicon diode in the following circuit requires a minimum current of 1 mA to be above the knee point (0.7 V) of its I–V characteristics. Also assume that the voltage across the diodes is independent of the current above the knee point.
(a) If VB = 5 V, what should be the maximum value of R so that the voltage is above the knee point?
(b) If VB = 5 V, what should be the value of R to establish a current of 5mA in the circuit ?
(c) What is the power dissipated in the resistance R and in the diode, when a current of 5 mA flows in the circuit at VB = 6 V.
(d) If R = 1 kΩ, what is the minimum voltage VB required to keep the diode above the knee point?
(a) Here,
Minimum current required by the silicon diode = 1mA= 10–3 A.
Minimum voltage across the diode is 0.7 V (so that it is above the knee point of the characteristic curve).
So, voltage drop across the resistance R = (5 – 0.7) V = 4.3 V.
The minimum current i = 1 mA .
Maximum value of resistance,
(b) Current through resistance R = 5 mA = 5 x 10–3 A.
Voltage drop across R = (5 – 0.7) V = 4.3 V
(c) Here, VB = 6 V.
Voltage V' across the resistance = 6 – 0.7 = 5.3 V
Current, i = 5 mA = 5 x 10–3 A
Power dissipated through the resistance R, P = i V’
= 5 x 10–3 x 5.3 W
= 26.5 x 10–3 W
= 26.5 mW
Power dissipated in the diode is P’' = i x 0.7 W
= 5 x 10–3 x 7 x 10–1 W
= 3.5 x 10–3 W
= 3.5 mW
(d) For keeping diode above the knee point, the minimum current required is 1 mA.
Voltage drop across R (= 1 kΩ), VR = 1 x 103 x 103
= 1 V
Minimum voltage drop across the diode = 0.7 V
Therefore,
Minimum voltage required, VB = (1 + 0.7) V = 1.7 V.
From the characteristic curve of a diode, we have
D.C. resistance,
A.C. resistance,
The following figure shows the V-I characteristics of a semiconductor diode
(i) Identify the semiconductor diode used.
(ii) Draw the circuit diagram to obtain the given characteristic of this device.
iii) How the characteristic of zener diode can be used as a voltage regulator?
(i) The semiconductor diode whose V-I characteristic is shown in the figure is Zener diode.
(ii) Circuit diagram to obtain the given characteristic is shown in Fig. 
(iii) The circuit of Zener diode used as voltage regulator is shown in Fig. 
The voltage to be regulated is applied across Zener diode. When input voltage increases, the current in Zener diode circuit also increases. As a result, voltage drop across series resistance Rs increases but, voltage drop across RL remain same. That is, the voltage drop across Zener diode remains the same.
Similarly, when voltage decreases, current in the zener diode circuit decreases and voltage drop across series Rs resistance decreases but, across the load resistance remains the same. This is how, voltage is regulated using a zener diode.
B.
Diffusion in forward biased, drift in reverse bias.C.
The majority carriers in n-type semiconductors are holesFor a transistor amplifier the voltage gain
C.
is low at high and low frequencies and constant at mid frequenciesB.
if either or both inputs are 1A.
electrons move froms base to collectorDeduce an expression for the conductivity of a p-type semiconductor.
Consider a block of semiconductor of length l and area of cross-section A, having electron density ne and hole density nh. Let, V be the the potential difference applied across the ends of the semiconductor. The magnitude of the applied electric field is given by,
E = ....... ( 1)
Due to the applied electric field, both electrons and holes move in a mutually opposite direction with drift velocities ve and vh and contribute current Ie and Ih.
Total current, I = Ie + Ih
Electrons and holes in the conduction band and valence band respectivel are moving in a random fashion. Therefore,
Ie = ne A e ve
Ih = nh A e vh
Thus,
Total current, I = ne A e ve + nh A e vh
= eA (neve + nh vh)
... (2)
Let R be the resistance of the semiconductor and is it's resistivity.
Then,
... (3)
Now, dividing 1 by 3, we get
...(4)
Therefore, from (2) and (4), we get
Now, mobility of electrons is defined as the drift velocity per unit electric field. Drift velocity is zero when, no electric field is applied.
mobility of electrons, =
Similarly,
We know, electrical conductivity is the reciprocal of resistivity.
So, electrical conductivity,
For a p-type semiconductor,
nh >> ne and nh = NA where, NA is the number density of acceptor atoms.
Hence, conductivity of a p-type semiconductor is given by,
is the required result.
The gap which is seperationg the valence band and the conduction band is known as fermi level.
When an intrinsic semiconductor is doped with impurity it becomes an extrinsic semiconductor.
Now, the fermi level will have to move away from it's mid-gap position inorder to conserve the number of particles as well as to maintain the electrical charge neutrality.
In an n-type semiconductor, energy gap decreases. Fermi-level shifts towards the conduction bandwhere, higher number of electrons are available for conduction. In a p-type semiconductor, energy band increases. Fermi levels shifts closer to the valence band because, holes are the majority carriers.
The circuit diagram of a common-emitter amplifier is given below: 
In a CE amplifier, the phase difference between the input and output signals is 180o.
Common emitter amplifier is preferred over common base amplifier because the current gain in CE amplifier is very high.
In CB amplifier the current gain is less than 1.
A semiconductor has equal electron and hole concentration of 2 x 108/m3. On doping with a certain impurity, the hole concentration increases to 4 x 1010/m3.
(i) What type of semiconductor is obtained on doping?
(ii) Calculate the new electron concentration of the semiconductor.
(iii) How does the energy gap vary with doping?
10 kHZ
10 MHz
1 GHz
1000 GHz
A.
10 kHZ
10 kHz cannot be radiated because the size of antenna is large. 1 GHz and 1000 GHz will penetrate.D.
Space waves.Reason: Propagation of waves in ultra high frequency range 0.3 - 3 GHz is possible by space waves. Because these waves can reflect from ionosphere.C.
(i), (ii) and (iii) but not (iv)(c) Reason: Digital signals connot utilise decimal system which represents a continuous set of values.
(i) The amplitude modulated waveform is given by superimposition of modulating signal on carrier wave and is shown below : 
(ii) We have,
Am = 1 V , Ac = 2V .
Therefore, modulation index is given by,
The AM wave is given by (Ac + Am sin ωmt) cos ωct where,
Ac is the amplitude of carrier wave and,
Am is the amplitude of message signal wave.
The maximum amplitude is M1 = Ac + Am = 10 V
while,
Minimum amplitude is M2 = Ac - Am = 2 V
Hence, the modulation index is,
When the minimum aplitude is M2 = 0. Then, we have clearly μ = 1, irrespective of M1.
The received signal is given by,
A1 cos (ωc + ωm) t
Carrier signal is given by,
Ac cos ωc t
where,
The carrier Ac cos ωct is available at the receiving station.
On multiplying the two signals, we get
If this signal is passed through a low-pass filter, it will pass through the high frequency signals and block the low frequency signals.
Thus, modulating signal, is given by
of a single frequency .
The three basic units of communication system are:
(i) Transmitter: It convert the message signal produced by the source of information into a form suitable for transmission.
(ii) Communication channel: The signal is passed through the channel for fulther processing of the information.
(iii) Receiver: It operates on the received signal in such a way that it is recognisable by the user of information.
Modulation is necessary for:
i) transmitting a low frequency signal to a distant place.
ii) avoiding mixing up of signals from different transmitters.
iii) so as to keep the height of antenna small.
iv) Modulation helps us to filter out the unwanted noises present in the signal and process out a more clearer form of information.
What is the length of the dipole antenna to transmit signals of frequency 200 MHz?
The range of transmission of signals by a TV tower can be increased by the following ways:
(i) By increasing the height of the transmission tower.
(ii) By increasing the height of the receiving antenna, so that it may directly intercept the signal from the transmitting antenna.
Frequency of the carrier wave, = 3 108 Hz
Wavelength,
The size of the antenna should be of the wavelength in case of a dipole antenna.
i.e., Length of the antenna = m
Two advantages of optical fibre:
(i) High bandwidth.
(ii) High data transmission capacity.
For frequency modulation, deviation in frequency is given by,
δ= v - vc
where, vc is constant frequency of carrier wave and v is instantaneous frequency of the FM wave at any time t.
(i) Transducer: It is a device which converts one form of energy into another. It converts physical variable like pressure, displacement, force, temperature etc. into it's corresponding electrical signal at it's output.
(ii) Repeater: A repeater is a combination of receiver and transmitter placed along the path of signal so as to extend the range of the communication system. Repeater picks up the signal from transmitter, amplifies it and re-transmits it to the receiver with a change in carrier frequency sometimes.
Mode of radio wave propagation by space waves, in which the wave travels in a straight line from transmitting antenna to the receiving antenna, is called line-of-sight (LOS) communication.
Two types of waves used for LOS communication are space wave and ground wave.
At frequencies above 40 MHz, LOS communication is essentially limited to line-of-sight paths.
The diagram below shows various communication systems that use space wave mode of propagation.
(i) AM Detector: The modulated wave is fed into a circuit for demodulation purpose. This circuit is called detector.
(ii) Envelope Detector: A diode followed by a suitable filter circuit is used for the detection of AM waves. The half wave rectification is done by the diode.
After these steps the desired original modulating signal is displayed at the output.
Amplitude modulation:
- The modulation technique is comparitively simple and instruments are cheap.
- The modulated signal's amplitude varies as per the modulating signal.
Frequency modulation:
- Modulation instruments and techniques are more costly and complex than AM.
- The modulated signal's frequency varies as per the modulating signal.
- It is less noisy and gives better quality transmission and has a larger bandwidth.
The noise in any signal is related to the amplitude variation of the signal emitted by a source. In FM, only frequency changes and amplitude remains constant. Therefore, FM signal is less susceptible to noise than an AM signal.
Explain, why high frequency carrier waves are needed for effective transmission of signals.
A message signal of 12 kHz and peak voltage 20 V is used to modulate a carrier wave of frequency 12 MHz and peak voltage 30 V. Calculate the (i) modulation index and (ii) side-band frequencies.
Optical detector is a device at the receiver end, which converts light into electrical signal so that the transmitted information may be decoded.
Characteristics of optical detector:
- Size is compatible with that of the optical fibre.
- High sensitivity at the desired optical wave-length.
- High response time for fast-speed data transmission/reception.
The efficiency of generating electron-hole pairs in a photodiode decides the quality of detector.
The parameters of carrier wave that vary in accordance with the modulating or message signal, in case of a sinusoidal wave are:
(a) Amplitude
(b) Frequency
(c) Phase.
And, in the case of pulsed wave the parameters varying are:
(a) Amplitude of pulse
(b) Width of pulse
(c) Position of pulse.
Frequency above 20 MHz is carried out with the help of satellites because
a) Higher frequencies being greater than critical frequency cannot be reflected by the ionosphere.
(b) The co-axial cables and wires are extremely radiating in nature at this high frequency which causes disruption in transmission.
Given,
Height of the ground transmitter = 300 m
Range of transmission = 100 km
Maximum distance covered by space wave propagation is,
As receiver-transmitter distance is 100 km, so space wave propagation is not possible for both 5 MHz and 100 MHz waves.
Critical frequency for ionospheric propagation is,
As frequency (5 MHz) of signal is less than the critical frequency, hence, this signal comes via ionospheric mode while signal of frequency 100 MHz is transmitted through the satellite mode.
Given, Eg = 0.73 eV
Energy of a photon is given by,
Therefore,
What is the total modulation index when a carrier wave is modulated by three sine waves for which modulation μ1 = 0.41, μ2 = 0.51 and μ3 = 0.45 respectively?
Given,
modulation index, μ1 = 0.41, μ2 = 0.51 and μ3 = 0.45
Total modulation index ,
Given,
Frequency of signal wave, s = 11.5 kHz
Frequency of carrier wave, = 3.45 MHz = 3450 kHz
Therefore,
The ability of a radio receiver to reject the unwanted noises and receive the desired signals is known as selectivity.
The ability of a receiver to amplify the desired weak signal is known as sensitivity.
Height of tv tower, h = 150 m
Height of the tower,
Increase in height of tower = 600 - 150 = 450 m.
Given,
We have,
is the required percentage modulation.
Electromagnetic waves are propagated through the following modes:
(i) ground wave propagation
(ii) sky wave propagation, and
(iii) space wave propagation.
For transmission of signal having frequency range from a few MHz up to 40 MHz, sky wave propagation mode is used.
Sky wave propagation:
The mode of wave propagation in which the radiowaves emitted from the transmitter antenna reach the receiving antenna after reflection by the ionosphere. 
An audio signal of frequency v0 is to be transmitted as an electromagnetic wave
(i) directly as such
(ii) through its use as the modulating signal on a carrier wave of frequency vc.
State the ratio of the size of the transmitting antenna, that can properly sense the time variation of the signal, in terms of v0 and vc.
Filter
Rectifier
FET
Oscillator
D.
Oscillator
Which of the following is not transducer?
B.
AmplifierD.
helps to guide the light signals through the coreB.
the frequency of modulated wave varies as amplitude of modulating waveD.
the transmitting length is very large and impracticableD.
Modulation and DemodulationB.
Photo diode in reverse biasDevice X is transducer and device Y is repeater.
In amplitude modulation, the amplitude of the carrier wave is varied in accordance with the amplitude of the audio frequency modulating signal.
In this case, the frequency of the amplitude modulated wave remains the same as that of the carrier wave. 
Limitation of amplitude modulation:
1. The amplitude of modulated signal obtaind can be transmitted only by amplifying it first and then fed to an antenna of appropriate size for radiation.
The size of the antenna matters.
The best mode of communication channel in this case is a twisted pair cable.
Twisted pair cables consists of pairs of plastic-coated copper wires that are twisted together.
Range is given by,
Given that height increase = 21%
A small bulb is placed at the bottom of a tank containing water to a depth of 80cm. What is the area of the surface of water through which light from the bulb can emerge out? Refractive index of water is 1.33. (Consider the bulb to be a point source.)
Here, the source of light (S) is 80 cm below the surface of water.
i.e., SO = 80 cm = 0.8 m
When angle of incidence,
Therefore, Area of the surface of water through which light from the bulb can emerge is area of the circle of radius,
The graph shown in the figure represents a plot of current versus voltage for a given semiconductor.
Identify the region, if any, over which the semiconductor has a negative resistance.
The part of the curve BC has a negative slope because with increase in current, voltage decreases.
Define the activity of a given radioactive substance. Write its S.I. unit.
The rate of decay of a radioactive substance is called activity of that substance. Activity is the negative of the rate of decay of the radioactive substance.
SI unit is Becquerel (Bq.)
Write the expression for the de Broglie wavelength associated with a charged particle having charge ‘q’ and mass ‘m’, when it is accelerated by a potential V.
De-broglie wavelength for a charged particle is given by,
Draw typical output characteristics of an n-p-n transistor in CE configuration. Show how these characteristics can be used to determine output resistance.
The output characteristics of an n-p-n transistor in CE configuration are shown below:
Output resistance is defined as the ratio of change in collector-emitter voltage to change in collector current at a constant base current. The reciprocal of the slope of the linear part of the output characteristics represents the output resistance.
A parallel beam of light of 500 nm falls on a narrow slit and the resulting diffraction pattern is observed on a screen 1 m away. It is observed that the first minimum is at a distance of 2.5 mm from the centre of the screen. Calculate the width of the slit.
Distance of nth minima from the centre of screen is given by, 
a is the width of the slit.
For first minimum, n=1
So, width of the slit is given by, 
A convex lens of focal length f1 is kept in contact with a concave lens of focal length f2. Find the focal length of the combination.
Focal length of convex lens = +f1
Now, using lens formula, we have
... (1)
For concave lens of formula, focal length is (-f2)
... (2)
Now, adding equations (1) and (2), we have
... (3)
Now, for the combined lens using the lens formula, we have,
... (4)
Therefore, from equations (3) and (4), we have
This is the required focal length of the combination of lenses.
In the block diagram of a simple modulator for obtaining an AM signal, shown in the figure, identify the boxes A and B. Write their functions. 
In the block diagram,
A is the square law device- This is a non-linear device and produces a non-linear output of message and carrier signals.
B is the band pass filter and it rejects high and low frequencies. It allows only a band of frequencies to pass through.
In the circuit shown in the figure, identify the equivalent gate of the circuit and make its truth table. 
A and B act as NOT gate which is then AND’ed.
The equivalent of these combination of gates is the OR gate.
Truth table of the combination is given by, 
Truth table of OR gate is as shown below:
Draw V – I characteristics of a p–n junction diode. Answer the following questions, giving reasons:
(i) Why is the current under reverse bias almost independent of the applied potential up to a critical voltage?
(ii) Why does the reverse current show a sudden increase at the critical voltage?
Name any semiconductor device which operates under the reverse bias in the breakdown region. 
V-I characteristics of p-n junction diode are as shown on the right side:
i) The current of order in reverse biasing is due to the drifting of minority charge carriers from one region to another through the junction. A small amount of applied voltage is sufficient to sweep the minority charge carriers through the junction. So reverse current is almost independent of critical voltage.
ii) During the critical voltage or most commonly called breakdown voltage, enormous covalent bonds break. As a result, large number of charge carriers increase. Therefore, current increases at critical voltage.
Zener diode operates under the reverse bias in the breakdown region.
Draw a labelled ray diagram of a refracting telescope. Define its magnifying power and write the expression for it.
Write two important limitations of a refracting telescope over a reflecting type telescope.
The labelled diagram of a refracting telescope is as shown below:
Magnification power is defined as the ratio of the angle () subtended by the final image on the eye to the angle () subtended by the object on eye.
The mathematical formula is given by,
Also, it is given by,
Limitations of refracting telescope over a reflecting telescope are:
(i) Image is not free from chromatic aberration and spherical aberration.
(ii) Aperture of the objective lens should be large for high resolving power.
Write Einstein’s photoelectric equation and point out any two characteristic properties of photons on which this equation is based. Briefly explain the three observed features which can be explained by this equation.
Einstein’s photoelectric equation is given by,
Characteristic property of photons:
i) Energy of photon is directly proportional to the frequency.
ii) Total energy and momentum of the system of two constituent particles remain constant in photon-electron collision.
Three observed features of photoelectric effect are:
i) When frequency of incident photon increases, kinetic energy of emitted electron increases. Kinetic energy does not have any effect on Intensity of radiations.
ii) When intensity of incident light increases, the number of incident photons increases. The increase in intensity will increase the number of ejected electrons. That is, photocurrent will increase with increase of intensity. Frequency has no effect on photocurrent.
iii) When the energy of incident photon is greater than work function, the photoelectron is immediately ejected. Thus, there is no time lag between incidence of light and emission of photoelectrons.
Name the type of waves which are used for line of sight (LOS) communication. What is the range of their frequencies?
A transmitting antenna at the top of a tower has a height of 20 m and the height of the receiving antenna is 45 m. Calculate the maximum distance between them for satisfactory communication in LOS mode. (Radius of the Earth = 6.4 × 106 m)
Types of waves are:
i) space waves
ii) radio waves
iii) microwave
The frequency range for LOS communication is 40 MHz.
Given,
Height of the transmitting antenna, h = 20 m
Height of receiving antenna = 45 m
So, maximum distance between transmitting antenna and receiving antenna is given by,
(a) What is linearly polarized light? Describe briefly using a diagram how sunlight is polarised.
(b) Unpolarized light is incident on a Polaroid. How would the intensity of transmitted light change when the polaroid is rotated? 
a) When an unpolarized light is transmitted through a polarizer, only one component is along the direction of propagation. This transmitted light is called as linearly polarized light.
The incident sunlight is unpolarized. Molecule in air acts as a dipole radiator. When the sunlight falls on a molecule, dipole molecule does not scatter energy along the dipole axis; however the electric field vector of light wave vibrates just in one direction perpendicular to the direction of the propagation. The light wave having direction of electric field vector in a plane is said to be linearly polarized.
b) When unpolarized light is incident on a Polaroid, the transmitted light has electric vibrations in the plane consisting of Polaroid axis and direction of wave propagation.
The plane of polarization will change when the Polaroid is rotated. However, the intensity of transmitted light will remain unchanged.
One day Chetan’s mother developed a severe stomach ache all of a sudden. She was rushed to the doctor who suggested for an immediate endoscopy test and gave an estimate of expenditure for the same. Chetan immediately contacted his class teacher and shared the information with her. The class teacher arranged for the money and rushed to the hospital. On realizing that Chetan belonged to a below average income group family, even the doctor offered concession for the test fee. The test was conducted successfully.
Answer the following questions based on the above information:
(a) Which principle in optics is made use of in endoscopy?
(b) Briefly explain the values reflected in the action taken by the teacher.
(c) In what way do you appreciate the response of the doctor on the given situation?
a) The principle of total internal reflection is made use of in endoscopy. If a light ray enters one end of an optic fiber coated with a material of low refractive index, it refracted and strikes the walls at angle greater than critical angle. Thus light rays show multiple reflections, without being absorbed at the side walls.
b) The values reflected from the character of teacher are caring nature, a helping attitude towards the kids. Being hospitable towards others makes one a better being and makes the society also a better place to live in.
c) Being sympathetic towards Chetan and his family’s condition, the doctor gave a reduction in the fee, which is highly appreciable. Such professional ethics of doctor in the society would be of immense help to a person belonging to average income groups.
Using Bohr’s postulates, derive the expression for the frequency of radiation emitted when electron in hydrogen atom undergoes transition from higher energy state (quantum number ni ) to the lower state, (nf ).
When electron in hydrogen atom jumps from energy state ni =4 to nf =3, 2, 1, identify the spectral series to which the emission lines belong.
According to Bohr’s postulates, in a hydrogen atom, a single electron revolves around a nucleus of charge +e. For an electron moving with a uniform speed in a circular orbit on a given radius, the centripetal force is provided by the Coulomb force of attraction between the electron and the nucleus.
Therefore, 
... (1) 
So, Kinetic Energy, K.E = 
Potential energy is given by, P.E = 
Therefore, total energy is given by, E = K.E + P.E = 
E = 
, is the total energy.
For nth orbit, E can be written as En,
... (2)
Now, using Bohr's postulate for quantization of angular momentum, we have
Putting this value of v in equation (1), we get
Now, putting value of rn in equation (2), we get
R is the rydberg constant.
For hydrogen atom Z =1,
If ni and nf are the quantum numbers of initial and final states and Ei & Ef are energies of electron in H-atom in initial and final state, we have 
That is, when electron jumps from ni = 4 to nf = 3.21 .
Radiation belongs to Paschen, Balmer and Lyman series.
(a) Draw the plot of binding energy per nucleon (BE/A) as a function of mass number A. Write two important conclusions that can be drawn regarding the nature of nuclear force.
(b) Use this graph to explain the release of energy in both the processes of nuclear fusion and fission.
(c) Write the basic nuclear process of neutron undergoing –decay. Why is the detection of neutrinos found very difficult?
Graphical representation of (BE/A) for nucleons with mass number A.
The variation of binding energy per nucleon VS. mass number is shown in the figure:
Characteristics of Nuclear force:
(i) Nuclear forces non-central and short ranged force.
(ii) Nuclear forces between proton-neutron and neutron-neutron are strong and attractive in nature.
b) When a heavy nucleus (A > 235 say) breaks into two lighter nuclei (nuclear fission), the binding energy per nucleon increases i.e, nucleons get more tightly bound. This implies that energy would be released in nuclear fission.
When two very light nuclei (A £10) join to form a heavy nucleus, the binding is energy per nucleon of fused heavier nucleus more than the binding energy per nucleon of lighter nuclei, so again energy would be released in nuclear fusion.
c) During the decay process of neutron, we have
Neutrinos show weak interaction with other particles. Hence, its detection is very different.
Here, a concave lens is placed in a medium of refractive index greater than the refractive index of the material of the lens.
So, the given lens will behave as a diverging lens.
When the audio frequency modulating signal wave is superimposed on a high frequency wave called carrier wave, the modulation process produce frequencies that are the sum and difference of carrier and modulating frequencies. These frequencies are side bands.
Lower side band frequency = 
Upper side band frequency = 
where,
is the carrier wave frequency and
is the modulating/audio signal frequency.
A proton and an α-particle have the same de-Broglie wavelength. Determine the ratio of
(i) their accelerating potentials
(ii) their speeds.
De-broglie wavelength of the particle is given by, 
;
where, V= Accelerating potential and v is the speed of the particle.
Given that, the de-broglie wavelength is same for both proton and a-particle.
Charge on 
particle = 2 qp;
Mass of 
-particle = 4 mp
Charge on proton= qp ;
Mass of proton = mp 
2 : 1 is the required ratio of the accelerating potential.
Also, 
4 : 1 is the required ratio of the speed of proton to speed of alpha-particle.
Show that the radius of the orbit in hydrogen atom varies as , where n is the principal quantum number of the atom.
The electrostatic force of attraction between nucleus and the electron is,
According to Bohr’s theory, a hydrogen atom consists of a nucleus with a positive charge e and a single electron of charge –e, which revolves around it in circular orbit of radius r.
To keep electron bound to its orbit, centripetal force on the electron becomes equal to the electrostatic attraction. Therefore, 
where,
m is the mass of electron, and
v is the speed of an orbit of radius r.
Now, using Bohr’s quantisation condition for angular momentum,
where, n is the principal quantum number.
|
Intrinsic semiconductor |
Extrinsic semiconductor |
|
1. It is a pure semiconductor material with no impurity atoms in it. |
1. The pure semiconductors doped with tetravalent or pentavalent impurity atoms are called extrinsic semiconductors. |
|
2. Their electrical conductivity is low. |
2. Their electrical conductivity is high. |
|
3. Number of free electrons in the conduction band is equal to the number of holes in the valence band. |
3. Number of holes is never equal to the number of electrons. No. of electrons is greater in n-type semiconductor and number of holes is greater in p-type semiconductor. |
Use the mirror equation to show that an object placed between f and 2f of a concave mirror produces a real image beyond 2f?
Mirror formula is given by,
where,
u is the distance of the object from the mirror,
v is the distance of image from the mirror, and
f is the focal-length of the mirror.
For a concave mirror, f < 0.
Object distance is negative for an object on the left side. i.e., u< 0.
Given, f < u <2f.
Subtracting throughout from 
, we get
Therefore, the image formed is negative and is on the left side.
Also, the inequality implies, 2f > v. i.e., |2f| < |v|.
That is, the real image is formed beyond 2f.
Let us consider three polaroids A, B and C where C is kept in between A and B.
Using Malu’s Law, the intensity of light emerging from the middle polaroid C is given by,
Therefore, intensity falls on the polaroid at the end (polaroid B) whose polarisation axis makes an angle of (
with the polarisation axis of the angle of middle polaroid.
Therefore, Intensity of light emerging from Polaroid B is, 
That is, transmitted intensity will be maximum when sin 2 
= 1 
(i) A giant refracting telescope has an objective lens of focal length 15 m. If an eye piece of focal length 1.0 cm is used, what is the angular magnification of the telescope?
(ii) If this telescope is used to view the moon, what is the diameter of the image of the moon formed by the objective lens ? The diameter of the moon is 3.48 × 106 m and the radius of lunar orbit is 3.8 × 108 m.
where,
fo is the focal length of the objective lens, and
fe is the focal length of the eye-piece.
ii)
Given, diameter of the moon = 3.48 × 106 m
Radius of the lunar orbit = 3.8 × 108 m
Diameter of the image of moon formed by the objective lens is given by, d = 
Write Einstein’s photoelectric equation and mention which important features in photoelectric effect can be explained with the help of this equation. The maximum kinetic energy of the photoelectrons gets doubled when the wavelength of light incident on the surface changes from 
to 
. Derive the expressions for the threshold wavelength 
and work function for the metal surface.
Einstein’s photoelectric equation is given as: 
where, h
is the energy of the photon,
, is the work function of the metal,
is the maximum kinetic energy of the emitted photoelectron.
Maximum kinetic energy is given by,
;
where,
is the threshold frequency of the metal surface.
Features of the equation of photo-electric effect are:
a) When a photon of suitable radiation falls on the metal surface, a photoelectron is ejected. If the intensity of the light is increased number of photons incident on the metal surface increases and hence, the number of photo-electrons ejected also increases. This implies, photo current is proportional to intensity and radiation.
b) If frequency is less than the threshold frequency, i.e., the, photoelectric emission does not take place for incident radiation.
c) When frequency of the incident light is greater than the threshold frequency, maximum K.E increases. This implies, maximum K.E of photoelectrons depends upon the frequency of incident light.
Einstein’s equation corresponding to wavelength 
Einstein’s equation corresponding to wavelength 
is, 
From equation (1) and (2), we have
where, 
is the threshold wavelength.
In the study of Geiger-Marsdon experiment on scattering of 
particles by a thin foil of gold, draw the trajectory of -particles in the coulomb field of target nucleus. Explain briefly how one gets the information on the size of the nucleus from this study. From the relation 
where Ro is constant and A is the mass number of the nucleus, show that nuclear matter density is independent of A.
Consider an alpha-particle with initial K.E = 
directed towards the center of nucleus of an atom.
The force that exists between nucleus and α-particle is Coulomb’s repulsive force. On account of this force, at the distance of closest approach 
, the particle stops and cannot go further closer to the nucleus. And, K.E gets converted to P.E.
That is, 
Hence, we can see that the size of the nucleus is approximately equal to the distance of closest approach, ro .
Now,
If ‘m’ is the average mass of the nucleon and r the radius of nucleus, then
Mass of nucleus = mA; where A is the mass number of the element.
Volume of the nucleus, 
This implies, 
Therefore, the nuclear density is independent of mass number A.
Distinguish between nuclear fission and fusion. Show how in both these processes energy is released.
Calculate the energy release in MeV in the deuterium-tritium fusion reaction:
|
Nuclear Fission |
Nuclear Fusion |
|
The process of breaking a heavy nucleus into two or more lighter nuclei. |
The phenomenon of combining or fusing two or more lighter nuclei to form a single nucleus. |
 |
 |
The given equation is, 
Mass of the reactant, 
= (2.014102 + 3.016049) u
= 5.030151u
Mass of the product, 
= (4.002603 + 1.008665) u
= 5.011268 u
Difference in energy, 
= (5.030151 – 5.011268) u
= 0.018883u
Therefore, Energy released, E = 
= 0.018883 × 931.5 MeV
= 17.589514 MeV
Draw a block diagram of a detector for AM signal and show, using necessary processes and the waveforms, how the original message signal is detected from the input AM waves.
Detection is the process of recovering modulating signal from the modulated carrier wave.
On receiving the message, the modulated signal is passed through a rectifier and an envelope signal is produced. This envelope signal is the message signal. The message signal is retrieved by passing it through an envelope detector. This way, modulating signal is obtained.
With what considerations in view, a photodiode is fabricated? State it’s working with the help of a suitable diagram.
Even though the current in the forward bias is known to be more than in the reverse bias, yet the photodiode works in reverse bias. What is the reason?A photodiode is used to observe the variation in current when the light intensity changes under reverse bias condition below the breakdown voltage.
A material of band gap ~1.5 eV or lower is chosen so that solar conversion efficiency is better. Hence, Ga-As or Si material is chosen for the fabrication of photo-diode.
Working: It is a p – n junction fabricated with a transparent window to allow light photons to fall on it. Upon absorption, these photons generate electron hole pairs. If the junction is reverse biased using an electrical circuit, these electron hole pair move in opposite directions so as to produce current in the circuit. This current is very small and is detected by the micro-ammeter connected to the circuit. 
The change in reverse saturation current is directly proportional to the change in incident light intensity when photo-diode is operated in reverse condition. This is not the case when the photo-diode is forward biased.
Draw a circuit diagram of a transistor amplifier in CE configuration.
Define the terms: (i) Input resistance and (ii) Current amplification factor. How are these determined using typical input and output characteristics?The circuit diagram of a transistor amplifier in CE configuration is given by, 
Input resistance-:
a)The ratio of change in base-emitter voltage to base current gives us the input resistance. We can calculate change in VBE and change in IB from the input characteristics. To find the input resistance, mark a point P on the input characteristic. Draw a tangent at point P. The reciprocal of slope AB will give us the input resistance. 
b) Current amplification factor: It is defined as the ratio of change in collector current to the change in base current. Output characteristics will help to calculate the current amplification factor.
Answer the following questions:
(a) In a double slit experiment using light of wavelength 600 nm, the angular width of the fringe formed on a distant screen is 0.1°. Find the spacing between the two slits.
(b) Light of wavelength 5000 Å propagating in air gets partly reflected from the surface of water. How will the wavelengths and frequencies of the reflected and refracted light be affected?Wavelength of light, 
Angular width of the fringe, 
Using the formula, 
Spacing between the slits, 
b) Wavelength of light, 
Frequency of reflected and refracted light is going to be the same.
Therefore,
Frequency, 
This is the required frequency of both refracted and reflected light.
Refractive index of water, 
Therefore, wavelength of the refracted light,
(a) Using Huygens’s construction of secondary wavelets explain how a diffraction pattern is obtained on a screen due to a narrow slit on which a monochromatic beam of light is incident normally.
(b) Show that the angular width of the first diffraction fringe is half that of the central fringe.
(c) Explain why the maxima at 
become weaker and weaker with increasing n.
a) Diffraction of light at a Single slit
A single narrow slit is illuminated by a monochromatic source of light. The diffraction pattern is obtained on the screen placed in front of the slits. There is a central bright region called as central maximum. All the waves reaching this region are in phase hence the intensity is maximum. On both side of central maximum, there are alternate dark and bright regions, the intensity becoming weaker away from the center. The intensity at any point P on the screen depends on the path difference between the waves arising from different parts of the wave-front at the slit.
According to the figure, the path difference (BP – AP) between the two edges of the slit can be calculated.
Path difference, BP – AP = NQ = 
Angle 
is zero at the central point C on the screen, therefore all path difference are zero and hence all the parts of slit are in phase.
Due to this, the intensity at C is maximum. If this path difference is, (the wavelength of light used), then P will be point of minimum intensity.
This is because the whole wave-front can be considered to be divided into two equal halves CA and CB and if the path difference between the secondary waves from A and B is 
, then the path difference between the secondary waves from A and C reaching P will be 
/2, and path difference between the secondary waves from B and C reaching P will again be 
/2. Also, for every point in the upper half AC, there is a corresponding point in the lower half CB for which the path difference between the secondary waves, reaching P is 
/2. Thus, destructive interference takes place at P and therefore, P is a point of first secondary minimum.
b)
Central bright lies between, 
Therefore, Angular width of central bright fringe = 
So, 1st diffraction fringe lies between 
Therefore,
Angular width of first diffraction fringe is, 
Hence proved.
c) As n increases, part of the slit contributing towards maximum decreases. Therefore, maxima gets weaker with increasing n.
A point object 'O' is kept in a medium of refractive index n1 in front of a convex spherical surface of radius of curvature R which separates the second medium of refractive index n2 from the first one, as shown in the figure.
Draw the ray diagram showing the image formation and deduce the relationship between the object distance and the image distance in terms of n1, n2 and R.
b) When the image formed above acts as a virtual object for a concave spherical surface separating the medium n2 from n1 (n2 > n1), draw this ray diagram and write the similar (similar to (a)) relation. Hence obtain the expression for the lens maker’s formula.
Ray diagram showing the image formation is given below:
The rays are incident from medium of refractive index n1 to another medium of refractive index n2.
NM is taken to be nearly equal to the length of the perpendicular from the point N on the principal axis.
We can have approximation for small angles. We get, 
b)
When the image formed is virtual and n2 > n1
A convex lens is placed in contact with a plane mirror. A point object at a distance of 20 cm on the axis of this combination has its image coinciding with itself. What is the focal length of the lens?
Given, convex lens placed in contact with a plane mirror.
Image of the object coincides with the object.
So, the rays refracted from the first lens and then reflected by the plane mirror will be retracing their path. This would happen when rays refracted by the convex lens fall normally on the mirror i.e., the refracted rays form a beam parallel to principal axis of the lens.
Hence, the object is at the focus of the convex lens.
Therefore, focal length, f = 20 cm
The figure given below shows the block diagram of a generalized communication system. Identify the element labelled 'X' and write its function.
The element labelled 'X' is called 'channel'.
Channel connects transmitter and receiver. The signal from the transmitter is carried to the receiver by the communication channel.
For a single slit of width "a", the first minimum of the interference pattern of a monochromatic light of wavelength λ occurs at an angle of 
. At the same angle of 
, we get a maximum for two narrow slits separated by a distance "a". Explain.
Width of the slit = a
The path difference between two secondary wavelets is given by,
N λ = a sin θ.
Since, θ is very small sin θ = θ.
So, for the first order diffraction n = 1, the angle is λ/a.
Now we know that θ must be very small θ = 0 (nearly) because of which the diffraction pattern is minimum.
Now for interference case, for two interfering waves of intensity I1 and I2 we must have two slits separated by a distance.
We have the resultant intensity, I = 
Since, θ = 0 (nearly) corresponding to angle λ/a so cos θ = 1 (nearly)
So,
We see the resultant intensity is sum of the two intensities, so there is a maxima corresponding to the angle λ/a.
This is why at the same angle of λ/a we get a maximum for two narrow slits separated by a distance "a".
Write the truth table for the combination of the gates shown. Name the gates used. 
From the logic gates given above, we can say that
R is the OR GATE
S is the AND GATE
|
A |
B |
Y' = A+B |
Y = Y'.A |
|
0 |
0 |
0 |
0 |
|
0 |
1 |
1 |
0 |
|
1 |
0 |
1 |
1 |
|
1 |
1 |
1 |
1 |
Identify the logic gates marked 'P' and 'Q' in the given circuit. Write the truth table for the combination.
P is NAND GATE and Q is the OR GATE
Assume:
X'=
|
A |
B |
X'= |
X =B + |
|
0 |
0 |
1 |
1 |
|
0 |
1 |
1 |
1 |
|
1 |
0 |
1 |
1 |
|
1 |
1 |
0 |
1 |
A proton and a deuteron are accelerated through the same accelerating potential. Which one of the two has
(a) greater value of de-Broglie wavelength associated with it, and
(b) less momentum?
Give reasons to justify your answer.De-Broglie wavelength of a particle depends upon its mass and charge for same accelerating potential, such that
Wavelength is inversely proportional to q and m.
Now, for proton and deuteron, we have 
where, e is the charge of an electron and mp is the mass of proton.
Thus, de-Broglie wavelength associated with proton is 2√ times of the de-Broglie wavelength of deuteron and hence it is more.
(b) Momentum is inversely proportional to the wavelength.
Mathematically it is given by,
;
where, h = plank's constant.
Wavelength of a proton is more than that of deuteron thus the momentum of a proton is lesser than that of deuteron.
(i) Monochromatic light of frequency 6.0 × 1014 Hz is produced by a laser. The power emitted is 2.0 × 10−3 W. Estimate the number of photons emitted per second on an average by the source.
(ii) Draw a plot showing the variation of photoelectric current versus the intensity of incident radiation on a given photosensitive surface.
i) Given,
Frequency of light,
Number of photons emitted per second on an average is equal to Power per unit energy of one photon.
Energy of one photon is given by, Energy = hν.
Therefore,
Number of photons emitted per second =
= 
(ii) The plot below shows the variation of photoelectric current Vs. Intensity of incident radiation on a given photosensitive surface. The plot shows a linear proportion.
A 12.5 eV electron beam is used to bombard gaseous hydrogen at room temperature. Upto which energy levels the hydrogen atoms would be excited?
Calculate the wavelengths of the first member of Lyman and first member of Balmer series.Amount of energy required to excite the electron = 12.5 eV
Energy of the electron in the nth state of an atom = 
; Z is the atomic number of the atom. [Z=1 for hydrogen atom]
Energy required to excite an atom from the initial state (ni) to the final state (nf) = 
This energy must be equal to or less than the energy of the incident electron beam.
… (1)
i=1 for ground state of hydrogen atom.
Energy of the electron in the ground state = −13.6 eV
Now, putting this in equation (1),
Since, the state cannot be a fractional number we have nf = 3
Therefore, hydrogen atom would be excited up to 3rd energy level.
b) Rydberg formula is given by, 
; 
is the wavelength and R is the Rydberg constant.
R = 1.097373157 × 10 7 m-1
For the first member of Lyman series, i=1; f=2
So, 
For the first member of the Balmer series, i=2; f=3
So, 
When Sunita, a class XII student, came to know that her parents are planning to rent out the top floor of their house to a mobile company she protested. She tried hard to convince her parents that this move would be a health hazard.
Ultimately her parents agreed:
(1) In what way can the setting up of transmission tower by a mobile company in a residential colony prove to be injurious to health?
(2) By objecting to this move of her parents, what value did Sunita display?1) Harmful electromagnetic waves such as microwaves which can cause severe health hazards like tumor and cancer are emitted by the transmitting tower. Added to that, the transmitting antenna operates on very high power. Hence, the chances of someone getting burnt are high. Therefore, TV tower should not be set up in residential areas.
(2) By objecting to this move of her parents, Sunita has displayed awareness towards the health and environment of her society.
(3) Range of the transmitting antenna, d = 
; h is the height of the antenna and R is the radius of the earth.
Given,
Height of the antenna, h = 20 m
R = 6400 km = 64 ×105 m
Therefore,
Range, d = 
=1600 m
(a) Draw a labelled ray diagram showing the formation of a final image by a compound microscope at least distance of distinct vision.
(b) The total magnification produced by a compound microscope is 20. The magnification produced by the eye piece is 5. The microscope is focused on a certain object. The distance between the objective and eyepiece is observed to be 14 cm. If least distance of distinct vision is 20 cm, calculate the focal length of the objective and the eye piece.a) The fig. below shows the formation of image by a compound microscope at least distance of distinct vision.
where, AB = object, A'B' = image formed by objective and A''B'' = image formed by eyepiece.
fo = focal length of objective,
uo = object distance from objective
vo = image distance from objective
D = distance of least distinct vision
L = length of the microscope
(b) Total magnification for least distance of clear vision is given by,
... (1)
where,
L is the separation between the eyepiece and the objective
fo is the focal length of the objective,
fe is the focal length of the eyepiece,
D is the least distance for clear vision,
me is the magnification of the eyepiece,
mo is the magnification of the objective.
Here, m =20, L =14 cm, D = 20 cm, me = 5
Magnification for the eyepiece:
Now, putting the value of mo and me in equation (1), we get
We have,
;is the required focal length of the objective lens.
(a) A mobile phone lies along the principal axis of a concave mirror. Show, with the help of a suitable diagram, the formation of its image. Explain why magnification is not uniform. (b) Suppose the lower half of the concave mirror's reflecting surface is covered with an opaque material. What effect this will have on the image of the object? Explain.
(a) Formation of image by the concave mirror is shown in the figure below:
The magnification of each part of the mobile phone is not uniform. The part of the mobile phone at centre of curvature will form image of the same size.
The part that lies between C and F will form an enlarged image beyond C as shown in the fig. above.
Since, the nature of the image is not the same at all sections of the principal axis, magnification varies.
(b) If lower half of the concave mirror is covered with an opaque material, the size of the image will not vary. But, since the reflecting surface has been reduced, the intensity of the image will be less. That is, brightness of the image will decrease.
a) Deduce the expression, N = N0 e−λt, for the law of radioactive decay.
(b) (i) Write symbolically the process expressing the β+ decay of 
. Also write the basic nuclear process underlying this decay.
, an isotope or isobar?
As per the law of radioactive decay, we have
where,
N = Number of nuclei in the sample
ΔN = Amount of nuclei undergoing decay
Δt = Time taken for decay
So,
where λ is Decay constant or disintegration constant.
(b)
(i) The β+ decay for is given below:
A proton is converted into neutron if, the unstable nucleus has excess protons than required for stability.
In the process, a positron e+ (or a β+) and a neutrino ν are created and emitted from the nucleus.
p → n + β+ + ν
This process is called beta plus decay.
(ii) The nucleus so formed is an isobar of 
because the mass number is same, whereas the atomic numbers are different.
(a) (i) 'Two independent monochromatic sources of light cannot produce a sustained interference pattern'. Give reason.
(ii) Light waves each of amplitude "a" and frequency "ω", emanating from two coherent light sources superpose at a point. If the displacements due to these waves is given by y1 = a cos ωt and y2 = a cos(ωt + ϕ) where ϕ is the phase difference between the two, obtain the expression for the resultant intensity at the point.
(b) In Young's double slit experiment, using monochromatic light of wavelength λ, the intensity of light at a point on the screen where path difference is λ, is K units. Find out the intensity of light at a point where path difference is λ/3.
(a)
(i) The condition for the sustained interference is that both the sources must be coherent (i.e. they must have the same wavelength and the same frequency, and they must have the same phase or constant phase difference).
Two sources are monochromatic if they have the same frequency and wavelength. Since they are independent, i.e. they have different phases with irregular difference, they are not coherent sources.
ii) 
Let the displacement of the waves from the sources S1 and S2 at point P on the screen at any time t be given by:
y1 = a cos ωt
y2 = a cos (ωt + Φ)
where, Φ is the constant phase difference between the two waves.
By the superposition principle, the resultant displacement at point P is given by:
y = y1 + y2
y = a cos ωt + a cos (ωt + Φ)
=2 a[cos cos 
y = 2 acos 
cos 
... (i)
... (2)
Then, equation (i) becomes:
y = A cos (ωt+
Now, we have:
... (3)
The intensity of light is directly proportional to the square of the amplitude of the wave. The intensity of light at point P on the screen is given by:
I = 4 a2 cos2 (
) ... (4)
(b) Wavelength of monochromatic light = 
Path difference = 
.
So, phase difference,
Intensity of light = K units
Intensity is given by, I = 
When path difference is 
Intensity of light, I’=
= 
(a) How does one demonstrate, using a suitable diagram, that unpolarized light when passed through a Polaroid gets polarized?
(b) A beam of unpolarized light is incident on a glass-air interface. Show, using a suitable ray diagram, that light reflected from the interface is totally polarized, when μ = tan iB, where μ is the refractive index of glass with respect to air and iB is the Brewster's angle.
Polarization of light is referred to as restricting the vibration of light in a perpendicular direction perpendicular to the direction of propagation of wave.
The vibration of particles of light which is parallel to the axis of crystal passes through the Polaroid on passing an unpolarized light. All other vibrations are absorbed and that is why intensity of emerging light is reduced.
The plane of vibration here is ABCD, in which the vibrations of the polarized light is confined and the plane KLMN is called plane of polarization. KLMN is perpendicular to the plane of vibration.
Reflected light is totally polarized, when unpolarized light is incident on the glass-air interface at the Brewster angle iB. This is known as Brewster’s law.
The reflected component OB and refracted component OC are mutually perpendicular to each other, when light is incident at Brewster’s angle.
Now, using Snell’s law, we have
i = iB and r = (900 – iB)
Therefore,
(a) State briefly the processes involved in the formation of p-n junction explaining clearly how the depletion region is formed.
(b) Using the necessary circuit diagrams, show how the V-I characteristics of a p-n junction are obtained in
(i) Forward biasing
(ii) Reverse biasing
How these characteristics are made use in rectification?
(a) The n-type semi-conductor has more concentration of electrons than that of a holes and p-type semi-conductor has more concentration of holes. Holes diffuse from p-side to n-side whereas electrons diffuse from n-side to p-side due to difference in concentration of charge carriers.
An ionized donor is left behind on n-side when electron diffuses from n side to p-side. The ionized donor (+ ve charge) is immobile as it is bound by the surrounding atoms. Therefore, a layer of positive charge is developed on the n-side of the junction. Similarly, a layer of negative charge is developed on the p-side. Hence, a space-charge region is formed on either side of the junction, which has immobile ions and is devoid of any charge carrier, called as depletion layer or depletion region.
(b)
For a p-n junction diode under forward bias, p-side is connected to the positive terminal and n-side is connected to the negative terminal.
When voltage is applied, electrons in n-region and holes in the p-region moves towards the p-n junction. Hence, there is decrease in the width of the depletion region thereby, offering less resistance. Diffusion of majority carriers takes place in the junction giving rise to a forward current.
The V-I characteristic of p-n junction in forward bias is shown below:
(ii) When p-n junction diode is reverse biased, the positive terminal of battery is connected to n-side and negative terminal to p-side.
The barrier height increases and the width of depletion region also increase as a result of reverse biasing. There is no conduction across the junction because of the lack of majority charge carriers. After applying a high reverse biased voltage, few minority carriers cross the junction. Hence, a current flows in reverse direction which is known as the reverse current.
The V-I characteristic of p-n junction diode in reverse bias is shown below:
p-n junction can be used for rectification purpose. Its working is based on the fact that, resistance of junction becomes low when forward biased and R becomes high when reverse biased.
(a) Differentiate between three segments of a transistor on the basis of their size and level of doping.
(b) How is a transistor biased to be in active state?
(c) With the help of necessary circuit diagram, describe briefly how n-p-n transistor in CE configuration amplifies a small sinusoidal input voltage. Write the expression for the ac current gain.On the basis of size and level of doping,
(a)
Emitter (E) - Emitter is heavily doped and is the left hand side thick layer of the transistor.
Base (B) - It is the central thin layer of the transistor, which is lightly doped.
Collector (C) - It is the right hand side thick layer of the transistor, which is moderately doped.
(b)
There are two conditions for a transistor to be into an active region.
1. The input circuit should be forward biased by using a low voltage battery.
2. The output circuit should be reverse biased by using a high voltage battery.
(c)
n-p-n Transistor as an amplifier:
The input circuit is forward biased by using a low voltage battery. Hence, the resistance of the input circuit is small. The output circuit is reverses biased using a high voltage battery. Hence, the resistance of the output circuit is large.
The operating point is fixed in the middle of its active region.
The output is taken between the collector and the ground.
Applying Kirchhoff’s law to the output loop:
VCC = VCE + ICRC
If Vc is the collector voltage then,
Vc = VCE − ICRC ... (A)
When the input signal voltage is fed to the emitter base circuit, it will change the emitter voltage and hence to the emitter current, which in turn will change the collector current. Due to this the collector voltage VC will vary in accordance with relation (A). This variation in collector voltage appears as the amplified output.
VBB = VBE + IBRB
Here, vi ≠ O
Then, VBB + vi = VBE + IBRB + ΔIB (RB + ri)
; is the required expression for current gain.
Under what condition does a biconvex lens of glass having a certain refractive index act as a plane glass sheet when immersed in a liquid?
When immersed in a liquid, a bi-convex lens of glass will behave as a plane glass,
when,
; 
is the refractive index of liquid and is the refractive index of glass.
State de-Broglie hypothesis.
According to hypothesis of de Broglie "The atomic particles of matter moving with a given velocity, can display the wave like properties.
i.e., 
A ray of light, incident on an equilateral glass prism ( 
) moves parallel to the base line of the prism inside it. Find the angle of incidence for this ray.
Since the prism is equilateral in shape, 
Angle of prism, A = 60o
Angle of refraction, r = A/2 = 30o
Now, using Snell’s law, 
Distinguish between ‘Analog and Digital signals’.
Difference between analog and digital signals is:
|
Analog signal |
Digital signal |
|
They are continuous signals i.e., continuous variation of voltage or current. |
These are the signals which can take only discrete values. In digital signal, high means 1 and low means 0. |
|
E.g. Sound of a human |
E.g., Temperature |
Mention the function of any two of the following used in communication system:
(i) Transducer (ii) Repeater
(iii) Transmitter (iv) Band pass Filteri) Transducer: A device which converts energy from one form to another form. It is used in microphone for converting sound energy into electrical energy and vice-versa.
ii) Repeater: It is a combination of a receiver and a transmitter. It is used in transmission system to regenerate the signal by picking up a signal, amplifying it and then transmitting it to the receiver.
iii) Transmitter: A device which processes the incoming message signal so as to make it suitable for transmission through a channel and for its subsequent reception.
iv) Bandpass filter: A device which allows only signals of a certain frequency range and blocks the rest of the frequencies. It is used in an electronic filter.
The current in the forward bias is known to be more (~mA) than the current in the reverse bias (~ A). What is the reason, then, to operate the photodiode in reverse bias?
The photodiode works in reverse biased condition although the current produced is less. In reverse bias, the width of depletion layer increases thereby reducing the capacitance across the junction. Hence, the response time for absorption of photons increases. Thus, the sensitivity of the photo-diode is increased which is certainly desired.
(a) Using Bohr’s second postulate of quantization of orbital angular momentum show that the circumference of the electron in the nth orbital state in hydrogen atom is n times the de Broglie wavelength associated with it.
(b) The electron in hydrogen atom is initially in the third excited state.
What is the maximum number of spectral lines which can be emitted when it finally moves to the ground state?i) According to Bohr’s second postulate, we have 
But, as per De- broglie hypothesis
Therefore, 
; where is the de- broglie wavelength.
ii) Given, electron in the hydrogen atom is in the third excited state.
For third excited state, n = 4
For ground state n= 1
Now, total number of possible spectral lines is given by,
The transition states are as shown in the figure above.
(i) What characteristic property of nuclear force explains the constancy of binding energy per nucleon (BE/A) in the range of mass number ‘A’ lying 30 < A < 170?
(ii) Show that the density of nucleus over a wide range of nuclei is constant- independent of mass number A.i) The characteristic property of nuclear force that explains the constancy of binding energy per nucleon is the saturation or short range nature of nuclear forces.
In heavy nuclei, nuclear size > range of nuclear force.
ii) Using the formula for radius of the nucleon, we have
Let, m be the mass of nucleon.
Therefore,
Density, 
Thus, we can see that density is independent of mass number A.
Write any two factor which justify the need for modulating a signal.
Draw a diagram showing an amplitude modulated wave by superposing a modulating signal over a sinusoidal carrier wave.Two factors justifying the need for modulation are:
(i) Practical size of antenna.
(ii) To avoid mixing up of signals from different transmitters i.e., avoid fluctuation and reduce the noise of signal.
The figures shows how modulated wave is formed by superimposing modulating signal over a carrier wave.
Write Einstein’s photoelectric equation. State clearly how this equation is obtained using the photon picture of electromagnetic radiation.
Write the three salient features observed in photoelectric effect which can be explained using this equation.Einstein’s photo electric equation is given by,
where, h is planck’s constant, 
is the frequency and 
is the work function.
The energy (h
) carried by a photon of frequency is absorbed by electrons on the surface to
i) overcome the work function of metal .
ii) impart maximum K.E. to the emitted electron (Kmax.)
Salient features observed in photoelectric effect are:
i) The stopping potential and maximum kinetic energy of emitted electrons varies linearly with the frequency of incident radiation.
ii) Photoelectric emission is possible when, 
where, 
is the cut-off frequency.
a) Why are coherent sources necessary to produce a sustained interference pattern?
(b) In Young’s double slit experiment using monochromatic light of wavelength 
, the intensity of light at a point on the screen where path difference is 
is K units. Find out the intensity of light at a point where path difference is 
.
a)
Coherent sources have almost the same wavelength. Coherent sources are necessary to ensure that the positions of maxima and minima do not change with time. Thereby, producing a sustained interference pattern.
b)
Wavelength of monochromatic light = 
Path difference = 
So, phase difference, 
Intensity of light = K units
Intensity is given by, I 
When path difference is 
, phase difference becomes 
Intensity of light I’ = 
, is the new intensity of light.
Use Huygen’s principle to explain the formation of diffraction pattern due to a single slit illuminated by a monochromatic source of light.
When the width of slit is made double the original width, how this affect the size and intensity of the central diffraction band?According to Huygen's principle "The net effect at any point due to a number of wavelets is equal to sum total of contribution of all wavelets with proper phase difference.
The point O is maxima because contribution from each half of the slit S1S2 is in phase, i.e., the path difference is zero.
At point P,
i) If 
.
ii) If 
, then point P would be maxima. At the point of maxima, intensity would decrease.
The width of the central maxima is given by, 
where a is the size of the slit; D is the distance between the screen and the slit.
When the width of the slit is made double the original width, then the size of central maxima will be reduced to half and intensity will be four times.
Define magnifying power of a telescope. Write its expression.
A small telescope has an objective lens of focal length 150 cm and an eye piece of focal length 5 cm. If this telescope is used to view a 100 m high tower 3 km away, find the height of the final image when it is formed 25 cm away from the eye piece.
a) Magnifying power of telescope is the ratio of the angle subtended at the eye by the image to the angle subtended at the unaided eye by the object.
Mathematically, we can write
where, fo is the focal length of the objective, fe is the focal length of the eye-piece and D is the least distance of distinct vision.
b) Using, the lens equation for objective lens,
Hence, magnification due to the objective lens is given by,
Now, using lens formula for eye-piece, we get
Therefore, magnification due to eyepiece me = 
Hence, total magnification, 
So, size of final image = 
Draw a simple circuit of a CE transistor amplifier. Explain its working. Show that the voltage gain AV, of the amplifier is given by 
, where 
is the current gain, RL is the load resistance and ri is the input resistance of the transistor. What is the significance of the negative sign in the expression for the voltage gain?
Circuit diagram of CE transistor amplifier, 
When an ac input signal VI (to be amplified) is superimposed on the bias VBB, the output, which is measured between collector and ground, increases.
We first assume that Vi = 0.
Applying Kirchoff’s law to the output loop,
VCC = VCE + IC RL
Similarly, the input loop gives,
VBB = VBE + IB RB
When Vi is not zero, we have
VBE + Vi = VBE + IB RB + 
Change in IB causes changes in Ic
Hence, 
As, 
The change in VCE is the output voltage Vo is, 
The voltage gain of the amplifier is,
Negative sign in the expression shows the output voltage and input voltage have phase difference of 
.
Draw the circuit diagram of a full wave rectifier using p-n junction diode. Explain its working and show the output, input waveforms.
(b) Show the output waveforms (Y) for the following inputs A and B of
(i) OR gate
a)
b) Output waveforms for the following inputs A and B of OR gate and NAND gate.
Show graphically, the variation of the de-Broglie wavelength (
) with the potential (V) through which an electron is accelerated from rest.
Wavelength is inversely proportional to the potential. 
In a transistor, doping level in base is increased slightly. How will it affect (i) collector current and (ii) base current?
When doping level is increased in base,
i) collector current decreases
ii) base current increases slightly.
When monochromatic light travels from one medium to another its wavelength changes but frequency remains the same, Explain.
When monochromatic light travels from one medium to another it’s wavelength and speed both changes and the frequency remains unchanged.
Two convex lenses of same focal length but of aperture A1 and A2 (A2 < A1), are used as the objective lenses in two astronomical telescopes having identical eyepieces. What is the ratio of their resolving power? Which telescope will you prefer and why? Give reason.
Resolving power is given by, R = 
; where A is the aperture. 
Magnification of telescope is given by, m = 
m is same for both.
Inorder to view the finer details of the object we use telescope of high resolving power.
Therefore, telescope having convex lens aperture A1 is used
Draw the output waveform at X, using the given inputs A and B for the logic circuit shown below. Also, identify the logic operation performed by this circuit. 
From the waveform given, we can say that 
which is the Boolean expression of OR.
The output waveform for OR gate is given by, 
Name the semiconductor device that can be used to regulate an unregulated dc power supply. With the help of I-V characteristics of this device, explain its working principle.
Zener diode can be used to regulate an unregulated dc power supply. 
Working principle of Zener diode as per the V-I characteristics:
In reverse breakdown region or zener region, a very small change in voltage across the zener diode produces very large change in current through the circuit but the voltage across the zener remains constant.
Draw the transfer characteristic curve of a base biased transistor in CE configuration. Explain clearly how the active region of the V0 versus Vi curve in a transistor is used as an amplifier.
The transfer characteristic curve of a base biased transistor in CE configuration is shown below: 
In the active region, a small increase of input voltage (Vi)results in a large increase in IC. The gradual increase in current is a linear phenomenon. This results in an increase in the voltage drop across RC.
(i) Define modulation index.
(ii) Why is the amplitude of modulating signal kept less than the amplitude of carrier wave?i) The ratio of the amplitude of modulating signal (Em) to the amplitude of carrier wave (Ec) is known as the modulation index of an amplitude modulated wave or amplitude modulation index.
Mathematically it is given by, 
For a modulated wave, we have 
ii) Keeping the amplitude of modulating signal less than the amplitude of carrier wave helps one to reduce the distortions in the signal.
An electron and a photon each have a wavelength 1.00 nm. Find
(i) their momenta,
(ii) the energy of the photon and
(iii) the kinetic energy of electron.Given, wavelength,
i) Momenta of electron = momenta of photon.
Therefore,
ii) Energy of photon, E = 
iii) Kinetic energy of electron, 
Draw a schematic diagram showing the (i) ground wave (ii) sky wave and (iii) space wave propagation modes for EM waves.
Write the frequency range for each of the following:
(i) Standard AM broadcast
(ii) Television
(iii) Satellite communicationThe diagram below illustrates the various modes of propagation for EM waves:
i) Standard AM broadcast used frequency of range 540-1600 kHz
ii) A frequency range of 54–890 MHz is used by television sets.
iii) Satellite communication uses 5.925–6.425 GHz as uplink frequency and 3.7–4.2 GHz downlink frequency.
Describe Young's double slit experiment to produce interference pattern due to a monochromatic source of light. Deduce the expression for the fringe width.
Conditions for Young’s double slit experiment are:
(i) The sources should be monochromatic and originating from common single source.
(ii) The amplitudes of the waves should be equal.
Let S1 and S2 be two coherent sources separated by a distance d.
Let the distance of the screen from the coherent sources be D. Let point M be the foot of the perpendicular drawn from O, which is the midpoint of S1 and S2 on the screen. Obviously point M is equidistant from S1 and S2. Therefore the path difference between the two waves at point M is zero. Thus the point M has the maximum intensity.
Now, Consider a point P on the screen at a distance y from M. Draw S1N perpendicular from S1 on S2 P. 
The path difference between two waves reaching at P from S1 and S2 is given by, 
Since, D >> d, 
Expression for Fringe Width:
Distance between any two consecutive bright fringes or any two consecutive dark fringes are called the fringe width. It is denoted by 
.
Let, 
be the distance of two consecutive fringes. Then, we have
So, fringe width is 
Fringe width is same for both bright and dark fringe.
Use Huygens principle to verify the laws of refraction.
Proof of Snell’s law of Refraction using Huygens wave theory is: 
As seen in the fig. above let XY be a surface separating the two media ‘1’ and ‘2’. Let v1 and v2 be the speeds of waves in these media.
A plane wavefront AB in the first medium is incident obliquely on the boundary surface XY and its end A touches the surface at A at time t = 0 while the other end B reaches the surface at point B after time-interval t.
Here, 
As the wavefront AB advances, it strikes the points between A and B¢ of boundary surface.
According to Huygens principle, secondary spherical wavelets emanate from these points, which travel with speed v1 in the first medium and speed v2 in the second medium.
Secondary wavelet starting from A, traverses a distance AA’ = v2t in second medium in time t. In the same time, point of wavefront traverses a distance in first medium and reaches B’, from where the secondary wavelet starts.
So, 
and AA’ = v2t.
Assuming A as centre, we draw a spherical arc of radius AA’ (= v2t) and draw tangent B’A’ on this arc from B’. As the incident wavefront AB advances, the secondary wavelets start from points between A and B’, one after the other and will touch A’B’ simultaneously.
According to Huygens principle A’B’ is the new position of wavefront AB in the second medium. Hence A’B’ will be the refracted wavefront.
First law: As AB, A’B’ and surface XY are in the plane of paper, therefore the perpendicular drawn on them will be in the same plane. As the lines drawn normal to wavefront denote the rays, therefore we may say that the incident ray, refracted ray and the normal at the point of incidence all lie in the same plane.
This is the first law of refraction.
Second law: Let the angle made by incident wavefront be i and angle made by the refracted wavefront A’B’ be r.
In 
, 
... (1)
Similarly, in right-angled triangle, <AA'B,
Therefore,
Now, dividing equation (1) by (2), we have 
The ratio of sine of angle of incidence and the sine of angle of refraction is a constant and is equal to the ratio of velocities of waves in the two media. This is the second law of refraction, and is called the Snell’s law.
(a) Describe briefly, with the help of suitable diagram, how the transverse nature of light can be demonstrated by the phenomenon of polarization.
(b) When unpolarized light passes from air to a transparent medium, under what condition does the reflected light get polarized?
If now the crystal A is kept fixed and B is gradually rotated in its own plane, the intensity of light emerging out of B decreases and becomes zero when the axis of B is perpendicular to that of A. If B is further rotated, the intensity begins to increase and becomes maximum when the axes of A and B are again parallel. Thus, we see that the intensity of light transmitted through B is maximum when axes of A and B are parallel and minimum when they are at right angles. From this experiment, it is obvious that light waves are transverse and not longitudinal; because, if they were longitudinal, the rotation of crystal B would not produce any change in the intensity of light.
b) When reflected and refracted rays are perpendicular to each other then, the reflected ray is totally plane polarized.
The energy levels of a hypothetical atom are shown below. Which of the shown transitions will result in the emission of a photon of wavelength 275 nm?
Which of these transitions correspond to emission of radiation of (i) maximum and (ii) minimum wavelength?
Given, wavelength of the photon, 
= 275 nm
Energy of the photon is given by, 
This corresponds to transition B as from the figure.
i) 
For maximum wavelength should be minimum.
Minimum energy corresponds to transition A.
ii) For minimum wavelength, should be maximum. Maximum energy corresponds to transition D.
State the law of radioactive decay.
Plot a graph showing the number (N) of undecayed nuclei as a function of time (t) for a given radioactive sample having half-life T½.
Depict in the plot the number of undecayed nuclei at (i) t = 3 T½ and (ii) t = 5 T½.Law of radioactive decay states that, the number of nuclei undergoing decay per unit time, at any instant, is proportional to the total number of nuclei present in the sample at that instant.
Number of undecayed nuclei at t = 3T1/2 is 
And, at t = 5T1/2 the number of undecayed nuclei is 
(a) Draw a ray diagram to show refraction of a ray of monochromatic light passing through a glass prism.
Deduce the expression for the refractive index of glass in terms of angle of prism and angle of minimum deviation.
(b) Explain briefly how the phenomenon of total internal reflection is used in fiber optics.
a) The phenomenon of refraction for a ray of monochromatic light passing through a glass prism is shown as below: 
Let, PQR be the principal section of the prism.
The refracting angle of prism is A. Monochromatic light EF is incident on face PQ at angle of incidence i1. This ray enters from a rarer to denser medium and hence is refracted towards the normal FN. Refracted ray is FG. Angle of refraction for this face is r1. Refracted ray FG becomes incident on face PR and is refracted away from the normal GN2 and emerges in the direction GH. The angle of incidence on this face is r2 (into prism) and angle of refraction (into air) is i2. At point O, incident and the emergent ray meet. The angle between these two rays is called the angle of deviation ‘
’. 
.
... (1)
The normal FN1 and GN2 on faces PQ and PR respectively, when produced meet at N.
Let, 
... (2) 
We can see that, for one angle of deviation we have two angles of incidence. But, angle of deviation is minimum for only one particular angle if incidence. Angle of minimum deviation is denoted by 
.
So let, i1 = i2 = i
And r1 = r2 = r
Therefore, we have (4) and (6), we have
b) An optical fiber is a device based on total internal reflection by which a light signal may be transmitted from one place to another with a negligible loss of energy. When a light ray is incident on one end at a small angle of incidence, it suffers refraction from air to quartz and strikes the quartz-coating interface at an angle more than the critical angle. Hence, total internal reflection happens and the ray of light strikes the opposite face again at an angle greater than critical angle. The phenomenon of total internal reflection takes place. Thus the ray within the fiber suffers multiple total internal reflections and finally strikes the other end at an angle less than critical angle for quartz-air interface and emerges in air.
(a) Obtain lens makers formula using the expression
Here the ray of light propagating from a rarer medium of refractive index (n1) to a denser medium of refractive index (n2) is incident on the convex side of spherical refracting surface of radius of curvature R.
(b) Draw a ray diagram to show the image formation by a concave mirror when the object is kept between its focus and the pole. Using this diagram, derive the magnification formula for the image formed.
Len’s Makers formula:
Consider a thick lens L. The refractive index of the material of lens is n2. The lens is placed in a medium of refractive index n1. R1 and R2 is the radius of curvature of the lens and poles are represented by P1 and P2. Thickness of the lens is given by t. O is a point object which is placed on the principal axis of the lens.
Object distance from pole P1 is u.
Image formed is at a distance of v’ from P1.
Now, using the refraction formula at spherical surface, we have 
... (1)
Image I’ acts as a virtual object for second surface and after refraction at second surface, final image is formed at I.
Distance of I from pole P2 of second surface is v.
Distance of virtual object I’ from pole P2 is (v’ – t).
For refraction at second surface, the ray is going from second medium (refractive index n2) to first medium (refractive index n1), therefore from refraction formula at spherical surface.
i.e., 
Thickness of the lens is negligibly small as compared to v’.
Therefore, 
... (2)
Now, adding equations (1) and (2), we have 
Now, using the lens formula, we get 
is the required Lens maker’s formula.
b) Ray diagram for the image formation of an object between focus (F) and pole (P) of a concave mirror is given by,
Magnification, m =
From fig. 
Also, 
Which of the following waves can be polarized (i) Heat waves (ii) Sound waves?
i) Heat waves can be polarized because they are transverse and electromagnetic in nature.
ii) Sound waves are longitudinal in nature and hence, cannot be polarized.
Write the relationship between angle of incidence ‘i’, prism ‘A’ and angle of minimum deviation for a triangular prism.
The relationship between i, A and 
for a triangular prism is given by,
The given graph shows the variation of photo-electric current (I) versus applied voltage (V) for two different photosensitive materials and for two different intensities of the incident radiation. Identify the pairs of curves that correspond to different materials but same intensity of incident radiation. 
The pair of curves (1, 3) and (2, 4) corresponds to different materials but having same intensity of incident radiation. The value of stopping potential for the curves (1, 2) and (3, 4) is same implying that they correspond to similar materials.
(a) Write the necessary conditions for the phenomenon of total internal reflections to occur.
(b) Write the relation between the refractive index and critical angle for a given pair of optical media.a) Conditions for total internal reflection to take place:
i) Light ray must travel from denser medium into rarer medium.
ii) The angle of incidence in denser medium must be greater than the critical angle for given pair of optical media.
b) Relation between refractive index and critical angle for a given pair is,
As per Snell’s law, 
In the given circuit diagram, a voltmeter ‘V’ is connected across a lamp ‘L’. How would (i) the brightness of the lamp and (ii) voltmeter reading ‘V’ be affected, if the value of resistance ‘R’ is decreased? Justify your answer.
The given figure is a Common Emitter configuration of an npn transistor. Input circuit is forward biased and collector circuit is reverse biased.
i)
If the value of the resistance R is reduced, the current in the forward biased input circuit increases. The emitter current IE and the collector current IC (= IE – IB) both increase. Hence, the brightness of the lamp increases.
ii)
Due to increase in IC, the potential drop across lamp L increases and hence the voltmeter reading V increases.
Block diagram of a receiver is shown in the figure:
(a) Identify ‘X’ and ‘Y’.
(b) Write their functions.
a) X represents Intermediate frequency stage and Y represents amplifier.
b) IF stage changes the electromagnetic wave of high frequency to a lower frequency for further detection in detector.
Function of amplifier is to enhance the power of the signals upto a required level.
Explain, with the help of a circuit diagram, the working of a photo-diode. Write briefly how it is used to detect the optical signals.
OR
Mention the important considerations required while fabricating a p-n junction diode to be used as a Light Emitting Diode (LED). What should be the order of band gap of an LED if it is required to emit light in the visible range?
In any diode, an electric field ‘E’ exists across the junction from n-side to p-side, when light with energy h greater than energy gap Eg illuminates the junction. Then electron-hole pairs are generated due to absorption of photons, in or near the depletion region of the diode. Due to existing electric field, electrons and holes get separated. The free electrons are collected on n-side and holes are collected on p-side, giving rise to an emf.
Due to this generated emf, an electric current of 
A order flows through the external resistance.
The figure above is the circuit diagram of photodiode.
Detection of optical signals:
Photodiode can be used to detect optical signals if we observe the change in current with change in light intensity when a reverse bias is applied.
OR
Important considerations in the fabrication of LED:
i) LED is heavily doped p-n junction.
ii) Reverse breakdown voltage of LED’s are very low, typically around 5 V.
The order of band gap of an LED to emit light in the visible range is about 3 eV to 1.8 eV.
Write three important factors which justify the need of modulating a message signal. Show diagrammatically how an amplitude modulate wave is obtained when a modulating signal is superimposed on a carrier wave.
Three factors for the need of modulating a message signal:
a) Size of the antenna or aerial should be 
/4.
b) Effective power radiated by an antenna should be close to 
c) Interference of signals from different transmitters. Inorder to avoid the interference of the signals, high frequency is required which can be acquired by modulation. 
a) In a typical nuclear reaction, e.g. , 
Although number of nucleons is conserved, yet energy is released. How? Explain.
b) Show that nuclear density in a given nucleus is independent of mass number A.a)
In a nuclear reaction,
In a nuclear reaction, the sum of the masses of the of the target nucleus and the bombarding particle may be greater or less than the sum of masses of the product nucleus and the outgoing particle. Therefore, from the law of conservation of mass-energy around 3.27 MeV is evolved or involved in a nuclear reaction. This energy is called the Q-value of the nuclear reaction.
Cause of the energy released:
i) Binding energy per nucleon of becomes more than the BE/A of 
ii) Mass defect between the reactant and the product nuclei.
b) The radius of nucleus of mass number A is given by R = 
Volume of the nucleus, 
Density of the matter in the nucleus is given by,
Hence, density is independent of mass number A.
(a) Why photoelectric effect cannot be explained on the basis of wave nature of light? Give reasons.
(b) Write the basis features of photon picture of electromagnetic radiation on which Einstein’s photoelectric equation is based.a) Wave theory of light cannot explain the photoelectric effect because of the following reasons:
i) Wave nature does not explain the instantaneous ejection of photoelectrons.
ii) It does not explain to us the threshold frequency for a metal surface.
iii) Wave nature do not tell us, why kinetic energy of the emitted electrons is independent of the intensity of light and depends upon its frequency.
Basic features of photon picture of electromagnetic radiation on which Einstein’s photoelectric equation is based upon is:
b) Features of photons:
i) Photons are particles of light having energy, E =
and momentum, p = 
where h is Planck constant.ii) Photons travel with the speed of light in vacuum, independent of the frame of reference.
iii) Intensity of light depends on the number of photons crossing unit area in a unit time.
Output characteristics of an n-p-n transistor in CE configuration is shown in the figure. Determine:
(a) Dynamic output resistance
(b) DC current gain and
(c) AC current gain at an operating point VCE=10V, when IB=30
A. 
Dynamic output resistance is given by,
For IB = 30 
A, = (12-8) = 4V and = (3.6-3.4) = 0.2 mA
Therefore, output resistance, 
ii) dc current gain is given by, 
At = 10 V and Ib = 30 
A
Ic = 3.5 mA
= 
= 117
iii) AC current gain is given by, 
At VCE = 10 V, = (3.5 -2.5) mA = 1 mA and
= (30 -20) A = 10 A
Using Bohr’s postulates, obtain the expression for the total energy of the electron in the stationary states of the hydrogen atom. Hence draw the energy level diagram showing how the line spectra corresponding to Balmer series occur due to transition between energy levels.
According to Bohr’s postulates, in a hydrogen atom, a single electron revolves around a nucleus of charge +e. For an electron moving with a uniform speed in a circular orbit on a given radius, the centripetal force is provided by the Coulomb force of attraction between the electron and the nucleus.
Therefore, 
... (1)
Potential energy is given by, P.E = 
Therefore, total energy is given by, E = K.E + P.E = 
E = 
For nth orbit, E can be written as En, 
... (2)
Now, using Bohr's postulate for quantization of angular momentum, we have
Putting this value of v in equation (1), we get
(a) In what way is diffraction from each slit related to the interference pattern in a double slit experiment?
(b) Two wavelength of sodium light 590 nm and 596 nm are used, in turn, to study the diffraction taking place at a single slit of aperture 2 x 10-4 m. The distance between the slit and the screen is 1.5.m. Calculate the separation between the positions of the first maxima of the diffraction pattern obtained in the two cases.= (a) When a plane wavefront of monochromatic light illuminates, the slit LN, each point in the slit LN becomes the source of secondary wavelets.
The secondary wavelets originating from different points superpose on each, while travelling towards the point C and point P; at angle q. However the superposition of the secondary wavelets produces a diffraction pattern of varying intensity, as shown in fig.
b) For maxima other than central maxima, 
For light of wavelength, 
= 590 nm
(a) Draw a ray diagram showing the image formation by a compound microscope.
Hence obtain expression for total magnification when the image is formed at infinity.
(b) Distinguish between myopia and hypermetropia. Show diagrammatically how these defects can be corrected.
OR
State Huygens principle. Using this principle draw a diagram to show how a plane wave front incident at the interface of the two media gets refracted when it propagates from a rarer to a denser medium. Hence verify Snell’s law of refraction.
(b) When monochromatic light travels from a rarer to a denser medium, explain the following, giving reasons:
(i) Is the frequency of reflected and reflected light same as the frequency of incident light?
(ii) Does the decrease in speed imply a reduction in the energy carried by light wave?a) The fig. below shows the formation of image by a compound microscope.
If the object AB is very close to the focus of the objective lens of focal length fo, then magnification Mo by the objective lens is given by, 
b) Difference between myopia and hypermetropia,
|
Myopia |
Hypermetropia |
|
1. The eye ball is shortened |
| 2. Person cannot see distant objects clearly. |
2. Person cannot see near objects clearly. |
A concave lens is used for the correction of myopic eye.
Hypermetropia is corrected using a convex lens.
OR
a) Law of Reflection: Let XY is a reflecting surface at which a wavefront is being incident obliquely. Let v be the speed of the wavefront and at time t = 0, the wavefront touches the surface XY at A. After time t, the point B of wavefront reaches the point B¢ of the surface. According to Huygens principle each point of wavefront acts as a source of secondary waves. When the point A of wavefront strikes the reflecting surface, then due to presence of reflecting surface, it cannot advance further; but the secondary wavelet originating from point A begins to spread in all directions in the first medium with speed v. As the wavefront AB advances further, its points A1, A2, A3… etc. strike the reflecting surface successively and send spherical secondary wavelets in the first medium.
First of all the secondary wavelet starts from point A and traverses distance AA’ (= vt) in first medium in time t. In the same time t, the point B of wavefront, after travelling a distance BB’, reaches point B’ (of the surface), from where the secondary wavelet now starts. Now taking A as centre we draw a spherical arc of radius AA’ (= vt) and draw tangent A’B’ on this arc from point B’. As the incident wavefront AB advances, the secondary wavelets starting from points between A and B’, one after the other and will touch A’B’ simultaneously. According to Huygens principle wavefront A’B’ represents the new position of AB, i.e., A’B’ is the reflected wavefront corresponding to incident wavefront AB.
Now, in right-angled ABB’ and AA’B’
i.e., incident wavefront AB and reflected wavefront A¢ B¢ make equal angles with the reflecting surface XY. As the rays are always normal to the wavefront, therefore the incident and the reflected rays make equal angles with the normal drawn on the surface XY.
i.e., angle of incidence i = angle of reflection r
This is the second law of reflection.
b) (i) If the radiation of certain frequency interact with the atoms/molecules of the matter, they start to vibrate with the same frequency under forced oscillations.
Thus, the frequency of the scattered light (under reflection and refraction) equals to the frequency of incident radiation.
(ii) No, energy carried by the wave depends on the amplitude of the wave, but not on the speed of the wave.
What is the function of a 'Repeater' used in communication system?
Repeater is used to increase the range of the transmission in communication systems with the help of a set of receiver and transmitter.
The line AB in the ray diagram represents a lens State whether the lens is convex or concave.
We can see that the lens slightly diverges the path of refracted light. Hence, the lens is concave in nature.
How does one explain, using de Broglie hypothesis, Bohr's second postulate of quantization of orbital angular momentum?
According to de-Broglie hypothesis, a stationary orbit is the one that contains an integral number of de-Broglie waves associated with the revolving electron.
Total distance covered by electron = Circumference of the orbit =
For the permissible orbit,
... (1)
Now, according to De-Broglie wavelength,
Now, putting this in equation (1), we have
= 
mvn rn = 
; which is the required Bohr’s second postulate of quantization of orbital angular momentum.
What is ground wave communication? Explain why this mode cannot be used for long distance communication using high frequencies.
Ground waves travel along the surface of the earth. In ground wave propagation, a large portion of wave energy is in space near the surface of the Earth. The propagation of the wave is guided along the Earth's surface and follows the curvature of the Earth.
High frequency wave propagation is not possible through ground waves for long-distance communication because while progressing, ground waves induce current in the ground and bend round the corner of the objects on the Earth due to which the energy of the ground waves of high frequency is almost absorbed by the surface of the Earth after travelling a small distance. Such signals are also absorbed by the obstacles (like mountains, tall buildings and trees) between the transmitter and the receiver. This loss in power of ground waves increases with the increase in frequency. Thus, ground wave communication is not suited for high frequency.
The equivalent wavelength of a moving electron has the same value as that of a photon of energy 6 x 10–17 J. Calculate the momentum of the electron.
Momentum of electron is given by, p = 
A ray of light passes through an equilateral glass prism such that the angle of incidence is equal to the angle of emergence and each of these angles is equal to 3/4 of angle of prism. Find the angle of deviation.
OR
Calculate the speed of light in a medium whose critical angle is 45°. Does critical angle for a given pair of media depend on the wavelength of incident light? Give reason.Given, i = e = ¾ A, …(1)
where,
A is the angle of prism.
Angle of deviation for a ray of light in a prism is given by,
Prism is an equilateral triangle.
That is A= 60o
OR
Using the snell's law we have
... (1)
where, C is the critical angle of the medium and 
is the refractive index of the medium.
Also, 
... (2)
Now, from equations (1) and (2), we have 
This is the speed of the light in the medium.
Refractive index of a medium is inversely proportional to the wavelength of light. So, critical angle also depends upon the wavelength of incident light.
Write two important considerations used while fabricating a Zener diode. Explain, with the help of a circuit diagram, the principle and working of a Zener diode as voltage regular.
Important considerations while fabricating Zener diode are:
i) Heavily doped p and n junctions are to be used for fabricating Zener diode.
ii) The breakdown voltage of the material must be noted down to avoid destruction of the device.
Principle: A properly doped p–n junction that is supposed to work in a breakdown region is called a Zener diode. Such a junction when reverse biased shows a sudden increase in current to a high value at a certain voltage known as the breakdown voltage or the Zener voltage.
Working: A Zener diode is connected across the fluctuating voltage source through a resistance R. A constant voltage supply is maintained across the load resistance RL.
When the input voltage increases, the resistance of the Zener diode decreases and hence the current through the diode increases. Thus, a large voltage drop is seen across the resistance R and the output voltage across RL remains at a constant value. When the input voltage decreases, the resistance of the Zener diode increases and hence the current through the diode decreases. Only a small voltage drop takes place now across the resistance R and the output voltage at RL remains constant. Thus, we get a constant voltage in spite of the fluctuating input voltage. In this way, a Zener diode acts as a voltage regulator.
Draw the necessary energy band diagrams to distinguish between conductors, semiconductors and insulators.
How does the change in temperature affect the behaviour of these materials? Explain briefly.
Energy band diagram is as given below:
Conductors:
i) The valence band is completely filled and the conduction band can have two possibilities—either it is partially filled with an extremely small energy gap between the valence and conduction bands or it is empty, with the two bands overlapping each other.
ii) On applying an even small electric field, conductors can conduct electricity.
Insulators:
i) For insulators, the energy gap between the conduction and valence bands is very large. Also, the conduction band is practically empty.
ii) When an electric field is applied across such a solid, the electrons find it difficult to acquire such a large amount of energy to reach the conduction band. Thus, the conduction band continues to be empty. That is why no current flows through insulators.
Semiconductors:
i) The energy band structure of semiconductors is similar to that of insulators, but in their case, the size of forbidden energy gap is much smaller than that of the insulators.
ii) When an electric field is applied to a semiconductor, the electrons in the valence band find it comparatively easier to shift to the conduction band. So, the conductivity of semiconductors lies between the conductivity of conductors and insulators.
(a) What are the three basic units in communication systems? Write briefly the function of each of these.
(b) Write any three applications of the Internet used in communication systems.a) A communication system consists of three basic units:
(i) Transmitter: This unit is used for transmitting the information after modifying it to a suitable form. It basically consists of a transducer that converts signal in any physical form to electrical signal for transmission. After that, the signal is modulated to transmit over long distances.
(ii) Communication channel: This unit carries the modulated signal from the transmitter to the receiver. Transmission lines act as a communication channel in case of telephony, whereas the free space serves the purpose of communication channel in case of the radio communication.
(iii) Receiver: This unit consists of an antenna, which receives the signal, followed by a demodulator, an amplifier and a transducer. The demodulator demodulates the modulated signal, the amplifier boosts up its intensity and the transducer converts it back again from electrical form to the needed physical form.
b) Application of internet used in communication system is:
i) E-banking: Financial transactions can be made through this mode.
ii) Internet surfing: Navigation over World Wide Web (www) from one webpage/website to another is called Internet surfing. It is an interesting way of searching and viewing information on any topic of interest.
iii) E-commerce: This is a mode of electronic commerce by which consumers can buy the products or the services over internet.
(a) Write the necessary conditions to obtain sustained interference fringes.
(b) In Young's double slit experiment, plot a graph showing the variation of fringe width versus the distance of the screen from the plane of the slits keeping other parameters same. What information can one obtain from the slope of the curve?
(c) What is the effect on the fringe width if the distance between the slits is reduced keeping other parameters same?
a) Necessary conditions for sustained interference fringes are:
i) Two sources must be coherent. They should emit continuous light waves of same wavelength or frequency.
ii) Two sources of light must be narrow.
iii) Sources of light must be monochromatic.
b) Fringe width is given by:
where, D is the distance between the slit and the screen.
Fringe width and distance of screen from the slit is a linear graph.
The slope of the curve gives is 
That is, when fringe width varies linearly with distance of screen from the slits, the ratio of wavelength to distance between the slits remain constant. That is why it is advised to take wavelengths of incident light nearly equal to the width of the slit.
c) If the distance between the slits is reduced by keeping other parameters same. Then, fringe width would become broader.
(a) Give two reasons to explain why reflecting telescopes are preferred over refracting type.
(b) Use mirror equation to show that convex mirror always produces a virtual image independent of the location of the object.a) Reflecting telescopes are preferred over refracting because of the following reasons:
i) There is no chromatic aberration as the objective is a mirror
ii) Spherical aberration is reduced in case of reflecting telescope by using mirror objective.
b) Focal length is always positive for convex mirror.
Consider an object placed on the left side of the mirror. That is u < 0.
Now, using the mirror formula,
Focal length is positive and object distance is negative.
Therefore, 
Hence, a virtual image is always formed at the back side of a mirror. That is, image formed by a convex mirror is virtual in nature.
Obtain the relation between the decay constant and half-life of a radioactive sample.
The half-life of a certain radioactive material against 
decay is 100 days. After how much time, will the undecayed fraction of the material be 6.25%?
In a radioactive sample, number of atoms at any instant is given by, 
Now, when t = T, where T is the half-life of the sample.
This implies, half-life of a radioactive substance is inversely proportional to decay constant.
Numerical:
We have,
Number of undecayed nuclei left = 6.25 % = 6.25/ 100 = 1/16
Let, t be the required time after which the undecayed fraction of the material will be 6.25%.
(a) Define the term 'intensity of radiation' in terms of photon picture of light.
(b) Two monochromatic beams, one red and the other blue, have the same intensity.
In which case:
(i) the number of photons per unit area per second is larger,
(ii) the maximum kinetic energy of the photoelectrons is more? Justify your answer.
a) The number of photons falling per unit area in unit time is defined as the intensity of radiation.
b)
i) Since, the two beams have the same intensity therefore, the number of photons emitted per unit are per unit time is the same.
ii) Maximum Kinetic energy of photoelectrons is given by, 
We know that the frequency of blue beam is more than that of the red beam. Therefore, maximum Kinetic energy of the blue beam will be more.
(a) Use Huygens' principle to show the propagation of a plane wavefront from a denser medium to a rarer medium. Hence find the ratio of the speeds of wavefront in the two media.
(b) (i) Why does an unpolarized light incident on a polaroid get linearly polarized ?
(ii) Derive the expression of Brewster's law when unpolarized light passing from a rarer to a denser medium gets polarized on reflection at the interface.
a) 
As seen in the fig. above let XY be a surface separating the two media ‘1’ and ‘2’. Let v1 and v2 be the speeds of waves in these media.
A plane wavefront AB in the first medium is incident obliquely on the boundary surface XY and its end A touches the surface at A at time t = 0 while the other end B reaches the surface at point B after time-interval t.
Here, 
.
As the wavefront AB advances, it strikes the points between A and B’ of boundary surface.
According to Huygens principle, secondary spherical wavelets emanate from these points, which travel with speed v1 in the first medium and speed v2 in the second medium.
Secondary wavelet starting from A, traverses a distance AA’ = v2t in second medium in time t. In the same time, point of wavefront traverses a distance in first medium and reaches B’, from where the secondary wavelet starts.
So, BB' = v1 t and AA’ = v2t.
Assuming A as centre, we draw a spherical arc of radius AA’ (= v2t) and draw tangent B’A’ on this arc from B’. As the incident wavefront AB advances, the secondary wavelets start from points between A and B’, one after the other and will touch A’B’ simultaneously.
According to Huygens principle A’B’ is the new position of wavefront AB in the second medium. Hence A’B’ will be the refracted wavefront.
Let the angle made by incident wavefront be i and angle made by the refracted wavefront A’B’ be r.
b)
Polarization of light is referred to as restricting the vibration of light in a perpendicular direction perpendicular to the direction of propagation of wave.
The vibration of particles of light which is parallel to the axis of crystal passes through the Polaroid on passing an unpolarized light. All other vibrations are absorbed and that is why intensity of emerging light is reduced.
The plane of vibration here is ABCD, in which the vibrations of the polarized light is confined and the plane KLMN is called plane of polarization. KLMN is perpendicular to the plane of vibration.
Reflected light is totally polarized, when unpolarized light is incident on the glass-air interface at the Brewster angle iB. This is known as Brewster’s law.
The reflected component OB and refracted component OC are mutually perpendicular to each other, when light is incident at Brewster’s angle.
i = iB and r = (900 – iB)
Therefore, 
, is the expression for Brewster’s law.
A biconvex lens with its two faces of equal radius of curvature R is made of a transparent medium of refractive index 
1. It is kept in contact with a medium of refractive index 2 as shown in the figure.
a) Find the equivalent focal length of the combination.
b) Obtain the condition when this combination acts as a diverging lens.
c) Draw the ray diagram for the case 
when the object is kept far away from the lens. Point out the nature of the image formed by the system.
a) Using the Len’s maker’s formula, we have
Let feq is the equivalent focal length of the combination, then
b)
For, the combination of lenses to behave as diverging lens, equivalent focal length < 0.
c)
For 
the combination of lenses will behave as converging lens. An object placed far from the lens will form image at the focus of the lens.
Image formed is real and diminished in nature.
Define intensity of radiation on the basis of photon picture of light. Write its S.I. unit.
The amount of light or photon energy incident per metre square per second is called the intensity of radiation. S.I unit is W/m2.
Why is it found experimentally difficult to detect neutrinos in nuclear 
-decay?
Neutrinos are chargeless, massless particles that hardly interact with matter. Hence, their detection during the experiment is very difficult.
A biconvex lens made of a transparent material of refractive index 1.25 is immersed in water of refractive index 1.33. Will the lens behave as a converging or a diverging lens? Give reason.
The lens will behave as a diverging lens when light rays enter from air to lens. The refractive of lens is lesser than the refractive index of the medium i.e., water. So. The lens will behave as a converging lens when light rays travel from lens to water.
Using Rutherford model of the atom, derive the expression for the total energy of the electron in hydrogen atom. What is the significance of total negative energy possessed by the electron?
OR
Using Bohr’s postulates of the atomic model derive the expression for radius of nth electron orbit. Hence obtain the expression for Bohr’s radius.According to Rutherford’s model, we have
Energy is negative implies that the electron –nucleus is a bound or attractive system.
OR
According to the Bohr’s atomic model, electrons revolve around the nucleus only in those orbits for which the angular momentum is an integral multiple of 
.
So, as per Bohr’s postulate, we have
This is the required expression for Bohr's radius.
Explain, with the help of a circuit diagram, the working of a p-n junction diode as a half-wave rectifier.
Circuit for p-n junction diode as half-wave rectifier is given below:
Two monochromatic rays of light are incident normally on the face AB of an isosceles right-angled prism ABC. The refractive indices of the glass prism for the two rays ‘1’ and ‘2’ are respectively 1.35 and 1.45. Trace the path of these rays after entering through the prism. 
The propagation of ray diagrams is as shown below:
Both the rays will fall on the side AC with angle of incidence (i) equal to 45°.
Critical angle of ray 1 is greater than that of angle of incidence. Hence, it will emerge from the prism. Critical angle of ray 2 is less than that of i. Hence, it will be internally reflected.
Write the functions of the following in communication systems:
(i) Transducer
(ii) RepeaterTransducer: Any device that converts one form of energy to another.
Repeater: A repeater accepts the signal from the transmitter amplifies and retransmits it to the receiver.
Draw a circuit diagram of n-p-n transistor amplifier in CE configuration. Under what condition does the transistor act as an amplifier?
The circuit diagram for n-p-n transistor in CE configuration is as shown below:
The transistor acts as an amplifier when the emitter-base junction is forward biased and the collector-emitter junction is reverse biased with a high voltage VCC.
(a) Using the phenomenon of polarization, show how transverse nature of light can be demonstrated.
(b) Two Polaroid P1 and P2 are placed with their pass axes perpendicular to each other. Unpolarised light of intensity Io is incident on P1. A third Polaroid P3 is kept in between P1 and P2 such that its pass axis makes an angle of 30o with that of P1. Determine the intensity of light transmitted through P1, P2 and P3.a)
i. Light from a source S is allowed to fall normally on the flat surface of a thin plate of a tourmaline crystal, cut parallel to its axis. Only a part of this light is transmitted through A.
ii. If now the plate A is rotated, the character of transmitted light remains unchanged.
iii. Now another similar plate B is placed at some distance from A such that the axis of B is parallel to that of A.
iv. If the light transmitted through A is passed through B, the light is almost completely transmitted through B and no change is observed in the light coming out of B. 
v. If now the crystal A is kept fixed and B is gradually rotated in its own plane, the intensity of light emerging out of B decreases and becomes zero when the axis of B is perpendicular to that of A.
vi. If B is further rotated, the intensity begins to increase and becomes maximum when the axes of A and B are again parallel. Thus, we see that the intensity of light transmitted through B is maximum when axes of A and B are parallel and minimum when they are at right angles.
From this experiment, it is obvious that light waves are transverse and not longitudinal; because, if they were longitudinal, the rotation of crystal B would not produce any change in the intensity of light.
Intensity of light transmitted through P1 = Io/2
Intensity of light transmitted through P3 = ( Io / 2) cos 2 30o = 3 Io / 8
Intensity of light transmitted through P2 = 3/8 Io cos2 60o = 3/32 Io
A convex lens of focal length 20 cm is placed coaxially with a convex mirror of radius of curvature 20 cm. The two are kept at 15 cm apart. A point object lies 60 cm in front of the convex lens. Draw a ray diagram to show the formation of the image by the combination. Determine the nature and position of the image formed.
For convex lens,
Focal length, f = + 20 cm
Object distance, u = - 60 cm
Now, using the lens formula, we have
Putting in the values, we get
Image distance, v = + 30 cm
Now, for convex mirror
Object distance, u = + (30- 15) cm = 15 cm
Focal length, f = + 20/2 cm = +10 cm
Object distance, u = + (30- 15) cm = 15 cm
Focal length, f = + 20/2 cm = +10 cm
Now, using the mirror formula we have
Putting all the values, we get
Image distance, v = + 30 cm
Therefore, the final image is formed at a distance of 30 cm from the convex mirror or at a distance of 45 cm from the convex lens to the right of the convex mirror.
Therefore, the final image formed is a virtual image.
Ray diagram for the image formation is given below:
Write any two distinguishing features between conductors, semiconductors and insulators on the basis of energy band diagrams.
Energy band diagram is given below:
Two distinguishing features of conductors, semiconductors and insulators are:
a) Aarti has incessantly displayed care towards the health of her sister. She has been a keen observer and concerned for others.
b) Using a radio-isotope, a doctor can understand the movement of a normal brain and a brain having tumor in it. A radio-isotope is made to enter the body along with the elements and a compound, whose absorption, functioning and distribution to the brain is studied. The isotope emits radiation and these radiations are detected for absorption and function of the compounds. So, a doctor can diagnose the brain tumor.
Write two basic modes of communication. Explain the process of amplitude modulation. Draw a schematic sketch showing how amplitude modulated signal is obtained by superposing a modulating signal over a sinusoidal carrier wave.
The two basic modes of communication are:
In amplitude modulation, the amplitude of the carrier wave is varied in accordance with the amplitude of the modulating wave.
Let, the carrier wave be given by,
Modulating signal is given by,
is the angular frequency of the message signal.
Therefore, modulated signal is given by,
Amplitude modulated signal is produced by superposing a modulating signal over a sinusoidal carrier wave. The process is illustrated in the fig. below.
An electron microscope uses electrons accelerated by a voltage of 50 kV. Determine the de-Broglie wavelength associated with the electrons. Taking other factors, such as numerical aperture etc. to be same, how does the resolving power of an electron microscope compare with that of an optical microscope which uses yellow light?
Given,
Accelerating voltage for electrons = 50 kV = 50 x 103 V
De-Broglie wavelength is given by, 
So, from the above formula, we can say that the resolving power of an electron microscope is much greater than that of optical microscope.
(a) In Young's double slit experiment, describe briefly how bright and dark fringes are obtained on the screen kept in front of a double slit. Hence obtain the expression for the fringe width.
(b) The ratio of the intensities at minima to the maxima in the Young's double slit experiment is 9: 25. Find the ratio of the widths of the two slits.
a)
Let S1 and S2 be two coherent sources separated by a distance d.
Let the distance of the screen from the coherent sources be D. Let point M be the foot of the perpendicular drawn from O, which is the midpoint of S1 and S2 on the screen. Obviously point M is equidistant from S1 and S2. Therefore the path difference between the two waves at point M is zero. Thus the point M has the maximum intensity.
Now, Consider a point P on the screen at a distance y from M. Draw S1N perpendicular from S1 on S2 P. 
The path difference between two waves reaching at P from S1 and S2 is given by, 
Since, D >> d, so 
is very small.
For dark fringe, path difference is an odd multiple of half wavelength.
So, we have 
Let, 
be the distance of two consecutive fringes. Then, we have
Fringe width is same for both bright and dark fringe.
b)
Give, ratio of minima to maxima intensity is given by, 9:25
That is, 
(a) Describe briefly how a diffraction pattern is obtained on a screen due to a single narrow slit illuminated by a monochromatic source of light. Hence obtain the conditions for the angular width of secondary maxima and secondary minima.
(b) Two wavelengths of sodium light of 590 nm and 596 nm are used in turn to study the diffraction taking place at a single slit of aperture 2 × 10−6 m. The distance between the slit and the screen is 1·5 m. Calculate the separation between the positions of first maxima of the diffraction pattern obtained in the two cases.
a) Diffraction of light at a Single slit
A single narrow slit is illuminated by a monochromatic source of light. The diffraction pattern is obtained on the screen placed in front of the slits. There is a central bright region called as central maximum. All the waves reaching this region are in phase hence the intensity is maximum. On both side of central maximum, there are alternate dark and bright regions, the intensity becoming weaker away from the center. The intensity at any point P on the screen depends on the path difference between the waves arising from different parts of the wave-front at the slit.
According to the figure, the path difference (BP – AP) between the two edges of the slit can be calculated.
Path difference, BP – AP = NQ = a sin 
Angle 
is zero at the central point C on the screen, therefore all path difference are zero and hence all the parts of slit are in phase. Due to this, the intensity at C is maximum. If this path difference is, (the wavelength of light used), then P will be point of minimum intensity. This is because the whole wave-front can be considered to be divided into two equal halves CA and CB and if the path difference between the secondary waves from A and B is , then the path difference between the secondary waves from A and C reaching P will be 
/2, and path difference between the secondary waves from B and C reaching P will again be 
/2. Also, for every point in the upper half AC, there is a corresponding point in the lower half CB for which the path difference between the secondary waves, reaching P is 
/2. Thus, destructive interference takes place at P and therefore, P is a point of first secondary minimum.
Width of secondary maximum is given by, 
Width of secondary minima is given by, 
Since angular width is independent of n. all secondary minima and maxima are of the same width.
b) Given,
Aperture of the slit, a = 2 10 -6 m
Distance between slit and screen = 1.5 m
Therefore, linear separation, between the first maxima (n=1) of the two wavelengths, on the screen, is given by,
Therefore, separation is given by, 
How does the angular separation between fringes in single-slit diffraction experiment change when the distance of separation between the slit and screen is doubled?
Angular separation is given by,
Angular separation would remain same when the distance of separation between the slit and the screen is doubled.
For the same value of angle incidence, the angles of refraction in three media A, B and C are 15°, 25° and 35° respectively. In which medium would the velocity of light be minimum?
As per Snell’s law we have,
For given i, v 
sin r
r is minimum in medium A, so velocity of light is minimum in medium A.
A proton and an electron have same kinetic energy. Which one has greater de-Broglie wavelength and why?
As per De-broglies formula, 
Kinetic energy of proton is equal to kinetic energy of proton.
Since, mass of proton > mass of electron,
This implies, 
That is, wavelength of electron is greater than the wavelength of proton.
An object AB is kept in front of a concave mirror as shown in the figure.
(i) Complete the ray diagram showing the image formation of the object.
(ii) How will the position and intensity of the image be affected if the lower half of the mirror’s reflecting surface is painted black?(i) Image formed will be inverted diminished between C and F.
ii) When the lower half of the mirror’s reflecting surface is painted black, the position of the image and its intensity will get reduced.
Draw a labelled ray diagram of a reflecting telescope. Mention its two advantages over the refracting telescope
Advantages of reflecting telescope over refracting telescope:
(i) It is free from chromatic aberration.
(ii) Its resolving power is greater than refracting telescope due to larger aperture of mirror.
Describe briefly with the help of a circuit diagram, how the flow of current carries in a p-n-p transistor is regulated with emitter-base junction forward biased and base-collector junction reverse biased.
The emitter-base junction is given a small forward bias, while base collector junction is given a large reverse bias.
Under the forward biasing of emitter-base region, the positive holes of P - region move towards the base. Due to thin base most of holes (about 98%) entering it pass onto collector while a very few of them (nearly 2%) combine with the electrons of base.
As soon as a hole combines with the electron, a fresh electron leaves the negative terminal of battery VEE and enters the base. This causes a very small base current I B. The holes entering the collector move under the aiding reverse bias towards terminal C. As a hole reaches terminal C, an electron leaves the negative terminal of battery VCC and neutralizes the hole.
This causes the collector current IC . Both these currents IB and IC combine to form the emitter current Ie.
i. e., IE = IB + IC
Obviously the holes are the charge carriers within the p-n-p transistor while the electrons are charge carriers in external circuit.
In the given block diagram of a receiver, identify the boxes labelled as X and Y and write their functions.
The element labelled 'X' is called 'channel'. Channel connects transmitter and receiver. The signal from the transmitter is carried to the receiver by the communication channel.
You are given three lenses L1, L2 and L3 each of focal length 20 cm. An object is kept at 40 cm in front of L1, as shown. The final real image is formed at the focus ‘I’ of L3. Find the separations between L1, L2 and L3.
Given,
f1 = f2 = f3 = 20 cm
In case of lens L1,
Object distance, u1 = -40 cm
Using Lens formula,
For lens L3,
So, lens L2 should produce the image at infinity. Thus, objective should be at focus. The image formed by lens L1 is at 40 cm on the right side of lens L1 which lies at 20 cm left of lens L2 i.e., focus of lens L2.
Hence, the distance between L1 and L2 = 40 + 20 = 60 cm.
Define the terms (i) ‘cut-off voltage’ and (ii) ‘threshold frequency’ in relation to the phenomenon of photoelectric effect.
Using Einstein’s photoelectric equation shows how the cut-off voltage and threshold frequency for a given photosensitive material can be determined with the help of a suitable plot/graph.i) Cut-off or stopping potential is that minimum value of negative potential at anode which just stops the photo electric current.
ii) The minimum frequency of light below which no photo electric emission will take place is known as threshold frequency.
Now using Einstein’s photoelectric equation, we have 
Clearly, Vo – 
graph is a straight line.
Mention three ‘different modes of propagation used in communication system. Explain with the help of a diagram how long distance communication can be achieved by ionospheric reflection of radio waves.
Three modes of propagation of electromagnetic waves:
(i) Ground waves
(ii) Sky waves
(iii) Space waves
Sky wave propagation is used for long distance communication. In this process, radio waves are reflected by ionosphere. When radio waves of frequency range 3 MHz to 30 MHz, emitted from transmitting antenna reach the receiving antenna after reflection from the ionosphere acts as a reflector for radio waves.
Draw a plot of potential energy of a pair of nucleons as a function of their separations. Mark the regions where the nuclear force is (i) attractive and (ii) repulsive. Write any two characteristic features of nuclear forces.
The below figure shows the plot of potential energy of a pair of nucleons as a function of their separation.
AB represents repulsive force and part BCD reprtesents attractive force.
Properties of nuclear forces are:
(1) Nuclear forces are attractive and stronger, then electrostatic force.
(2) Nuclear forces are charge-independent.
In a Geiger–Marsden experiment, calculate the distance of closest approach to the nucleus of Z =80, when an a-particle of 8 MeV energy impinges on it before it comes momentarily to rest and reverses its direction.
How will the distance of closest approach be affected when the kinetic energy of the a-particle is doubled?
Let ro be the distance of closest approach where the kinetic energy of the alpha-particle is converted into it’s potential energy.
Given, Z = 80, Ek = 8 MeV
Since, 
Kinetic energy of the particle becomes doubled if the distance of closest approach becomes half.
The ground state energy of hydrogen atom is – 13.6 eV. If an electron makes a transition from an energy level – 0.85 eV to –3.4 eV, calculate the wavelength of the spectral line emitted. To which series of hydrogen spectrum does this wavelength belong?
Using the formula,
For, n=1; E1 = - 13.6 eV
During the electron transmission, EA = - 0.85 eV to EB = -3.4 eV
So, from equation (i), we have
Therefore, electron transition takes place from n=4 to n=2 which corresponds to Balmer series.
We know,
Here,
nA = 4 ; nB = 2 ; R = 1.097 x 107 m-1
Then, 
a) In Young’s double slit experiment, derive the condition for (i) constructive interference and
(ii) Destructive interference at a point on the screen.
(b) A beam of light consisting of two wavelengths, 800 nm and 600 nm is used to obtain the interference fringes in a Young’s double slit experiment on a screen placed 1.4 m away. If the two slits are separated by 0.28 mm, calculate the least distance from the central bright maximum where the bright fringes of the two wavelengths coincide.
a)
Conditions of constructive interference and destructive interference.
Consider two coherent waves travelling in the same direction along a straight line.
Frequency of each wave is given by 
,
Amplitude of electric field vectors are a1 and a2 rspectively.
Wave equation is represented by, 
Intensity of the wave is proportional to the amplitude of the wave.
Thus, Intensity of the resultant wave is given by, 
Constructive interference: For maximum intensity at any point, cos = +1
So, maximum intensity is, 
Path difference is, 
Constructive interference is obtained when the path difference between the waves is an integral multiple of 
Destructive Interference: For minimum intensity at any point, cos = -1
Phase difference is given by, 
Path difference is, 
In destructive interference, path difference is odd multiple of 
.
b)
Given,
d = 0.28 mm = 0.28 x 10-3 m
As, 
If n1 = n then, n2 = n+1
(a) How does an unpolarized light incident on a Polaroid get polarized?
Describe briefly, with the help of a necessary diagram, the polarization of light by reflection from a transparent medium.
(b) Two Polaroid’s ‘A’ and ‘B’ are kept in crossed position. How should a third Polaroid ‘C’ be placed between them so that the intensity of polarized light transmitted by Polaroid B reduces to 1/8th of the intensity of unpolarized light incident on A?a) When an unpolarized light falls on a polaroid, only those electric vectors that are oscillating along a direction perpendicular to the aligned molecules will pass through. Thus, incident light gets linearly polarized.
Electric vectors which are along the direction of the aligned molecules gets absorbed.
Whenever unpolarized light is incident on the boundary between two transparent media, the reflected light gets partially or completely polarized. When reflected light is perpendicular to the refracted light, the reflected light is a completely polarized light.
b) Let the angle between A and C be 
.
Intensity of light passing through A = Io/2
Intensity of light passing through C = Io/2 cos2 
Intensity of light passing through B = Io/2 cos2 
. cos2 (90 - 
)
Given that, 
Thus, the third polaroid is placed at an angle of 45o.
(a) Describe briefly, with the help of a diagram, the role of the two important processes involved in the formation of a p-n junction.
(b) Name the device which is used as a voltage-regulator. Draw the necessary circuit diagram and explain its working.
OR
(a) Explain briefly the principle on which a transistor-amplifier works as an oscillator. Draw the necessary circuit diagram and explain its working.
(b) Identify the equivalent gate for the following circuit and write its truth table.
a) Important process occurring during the formation of p-n junction are:
1) Diffusion: When p-n junction is formed due to concentration gradient, the holes diffuse from p-side to n-side and electrons diffuse from n-side to p-side. This motion of charge carriers gives rise to diffusion current across the junction.
2) Drift: Drift of charge carriers occurs due to the applied electric field. As a result of the potential barrier, an electric field from n-region to p-region is developed across the junction. This electric field causes motion of electrons on p-side of the junction to n-side of the junction. Holes are moved from n-side to p-side. Thus, a drift current starts which is opposite to the direction of diffusion current.
b) Zener diode is used as voltage regulator. 
There is a voltage drop across Rs whenever there is decrease or increase in the input voltage. Change in voltage across Rs does not affect the voltage across Zener diode.
OR
a) The below fig. is that of a transistor amplifier.
The transistor amplifier works on the principle of mutual induction.
The above circuit contains tuned circuit made of variable capacitor C and an inductor L in the collector circuit and hence is named as tuned collector oscillator. The feedback coil L¢ connected to base circuit is mutually coupled with coil L.
The biasing is provided by emitter resistance RE and potential divider arrangement consisting of resistances R1 and R2. The capacitor C1 connected in the base circuit provides low reactance path to the oscillations and the capacitor CE is the emitter by-pass capacitor so that the resistor RE has no effect on the ac operation of the circuit.
Closing switch S will switch on the collector supply voltage. So, collector current starts increasing and capacitor C is charged. When the capacitor attains maximum charge, it discharges through coil L , setting up oscillations of natural frequency which is given by, 
These oscillations induce a small voltage in coil L’ by mutual induction. This induced voltage is the feedback voltage; its frequency is same as that of resonant LC circuit but its magnitude depends on the number of turns in L’ and coupling between L and L’ . The feedback voltage is applied between the base and emitter and appears in the amplified form in the collector circuit. A part of this amplifier energy is used to meet losses taking place in oscillatory circuit to maintain oscillations in tank circuit and the balance is radiated out in the form of electromagnetic waves.
b) The Boolean expression for the output is given by, 
This is the expression for AND gate.
Symbol for AND gate is given by, 
Truth table is given by, 
Write any two characteristic properties of nuclear force.
(i) Nuclear forces are short range attractive forces.
(ii) Nuclear forces are charge – independent.
What happens to the width of depletion layer of a p-n junction when it is (i) forward biased, (ii) reverse biased?
(i) When forward biased, the width of depletion layer decreases.
(ii) When reverse biased, the width of depletion layer increases.
Define the term ‘stopping potential’ in relation to photoelectric effect.
Stopping potential is the minimum retarding or negative potential of anode of a photoelectric tube for which photoelectric current stops or becomes zero.
What is sky wave communication? Why is this mode of propagation restricted to the frequencies only up to few MHz?
Sky wave propagation is a mode of propagation in which communication of radio waves in frequency range 30 MHz–40 MHz takes place due to reflection from the ionosphere.
For frequencies higher than few MHz, the sky waves penetrate the ionosphere and are not reflected back.
Using the curve for the binding energy per nucleon as a function of mass number A, state clearly how the release in energy in the processes of nuclear fission and nuclear fusion can be explained.
The binding energy per nucleon increases when a heavy nucleus breaks into two lighter nuclei. That is, energy would be released during the process of nuclear fusion.
When light nuclei join to form a heavy nucleus, the binding energy per nucleon of fused heavier nucleus is more than the binding energy per nucleon of lighter nuclei. Again, there is release of energy in this processA convex lens made up of glass of refractive index 1.5 is dipped, in turn, in (i) a medium of refractive index 1.65, (ii) a medium of refractive index 1.33.
(a) Will it behave as a converging or a diverging lens in the two cases?
(b) How will its focal length change in the two media?Given, refractive index of glass, ng = 1.5
Therefore, focal length of lens in liquid, 
a) i) ng = 1.5, nl
So fi and fa are of opposite sign.
Therefore, convex in liquid 1 behaves as a diverging lens.
ii) Refractive index of medium, nl = 1.33
Therefore, fl and fa are of the same sign.
So, convex lens in liquid l behaves a s a convergent lens.
b)
i) Focal length, f1 is, 
Focal length becomes negative and it’s magnitude increases.
ii) Focal length f2 is given by, 
Focal length increases in magnitude and is positive.
Draw a plot showing the variation of photoelectric current with collector plate potential for two different frequencies, v1 >v2 , of incident radiation having the same intensity. In which case will the stopping potential be higher? Justify your answer.
The plots showing the variation of photoelectric current with collector potential is shown below.
Stopping potential is directly proportional to the frequency of incident radiation.
Write briefly any two factors which demonstrate the need for modulating a signal.
Draw a suitable diagram to show amplitude modulation using a sinusoidal signal as the modulating signal.
Two factors justifying the need for modulation are:
(i) Practical size of antenna.
(ii) To avoid mixing up of signals from different transmitters i.e., avoid fluctuation and reduce the noise of signal.
Use the mirror equation to show that:
(a) an object placed between f and 2f of a concave mirror produces a real image beyond 2f.
(b) a convex mirror always produces a virtual image independent of the location of the object.
(c) an object placed between the pole and focus of a concave mirror produces a virtual and enlarged image.
Mirror equation is given by,
a)
For a concave mirror, f is negative, i.e., f < 0
For a real object i.e., which is on the left side of the mirror,
For u between f and 2f implies that 1/u lies between 1/f and 1/2f
i.e., 
Implies, v is negative and greater than 2f. Therefore, image lies beyond 2f and it is real.
b)
Focal length is positive for convex mirror, i.e., f > 0.
For a real object on the left side of the mirror, u is negative.
That is, 
Since u is negative and f is positive so, 1/v should also be positive, so v must be positive.
Hence, image is virtual and lies behind the mirror.
c)
Using the mirror equation, we have
That is, the image is enlarged.
(a) Using de Broglie’s hypothesis, explain with the help of a suitable diagram, Bohr’s second postulate of quantization of energy levels in a hydrogen atom.
(b) The ground state energy of hydrogen atom is –13.6 eV. What are the kinetic and potential energies of the electron in this state?a) According to De- broglie hypothesis
…(i)
As per De-broglie’s condition of stationary orbits, the stationary orbits are those which contain complete de-Broglie wavelength.
... (ii)
Putting value of from equation (i) into (ii)
which is the required postulate for Bohr’s quantization of energy levels.
b) Given, ground state energy of hydrogen atom = -13.6 eV
Kinetic energy is given by,
Potential energy is given by,
So, total energy is given by,
Now, from equations (i), (ii) and (iii), we can see that
K.E = - E and U = 2E
We have, E = -13.6 eV
So,
K.E = 13.6 eV
P.E, U = 2 x (-13.6) = -27.2 eV
You are given a circuit below. Write its truth table. Hence, identify the logic operation carried out by this circuit. Draw the logic symbol of the gate it corresponds to. 
The logical expression for output is given by,
which is the equivalent of AND gate.
Logic symbol for AND gate is given by, 
Truth table for AND gate is given by, 
A compound microscope uses an objective lens of focal length 4 cm and eyepiece lens of focal length 10 cm. An object is placed at 6 cm from the objective lens. Calculate the magnifying power of the compound microscope. Also calculate the length of the microscope.
OR
A giant refracting telescope at an observatory has an objective lens of focal length 15 m. If an eyepiece lens of focal length 1.0 cm is used, find the angular magnification of the telescope.
If this telescope is used to view the moon, what is the diameter of the image of the moon formed by the objective lens? The diameter of the moon is 3.42 × 106 m and the radius of the lunar orbit is 3.8 × 108 m.
Focal length of the objective lens, fo = 4 cm
Focal length of the eyepiece, fe = 10 cm
Object distance, uo = -6 cm
Magnifying power of a microscope is given by, 
Now, using the lens formula, we have
Negative sign shows that the image is inverted.
Length of the microscope is given by,
For eyepiece of the microscope,
We have, ve = D = -25 cm
Here, L is the length of the microscope.
OR
Angular magnification is given by, 
where, fo is the focal length of the objective lens, and
fe is the focal length of the eye-piece.
Given, diameter of the moon = 3.48 × 106 m
Radius of the lunar orbit = 3.8 × 108 m
Diameter of the image of moon formed by the objective lens is given by, d = 
State the importance of coherent sources in the phenomenon of interference.
In Young’s double slit experiment to produce interference pattern, obtain the conditions for constructive and destructive interference. Hence, deduce the expression for the fringe width.
How does the fringe width get affected, if the entire experimental apparatus of Young is immersed in water?
If coherent sources are not taken, the phase difference between two interfering waves, will change continuously and a sustained interference pattern will not be obtained. Thus, coherent sources provide sustained interference pattern.
Let S1 and S2 be two coherent sources separated by a distance d.
Let the distance of the screen from the coherent sources be D. Let point M be the foot of the perpendicular drawn from O, which is the midpoint of S1 and S2 on the screen. Obviously point M is equidistant from S1 and S2. Therefore the path difference between the two waves at point M is zero. Thus the point M has the maximum intensity.
Now, Consider a point P on the screen at a distance y from M. Draw S1N perpendicular from S1 on S2 P. 
The path difference between two waves reaching at P from S1 and S2 is given by, 
Expression for Fringe Width:
Distance between any two consecutive bright fringes or any two consecutive dark fringes are called the fringe width. It is denoted by 
.
Let, 
be the distance of two consecutive fringes. Then, we have
So, fringe width = 
Fringe width is same for both bright and dark fringe.
When the entire apparatus is immersed in water, the fringe width decreases.
(a) State Huygens principle. Using this principle explain how a diffraction pattern is obtained on a screen due to a narrow slit on which a narrow beam coming from a monochromatic source of light is incident normally.
(b) Show that the angular width of the first diffraction fringe is half of that of the central fringe.
(c) If a monochromatic source of light is replaced by white light, what change would you observe in the diffraction pattern?
Huygen’s principle:
(i) Every point on a given wavefront may be regarded as a source of new disturbance.
(ii) The new disturbances from each point spread out in all directions with the velocity of light and are called the secondary wavelets.
(iii) The surface of tangency to the secondary wavelets in forward direction at any instant gives the new position of the wavefront at that time.
a) Diffraction of light at a Single slit
A single narrow slit is illuminated by a monochromatic source of light. The diffraction pattern is obtained on the screen placed in front of the slits. There is a central bright region called as central maximum. All the waves reaching this region are in phase hence the intensity is maximum. On both side of central maximum, there are alternate dark and bright regions, the intensity becoming weaker away from the center. The intensity at any point P on the screen depends on the path difference between the waves arising from different parts of the wave-front at the slit.
According to the figure, the path difference (BP – AP) between the two edges of the slit can be calculated.
Path difference, BP – AP = NQ = a sin 
Angle is zero at the central point C on the screen, therefore all path difference are zero and hence all the parts of slit are in phase. Due to this, the intensity at C is maximum. If this path difference is 
, (the wavelength of light used), then P will be point of minimum intensity. This is because the whole wave-front can be considered to be divided into two equal halves CA and CB and if the path difference between the secondary waves from A and B is 
, then the path difference between the secondary waves from A and C reaching P will be 
/2, and path difference between the secondary waves from B and C reaching P will again be 
/2. Also, for every point in the upper half AC, there is a corresponding point in the lower half CB for which the path difference between the secondary waves, reaching P is 
/2. Thus, destructive interference takes place at P and therefore, P is a point of first secondary minimum.
b)
Central bright lies between 
Therefore, Angular width of central bright fringe = 
So, 1st diffraction fringe lies between 
and 
Therefore,
Angular width of first diffraction fringe is, 
So, 
c) When monochromatic light is replaced by white light, each diffraction band splits into a number of colored bands. Angular width of violet being the least and that of red light is maximum.
Explain the terms (i) Attenuation and (ii) Demodulation used in Communication System.
, where V is the is accelerating potential for two particles A and B, carrying the same charge but different masses m1 and m2 (m1 > m2). Which one of the two represents a particle of smaller mass and why? Given,
Charge (q) is the same but the mass of both particles is different (m1 > m2).
The De-broglie wavelength given by, 
The slope of the graph of λ versus 
= 
The slope of the smaller mass is larger; therefore, plat A in the above graph represents mass m2.
The B.E. of the nucleus of mass number 240, B1 = 7.6 x 240 = 1824 MeV
The B.E of each product nucleus, B2 = 8.5 x 120 - 1020 MeV
Then, the energy released as the nucleus breaks is given by,
E = 2B2 - B1 = 2 x 1020 - 1824 = 216 MeV
OR
Given:
B.E of 
B.E of 
Energy in the fusion reaction is given by, 
The three characteristics of photoelectric effect, which cannot be explained on the basis of wave theory of light are:
1. For a given metal and frequency of incident radiation, the number of photoelectrons ejected per second is directly proportional to the intensity of the incident light.
2. For a given metal, a certain minimum frequency of the incident radiation below which no emission of photoelectrons take place. The is the threshold frequency.
3. Above the threshold frequency, the maximum kinetic energy of the emitted photoelectron is independent of the intensity of the incident light and is dependent only upon the frequency or wavelength of the incident light.
(a) Explain any two factors that justify the need of modulating a low-frequency signal.
(b) Write two advantages of frequency modulation over amplitude modulation.
Modulating low-frequency signals are required because:
i) Size of the antenna: Antenna is required for both transmission and reception of the signal whose size (at least 
) should be comparable to the wavelength of the signal so that time variation of the signal can be properly sensed by the antenna. For a low-frequency signal, the wavelength is large. Therefore, a large antenna of such a huge height is practically impossible to construct and operate.
Hence, the need for high-frequency transmission arises.
ii) Overlapping signals from different transmitters: When an information is transmitted using different transmitters, the signals get overlapped and the information is lost. Hence, high-frequency transmission is required. Each band should be allocated to each frequency range in order to avoid mixing of signals.
iii) Higher energy: High-frequency signals have high energy and therefore, even after loss due to attenuation, signals can be carried over longer distances.
b)
Advantage of frequency modulation over amplitude modulation are:
i) FM is more efficient than AM because the amplitude of an FM wave is constant, irrespective of its modulation index. Thus, the transmitted power is constant. Also, in AM the transmitted power goes waste in the transmitted carrier.
ii) FM reception is quite immune to noise in comparison to AM reception. Noise is a form of amplitude variation in the transmitted signal. Thus, using amplitude limiters in the FM receivers, the noise is eliminated.
i) The function of three segments of a transistor are:
Emitter: Emits the majority charge carriers
Collector: Collects the majority charge carriers
Base: Base provides the interaction between the collector and the base
ii) An n-p-n transistor is as shown below:
Input characteristics: The variation between the base current and the base-emitter voltage is obtained. Input characteristics is used to find input dynamic resistance of the transistor as it is represented by the slope.
Output characteristics: A graph representing the variation of the collector current and emitter voltage is obtained, keeping base current fixed. The slope fo the output characteristic graph gives us the output dynamic resistance.
a)
Given,
Radius of curvature, R = 20 cm
So, focal length, f = R/2 = - 10 cm
Since the image obtained is real, therefore magnification of the image, m = -2
Now, using the formula, 
Therefore, the distance of the object is 15 cm in front of the mirror and the position of the image is 30 cm, formed in front of the mirror.
b)
For a convex mirror,
Focal length, f > 0
Position of the object, u < 0
Using mirror formula, we have
That is, the image formed by a convex lens is always behind the mirror and hence is virtual.
Bohr's Quantisation Rule:
According to Bohr, an electron can revolve only in certain discrete, non-radiating orbits for which the total angular momentum of the revolving electron is an integral multiple of 
; where h is the Planck's constant.
That is, 
b)
Using Rydberg's formula for spectra of hydrogen atom, we have
Hence, the relation between 3 wavelengths from the energy-level diagram is obtained.
A reflecting telescope is as shown below: 
In a reflecting telescope, an image is formed by reflection from a curved mirror. The image formed is then magnified by a secondary mirror.
Advantages of reflecting telescope over a refracting telescope are:
1. There is no chromatic aberration for reflecting telescopes as the objective is a mirror.
2. Spherical aberration is reduced in the case of reflecting telescopes by using mirror objective in the form of a paraboloid.
i)
The values displayed by Meeta and her father are:
Meeta: Curious mind
Meeta's Father: knowledge and patience
ii)
The answer that Meeta's father had given would be the advantages of using LED lights over a single bulb.
Advantages of LED:
a) LED's consume very less power as compared to an incandescent bulb.
b) The cost of tiny LED is much less than of a bulb. This reduces the maintenance cost of LED bulbs.
c) The working of the traffic will remain unhindered even if one of the bulbs is not working.
iii)
The tiny lights are called LED (Light Emitting Diode) and they work on the principle of de-excitation of electrons in a forward biased semiconductor upon passing electricity through them.
Expression for fringe width in Young's Double Slit Experiment
Let S1 and S2 be two slits separated by a distance d.
GG' is the screen at a distance D from the slits S1 and S2.
Point C is equidistant from both of the slits.
The intensity of light will be maximum at this point because the path difference of the waves reaching this point will be zero.
At point P, the path difference between the rays coming from the slits is given by,
S1 = S2 P - S1 P
Now, S1 S2 = d, EF = d, and S2 F = D
In 
S2P = 
For constructive interference, the path difference is an integral multiple of wavelengths, that is, path difference is n
.
Graph of intensity distribution in young's double slit experiment is, 
ii)
Three distinguishing features observed in Young's Double Slit experiment as compared to single slit diffraction pattern is,
1. In the interference pattern, all the bright fringes have the same intensity. The bright fringes are not of the same intensity in a diffraction pattern.
2. In interference pattern, the dark fringes have zero or small intensity so that the bright and dark fringes can be easily distinguished. While in diffraction pattern, all the dark fringes are not of zero intensity.
3. In interference pattern, the width of all fringes are almost the same, whereas in diffraction pattern, the fringes are of different widths.
i)
If the angle of incidence is increased gradually, then the angle of deviation first decreases, attains a minimum value (
) and then again starts increasing.
; where a, b and c are constants. 
Name the essential components of a communication system.
Essential components of communication system:
1. Transmitter
2. Channel
3. Receiver
Why does sun appear red at sunrise and sunset?
Sun appears red at sunrise and sunset because of least scattering of red light as it has the lowest wavelength.
As per Rayleigh scattering, the amount of light scattered 
.
Define modulation index. Why is it kept low ? What is the role of a bandpass filter?
Modulation index is the ratio of the amplitude of modulating signal to that of carrier wave.
Modulation index is kept low in order to avoid distortion. The low frequency modulating signal is mixed with high-frequency carrier wave, the distortion is restricted due the high-frequency carrier wave for modulation index lying between 0 and 1.
Bandpass filter rejects low and high frequencies and allows a band of frequencies to pass through.
A ray PQ incident normally on the refracting face BA is refracted in the prism BAC made of material of refractive index 1.5. Complete the path of ray through the prism. From which face will the ray emerge ? Justify your answer.
Here,
sin ic = 
The light will emerge out from face AC.
Calculate the de-Broglie wavelength of the electron orbitting in the n=2 state of hydrogen atom.
Given,
n =2
K.E for the second state, Ek,
De-Broglie wavelength, 
Define ionisation energy.
How would the ionisation energy change when electron in hydrogen atom is replaced by a particle of mass 200 times that of the electron but having the same charge ?
The minimum energy, required to free the electron from the ground state of the hydrogen atom, is known as ionization energy.
Ionisation energy is given by,
Since, ionisation energy is directly proportional to mass, the Ionization energy will become 200 times.
Calculate the shortest wavelength of the spectral lines emitted in Balmer series. [ Given Rydberg constant, R = 107 m-1]
The formula is,
For shortest wavelength,
Therefore,
(i) State law of Malus.
(ii) Draw a graph showing the variation of intensity (I) of polarised light transmitted by an analyser with angle (
) between polarizer and analyser.
(iii) What is the value of refractive index of a medium of polarising angle 60o?
i) Law of Malus states that when a completely plane polarised light beam is incident on an analyser, the intensity of the emergent light varies as the square of the cosine of the angle between the plane of transmission of the analyser and the polariser.
I = Io cos2
ii) The variation of intensity (I) of polarised light transmitted by an analyser with angle 
.
Sketch the graphs showing variation of stopping potential with frequency of incident radiations for two photosensitive materials A and B having threshold frequencies 
A > 
B.
(i) In which case is the stopping potential more and why?
(ii) Does the slope of the graph depend on the nature of the material used? Explain.
Graph showing the variation of stopping potential with frequency of incident radiations:
From the above graph, we can observe that,
i) The stopping potential is inversely proportional to the threshold frequency. Hence, the stopping potential is higher for metal B.
ii)The slope of the graph does not depend on the nature of the material used:
We know that,
Dividing the equation throughout by e, we get
On comparing the above equation with the straight line equation, we get
Slope of the graph = 
That is, the slope of the graph does not depend on the nature of the material used.
a) Write the basic nuclear process involved in the emission of beta plus in a symbolic form, by a radioactive nucleus.
b) In the reactions given below:
Find the values of x, y and z and a, b and c.
a)
The basic nuclear process involved in the emission of beta plus in a symbolic form, by a radioactive nucleus,
In a beta-plus decay, a proton transforms into a neutron within the nucleus.
b)
i) 
The corresponding y and z are 5 and 11 respectively.
Here, x is the positron.
ii) 
The corresponding values of a, b and c are 10, 2 and 4 respectively.
(i) Explain with the help of a diagram the formation of depletion region and barrier potential in a p-n junction.
(ii) Draw the circuit diagram of a half wave rectifier and explain its working.
i)
(i) Which mode of propagation is used by shortwave broadcast services having frequency range from a few MHz upto 30 MHz ? Explain diagrammatically how long distance communication can be achieved by this mode.
(ii) Why is there an upper limit to frequency of waves used in this mode?
i) Sky-wave propagation is used for long distance communication. Long distance communication is achieved by reflection of radio waves by the ionosphere, back towards the earth. This ionosphere layer acts as a reflector for certain range of frequencies from few MHz to 30 MHz.
For a CE-transistor amplifier, the audio signal voltage across the collector resistance of 2 k
is 2 V. Suppose the current amplification factor of the transistor is 100, find the input signal voltage and base current, if the base resistance is 1 kV.
Given,
Output voltage, Vo = 2 V
Output resistance, Ro = 2 k
Base resistance, Ri = 1 k
Current amplification factor, 
Then input signal voltage is,
Now, collector current is,
IC = 
Therefore,
Base current, IB = 
Define the term wave front. State Huygen’s principle.
Consider a plane wave front incident on a thin convex lens. Draw a proper diagram to show how the incident wave front traverses through the lens and after refraction focusses on the focal point of the lens, giving the shape of the emergent wave front.
OR
Explain the following, giving reasons :
(i) When monochromatic light is incident on a surface separating two media, the reflected and refracted light both have the same frequency as the incident frequency.
ii) When light travels from a rarer to a denser medium, the speed decreases. Does this decrease in speed imply a reduction in the energy carried by the wave?
(iii) In the wave picture of light, intensity of light is determined by the square of the amplitude of the wave. What determines the intensity in the photon picture of light?
Wavefront is defined as the locus of all the points in space that reach a particular distance by a propagating wave at the same instant.
Huygen's principle:
i) Each point of the wavefront is the source of a secondary disturbance and the wavelets emanating from these points spread out in all directions. These travel with the same velocity as that of the original wavefront.
ii) The shape and position of the wave-front, after time 't', is given by the tangential envelope to the secondary wavelets.
A plane wavefront is incident on a thin convex lens:
OR
i) Reflection and refraction arise through interaction of incident light with atomic constituents of matter which vibrate with the same frequency as that of the incident light. Hence, frequency remains unchanged.
ii) No, energy carried by a wave depends on the amplitude of the wave, not on the speed of wave propagation.
iii) For a given frequency, intensity of light in the photon picture is determined by the number of photon incident normally on crossing unit area per unit time.
(i) Derive the mathematical relation between refractive indices n1 and n2 of two radii and radius of curvature R for refraction at a convex spherical surface. Consider the object to be a point since lying on the principle axis in rarer medium of refractive index n1 and a real image formed in the denser medium of refractive index n2. Hence, derive lens maker’s formula.
(ii) Light from a point source in air falls on a convex spherical glass surface of refractive index 1.5 and radius of curvature 20 cm. The distance of light source from the glass surface is 100 cm. At what position is the image formed ?
The figure given above shows the geometry of formation of real image I of an object O and the principal axis of a spherical surface with centre of curvature c and radius of curvature R.
(a) Draw a labelled ray diagram to obtain the real image formed by an astronomical telescope in normal adjustment position. Define its magnifying power.
(b) You are given three lenses of power 0.5 D, 4 D and 10 D to design a telescope.
(i) Which lenses should he used as objective and eyepiece ? Justify your answer.
ii) Why is the aperture of the objective preferred to be large?
a) The image formed by an astronomical telescope in normal adjustment position is given below,
are very small,
Define the distance of closest approach. A α-particle of kinetic energy 'K' is bombarded on a thin gold foil. The distance of the closest approach is 'r'. What will be the distance of closest approach for a α-particle of double the kinetic energy?
It is the distance of charged particle from the centre of the nucleus, at
which the whole of the initial kinetic energy of the (far off) charged
particle gets converted into the electric potential energy of the system.
Distance of closest approach (rc) is given by
(rc) is inversely proportional to K. Thus,
‘K’ is doubled therefore rc becomes r/2
Write two important limitations of Rutherford nuclear model of the atom.
1. According to Rutherford model, an electron orbiting around the nucleus continuously radiates energy due to the acceleration; hence the atom will not remain stable.
2. As electron spirals inwards; its angular velocity and frequency change continuously; therefore it will emit a continuous spectrum.
Radius of ground state of hydrogen atom =0.53A° = 0.53 x10-10m
According to de Broglie relation 2πr = nλ
For ground state n=1
2 x 3.14 x 0.53 x 10-10= 1 x λ
therefore, λ = 3.32 x10-10 m
= 3.32 A°
Magnifying power is defined as the angle subtended at the eye by the image
to the angle subtended (at the unaided eye) by the object.
To increase the magnifying power both the objective and eyepiece must have short focal lengths such as 
Which basic mode of communication is used in satellite communication? What type of wave propagation is used in this mode? Write, giving a reason, the frequency range used in this mode of propagation.
There are two basic modes of communication which are used in satellite communication:
(i) point-to-point and broadcast- In the broadcast mode, there are a large number of receivers corresponding to a single transmitter. Satellite communication is an example of the broadcast mode of communication.
(ii) Space wave mode of propagation: In the frequency range 30 MHz to 1000 Mhz, the wavelengths are in the range of 30 cm to 10 m. At these wavelengths the diffraction of waves is high and the waves lose their directional property. So frequency range of 5.925 - 6.425 GHz is used in satellite communication.
Diffusion and Drift are the two processes that take place in the formation of a p-n junction.
In an n-type semiconductor, the concentration of electrons (number of electrons per unit volume) is more compared to the concentration of holes. Similarly, in a P -type semiconductor, the concentration of holes is more than the concentration of electrons. During the formation of p-n junction.
Due to the diffusion of electrons and holes across the junction a region of
(immobile) positive charge is created on the n-side and a region of(immobile) negative charge is created on the p-side, near the junction; this is called depletion region.
Barrier potential is formed due to loss of electrons from n-region and gain of electrons by p-region. Its polarity is such that it opposes the movement of charge carriers across the junction.
(i) Obtain the expression for the cyclotron frequency.
(ii) A deuteron and a proton are accelerated by the cyclotron. Can both be accelerated with the same oscillator frequency? Give reason to justify your answer.
When a charged particle (q) moves inside a magnetic field B with velocity v, it experiences a force F = q(v xB) when v is perpendicular to B, the force F on the charged acts as the centripetal force and makes it move along a circular path.
Let m be the mass of particle and r is radius of circular path
Time period of the circular motion of a charged particle is given by
ii) No, The mass of the two particles, i.e deuteron and proton, is different.Since (cyclotron) frequency depends inversely on the mass, they cannot be accelerated by the same oscillator frequency.
How does one explain the emission of electrons from a photosensitive surface with the help of Einstein's photoelectric equation?
Einstein’s Photoelectric equation is
When a photon of energy is incident on the metal, some part of this
energy is utilized as work function to eject the electron and remaining
energy appears as the kinetic energy of the emitted electron.
The work function of the following metals is given : Na 2.75 ev, K = 2.3 eV, Mo = 4.17 eV and Ni = 5.15 eV. Which of these metals will not cause photoelectric emission for radiation of wavelength 3300 Å from a laser source placed 1 m away from these metals? What happens if the laser source is brought nearer and placed 50 cm away?
The work function of Mo and Ni is more than the energy of the incident
photons; so photoelectric emission will not take place from these metals.
Kinetic energy of photo electrons will not change, only photoelectric current
will change.
Define the term 'amplitude modulation'. Explain any two factors which justify the need for modulating a low-frequency base band signal.
It is the process of superposition of information/message signal over a carrier wave in such a way that the amplitude of carrier wave is varied according to the information signal/message signal.
Direct transmission, of the low frequency base band information signal, is not possible due to the following reasons;
(i) Size of Antenna: For transmitting a signal, minimum height of the antenna should be λ/4; with the help of modulation wavelength of signal decreases, hence the height of antenna becomes manageable.
(ii) Effective power radiated by an antenna: Effective power radiated by an antenna varies inversely as λ2, hence effective power radiated into the space, by the antenna, increases.
(iii)To avoid mixing up of signals from different transmitters.
A radioactive nucleus 'A' undergoes a series of decays as given below:
The mass number and atomic number of A2 are 176 and 71 respectively. Determine the mass and atomic number of A4 and A.
The mass number and atomic number of A4 is 172 and 69, respectively.
The mass number and atomic number of A is 180 and 74, respectively.
Write the basic nuclear process underlying β+ and β- decays.
Basic process underlying β+ and β- decay are
During a weak interaction an atomic nucleus converts into a nucleus with one higher atomic number while emitting one electron and an electron antineutrino this is called beta minus decay.
During a weak interaction an atomic nucleus converts into a nucleus with one Lower atomic number while emitting one electron and an electron neutrino this is called beta minus decay.
For a CE-transistor amplifier, the audio signal voltage across the collector resistance of 2 kΩ is 2 V. Given the current amplification factor of the transistor is 100, find the input signal voltage and base current, if the base resistance is 1 kΩ.
Given, Rc = 2000Ω
RB = 1000 Ω,
β = 100,
voltage gain can given as,
Mrs Rashmi Singh broke her reading glasses. When she went to the shopkeeper to order new specs, he suggested that she should get spectacles with plastic lenses instead of glass lenses. On getting the new spectacles, she found that the new ones were thicker than the earlier ones. She asked this question to the shopkeeper but he could not offer a satisfactory explanation for this. At home, Mrs Singh raised the same question to her daughter Anuja who explained why plastic lenses were thicker.
(a) Write two qualities displayed each by Anuja and her mother.
(b) How do you explain this fact using lens maker's formula?
(a) Two qualities displayed each by Anuja and her mother:
Anuja has good knowledge of lens and she is very co-operative.
Mrs Rashmi is curious to know about lenses and science behind it.
(b)The lens maker formula is given as 
where f is the focal length, μ is the refractive index and R1 and R2 is the radius of curvature of the lens.
Since μg>μp
where g stands for glass and p stands for plastic
Therefore, we get (μ-1)> (μp-1)
Now, using the lens maker formula, we see that focal length is inversely proportioned to (μ-1)
Hence, fp>fg
Thus, in the case of the plastic lens, the thickness of the lens should be increased to keep the same focal length as that of the glass lens to give the same power.
(i) A ray of light incident on face AB of an equilateral glass prism shows the minimum deviation of 30°. Calculate the speed of light through the prism.
(ii) Find the angle of incidence at face AB so that the emergent ray grazes along the face AC.
At the minimum deviation, the refracted ray inside the prism becomes parallel to its base.
Angle of minimum deviation is given as Dm = 30°
Since, the prism is equilateral, So, A = 60°
Refractive index of the prism
We know that μ = v1/v2
Hence the speed of light in prism would be 1/√2 times the speed of light in air i.e = 3 x108 /1.414 = 2.121 x108 m/s
(ii)
From Snell's law, we know that 
For the emergent ray to graze at the face AC, the angle of refraction should be 90
So, applying snell's law at face AC, we get
From figure, we can see that angle of refraction at face AB is 15
So applying Snell's law we get
Sin iAB = sin rAB x μ12
or iAB = sin-1 (√sin 15°)
In a beam of unpolarized light, the vibrations of light vectors are in all directions in a plane perpendicular to the direction of propagation. 
In the linearly polarised light, the vibrations of light take place in a particular direction, perpendicular to the direction of wave motion.
Polarized light can be distinguished, from unpolarized light, when it is allowed to pass through a Polaroid. Polarized light does can show the change in its intensity, on passing through a Polaroid; intensity remains same in case of unpolarized light.
When the unpolarised light wave is incident on a polaroid, then the electric vectors along the direction of its aligned molecules, get absorbed; the electric vector, oscillating along a direction perpendicular to the aligned molecules, pass through. This light is called linearly polarized light.
A narrow beam of unpolarized light of intensity I0 is incident on a Polaroid P1. The light transmitted by it is then incident on a second polaroid P2 with its pass axis making the angle of 60° relatives to the pass axis of P1. Find the intensity of the light transmitted by P2.
According to Malus’ Law:
Explain two features to distinguish between the interference pattern in Young's double slit experiment with the diffraction pattern obtained due to a single slit.
| Interference pattern | Diffraction pattern |
| 1) All fringes are of equal width. | 1) The width of central maxima is twice the width of higher order band. |
| 2) Intensity of all bright bands is equal. | 2) The intensity goes on decreasing for a higher order of diffraction bands. |
Angular width of central maximum
Linear width of central maxima in the diffraction pattern
Let ‘n’ be the number of interference fringes which can be accommodated in the central maxima
If one of two identical slits producing interference in Young’s experiment is convered with glass, so that the light intensity passing through it is reduced to 50%, find the ratio of the maximum and minimum intensity of the fringe in the interference pattern.
As we know that intensity is directly proportional to the square of the amplitude.
What kind of fringes do you expect to observe if white light is used instead of monochromatic light?
If the White light is used instead of monochromatic light then we see a sequence in which central fringe will be white and remaining will be coloured in VIBGYOR sequence.
Define a wavefront. Using Huygens’ principle, verify the laws of reflection at a plane surface.
A wavefront is an imaginary surface over which an optical wave has a constant phase. The shape of a wavefront is usually determined by the geometry of the source.
Huygen's principle:
(i) Every point on a given wavefront acts as a fresh source of secondary wavelets which travel in all directions with the speed of light.
(ii) The forward envelope of these secondary wavelets gives the new wavefront at any instant.
Laws of reflection by Huygen's principle:
Let, PQ be reflecting surface and a plane wavefront AB is moving through the medium (air) towards the surface PQ to meet at the point B. 
Let, c be the velocity of light and t be the time taken by the wave to reach A' from A.
Then, AA' = ct.
Using Huygen's principle, secondary wavelets starts from B and cover a distance ct in time t and reaches B'.
To obtain new wavefront, draw a circle with point B as centre and ct (AA' = BB') as radius. Draw a tangent A'B' from the point A'.
Then, A'B' represents the reflected wavelets travelling at right angle. Therefore, incident wavefront AB, reflected wavefront A'B' and normal lies in the same plane.
Consider ∆ABA' and B'BA'
AA' = BB' = ct [∵ AA' = BB' = BD = radii of same circle]
BA' = BA' [common]
∠BAA' = ∆BB'A' [each 90°]
∴ ∆ABA' ≅ ∠DBA' [by R.H.S]
∠ABA' = ∠B'A'B [corresponding parts of congruent triangles]
∴ incident angle i = reflected angle r
i.e., ∠i = ∠r
In a single slit diffraction experiment, the width of the slit is made double the original width. How does this affect the size and intensity of the central diffraction band ? Explain.
Size will be halved, intensity will be increased to four times.
When a tiny circular obstacle is placed in the path of light from a distant source, a bright spot is seen at the centre of the obstacle. Explain why
A bright spot is seen at the centre of the shadow of objective because wave diffracted from the edge of the circular obstacle interface constructively at the centre of the shadow producing a bright spot.
The temperature dependence of resistances of Cu and undoped Si in the temperature range 300-400 K is best described by:
Linear increase for Cu, linear increase for Si.
Linear increase for Cu, exponential increase for Si.
Linear increase for Cu, exponential decrease for Si.
Linear decrease for Cu, linear decrease for Si.
C.
Linear increase for Cu, exponential decrease for Si.
As we know Cu is a conductor, so increase in temperature, resistance will increase. Then, Si is a semiconductor, so with the increase in temperature, resistance will decrease.
Arrange the following electromagnetic radiations per quantum in the order of increasing energy:
A: Blue light
B: Yellow light
C: X-ray
D: Radiowave
D, B, A, C
A, B, D, C
C, A, B, D
B, A, D, C
A.
D, B, A, C
As, we know energy liberated, E = hc/λ
i.e 
So, lesser the wavelength greater will be energy liberated by electromagnetic radiations per quantum.
As, order of wavelength is given by
X- rays, VIBGYOR, Radio waves
therefore, the order of electromagnetic radiations per quantum.
D<B<A<C
An observer looks at a distant tree of height 10 m with a telescope of magnifying power of 20. To the observer the tree appears:
10 times taller.
10 times nearer.
20 times taller.
20 times nearer
C.
20 times taller.
The height of image depends upon magnifying rows to see 20 times taller object, angular magnification should be 20 and we observe angular magnification option (c) would not be very correct as the telescope can be adjusted to form the image anywhere between infinity and least distance for distinct vision.Suppose that the image is formed at infinity. Then. the observer will have to focus the eyes at infinity to observe the image. Hence, it is incorrect to say that the image will appear nearer to the observer.
The box of a pinhole camera, of length L, has a hole of radius a. It is assumed that when the hole is illuminated by a parallel beam of light of wavelength λ the spread of the spot (obtained on the opposite wall of the camera) is the sum of its geometrical spread and the spread due to diffraction. The spot would then have its minimum size (say bmin) when:
C.
In diffraction, first minima, we have
sin θ = λ/a
So , size of a spot
b= 2a +2Lλ/a (i)
Then, minimum size of a spot, we get
Radiation of wavelength λ is incident on a photocell. The fastest emitted electron has to speed v. If the wavelength is changed to 3λ/4, the speed of the fastest emitted electron will be:
= v(4/3)1/2
=v(3/4)1/2
A.
According to the law of conservation of energy, i.e. Energy of a photon (hv) = work function (Φ) + Kinetic energy of the photoelectron (mv2/2)
according to Einstein's photoelectric emission of light,
E =(KE)max + Φ
As, hc/λ = (KE)max + Φ
If the wavelength of radiation is changed to 3λ/4
then,
Half-lives of two radioactive elements A and B are 20 minutes and 40 minutes, respectively. Initially, the samples have equal number of nuclei. After 80 minutes, the ratio of decayed numbers of A and B nuclei will be:
1: 16
4 : 1
1: 4
5: 4
D.
5: 4
Given 80 min = 4 half-lives of A = 2 half-lives of B.
Let the initial number of nuclei in each sample be N.
For radioactive element A,
NA after 80 min = N/24
⇒ Number of A nuclides decayed =
If a, b, c, d are inputs to a gate and x is its output, then, as per the following time graph, the gate is:
NOT
AND
OR
NAND
C.
OR
The output of OR gate is 0 when all inputs are 0 and output is 1 when at least one of the input is 1.
Observing output x it is 0 when all inputs are 0 and it is 1 when at least one of the inputs is 1.
therefore, the gate is OR
Choose the correct statement:
In amplitude modulation, the amplitude of the high-frequency carrier wave is made to vary in proportion to the amplitude of the audio signal.
In amplitude modulation, the frequency of the high-frequency carrier wave is made to vary in proportion to the amplitude of the audio signal.
In frequency modulation, the amplitude of the high-frequency carrier wave is made to vary in proportion to the amplitude of the audio signal.
In frequency modulation, the amplitude of the high-frequency carrier wave is made to vary in proportion to the frequency of the audio signal.
B.
In amplitude modulation, the frequency of the high-frequency carrier wave is made to vary in proportion to the amplitude of the audio signal.
As, we know, an amplitude modulated wave, the bandwidth is twice the frequency of modulating the signal. Therefore, amplitude modulation (AM), the frequency of the high-frequency carrier wave is made to vary in proportion to the amplitude of the audio signal.
A pipe open at both ends has a fundamental frequency f in the air. The pipe is dipped vertically in water so that half of it is in water. The fundamental frequency of the air column is now:
f/2
3f/4
2f
f
D.
f
For open ends, fundamental frequency f in air
we have
When a pipe is dipped vertically in the water, so that half of it is in water, we have
Thus, the fundamental frequency of the air column is now,
f=f'
In an experiment for determination of refractive index of glass of a prism by i− δ, plot, it was found that a ray incident at angle 35o, suffers a deviation of 408 and that it emerges at angle 79o ⋅ Ιn that case which of the following is closest to the maximum possible value of the refractive index?
1.5
1.6
1.7
1.8
A.
1.5
C.
1.7
If μ is the refractive index of the material of prism, the from shell's law
This angle is greater than the 40o and deviation angle already given. For greaterμ deviation will be even higher. Hence μ of the given prism should be lesser than 1.5. Hence, the closest option will be 1.5
Identify the semiconductor devices whose characteristics are given below, in the order (a), (b), (c), (d):
A.
Zener diode works in breakdown region
So, simple diode → (a)
zener diode (b)
solar cell → (c)
Light dependent resistance → (d)
A signal of 5 kHz frequency is amplitude modulated on a carrier wave of frequency 2 MHz. The frequencies of the resultant signal is/are:
2005 kHz, and 1995 kHz
2005 kHz, 2000 kHz and 1995 kHz
2000 kHz and 1995 kHz
2 MHz only
C.
2000 kHz and 1995 kHz
Frequency associated with AM are
fc - fm, f, fc + fm
according to the question
fc = 2 MHz = 2000 kHz
Match List – I (Fundamental Experiment) with List – II (its conclusion) and select the correct option from the choices given below the list:
List I |
List II |
| (A) Franck-Hertz Experiment | (i) Particle nature of light |
| (B) Photo-electric experiment | (ii) Discrete energy levels of atom |
| (C) Davison – Germer Experiment | (iii) Wave nature of electron |
| (iv) Structure of atom |
| A | B | C |
| 1 | 4 | 3 |
| A | B | C |
| 2 | 4 | 3 |
| A | B | C |
| 2 | 1 | 3 |
| A | B | C |
| 4 | 3 | 2 |
C.
| A | B | C |
| 2 | 1 | 3 |
a) Frank-Hertz experiments are associated with the discrete energy level of the atom.
b) The photo-electric experiment is associated with particle nature of light.
c) Davisson -Germer experiment is associated with wave nature of the electron.
As an electron makes a transition from an excited state to the ground state of a hydrogen – like atom/ion:
kinetic energy, potential energy and total energy decrease
kinetic energy decreases, potential energy increases but total energy remains same
kinetic energy and total energy decrease but potential energy increases
its kinetic energy increases but the potential energy and total energy decrease
A.
kinetic energy, potential energy and total energy decrease
As we know that kinetic energy of an electron is
KE ∝ (Z/n)2
when the electron makes the transition from an excited state of the ground state then n, decreases and KE increases. We know that PE is lowest for the ground state. As TE=- KE and TE also decreases.
Assuming human pupil to have a radius of 0.25 cm and a comfortable viewing distance of 25 cm, the minimum separation between two objects that human eye can resolve at 500 nm wavelength is:
1 μm
30 μm
100 μm
300 μm
B.
30 μm
We can write resolving angle of necked eye as,
Where D is the diameter of the eye lens.
On a hot summer night, the refractive index of air is smallest near the ground and increases with height from the ground. When a light beam is directed horizontally, the Huygens’ principle leads us to conclude that as it travels, the light beam:
goes horizontally without any deflection
bends downwards
bends upwards
becomes narrower
D.
becomes narrower
According to Shell's law,
μ sinθ = constant
sin θ ∝ 1/μ
as μ increases, θ decreases.
A train is moving on a straight track with speed 20 ms-1 .It is blowing its whistle at the frequency of 1000 Hz. The percentage change in the frequency heard by a person standing near the track as the train passes him is (speed of sound = 320 ms-1 ) close to :
6%
12%
18%
24%
B.
12%
Apparent frequency heard by the person before crossing the train.
Similarly, apparent frequency heard, after crossing the trains
Monochromatic light is incident on a glass prism of angle A. If the refractive index of the material oft the prism is μ, a ray incident at an angle θ, on the face AB would get transmitted through the face AC of the prism provided
A.
The Ray will get transmitted through face AC if iAC < iC
Consider the ray diagram is shown below θ.
A ray of light incident on face AB at angle
r1 = Angle of refraction on face AB
r2 = Angle of incidence at face AC
For transmission of light through face AC
iAC < iC or A-ri < iC
or Sin (A-r1)1)< 1/μ
Now, applying snell's law at the face AB
1 x sin θ = μ sin r1 or sinr1 = sinθ/μ
A thin convex lens made from crown glass 
has focal length f. When it is measured in two different liquids having refractive indices 4/3 and 5/3, it has the focal lengths f1 and f2 respectively. The correct
f1 = f2 <f
f1 >f and f2 becomes negative
f2 >f and f1 becomes negative
f1 and f2 both become negative
B.
f1 >f and f2 becomes negative
according to lens maker's formula, when the lens in the air.
In case of liquid, where refractive index is 4/3 and 5/3
we get
Focal length in first liquid
⇒ f1 is positive.
Focal length in second liquid
A green light is an incident from the water to the air – water interface at the critical angle (θ). Select the correct statement
The entire spectrum of visible light will come out of the water at various angles to the normal.
The spectrum of visible light whose frequency is less than that of green light will come out to the air medium
The spectrum of visible light whose frequency is more than that of green light will come out to the air medium.
The entire spectrum of visible light will come out of the water at an angle of 900 to the normal.
D.
The entire spectrum of visible light will come out of the water at an angle of 900 to the normal.
As the frequency of visible light increases refractive index increases. With the increase of refractive index critical angle decreases. So that light having a frequency greater than green will get total internal reflection and the light having the frequency less than green will pass to air.
Hydrogen (1H1), deuterium (1H2), singly ionised helium (2He4+) and doubly ionised lithium (3Li8)2+ all have one electron around the nucleus. Consider an electron transition from n =2 to n=1. If the wavelengths of emitted radiation are λ1,λ2,λ3 andλ4, respectively for four elements, then approximately which one of the following is correct?
4λ1=2λ2=2λ3 =λ4
λ1=2λ2=2λ3 =λ4
λ1=λ2=4λ3 =9λ4
λ1=2λ2=3λ3 =4λ4
C.
λ1=λ2=4λ3 =9λ4
For hydrogen atom, we get
A circular loop of radius 0.3 cm lies parallel to a much bigger circular loop of radius 20 cm. The centre of the small loop is on the axis of the bigger loop. The distance between their centres is 15 cm. If a current of 2.0 A flows through the smaller loop, then the flux linked with bigger loop is
9.1 x 10-11 Wb
6 x 10-11 Wb
3.3 x 10-11 Wb
6.6 x 10-9 Wb
A.
9.1 x 10-11 Wb
A diode detector is used to detect an amplitude modulated wave of 60% modulation by using a condenser of capacity 250 pico farad in parallel with a load resistance 100 kilo ohm. Find the maximum modulated frequency which could be detected by it.
10.62 MHz
10.62 kHz
5.31 mHz
5.31 kHz
B.
10.62 kHz
The frequency is given as
μ =0.6
R = 100 k = 100 x1000 Ω
c = 250 pico farad = 250 x 10-12 F
So,
The anode voltage of a photocell is kept fixed. The wavelength λ of the light falling on the cathode is gradually changed. The plate current I of the photocell varies as follows:
D.
As λ is increased, there will be a value of λ above which photoelectrons will cease to come out. So, photocurrent will be zero.
The I-V characteristic of an LED is
A.
For same value of current, higher value of voltage is required for higher frequency.
In a hydrogen like atom, electron makes the transition from an energy level with quantum number n to another with a quantum number (n – 1). If n >> 1, the frequency of radiation emitted is proportional to
1/n
1/n2
1/n3/2
1/n3
D.
1/n3
An energy gap, ΔE = hv
Here, h is Planck's constant
therefore,
Frequency
Proton, Deuteron and alpha particle of the same kinetic energy is moving in circular trajectories in a constant magnetic field. The radii of the proton, deuteron and alpha particle are respectively rp, rd and rα. Which one of the following relations is correct?
rα = rp= rd
rα = rp< rd
rα > rd> rp
rα = rd> rp
B.
rα = rp< rd
For charged particle moving with a speed v, in magnetic field B, on a circular track of radius
An object 2.4 m in front of a lens forms a sharp image on a film 12 cm behind the lens. A glass plate 1cm thick, of refractive index 1.50 is interposed between lens and film with its plane faces parallel to film. At what distance (from lens) should object be shifted to be in sharp focus on film?
7.2 m
2.4 m
3.2 m
5.6 m
D.
5.6 m
Case I: u = –240cm, v = 12, by Lens formula
A cylindrical tube, open at both ends, has a fundamental frequency, f, in the air. The tube is dipped vertically in water so that half of it is in water. The fundamental frequency of the air-column is now
f
f/2
3f/4
2f
A.
f
Initially for open organ pipe, fundamental frequency,
But when it is half dipped in water, then it becomes closed organ pipe of length l/2. In this case fundamental frequency.
A radar has a power of 1 Kw and is operating at a frequency of 10 GHz. It is located on a mountain top of height 500 m. The maximum distance upto which it can detect object located on the surface of the earth (Radius of earth = 6.4 x 106 m) is
80 km
16 km
40km
64 km
A.
80 km
Maximum distance on earth where object can be detected is d, then
(h + R)2 = d2 + R2
⇒ d2 = h2 + 2Rh
Since, h<<R,
⇒d2 =2hR
A charge Q is uniformly distributed over the surface of non conducting disc of radius R. The disc rotates about an axis perpendicular to its plane and passing through its centre with an angular velocity ω. As a result of this rotation, a magnetic field of induction B is obtained at the centre of the disc. If we keep both the amount of charge placed on the disc and its angular velocity to be constant and vary the radius of the disc than the variation of the magnetic induction at the centre of the disc will be represented by the figure
A.
Consider ring like the element of the disc of radius r and thickness dr.
If σ is charge per unit area, then charge on the element
dq = σ(2πr dr)
current ‘i’ associated with rotating charge dq is
Magnetic field dB at center due to element
So if Q and w are unchanged then
This question has statement 1 and statement 2. Of the four choices given after the statements, choose the one that best describes the two statements
Statement 1: Davisson – Germer experiment established the wave nature of electrons.
Statement 2: If electrons have wave nature, they can interfere and show diffraction.
Statement 1 is false, Statement 2 is true
Statement 1 is true, Statement 2 is false
Statement 1 is true, Statement 2 is the correct explanation for statement 1
Statement 1 is true, Statement 2 is true, Statement 2 is not the correct explanation for statement 1
C.
Statement 1 is true, Statement 2 is the correct explanation for statement 1
Davisson – Germer experiment showed that electron beams can undergo diffraction when passed through atomic crystals. This shows the wave nature of electrons as waves can exhibit interference and diffraction
In Young's double slit experiment, one of the slit is wider than other, so that the amplitude of the light from one slit is double of that from other slits. If Im be the maximum intensity, the resultant intensity I when they interfere at phase difference φ is given by
D.
Let A1 = A0, A2 = 2A0
If amplitude of resultant wave is A then
For maximum intensity
The hydrogen atom is excited from ground state to another state with the principal quantum number equal to 4. Then the number of spectral lines in the emission spectra will be
2
3
3
4
D.
4
Number of spectral lines from a state n to ground state is
Let the xz-plane be the boundary between two transparent media. Medium 1 in z ≥ 0 has a refractive index of √2 and medium 2 with z<0 has a refractive index of √3. A ray of light in medium 1 given by the vector
is incident on the plane of separation. The angle of refraction in medium 2 is
45°
60°
75°
30°
A.
45°
As a refractive index for z>0 and z≤0 is different xy-plane should be the boundary between two media.
Angle of incidence,
A car is fitted with a convex sideñview mirror of focal length 20 cm.Asecond car 2.8 m behind the first car is overtaking the first car at a relative speed of 15 m/s. The speed of the image of the second car as seen in the mirror of the first one is
1/15 m/s
10 m/s
15 m/s
1/10 m/s
A.
1/15 m/s
A nucleus of mass M + Δm is at rest and decays into two daughter nuclei of equal mass M/2 each. The speed of light is c.
The binding energy per nucleon for the parent nucleus is E1 and that for the daughter nuclei is E2. Then
E1 = 2E2
E2 = 2E1
E1 > E2
E2 > E1
D.
E2 > E1
After decay, the daughter nuclei will be more stable hence, binding energy per nucleon will be more than that of their parent nucleus.
A nucleus of mass M + Δm is at rest and decays into two daughter nuclei of equal mass M/2 each. Speed of light is c.
The speed of daughter nuclei is
B.
Conserving the momentum
Now, from energy conservation and mass -energy equivalence
A radioactive nucleus (initial mass number A and atomic number Z) emits 3 α–particles and 2 positions. The ratio of number of neutrons to that of protons in the final nucleus will be
B.
In positive beta decay a, proton is transformed into a neutron and a positron is emitted
p+ → n0 + e+
Number of neutrons initially was A-Z
Number of neutrons after decay (A-Z) -3 x 2 (due to alpha particles) + 2 x 1 (due to positive beta decay)
The number of protons will reduce by 8. so, the ratio number of neutrons to that of protons = 
The combination of gates shown below yields
OR gate
NOT gate
XOR gate
NAND gate
A.
OR gate
| A | B | X |
| 0 | 0 | 0 |
| 0 | 1 | 1 |
| 1 | 0 | 1 |
| 1 | 1 | 1 |
If a source of power 4kW produces 1020 photons/second, the radiation belongs to a part of the spectrum called
X -rays
ultraviolet rays
microwaves
γ rays
A.
X -rays
As power of source = 4 x 103 = 1020 x hv
(∴ P/E = Number of photons)
The potential energy function for the force between two atoms in a diatomic molecule is approximately given by 
where a and b are constants and x is the distance between the atoms. If the dissociation energy of the molecule is D = [U(x = ∞) – Uat equilibrium], D is
b2/2a
b2/12a
b2/4a
b2/6a
C.
b2/4a
The potential energy for diatomic molecule is 
The speed of light in the medium is
maximum on the axis of the beam
minimum on the axis of the beam
the same everywhere in the beam
directly proportional to the intensity I.
B.
minimum on the axis of the beam
The speed of light at the surface or interface of the medium is maximum while on the axis of the medium is minimum
An initially parallel cylindrical beam travels in a medium of refractive index μ(I) = μ0 +μ2I , where μ0 and μ2 are positive constants and I is the intensity of the light beam. The intensity of the beam is decreasing with increasing radius.
As the beam enters the medium, it will
travel as a cylindrical beam
diverge
converge
diverge near the axis and converge near the periphery.
C.
converge
At intensity is maximum at the axis, therefore, μ will be maximum an speed will be minimum on the axis of the beam. So, the beam will converge.
An initially parallel cylindrical beam travels in a medium of refractive index μ(I) = μ0 +μ2I , where μ0 and μ2 are positive constants and I is the intensity of the light beam. The intensity of the beam is decreasing with increasing radius.
The initial shape of the wavefront of the beam is
convex
concave
convex near the axis and concave near the periphery
Planar
D.
Planar
For a parallel cylindrical beam, wave length will be planar.
A diverging lens with the magnitude of focal length 25 cm is placed at a distance of 15 cm from a converging lens of the magnitude of focal length 20 cm. A beam of parallel light falls on the diverging lens. The final image formed is
real and at a distance of 40 cm from the divergent lens
real and at a distance of 6 cm from the convergent lens
real and at a distance of 40 cm from the convergent lens
virtual and at a distance of 40 cm from the convergent lens
C.
real and at a distance of 40 cm from the convergent lens
As parallel beam incident on diverging lens if forms virtual image at v1 = –25 cm from the diverging lens which works as an object for the converging lens (f = 20 cm)
So for converging lens u = -40 cm, f = 20 cm
therefore final image
V = 40 cm from the converging lens.
In a common emitter amplifier circuit using an n-p-n transistor, the phase difference between the input and the output voltages will be
135°
180°
45°
90°
B.
180°
In common emitter amplifier circuit input and output voltage are out of phase. When the input voltage is increased then ib is increased, ic also increases so the voltage drop across Rc is increased. However, increase in voltage across RC is in opposite sense.
In a Young's double slit experiment, slits are separated by 0.5 mm, and the screen is placed 150 cm away. A beam of light consisting of two wavelengths, 650 nm and 520 nm, is used to obtain interference fringes on the screen. The least distance from the common central maximum to the point where the bright fringes due to both the wavelengths coincide is
9.75 mm
15.6 mm
1.56 mm
7.8 mm
D.
7.8 mm
For common maxima
n1λ1 = n2λ2
n1 × 650 = n2 × 520
The refractive index of glass is 1.520 for red light and 1.525 for blue light. Let D1 and D2 be an of minimum deviation for red and blue light respectively in a prism of this glass. Then
D1 > D2
D1 < D2
D1 = D2
D1 can be less than or greater than depending upon the angle of prism
B.
D1 < D2
D = (µ − 1)A
D2 > D1
When an unpolarized light of intensity I0 is incident on a polarizing sheet, the intensity of the light which does not get transmitted is
I0 /2
I0 /4
zero
I0
A.
I0 /2
When unpolarised light of intensity Io is incident on a polarizing sheet, only Io/2 is transmitted.
Unpolarized light of intensity I passes through an ideal polarizer A. Another identical polarizer B is placed behind A. The intensity of light beyond B is found to be I/2.Now another identical polarizer C is placed between A and B. The intensity beyond B is now found to be I/8. The angle between polarizer A and C is :
60°
0°
30°
45°
D.
45°
Axis of transmission of A and B are parallel
After introducing polariser C between A and B,
To get output for the following circuit, the correct choice for the input is,
A=1, B=0, C = 0
A=1, B=1, C = 0
A=1, B=0, C=1
A=0, B=1, C=0
C.
A=1, B=0, C=1
Consider the figure below,
The resultant boolean expression of the above logic circuit is given by,
Y = (A+B).C
Using the inputs given in the options,
If A=0, B=0,C=0, we have
Y = (0+0).0
i.e., Y = 0
If A=1, B=1, C=0, then we have
Y = (1+1).0
i.e., Y = 1.0 = 0
If A=1, B=0, C=1, then
Y = (1+0).1
i.e., Y = 1.1 = 1
If A=0, B = 1, C =0, then
Y = (0+1).0
i.e., Y = 1.0
Y = 0
Therefore, output Y = 1 only when inputs A=1, B=0 and C=1.
When a metallic surface is illuminated with radiation of wavelength 
, the stopping potential is V. If the same surface is illuminated with radiation of wavelength 2 
, the stopping potential is V/4. The threshold wavelength for the metallic surface is,
3
4
C.
3
When a metallic surface is illuminated with radiation of wavelength 
, the stopping potential is V.
Photoelectric equation can be written as,
... (i)
Now, when the same surface is illuminated with radiation of wavelength 2
, the stopping potential is V/4. So, photoelectric equation can be written as,
From equations (i) and (ii), we get
The intensity at the maximum in a Young's double slit experiment is Io. Distance between two slits is d = 
, where 
is the wavelength of light used in the experiment. What will be the intensity infront of one of the slits on the screen placed at a distance D = 10 d?
3
Io
C.
is the wavelength of the light used in the experiment.
corresponds to phase difference 2
.
, phase difference is
Given the value of rydberg constant is 10 m-1, the wave number of the last line of the Balmer series in hydrogen spectrum will be:
0.5 x 107 m-1
0.25 x 107 m-1
2.5 x107 m-1
0.025 x 104 m-1
B.
0.25 x 107 m-1
Rydberg constant, r = 107 m-1
For last line in Balmer series,
; n1 = 2
The ratio of escape velocity at earth (Ve) to the escape velocity at a planet (vp) whose radius and mean density are twice as that of earth is,
1:
1:4
1:
1:2
A.
1:
Escape velocity of earth can be given as,
ve = 
= R
, where 
is the density of the earth.
... (i)
Given that the radius and mean density of planet are twice as that of earth.
So, escape velocity at planet will be,
A siren emitting a sound of frequency 800 Hz moves away from an observer towards a cliff at a speed of 15 m/s. Then, the frequency of sound that the observer hears in the echo reflected from the cliff is (Take, velocity of sound in air = 330 m/s)
800 Hz
838 Hz
885 Hz
765 Hz
B.
838 Hz
The situation can be illustrated as follows:
Frequency of sound that the observer hear in the echo reflected from the cliff is given by,
In a diffraction pattern due to single slit of width a, the first minimum is obsrved at an angle 30o when light of wavelength 
is incident on the slit. The first secondary maximum is observed at an angle of:
sin-1 (2/3)
sin-1 (1/2)
sin-1 (3/4)
sin-1 (1/4)
C.
sin-1 (3/4)
Given that, first minimum is observed at an angle of 30o in a diffraction pattern due to a single slit of width a.
i.e., n =1, 
According to the Bragg's law of diffraction,
For 1st secondary maxima,
... (2)
Putting the value of a from Eqn. (i) to Eqn. (ii), we get
A black body is at a temperature of 5760 K. The energy of radiation emitted by the body at wavelength 250 nm is U1, at wavelength 500 nm is U2 and that at 1000 nm is U3. Wien's constant, b = 2.88 x 106 nmK. Which of the following is correct?
U3 = 0
U1 > U2
U2 > U1
U1 = 0
C.
U2 > U1
Given, temperature, T1 = 5760 K
Given that energy of radiation emitted by the body at wavelength 250 nm in U1, at wavelength 500 nm is U2 and that at 1000 nm is U3.
Now, according to Wein's law, we get
where, b = Wien's constant = 2.88 x 106 nmK
is the wavelength corresponding to maximum energy, so U2 > U1.
Consider the junction diode as ideal. The value of current flowing through AB is,
10-2 A
10-1 A
10-3 A
0 A
A.
10-2 A
Let, I be the current through the diode.
From the given condition,
An astronomical telescope has objective eyepiece of focal lengths 40 cm and 4 cm respectively. To view an object 200 cm away from the objective lens, the lenses must be separated by a distance,
46.0 cm
50.0 cm
54.0 cm
37.3 cm
C.
54.0 cm
According to a question,
Focal length of the objective lens, Fo = +40 cm
Focal length of eyepiece, Fe = 4 cm
Object distance for objective lens (uo) = -200 cm
Applying lens formula for objective lens,
Image will be formed at the focus of a eyepiece lens.
So, for normal adjustment distance between objectives and eyepiece (length of tube) will be,
A p-n-p transistor is connected in a common emitter configuration in a given amplifier. A load resistance of 800 ohm is connected in the collector circuit and the voltage drop across it is 0.8 V. If the current amplification factor is 0.96 and the input resistance of the circuit is 192 ohm, the voltage gain and the power gain of the amplifier will respectively be,
3.69,3.84
4,4
4,3.69
4, 3.84
D.
4, 3.84
Given,
Resistance across load, RL = 800 ohm
Voltage drop across load, VL = 0.8 V
Input resistance of the circuit, Ri = 192 ohm
Collector current is given by,
Therefore, Current amplification = 
Voltage gain, AV is,
Power gain,
AP = 
= (0.96)2 x 
AP = 3.84
Match the corresponding entries of Column 1 with Column 2. [ where m is the magnification produced by the mirror]
| Column 1 | Column 2 |
| A. m = -2 | a. convex mirror |
| B. m = -1/2 | b. Concave mirror |
| C. m = +2 | c. Real image |
| D. m = +1/2 | d. Virtual image |
A --> a and c; B --> a and d; C --> a and b ; D --> c and d
A --> a and d; B --> b and c; C --> b and d; D --> b and c
A --> c and d; B --> b and d; C --> b and c; D --> a and d
A --> b and c; B --> b and c; C --> b and d; D --> a and d
D.
A --> b and c; B --> b and c; C --> b and d; D --> a and d
A concave mirror forms real and virtual image, whose magnification can be negative or positive depending upon the position of the object.
If object is placed between focus and pole the image obtained will be virtual and its magnification will be positive. In all other cases concave mirror forms real images whose magnification will be negative.
A convex mirror always forms a virtual image whose magnification will always be positive.
On observing light from three different stars P, Q and R, It was found that intensity of violet colour is maximum in the spectrum of P, the intensity of green colour is maximum in the spectrum of R and the intensity of red colour is maximum in the spectrum of Q. If TP, TQ and TR are the respective absolute temperature of P, Q and R, then it can conclude from the above observations that
TP > TQ>TR
TP > TR> TQ
TP > TR < TQ
TP < TQ < TR
B.
TP > TR> TQ
We know from Weins displacement law
A radiation of energy 'E' falls normally on a perfectly reflecting surface. The momentum transferred to the surface is
(c = velocity of light)
B.
The radiation energy is given by 
In a double slit experiment, the two slits are 1 mm apart and the screen is placed in 1 m away. A monochromatic light of wavelength 500nm is used. What will be the width of each slit for obtaining ten maxima of double slit within the central maxima of double slit within the central maxima of single slit pattern?
0.2mm
0.1 mm
0.5 mm
0.02 mm
A.
0.2mm
Given d =1 mm = 1 x 10-3 m
D =1m
Consider 3rd orbit of He+ (Helium), using non-relativistic approach, the speed of electron in this orbit will be (given K= 9 x 109 constant, Z=2 and h (Planck's constant = 6.6 x 10-34 Js-1
2.92 x 106 m/s
1.46 x 106 m/s
0.73 x 106 m/s
3.0 x 108 m/s
B.
1.46 x 106 m/s
Energy of electron in the 3rd orbit of He+ is
For a parallel beam of monochromatic light of wavelength 'λ'diffraction is produced by single slit whose width 'a' is of the order of the wavelength of the light. If 'D' is the distance of the screen from the slit the width of the central maxima will be
2Dλ/a
Dλ/a
Da/λ
2Da/λ
A.
2Dλ/a
For the condition of maxima
An electron of mass m and a photon have the same energy E. The ratio of de-Broglie wavelength associated with them is,
D.
Given that electron has a mass m.
De-Broglie wavelength for an electron will be given as,
where,
h is the Planck's constant, and
p is the linear momentum of electron
Kinetic energy of electron is given by, E = 
From equation (i) and (ii), we have
Energy of a photon can be given as,
Hence, 
is the de-Broglie wavelength of photon.
Now, dividing equation (iii) by (iv), we get
Which of the following figures represent the variation of particle momentum and the associated de-Broglie wavelength?
B.
The de-Broglie wavelength is given by
This equation is in the form of yx =c, which is the equation of a rectangular hyperbola.
If in a p-n junction, a square input signal of 10 V is applied as shown,
then the output across RL will be
D.
As it is forward biased so it takes positive value. Hence, option (d) is correct.
Which logic gate is represented by this following combination of logic gates?
OR
NAND
AND
NOR
C.
AND
The truth table for the given circuit is
|
A |
B |
y1 |
y2 |
y= y1 +y2 |
|
0 |
0 |
1 |
1 |
0 |
|
0 |
1 |
1 |
0 |
0 |
|
1 |
0 |
0 |
1 |
0 |
|
1 |
1 |
0 |
0 |
1 |
When an alpha particle of mass m moving with velocity v bombards on a heavy nucleus of charge Ze, its distance of closest approach from the nucleus depends on mass as,
m
D.
When an alpha particle moving with velocity v bombards on a heavy nucleus of charge Ze, then there will be no loss of energy.
Initial Kinetic energy of the alpha particle = Potential energy of alpha particle at closest approach.
That is,
This is the required distance of closest approach to alpha particle from the nucleus.
The angle of incidence for a ray of light at a refracting surface of a prism is 45o. The angle of prism is 60o. If the ray suffers minimum deviation through the prism, the angle of minimum deviation and refractive index of the material of the prism respectively, are
30o, 
45o ; 
30o; 
45o; 
A.
30o,
Consider a ray of light PQ incident on the surface AB and moves along RS, after passing through the prism ABC.
Given that, incident ray suffers minimum deviation. therefore, the ray inside the prism must be parallel to the base BC of the prism.
is the refractive index of the material of the prism.
For the angle of minimum deviation of a prism to be equal to its refracting angle, the prism must be made of a material whose refractive index
Lies between 
Lies between 2 and 
is less than 1
is greater than 2
B.
Lies between 2 and
In given situation value of A varies from 0 to 90o. so,
A rod of length 10 cm lies along the principal axis of a concave mirror of focal length 10 cm in such way that it end closer to pole is 20 cm away from the mirror. The length of the image is
10 cm
15 cm
2.5 cm
5 cm
D.
5 cm
By mirror formula, image distance of A 
The transition from the state n = 3 to n=1 in a hydrogen-like atom results in ultraviolet radiation. Infrared radiation will be obtained in the transition from.
2→ 1
3 →2
4 → 2
4 →3
D.
4 →3
Infrared radiation is found in Paschen, Brackett and pfund series nd it is obtained when electron transition occurs from high energy level to minimum third level.
To get an output Y = 1 in given circuit which of the following input will be correct
|
A |
B |
C |
|
1 |
0 |
0 |
|
A |
B |
C |
|
1 |
0 |
1 |
|
A |
B |
C |
|
1 |
1 |
0 |
|
A |
B |
C |
|
0 |
1 |
0 |
B.
|
A |
B |
C |
|
1 |
0 |
1 |
Here Y = (A+B).C
Truth Table
|
A |
B |
C |
Y = (A+B).C |
|
0 |
0 |
0 |
0 |
|
1 |
0 |
0 |
0 |
|
0 |
1 |
0 |
0 |
|
0 |
1 |
0 |
0 |
|
0 |
1 |
1 |
1 |
|
1 |
0 |
1 |
1 |
|
1 |
1 |
1 |
1 |
An electron in hydrogen atom first jumps from third excited state to second excited state and then from second excited to the first excited state. The ratio of the wavelengths λ1:λ2 emitted in the two cases is
7/5
27/20
27/5
20/7
D.
20/7
Here, for wavelength λ1
n1 = 4 and n2 =3
and for λ2, n1 = 3 and n2 = 2
When a biconvex lens of glass having refractive index 1.47 is dipped in a liquid, it acts as a plane sheet of glass. This implies that the liquid must have refractive index
equal to that of glass
less than one
greater than that of glass
less than that of glass
A.
equal to that of glass
If biconvex lens behaves like a plane sheet, the ray will pass undeviated through it only when the medium has a same refractive index as that of a biconvex lens.
A mixture consists of two radioactive materials A1 and A2 with half-lives of 20 s and 10 s respectively. Initially, the mixture has 40 g of A1 and 160 g of A2. The amount of the two in the mixture will become equal after
60 s
80 s
20 s
40 s
D.
40 s
For 40 g amount
So, after 40 A1 and A2 remains same.
C and si both have a same lattice structure, having 4 bonding electrons in each. However, C is insulator whereas Si is an intrinsic semiconductor. This is because
in the case of C, the valence band is not completely filled at absolute zero temperature
in a case C, the conduction band is partly filled even at absolute zero temperature
the four bonding electrons in the case of C lie in the second orbit, whereas in case of Si they lie in the third
the four bonding electrons in the case of C lie in the third orbit, whereas for Si they lie in the fourth orbit
C.
the four bonding electrons in the case of C lie in the second orbit, whereas in case of Si they lie in the third
The four bonding electrons in the case of C lie in the second orbit, whereas in the case of Si they lie in the third orbit so loosely bounded valency electron in Si as compared to C.
Two sources of sound placed close to each other, are emitting progressive waves given by
y1 = 4sin 600π t and y2 = 5 sin 608πt
An observer located near these two sources of sound will hear
4 beats per second with intensity ratio 25: 16 between waxing and waning
8 beats per second with intensity ratio 25:16 between waxing and waning
8 beats per second with intensity ratio 81:1 between waxing and waning
4 beats per second with intensity ratio 81:1 between waxing and waning
D.
4 beats per second with intensity ratio 81:1 between waxing and waning
Given, y1 = 4sin 600π t and y2 = 5 sin 608πt
Comparing with general equation
y = a sin 2πft
we get, f2 = 300 Hz and f2 304 Hz
Number of beats = f2-f1 = 4s-1
The magnifying power of a telescope is 9. When it is adjusted for parallel rays the distance between the objective and eyepiece is 20 cm. The focal length of lenses are
10 cm, 10 cm
15 cm, 5 cm
18 cm, 2cm
11 cm, 9 cm
C.
18 cm, 2cm
Given M = fo/fe =9 and fo+fe = 20
fo = 9fe
So, 9fe+fe = 20
fe = 2cm
fo = 9 x2
fo =18
Monochromatic radiation emitted when electron on hydrogen atom jumps from first excited to the ground state irradiates a photosensitive material. The stopping potential is measured to be 3.57 V. The threshold frequency of the material is
4 x 1015 Hz
5 x 1015 Hz
1.6 x1015 Hz
2.5 x 1015 Hz
C.
1.6 x1015 Hz
Energy released from emission of electron
E = (-3.4)-(-13.6) = 10.2 eV
From photo electric equation work function
The figure shows a logic circuit with two inputs A and B and the output C. The voltage waveforms across A, B and C are as given. The logic circuit gate is
OR Gate
NOR Gate
AND Gate
NAND Gate
A.
OR Gate
From the given waveforms, the following truth table can be made.
|
Inputs |
Output |
|
|
A |
B |
C |
|
0 |
0 |
0 |
|
1 |
0 |
1 |
|
1 |
1 |
1 |
|
0 |
1 |
1 |
Transfer characteristic [output voltage (Vo) vs input voltage (Vi)] for a base biased transistor in CE configuration is as shown in the figure. For using transistor as a switch, it is used
in region III
both in region (I) and (III)
in region (II)
in region
B.
both in region (I) and (III)
For using a transistor as a switch, it is used in cut-off state and saturation state only.
Light with an energy flux of 25 x 104 W/m2 falls on a perfectly reflecting surfaces at normal incidence. If the surface area is 15 cm2, the average force exerted on the surface is,
1.25 x 10-6 N
2.50 x 10-6 N
1.20 x 10-6 N
3.0 x 10-6 N
A.
1.25 x 10-6 N
Energy flux = 
Force on unit area = momentum transferred in unit time on unit area = 
Force on the 15 x 10-4 m2 area,
= 8.3 x 10-4 N/m2 x 15 x 10-4 m2
= 124.5 x 10-8 N
= 1.25 x 10-6 N
A beam of light of 
= 600 nm from a distant source falls on a single slit 1 mm wide and the resulting diffraction pattern is observed on a screen 2 m away. The distance between first dark fringes on either side of the central bright fringe is,
1.2 cm
1.2 mm
2.4 cm
2.4 mm
D.
2.4 mm
and when 
is counted in rad, tan 
In the young's double slit experiment, the intensity of light at a point on the screen (where the path difference is 
) is K. The intensity at a point where the path difference is 
will be,
K
K/4
K/2
zero
C.
K/2
For net intensity,
I' = 4 Io cos2 
For the first case,
K = 4 Io cos2 
K = 4 Io ... (i)
For the second case,
... (ii)
Comparing equations (i) and (ii), we have
K' = K/2
If the focal length of the objective lens is increased, then the magnifying power of,
microscope will increase but that of telescope will decrease
microscope and telescope both will increase
microscope and telescope both will decrease
microscope will decrease but that of telescope will increase
D.
microscope will decrease but that of telescope will increase
For microscope, m = 
For telescope, m = 
m 
fo
Therefore, the magnifying power of microscope will decrease but the magnifying power of telescope will increase.
The angle of a prism is A. One of its refracting surfaces is silvered. Light rays falling at an angle of incidence 2A on the first surface returns back through the same path after suffering reflection at the silvered surface. The refractive index 
, of the prism is,
2 sin A
2 cos A
1/2 cos A
tan A
B.
2 cos A
When the energy of the incident radiation is increased by 20%, the kinetic energy of the photoelectrons emitted from a metal surface increased from 0.5 eV to 0.8 eV. The work function of the metal is,
0.65 eV
1.0 eV
1.3 eV
1.5 eV
B.
1.0 eV
For photoelectric equation,
(KE)min = 
For the first condition,
0.5 = E - 
... (i)
For the second condition,
... (ii)
From equations (i) and (ii),
and 
-0.3 = -0.2 E
E = 
From expression (i),
0.5 = 1.5 - 
If the kinetic energy of the particle is increased to 16 times its previous value, the percentage change in the de-Broglie wavelength of the particle is,
25
75
60
50
B.
75
For de-Broglie wavelength,
... (i)
... (ii)
From equations (i) and (ii),
There is 75% change in the wavelength.
Hydrogen atom in ground state is excited by a monochromatic radiation of 
. The number of spectral lines in the resulting spectrum emitted will be,
3
2
6
10
B.
2
Energy provided to the ground state electron,
The electron will jump from n= 3 to n=1.
The number of possible field lines from n=3 to n=1 is 2.
The binding energy per nucleon of 
nuclei are 5.60 MeV and 7.06 MeV, respectively. In the nuclear reaction, 
, the value of energy Q released is,
19.6 MeV
-2.4 MeV
8.4 MeV
17.3 MeV
D.
17.3 MeV
The binding energy for 
is around zero and also not given in the question so we can ignore it,
Two ideal diodes are connected to a battery as shown in the circuit. The current supplied by the battery is
0.75 A
zero
0.25
0.5 A
D.
0.5 A
Here in circuit D1 is forward bias and D2 is reverse bias. Current flow only in diode D1
The current supplied by the battery
i = 5/10 = 0.5 A
A radio isotope X with a half life 1.4 x 109 yr decays of Y which is stable. A sample of the rock from a cave was found to contain X and Y in the ratio 1:7. The age of the rock is,
1.96 x 109 yr
3.92 x 109 yr
4.20 x 109 yr
8.40 x 109 yr
C.
4.20 x 109 yr
Ratio of X:Y is given = 1:7
That is,
Let, the initial total mass is m.
Therefore, time taken to become 1/8 unstable part
= 3 x T1/2
= 3 x 1.4 x 109
= 4.2 x 109 y
The given graph represents V-I characteristic for a semiconductor device. Which of the following statement is correct?
It is V-I characteristic for solar cell where point A represents open circuit voltage and point B short circuit current.
It is for a solar cell and points A and B represent open circuit voltage and current, respectively.
It is for a photodiode and points A and B represent open circuit voltage and current, respectively.
It is for a LED and points A and b represents open circuit voltage and short circuit current respectively.
A.
It is V-I characteristic for solar cell where point A represents open circuit voltage and point B short circuit current.
V-I characteristics of a solar cell is shown where, A represents open circuit voltage and B shows short circuit current.
The barrier potential of a p-n junction diode depends on:
i) type of semiconductor material
ii) amount of doping
iii) temperature
Which one of the following is correct?
(i) and (ii) only
(ii) only
(ii) and (iii) only
(i), (ii) and (iii)
D.
(i), (ii) and (iii)
Barrier potential depends on the material used to make p-n junction diode (whether it is Si or Ge).
It should also depend on the amount of doping due to which the number of majority charge carriers will change. Also, it depends on temperature due to which the number of minority carriers will change.
The output (X) of the logic circuit shown in a figure will be.
X = A. B
C.
X = A. B
In an n-type semiconductor, which of the following statement is true?
Electrons are majority carriers and trivalent atoms are dopants
Electrons are minority carriers and pentavalent atoms are dopants
Holes are minority carriers and pentavalent atoms are dopants
Holes are majority carriers and trivalent atoms are dopants
C.
Holes are minority carriers and pentavalent atoms are dopants
The n-type semiconductor can be produced by doping impurity atoms of valence 5 i.e., pentavalent atoms, i.e., phosphorous.
A parallel beam of fast moving electrons is incident normally on a narrow slit. A fluorescent screen is placed at a large distance from the slit. If the speed of the electrons is increased, then which of the following statements is correct ?
Diffraction pattern is not observed on the screen in the case of electrons.
The angular width of the central maximum of the diffraction pattern will increase
The angular width of the central maximum will decrease
The angular width of the central maximum will be unaffected
B.
The angular width of the central maximum of the diffraction pattern will increase
In young's double-slit experiment, the slits are 2 mm apart and are illuminated by photons of two wavelengths λ1 = 12000 A and λ2 =10000 A. At what minimum distance from the common central bright fringe on the screen 2 m from the slit will a bright fringe from one interference pattern coincide with a bright fringes from the other?
8 mm
6 mm
4 mm
3 mm
B.
6 mm
Given λ1 = 12000 Ao and λ2 = 10000 Ao
D = 2 cm and d=2 mm= 2x 10-3
For a normal eye, the cornea of eye provides a converging power of 40 D and the least converging power of the eye lens behind the cornea is 20 D. Using this information, the distance between the retina and the cornea eye lens can be estimated to be
5 cm
2.5 cm
1.67 cm
1.5 cm
C.
1.67 cm
Given p1 = 40 D, and P2 = 20 D
We have Peff = 40D +20D = 60 D
⇒ f= 100/peff = 100/60
= 1.67 cm
In the Davisson and Germer experiment, the velocity of electrons emitted from the electron gun can be increased by
Increasing the filament current
decreasing the filament current
decreasing the potential difference between the anode and filament
increasing the potential difference between the anode and filament
D.
increasing the potential difference between the anode and filament
In the Davisson and Germer experiment, the velocity of the electron emitted from the electron gun can be increased by increasing the potential difference between the anode and filament.
The power obtained in a reactor using U235 disintegration is 1000kW. The mass decay of U235 per hour is
20 μg
40 μg
1 μg
10 μg
B.
40 μg
Let assume power P = 1000 W
Energy per hour = 1000 x 3600 J
Energy per fission = 200 MeV
= 200 x 1.6 x 10-13 J
therefore number per hour 
The nearest value is 40 μg, Hence option b is correct.
In photoelectric emission process from a metal of work function 1.8 eV, the kinetic energy of most energetic electrons is 0.5 eV. The corresponding stopping potential is
1.3 V
0.5 V
2.3 V
1.8 V
B.
0.5 V
stopping potential = Maximum KE
eV = KEmax
Symbolic representation of four logic gates are shown as,
Pick out which ones are for AND, NAND and NOT gates, respectively.
(iii), (ii) and (i)
(iii), (ii) and (i)
(ii), (iv), and (iii)
(ii), (iii) and (iv)
C.
(ii), (iv), and (iii)
In symbol given in problem are
(i) OR gate
(ii) AND gate
(iii) Not gate
(iv) NAND gate
If a small amount of antimony is added to germanium crystal
the antimony becomes an acceptor atom
there will be more free electrons than holes in the semiconductor
its resistance is increased
it becomes a p - type semiconductor
B.
there will be more free electrons than holes in the semiconductor
When a small amount of antimony is added to germanium crystal, the crystal becomes n- type semiconductor because antimony is a pentavalent substrate. It excess free electrons.
Fusion reaction takes place at high temperature because
atoms get ionised at high temperature
kinetic energy is high enough to overcome the coulomb repulsion between nuclei
molecules break up at high temperature
nuclei break up at high temperature
B.
kinetic energy is high enough to overcome the coulomb repulsion between nuclei
Fusion reaction takes place at high temperature because kinetic energy is high enough to overcome the coulomb repulsion between nuclei.
Photoelectric emission occurs only when the incident light has more than a certain minimum
wavelength
intensity
frequency
power
C.
frequency
By the concept of threshold minimum frequency needed for photoelectric emission.
Electrons used in an electron microscope are accelerated by a voltage of 25 kV. If the voltage is increased to 100 kV then the de - Broglie wavelength associated with the electrons would
decrease by 2 times
decrease by 4 times
increase by 4 times
increase by 2 times
A.
decrease by 2 times
We have
A biconvex lens has a radius of curvature of magnitude 20 cm. Which one of the following options describes best the image formed on an object of height 2 cm placed 30 cm from the lens?
Virtual, upright, height = 0.5 cm
Real, invented, height = 4 cm
Real, inverted, height = 1 cm
Virtual, upright, height = 1 cm
B.
Real, invented, height = 4 cm
Which of the following is not due to total internal reflection ?
Difference between apparent and real depth of a pond
Mirage on hot summer days
Brilliance of diamond
Working of optical fibre
A.
Difference between apparent and real depth of a pond
Real and apparent depth are explains on the basis of refraction only. The concept of TIR is not involve here.
In forward biasing of the p - n junction
the positive terminal of the battery is connected to n - side and the depletion region becomes thin
the positive terminal of the battery is connected to n- side and the depletion region becomes thick
the positive terminal of the battery is connected to p - side and the depletion region become thin
the positive terminals of the battery is connected to p- side and the depletion region becomes thick
C.
the positive terminal of the battery is connected to p - side and the depletion region become thin
In forwarding biasing of p - n junction the positive terminal of the battery is connected to p -side and the depletion region becomes thin
In the spectrum of hydrogen, the ratio of the longest wavelength in the Lyman series to the longest wavelength in the Balmer series is
4/9
9/4
27/5
5/27
D.
5/27
In hydrogen atom, wavelength of characteristic spectrum
A photoelectric surface is illuminated successively by monochromatic light of wavelength λ and λ/2. If the maximum kinetic energy of the emitted photoelectrons in the second case is 3 times that in the first case, the work function of the surface of the material is
(h = planck's constant, c= speed of light)
hc/ 2λ
hc/λ
2 hc/λ
hc/3λ
A.
hc/ 2λ
According to Einstein photoelectric equation,
E = Kmax + Φ
Where Kmax is the maximum kinetic energy of emitted electron and Φ is work function of electrons.
Kmax = E - Φ = hv - Φ
Kmax = 
Similarly, in the second case, maximum kinetic energy of emitted electron is 3 times that in the first case, we get
3Kmax 
solving EQs (i) and (ii), we get work function of an emitted electron from a metal surface.
Φ = hc/2λ
Light of wavelength 500 nm is incident on metal with work function 2.28 eV. The de - Broglie wavelength of the emitted electrons is
<2.8 x 10-10 m
<2.8 x 10-9 m
> equal to 2.8 x 10-9 m
< equal to 2.8 x 10-12 m
B.
<2.8 x 10-9 m
As, energy of photon, E = hv
The input signal given to a CE amplifier having a voltage gain of 150 is 
The corresponding output signal will be
D.
Input signal of a CE amplifier,
A nucleus of uranium decays at rest into nuclei of thorium and helium. Then,
the helium nucleus has more kinetic energy than the thorium nucleus
the helium nucleus has less momentum than the thorium nucleus
the helium nucleus has more momentum than the thorium nucleus
the helium nucleus has less kinetic energy than the thorium nucleus
A.
the helium nucleus has more kinetic energy than the thorium nucleus
Two slits in Young's experiment have widths in the ratio 1:25.The ratio of intensity at the maxima and minima in the interference pattern Imax/Imin is
9/4
121/49
49/121
4/9
A.
9/4
Given, YDSE experiment, having two slits of width are in the ratio of 1:25
So, ratio of intensity
In an astronomical telescope in normal adjustment, a straight black line of length L is drawn on inside part of the objecitive lens. The eye - piece forms a real image of this line. The length of this image is I. The magnification of the telescope is
A.
A beam of light consisting of red, green and blue colours is incident on a right-angled prism. The refractive index of the material of the prism for the above red, green and blue wavelength are 1.39, 1.44 nd 1.47
The Prism will
separate the blue colour part from the red and green colours
separate all the three colours from one another
not separate the three colours at all
separate the red colour part from the green and blue colours
D.
separate the red colour part from the green and blue colours
For refractive index of a index.
On the first minimum adjacent to the central maximum of a single slit diffraction pattern, the phase difference between the Huygens wavelet from the edge of the slit and the wavelet from the midpoint of the slit is
C.
π radianFor the first minima at P, a sin θ = nλ
where , N = 1
In the following figure, the diodes which are forward biased, are
B.
For forward bias p-type should be higher potential and n-type at a lower potential.
Pure Si at 500 K has an equal number of the electron (ne) and hole (nh) concentrations of 1.5 x 1016 m-3.Doping by indium increases nh to 4.5 x 1022 m -3.The doped semiconductor is of
n- type with electron concentration ne = 5 x 1022 m -3
P- type with electron concentration ne = 2.5 x 1010 m -3
n- type with electron concentration ne = 2.5 x 1023 m -3
p- type with electron concentration ne = 5 x 109 m -3
D.
p- type with electron concentration ne = 5 x 109 m -3
An electron in the hydrogen atom jumps from excited state n to the ground state. The wavlength so emitted illuminates a photo -sensitive material having work function 2.75 eV. If the stopping potential of the photo-electron is 10 V, the value of n is
3
4
5
2
A.
3
E = KEmax + W
eVo + W
= 10 + 2.75
E = 12.75 eV
The difference of 4 and 1 energy level is 12.75 eV. So, the higher energy level is 4 to the ground and excited state is n=3.
Two radioactive nuclei P and Q, in a given sample decay into a stable nucleous R. At time t =0, number of P species are 4No and that of Q are No.Half -life of P (for conversion to R) is 1 min where asthat of Q is 12 min. Initially there are no nuclei of R present in the sample. When number of nuclei of R presnent in the sample would be
3No
9No / 2
5No/2
2No
B.
9No / 2
Initially P → 4No
Q → No
Half life TP →1 min
TQ = 2 min
Let after time t number of nuclei of P and Q are equal i.e,
Out of the following which one is not a possible energy for a photon to be emitted by hydrogen atom according to Bohr's atomic model?
1.9 eV
11.1 eV
13.6 eV
0.65 eV
B.
11.1 eV
A conversing beam of rays is incident on a diverging lens. Having passed though the lens the rays intersect at a point 15 cm from the lens on the opposite side. If the lens is removed the point where the rays meets will moves 5 cm closer to the lens. The focal length of the lens is
-10 cm
20 cm
-30 cm
5 cm
C.
-30 cm
Given u = 10 cm, v = 15 cm
The mass of a 
nucleus is 0.042u less than the sum of the masses of all its nucleons. The binding energy per nucleon is 
nucleus is nearly
46 MeV
5.6 MeV
3.9 MeV
23 MeV
B.
5.6 MeV
If, m = 1 u, c= 3 x 108 ms-1, then
E = 931 MeV i.e. 1 u = 931 MeV
Binding energy = 0.042 x 931 = 39.10 MeV
therefore, Binding energy per nucleon
= 39.10/7 = 5.58 = 5.6 MeV
Which one of the following statement is false?
Pure Si doped with trivalent impurities given a p- type semiconductor
Majority carriers in a p- type semiconductor are holes
Minority carriers in a p- type semiconductor are electron
The resistance of intrinsic semiconductor decreases with increase of temperature
A.
Pure Si doped with trivalent impurities given a p- type semiconductor
p-type semiconductor is obtained by adding a small amount of trivalent impurity to a pure sample of semiconductor (Ge).
Majority charge carriers - holes
Minority charge carriers - electrons
n - type semiconductor are obtained by adding a small amount of pentavalent impurity to a pure sample of semiconductor (Ge)
Majority charge carriers - electrons
The resistance of intrinsic semiconductors decreases with the increase of temperature.
A ray of light travelling in a transparent medium of refractive index μ falls, on a surface separating the medium from the air at an angle of incidence of 45o. For which of the following value of μ the ray can undergo total internal reflection?
μ =1.33
μ = 1.40
μ = 1.50
μ = 1.25
C.
μ = 1.50
For total internal reflection
i > c
⇒ sin i > sin c
⇒ sin 45o > 1/ μ
μ > 21/2
μ > 1.4
The device that can act as a complete electronic circuit is
Junction diode
Integrated circuit
Junction transistor
Zener diode
B.
Integrated circuit
Integrated circuits are miniature electronic circuits produced within a single crystal of semiconductors such as silicon. They contain a million or so transistors and resistors or capacitors. They are widely used in memory circuits, microcomputers, pocket calculators and electronic watches on account of their low cost and bulk, reliability into specific regions of the semiconductor crystals.
The energy of a hydrogen atom in the ground state is -13.6 eV. The energy of a He+ ion in the first excited state will be
-13.6 eV
-27.2 eV
-54.4eV
-6.8 eV
A.
-13.6 eV
Energy E of an atom with principal quantum number n is given n 
A source S1 is producing, 1015 photons of wavelength 500 A. Another source S2 is producing 1.02 x 1015 photons per second of wavelength 5100 A. Then, power of S2) (power of S1) is equal to
1.00
1.02
1.04 A
0.98
A.
1.00
Number of photons emitted per seconds 
The potential difference that must be applied to stop the fastest photoelectrons emitted by a nickel surface, having work function 5.01 eV, when ultraviolet light of 200 nm falls on it, must be
2.4 V
-1.2 V
-2.4 V
1.2 V
D.
1.2 V
Energy of incident light E (eV) = 12375/2000 = 6.2 eV (200 nm = 2000 A)
According to the relation E = Wo +eVo
To get an output Y = 1 from the circuit shown below, the input must be
|
A |
B |
C |
|
0 |
1 |
0 |
|
A |
B |
C |
|
0 |
0 |
1 |
|
A
|
B
|
C
|
|
1
|
0
|
1
|
|
A
|
B
|
C
|
|
1
|
0
|
0
|
C.
|
A
|
B
|
C
|
|
1
|
0
|
1
|
Gate I is OR gate Y' = A + B
Gate II is AND gate Y = Y'.C
therefore,
A = 1, B = 0, C = 1 will give Y = 1
A ray of light is incident on a 60o prism at the minimum deviation position. The angle of refraction at the first face (ie, incident face) of the prism is
zero
30o
45o
60o
B.
30o
The refracting angle of prism
A = r1 + r2
For minimum deviation
For transistor action
A) Base, emitter and collector regions should have similar size and doping concentrations.
B) The base region must be very thin and lightly doped.
C) The emitter-base junction is forward biased and base -collector junction is reverse biased
D) Both the emitter-base junction as well as the base-collector junction are forward biased.
(D) and (A)
(A) and (B)
(B) and (C)
(C) and (D)
C.
(B) and (C)
A transistor is mostly used in the active region of operation ie, emitter-base junction is forward biased and collector-base junction is reverse biased. The base region must be very thin and lightly doped.
The decay constant of a ratio isotope is λ. If A1 and A2 are its activities at times t1 and t2 respectively, te number of nuclei which have decayed during the time (t1-t2)
A1t1 - A2t2
A1 - A2
(A1 - A2)/λ
λ(A1 - A2)
C.
(A1 - A2)/λ
A1 = λN1
A2 = λN2
N1 -N2 = [A1 -A2/ λ]
When monochromatic radiation of intensity I falls on a metal surface, the number of photoelectrons and their maximum kinetic energy and N and T respectively. If the intensity of radiation is 2I, the number of emitted electrons and their maximum kinetic energy are respectively
N and 2T
2N and T
2N and 2T
B.
2N and T
i) Kinetic energy of photoelectrons depends on the frequency of incident radiations and is independent of the intensity of illumination.
ii) A number of photoelectrons depend upon the intensity.
So, the number of emitted electrons and their maximum kinetic energy are 2N and T respectively.
The speed of light in media M1 and M2 is 1.5 x 108 m /s respectively. A ray of light enters from medium M1 and M2 at incidence angle i. If the ray suffers total internal reflection the value of i is
equal to sin-1 (2/3)
equal to or less than sin-1 (3/5)
equal to or greater than sin-1 (3/4)
less than sin-1 (2/3)
C.
equal to or greater than sin-1 (3/4)
In total internal reflection, the angle of incidence (i) must be greater than critical angle C
In the nuclear decay given below,
the particle emitted in the sequence are
β, α, γ
γ, β, α,
β,γ, α
α, β,γ
A.
β, α, γ
In a nuclear reaction conservation of charge number and mass, the number must hold good.
Alpha particles are positively charged particles with charge +2e and mass 4 m. Emission of an α particle reduces the mass of the radionuclide by 4 and its atomic number by 2. β- particles are negatively charged particles with rest mass as well as charge same as that of electrons. γ particles carry no charge and mass.
Radioactive transition will be as follows.
A p-n photodiode is fabricated from a semiconductor with a band gap of 2.5 eV. It ca detect a signal of wavelength
6000 Ao
4000 nm
6000 nm
4000 Ao
D.
4000 Ao
Only signals having wavelength less than threshold wavelength will be detected.
Energy E = hv = hc/λ
λ = hc / E
substituting the values of h, c and E in the above equation
If a diamagnetic substance is brought near the north or the south pole of a bar magnet, it is
repelled by both the poles
repelled by the north pole and attracted by the south pole
attracted by the north pole and repelled by the south pole
attracted by both the poles
A.
repelled by both the poles
The diamagnetic substance is weakly magnetised in a direction opposite to that of applied magnetic field. These are repelled in an external magnetic field ie, have a tendency to move from high to low field region.
The number of beta particles emitted by radioactive substance is twice the number of alpha particle emitted by it. The resulting daughter is an
isobar of parent
isomer of parent
isotone of parent
isotope of parent
D.
isotope of parent
Let the radioactive transition is given by
The atoms of an element having a same atomic number but a differnent mass number are called isotopes.
So, 
A transistor is operated in common-emitter configuration at Vc = 2 volts such that a change in the base current from 100 μ A to 200 μ A produces a change in the collector current from 5 mA to 10 mA. The current gain is
75
100
150
50
D.
50
For a transistor
IE = IB + IC
where
In a Rutherford scattering experiment when a projectile of charge Z1 and mass M1 approaches a target nucleus of charge Z2 and mass M2, the distance of closest approach is ro.The energy of the projectile is
directly proportional to M1 x M2
directly proportional to Z1Z2
inversely proportional to Z1
directly proportional to mass M1
B.
directly proportional to Z1Z2
The particle of mass M1 and charge z1 possess initial velocity u when it at a large distance from the nucleus of an atom having atomic number z2. At the distance of closest approach, the kinetic energy of the particle is completely converted to potential energy. Mathematically,
so the energy of the particle is directly proportional to z1z2
The ionisation energy of the electron in the hydrogen atom in its ground state is 13.6 eV. The atoms are excited to higher energy levels to emit radiations of 6 wavelengths. Maximum wavelength of emitted radiation corresponds to the transition between
n = 3 to n =2 states
n = 3 to n = 1 states
n = 2 to n = 1 states
n = 4 to n =3 states
D.
n = 4 to n =3 states
Number of spectral lines obtained due to transition of electron from nth orbit to lower orbit is N = n(n-1)/2 and for maximum wavelength, the difference between the orbits of the series should be minimum
Number of spectral lines N = n (n-1)/2
= n (n-1)/2 = 6
= n2-n-12 = 0
(n-4)(n+3) = 0
n=4
Now as the first line of the series has the maximum wavelength, therefore, electrons jump from the 4th orbit to the third orbit.
The mean free path of electrons in a metal is 4 x 10-8 m. The electric field which can give on an average 2 eV energy to an electron in the metal will be in unit of V m-1
8 x 107
5 x 10-11
8 x 10-11
5 x107
D.
5 x107
Energy = 2eV
eVo = 2eV
⇒ Vo = 2
Now, electric field E = 2 / 4 x 10-8 = 0.5 x 108 = 5 x 107 Vm-1
The number of photoelectrons emitted from light of a frequency v (higher than the threshold frequency vo) is proportional to
v-vo
threshold frequency (vo)
the intensity of light
the frequency of light (v)
C.
the intensity of light
The number of photoelectrons emitted is directly proportional to the intensity of light.
The figure shows a plot of photocurrent versus anode potential for a photosensitive surface for three different radiations. Which one of the following is a correct statement?
Curves a and b represent incident radiations of different frequencies and different intesities
Curves a and b represent incident radiations of same frequency but of different intesities
Curves b and c represent incident radiations of different frequencies and different intensities
Curves b and c represent incident radiations of same frequency having same intensity
B.
Curves a and b represent incident radiations of same frequency but of different intesities
The work function of a surface of a photosensitive material is 6.2 eV. The wavelength of the incident radiation for which the stopping potential is 5 V lies in the
ultraviolet region
visible region
infrared region
X-ray region
A.
ultraviolet region
According to laws of photoelectric effect
KEmax = E - Φ
where Φ is work function and KEmax is the maximum kinetic energy of photoelectron.
hv = eVo + Φ
or
hv = 5eV + 6.2 eV = 11.2 eV
therefore, λ = (12400/11.2) = 1000 A
Hence, the radiation lies in ultraviolet region.
The ground state energy of hydrogen atom is -13.6 eV. When its electron is in the first excited state, its excitation energy is
3.4 eV
6.8 eV
10.2 eV
zero
C.
10.2 eV
Excitation energy is defined as the energy required to take the electron from ground level orbit to any higher order orbit (ie, n = 2,3,4...)
Given, ground state energy of hydrogen atom
E1 = - 13.6/(2)2 eV
Energy of electron in first excited state (ie, n = 2)
E2 = - 13.6/ (2)2 eV
Therefore, excitation energy
ΔE = E2-E1
=(-13.6/4) - (-13.6)
= - 3.4 + 13.6 = 10.2 eV
A p-n photodiode is made of a material with a band gap of 2.0 eV. The minimum frequency of the radiation that can be absorbed by the material is nearly
10 x 1014 Hz
5 x 1014 Hz
1 x 1014 Hz
20 x 1014 Hz
B.
5 x 1014 Hz
the p-n photodiode is a semiconductor diode that produces a significant current when illuminated. It is reversed biased but is operated below the breakdown voltage.
Energy of radiation = band gap energy
ie, hv = 2.0 eV
or v= 2.0 x 1.6 x 10-19 / 6.6 x 10-34 = 5 x 1014 Hz
Two radioactive materials X1 and X2 have decay constant 5 λ respectively. If initially, they have the same number of nuclei, then the ratio of the number of nuclei of X1 to that of X2 will be 1/e after a time
λ
λ/2
1/4λ
e/λ
C.
1/4λ
If N is the number of radioactive nuclei present at some instant then
Two thin lenses of focal lengths f1 and f2 are in contact and coaxial. The power of the combination is
D.
If two thin lenses of focal lengths f1, f2 are placed in contact coaxially, then equivalent focal length of combination is 
A particle of mass 1 mg has the same wavelength as an electron moving with a velocity of 3 x 106 ms-1. The velocity of the particle is.
(mass of electron = 9.1 x 10-31 kg)
2.7 x 10-18 ms-1
9 x 10-2 ms-1
3 x 10-31 ms-1
2.7 x 10-21 ms-1
A.
2.7 x 10-18 ms-1
wavelength of a particle is given by
If M (A, Z), MP and Mn denote the masses of the nucleus 
proton and neutron respectively in units of u (1 u = 931.5 MeV/c2) and BE represents its binding energy in MeV, then
M (A,Z) = ZMp + (A-Z)Mn - BE/c2
M (A,Z) = ZMp + (A-Z)Mn + BE
M (A,Z) = ZMp + (A-Z)Mn - BF
M (A,Z) = ZMp + (A-Z)Mn + BE/c2
A.
M (A,Z) = ZMp + (A-Z)Mn - BE/c2
Two periodic waves of intensities I2 and I2 pass through a region at the same time in the same direction. The sum of the maximum and minimum intensities is
I1 + I2
2(I1 +I2)
D.
2(I1 +I2)
Resultant intensity of two periodic waves is given by
where δ is the phase difference between the waves. For minimum intensity
δ = 2nπ; n = 0,1,2,...... etc.
Therefore, for zero order maxima, cos δ =-1
A boy is trying to start a fire by focusing sunlight on a piece of paper using an equal convex lens of focal length 10 cm. The diameter of the sun is 1.39 x 109 m and its mean distance from the earth is 1.5 x 1011 m. What is the diameter of the sun's image on the paper?
9.2 x 10-4 m
6.5 x 10-4 m
6.5 x 10-5 m
12.4 x 10-4 m
A.
9.2 x 10-4 m
For the relation
Two points are located at a distance of 10 m and 15 m from the source of oscillation. The period of oscillation is 0.05 s and the velocity of the wave is 300 m/s. What is the phase difference between the oscillation of two points?
π/3
2π/3
π
π/6
B.
2π/3
Phase difference = (2π/λ) x path difference
path difference between two points,
Δ x = 15-10 = 5 m
Time period, T = 0.05 s
⇒ frequency v = 1/T
= 1/0.05 = 20 Hz
Velocity, v= 300 m/s
therefore, wavelength, λ = v/V = 300/20 = 15 m
Hence, phase difference
Δ Φ = (2π /λ) x Δ x
= (2π /15) x 5 = (2π /3)
Two nuclei have their mass number in the ratio of 1:3. The ratio of their nuclear densities would be
1:3
3:1
(3)1/3 : 1
1:1
D.
1:1
Density of nuclear matter is independent of mass number, so the required ratio is 1:1
Alternative :
A1 : A2 = 1:3
Their radii will be in the ratio
Their nuclear densities will be the same.
Tips: -
The Circuit is equivalent to 
AND gate
NAND gate
NOR gate
OR Gate
C.
NOR gate
The gate circuit can be shown by giving two inputs A and B.
Output of NOR gate,
Which is output of NOR gate.
In radioactive decay process, the negatively charged emitted beta particles are:
the electrons present inside the nucleus
the electrons produced as a result of the decay of neutrons inside the nucleus
the electrons produced as a result of collisions between atoms
the electrons produced as a result of collisions between atoms
B.
the electrons produced as a result of the decay of neutrons inside the nucleus
Beta decay can involve the emission of either electrons or positirons. The electrons or positrons emitted in a beta-decay do not exist inside the nucleus. They are only createdat the time of emission, just as photons are created when an atom makes a transition from higher to a lower energy state.
In negative Beta decay a neutron in the nucleus is transformed into a proton, an electron and an antineutrino. Hence, in radioactive decay process, the negatively charged emitted beta process are the electrons produced as a result of the decay of neutrons present inside the nucleus.
A nucleus has 
mass represented by M (A, Z). If Mp and Mn denote the mass of proton and neutron respectively and BE the binding energy (in MeV), then:
BE = [M(A,Z)-ZMp - (A-Z)Mn]c2
BE = [ZMp + (A-Z)Mn -M(A,Z)]c2
BE = [ZMp + AMn - M (A,Z)]c2
BE = M (A,Z)- ZMp - (A-Z) Mn
B.
BE = [ZMp + (A-Z)Mn -M(A,Z)]c2
In the case of formation of a nucleus, the evolution of energy equal to the binding energy of the nucleus takes place due to the disappearance of a fraction of the total mass. If the quantity of mass disappearing is ΔM, then the binding energy is
BE = ΔMc2
From the above discussion, it is clear that the mass of the nucleus must be less than the sum of the masses of the constituent neutrons and protons. We can then write.
If the nucleus 
has a nuclear radius of about 3.6 fm, then 
would have its radius approximately as:
6.0 fm
9.6 fm
12.0 fm
4.8 fm
A.
6.0 fm
If R is the radius of the nucleus, the corresponding volume 4πR3/3 has been found to be proportional to A.
This relationship is expressed in inverse form as
R = Ro A1/3
The value of Ro is 1.2 x 10-15 m, ie, 1.2 fm
The total energy of an electron in the ground state of a hydrogen atom is -13.6 eV. The kinetic energy of an electron in the first excited state is:
3.4 eV
6.8 eV
13.6 eV
1.7 eV
A.
3.4 eV
The total energy of an electron in the orbit is equal to negative of its kinetic energy.
The energy of hydrogen atom when the electron revolves in nth orbit is
E = -13.6/n2 eV
In the ground state: n =1
E = - -13.6/12 = -13.6 eV
For n = 2, E = -13.6/22 = -3.4 eV
so, the kinetic energy of an electron in the first excited state (i,e, for n = 2 ) is
K = - E = - (-3.4) - 3.4 eV
Monochromatic light of frequency 6.0 x 1014 Hz is produced by a laser. The power emitted is 2 x 10-3 W. The number of photons emitted, on the average, by the source per second is:
5 x 1015
5 x 1016
5 x 1017
5 x104
A.
5 x 1015
Photons are the packets of energy. Power emitted ,
P = 2 x 10-3 W
Energy of photon,
E = hv
= 6.6 x 10-34 x 6 x 1014 J
h being planck's constant.
Numer of photons emitted per second
n = P/E
= 2 x 10-3 / 6.6 x 10-34 x 6 x 1014
= 5 x 1015
The frequency of a light wave in the material is 2 x 10 Hz and wavelength is 5000 A. the refractive index of material will be:
1.40
1.50
3.00
1.33
C.
3.00
The velocity of light waves in the material is
ν = vλ ... (i)
Refractive index of material is
μ = c/v .... (ii)
where c is the speed of light in vacuum or air
Where c is speed of light in vacuum or air
or μ = c/vλ (iii)
Given, v = 2 x 1014 Hz,
λ = 5000 A = 5000 x 10-10 m,
c = 3x 108 m/s
Hence from Eq. (iii), we get
μ = 3 x 108 / 2 x 1014 x 500 x 10-10 = 3.00
A 5 W source emits monochromatic light of wavelength 5000 A. When placed 0.5 m away, it liberates photoelectrons from a photosensitive metallic surface. when the source si moved to a distance of 1.0 m, the number of photoelectrons liberated will be reduced by a factor of:
4
8
16
2
A.
4
The intensity of light is inversely proportional to the square of the distance.
Now, since number of photoelectric emitted per second is directly proportional intensity, so number of electrons emitted would decrease by factor of 4.
Two radioactive substance A and B have decay constants 5λ and λ respectively. At t = 0 they have the same number of nuclei. The ratio of a number of nuclei of A t\o those of B will be 
after a time interval:
1/ 4λ
4λ
2λ
1/2λ
D.
1/2λ
Number of nuclei remained after time t can be written as
N = Noe-λt
where No is initial number of nuclei of both the substances.
Monochromatic light of frequency 6.0 x 1014 Hz is produced by a laser. The power emitted is 2 x 10-3 W The number of photons emitted, on the average, by the source per second is:
5 x 1015
5 x 106
5 x 1017
5 x 1014
A.
5 x 1015
Photons are the packest of energy.
Power emitted,
P = 2 x 10-3 W
Energy of photon,
E = hv
= 6.6 x 10-34 x 6 x 1014 J
h being Planck's constant
Number of photons emitted per second
n = P/E
=
= 5 x 1015
The binding energy of deuteron is 2.2 MeV and that of 
is 28 MeV. If two deuterons are fused to form one 
then the energy released is
25.8 MeV
23.6 MeV
19.2 MeV
30.2 MeV
B.
23.6 MeV
The reaction can be written as:
The energy released in the reaction is difference of binding energies of daughter and parent nuclei.
Hence, energy released.
= binding energy of 
- 2 x binding energy of 
= 28 - 2 x 2.2 = 23.6 MeV
In a radioactive material the activity at time t1 is R1 and at a later time t2, it is R2. If the dacay constant of the material is λ, then
R1 = R2 e-λ(t1 -t2 )
R1 = R2 eλ(t1 -t2 )
R1 = R2 e(t2 /t1 )
R1 = R2
A.
R1 = R2 e-λ(t1 -t2 )
The decay rate R of radioactive materials the number of decays per second.
From radioactive decay law.
Thus, 
or 
...(i)
where 
is the activity of the radioactive material at time t = 0.
At time t1, 
...(ii)
At time t2, 
...(iii)
Dividing Eq. (ii) by (iii), we have,
Ionization potential of hydrogen atom is 13.6 eV. Hydrogen atoms in the ground state are excited by monochromatic radiation of photon energy 12.1 eV. According to Bohr's theory, the spectral lines emitted by hydrogen will be
two
three
four
one
B.
three
Ionization energy corresponding to ionization potential = -13.6 eV.
Photon energy incident = 12.1 eV
So, the energy of electron in excited state
= -13.6 + 12.1 = -1.5 eV
i.e., 
i.e., energy of electron in excited state corresponds to third orbit.
The possible spectral lines are when electron jumps from orbit 3rd to 2nd; 3rd to 1st and 2nd to 1st. Thus, 3 spectral lines are emitted.
A microscope is focussed on a mark on a piece of paper and then a slab of glass of thickness 3 cm and refractive index 1.5 is placed over the mark. How should the microscope be moved to get the mark in focus again?
1 cm upward
4.5 cm downward
1 cm downward
2 cm upward
A.
1 cm upward
Apparent depth of mark as seen through a glass slab of thickness x and refractive index 
is
As image appears to be raised by 1 cm, therefore, microscope must be moved upward by 1 cm.
A photo-cell employs photoelectric effect to convert
change in the frequency of light into a change in electric voltage
change in the intensity of illumination into a change in photoelectric current
change in the intensity of illumination into a change in the work function of the photocathode
change in the frequency of light into a change in the electric current
B.
change in the intensity of illumination into a change in photoelectric current
In photoelectric effect when monochromatic radiations of suitable frequency fall on the photo-sensitive plate called cathode, the photoelectrons are emitted which get accelerated towards anode. These electrons flow in the outer circuit resulting in the photoelectric current.
Using the incident radiations of a fixed frequency, it is found that the photoelectric current increases linearly with the intensity of incident light as shown in figure. Hence, a photocell employs photoelectric effect to convert change in the intensity of illumination into a change in photoelectric current.
When photons of energy hv fall on an aluminium plate (of work function E0), photoelectrons of maximum kinetic energy K are ejected. If the frequency of the radiation is doubled, the maximum kinetic energy of the ejected photoelectrons will be
K+E0
2K
K
K+hv
D.
K+hv
The energy of photon is used in liberating the electron from metal surface and in imparting the kinetic energy to emitted photoelectron.
According to Einstein's photoelectric effect energy of photon = KE of photoelectron + work function of metal
i.e., 
or 
Now, we have given,
Therefore, 
...(ii)
From Eqs. (i) and (ii), we have
Putting 
The momentum of a photon of energy 1 MeV in kg m/s, will be
D.
Energy of photon is given by
...(i)
Where h is Planck's constant, c the velocity of light and 
its wavelength.
de-Broglie wavelength is given by
...(ii)
p being momentum of photon.
From Eqs. (i) and (ii), we can have
Hence, after putting numerical values, we obtain
The radius of germanium (Ge) nuclide is measured to be twice the radius of 
The number of nucleons in Ge are
73
74
75
72
D.
72
Let radius of 
be nucleus be r. Then radius of germanium (Ge) nucleus will be 2r.
Radius of a nucleus is given by:
Thus, in germanium (Ge) nucleus number of nucleons is 72.
A convex lens and a concave lens, each having same focal length of 25 cm, are put in contact to form a combination of lenses. The power in diopters of the combination is
25
50
infinite
zero
D.
zero
Focal length of combination of lenses placed in contact is
For convex lens, 
For concave lens, 
Hence, 
Hence, power of combination
Unpolarised light is incident from air on a plane surface of a material of refractive index 'µ'. At a particular angle of incidence 'i', it is found that the reflected and refracted rays are perpendicular to each other. Which of the following options is correct for this situation?
Reflected light is polarised with its electric vector parallel to the plane of incidence
Reflected light is polarised with its electric vector perpendicular to the plane of incidence
i = tan-1
i = sin-1
B.
Reflected light is polarised with its electric vector perpendicular to the plane of incidence
When reflected light rays and refracted rays are perpendicular, reflected light is polarised with electric field vector perpendicular to the plane of incidence.
Also, tan i = µ (i = Brewester angle)
An em wave is propagating in a medium with a velocity . The instataneous oscillating electric field of this em wave is along +y axis. Then the direction of oscillating magnetic field of the em wave will be along
-z direction
+z direction
-x direction
-y-direction
B.
+z direction
As we know,
(Therefore , electric field vector is along + y axis)
i.e., direction of magnetic field vector is along + z direction.
The maximum numbers of possible interference maxima for slit separation equal to twice the wavelength in Young's double slit experiment is
infinite
five
three
zero
B.
five
The condition of interference maxima is
The magnitude of sin θ lies between 0 and 1
When n = 0, sinθ = 0 ⇒θ = 0°
When n = 1, sin θ = 1/2 ⇒ θ =30°
When n = 2, sin θ = 1⇒ θ =90°
Thus, there is central maximum ( θ = 0°), on other side of it maxima lie at θ = 30° and θ = 90°, so maximum number of possible interference maxima is 5.
Light of wavelength 6000 A0 is reflected at nearly normal incidence from a soap film of refractive index 1.4. The least thickness of the film that will appear black is
200 A0
2000 A0
1000 A0
Infinity
B.
2000 A0
for the minimum in reflected light
2μtcosr =nλ
Light travels faster in air than that in glass .This is accordance with
wave theory of light
corpuscular theory of light
Neither(a) nor(b)
Both (a) and (b)
A.
wave theory of light
Light travels faster in air than in glass. This is accordance with wave theory of light.
In Young's double slit experiment, light of wavelength 6000 A0 is used to produced fringes of width 0.8 mm at a distance of 2.5 m. If the whole appart is dropped in liquid of refractive index 1.6, then the fringe width will be
0.2mm
0.4mm
0.5mm
0.6mm
C.
0.5mm
The Young's double slit experiment belongs to a general class of "double path "experiments, in which a wave is split into two separate waves that later combine into a single wave.
In forward biased if the p-n junctions diode is forward biased, then width of potential barrier in p-n junction diode
remains constant
increase
decrease
None of these
C.
decrease
In forward bias the terminal of the same polarities are connected between the diode and the supply. Hence by repulsion excess number of holes or electron come out and width of potential barrier can be decreased.
If the energy released in the fission of one nucleus is 3.2x10-11J, then number of nuclei required per second in a power plant of 16 kW is
D.
The number of nuclei required per second is
n==
The enregy of a photon corresponding to the visible light maximum wavelength is nearly
7ev
3.2eV
1eV
1.6eV
D.
1.6eV
In the visible light λmax= 7800Ao
In junction diode, the holes are because of
Missing electrons
extra electrons
protons
neutrons
A.
Missing electrons
In semiconductor, when an electron leaves its place, a positive charge is left behind and it is known as hole. Hence, holes are created because of missing electrons.
The half-life of a radioactive substance is 3.6 days. How much of 20 mg of this radioactive substance will remain after 36 days?
0.0019 mg
1.019 mg
1.109 mg
0.019 mg
D.
0.019 mg
Half time T1/2 = 3.6 days
Initial quantity N0 =20 mg
Total time = 36 days
The number of half lives
The energy in MeV is released due to transformation of 1kg mass completely into energy is (c=3×108 m/s)
7.625×10MeV
10.5×1029MeV
2.8×10-28MeV
5.625×1029MeV
D.
5.625×1029MeV
According to Einstein's mass energy equivalence
E=mc2 =1×(3×108)2 =9×1016 J
=
20 kV potential is applied across X-ray tube, the minimum wavelength of X-ray emitted will be
0.62 Ao
0.37 Ao
1.62 Ao
1.31 Ao
A.
0.62 Ao
(i)The minimum energy required to eject an electron from the surface is
called the photoelectric work function of the metal.
(ii)Energy greater than the work function of the metal (Φo)
required for electron emission from the metal surface can be supplied by suitably heating
(or) applying strong electric field or irradiating it by suitable frequency.
In the CB mode of a transistor, when the collector voltage is changed by 0.5 volt, the collector current changes by 0.05 mA. The output resistance will be
10 kΩ
20 kΩ
5 kΩ
2.5 kΩ
A.
10 kΩ
Here: ΔVc = 0.5 volt
ΔIc =0.05 mA=0.05×10-3
Output resistance is defined as the ratio of change in output voltage or collector voltage (VCB) to the corresponding change in output current or collector current (IC), with the input current or emitter current (IE) kept at constant.
A student can distinctly see the object upto a distance 15 cm. He wants to see the blackboard at a distance of 3m. Focal length and power of lens used respectively will be
-4.8cm, -3.3D
-5.8cm, -4.3D
-7.5cm, -6.3D
-15.8cm, -6.33D
D.
-15.8cm, -6.33D
The rays emanating from a point actually meet at another point after reflection and/or refraction. That point is called the image of the first point. The image is real if the rays actually converge to the point, it is virtual if the rays do not actually meet but appear to diverge from the point from the point when produces backwards.
The student should use a lens which forms image at a distance of 15 cm of the object placed at 3m i.e object distance u= -3 m =-3000 cm, image distance v =-15 cm
A thin glass prism (μ =15) in the position of minimum deviation deviates the monochromatic light ray by 10o, the refracting angle of prism is
20o
10o
30o
45o
A.
20o
When a narrow beam of sunlight, is incident on a glass prism the emergent light is seen to be consisting of several colours. There is actually a continuous variation of colour, but broadly in seven colours (acronym VIBGYOR)
The phenomenon of splitting of light into different colours known as dispersion.
Two thin lenses of powers 12 D and-2D respectively are placed in contact, the power, focal length and nature
respectively will be
8D, 0.8m, convex
14D, 0.5m, convex
5D, 0.2m, convex
10D, 0.1m, convex
D.
10D, 0.1m, convex
Convex lens refers to the lens which merges the light at a particular point, that travels through it.
Interference was observed in an interference chamber when air was present. Now, the chamber is evacuated and if the same light is used, a careful observation will show
no interference
interference with bright band
interference with dark bands
interference in which breadth of the fringe will be slightly increased
D.
interference in which breadth of the fringe will be slightly increased
When chamber is evacuated, the refractive index (µ) decreases. Therefore, wavelength increases.
Now, fringe width is given by
β ∝ λ
Therefore, as wavelength increases, the fringe width will increase.
The tip of a needle does not give a sharp image on a screen. This is due to
polarisation
interference
diffraction
None of these
C.
diffraction
As the size of tip of the needle is very small ( because its a point ) and it comes under the range of wavelength of light, so the ray coming from back side of needle ( i.e consider needle's tip just blocking a ray of light) will bend across the tip.
And hence on screen, that ray will also give brightness near so formed image of needle's tip, hence reducing the original sharpness of the needle's tip.
The tip of a needle does not give a sharp image, it is due to diffraction.
Two coherent light beams of intensities I and 41 are superposed. The maximum and minimum possible intensities in the resulting beam are
5I and I
5I and 3I
9I and I
9I and 3I
C.
9I and I
Given:- I1 = I, I2 = 4I
Resultant intensity =
where δ is the phase difference between two waves.
intensity will be maximum when δ = 0 ....(cos 0o = 1)
Imax =
=
=
Imax = 9I
Intensity will be minimum when δ = 0
and Imin =
= .....(cos 180o = -1)
Imin = I
If prism angle α = 1o , μ= 1.54, distance between screen and prism (D) = 0. 7 m, distance between prism and source nm, then in Fresnal biprism find the value of β (fringe width).
10-4 m
10-3 mm
10-4 × m
B.
10-3 mm
a = 0.3
μ = 1.54
We know that
d = 2a ( μ - 1 )
= 2 × 0.3 ( 1.54 - 1 ) × 1
= 2 × 0.3 × 0.54
d = 0.324
Fringe width F =
=
=
F = 10-3 mm
The magnification produced by a astronomical telescope for normal adjustment is 10 and the length of the telescope is 1.1 m. The magnification, when the image is formed atleast distance of distinct vision is
6
18
16
14
D.
14
Given:-
m = 10, length of telescope= 1.1 m
We know that,
Magnification m =
10 =
f0 = 10 fe
∴ fe + f0 = 1.1 m
⇒ fe + 10 fe = 1.1 mm ....[ f0 = 10fe ]
⇒ fe ( 1 + 10 ) = 1.1 m
⇒ fe = 0.1 m or 10 cm
Magnification least distance of distinct vision,
Mb =
= 10 .....[ since D = 25 cm ]
= 10 ×
Mb = 14
Two slits are separated by a distance of 0.5 mm and illuminated with light of wavelength λ = 6000 Ao. If the screen is placed at 2.5 m from the slits. The distance of the third bright fringe from the centre will be
1.5 mm
3 mm
6 mm
9 mm
D.
9 mm
Distance of nth bright fringe from the centre,
xn =
where n = 0, ± 1,±2,.....
The third bright fringe width from the centre,
x3 =
= 9 × 10-3 m
x3 = 9 mm
Assertion: A ray of light is incident from outside on a glass sphere surrounded by air. This ray may suffer total internal reflection at second interface.
Reason: If a ray of light goes from denser to rarer medium, it bends away from the normal.
If both assertion and reason are true and reason is the correct explanation of assertion.
If both assertion and reason are true but reason is not the correct explanation of assertion.
If assertion is true but reason is false.
If both assertion and reason are false.
B.
If both assertion and reason are true but reason is not the correct explanation of assertion.
When a ray of light enters a spherical glass sphere, it is first refracted at first interface and then it strike the inner surface of sphere (second interface) and get totally internally reflected if the angle between the refracted ray and normal to the drop surface is greater than the critical angle (42°, in this case).
Assertion: Two point coherent sources of light S1 and S2 are placed on a line as shown. P and Q are two points on that line. If at point P maximum intensity is observed then maximum intensity should also be observed at Q.
Reason: In the figure of assertion the distance is equal to distance .
If both assertion and reason are true and reason is the correct explanation of assertion.
If both assertion and reason are true but reason is not the correct explanation of assertion.
If assertion is true but reason is false.
If both assertion and reason are false.
B.
If both assertion and reason are true but reason is not the correct explanation of assertion.
If maximum intensity is observed at P then for maximum intensity to be also observed at Q, S1 and S2 must have phase difference of 2 m (where m is an integer).
In a Fraunhofer diffraction at single slit of width d with incident light of wavelength 5500 Ao, the first minimum is observed, at angle 30°. The first secondary maximum is observed at an angle θ =
sin-1
sin-1
sin-1
sin-1
C.
sin-1
Slit width = d
λ = 5500 Ao
Ao = 5.5 × 10-7 m
θn = 30o
For first secondary minima
d sin θn = λ
⇒ d =
⇒ d = 11 × 10-7 m
For first secondary maxima
d sin θn =
⇒ sinθn =
=
⇒ sinθn =
⇒ θn = sin-1 (3/4 )
The intensity ratio of the maxima and minima in an interference pattern produced by two coherent sources of light is 9 : 1. The intensities of the used light sources are in ratio
3 : 1
4 : 1
9 : 1
10 : 1
B.
4 : 1
Given:- The intensity ratio interference pattern produced by two coherent sources of light 9:1.
⇒ 3 ( a b ) = a + b
⇒ 3a 3b = a + b
⇒ 2a = 4b
⇒ a = 2b
∴
∴
The two coherent sources with intensity ratio β produce interference. The fringe visibility will be
2 β
A.
= β
∴
Imax = ( a + b )2 and Imin = ( a b )2
Fringe visibility is given by
V =
=
V =
V =
V =
Assertion: In YDSE bright and dark fringe are equally spaced.
Reason: It only depends upon phase difference.
If both assertion and reason are true and reason is the correct explanation of assertion.
If both assertion and reason are true but reason is not the correct explanation of assertion.
If assertion is true but reason is false.
If both assertion and reason are false.
C.
If assertion is true but reason is false.
Fringe width is given by
β =
where, D= distance between slits and screen,
d = distance between coherent sources of light
and A = wavelength of incident light.
The assertion is true but reason is false. Young's Double-slit Experiment depends on distance between slits and screen (D), the wavelength of light used (λ), distance between the two slits (d).
In a certain double slit experimental arrangement, interference fringes of width 1 mm each are observed when light of wavelength 5000 Ao is used. Keeping the setup unaltered, if the source is replaced by another of wavelength 6000 Ao, the fringe width will be
1.2 mm
0.5 mm
1 mm
1.5 mm
A.
1.2 mm
Fringe width
β =
Since, D and d are unaltered β ∝ λ
∴
β' = β ×
= 1 ×
⇒ β = 1.2 mm
Assertion: In a common emitter transistor amplifier, the input current is much less than the output current than output impedance.
Reason: The common emitter transistor amplifier has higher input impedance than output impedance.
If both assertion and reason are true and reason is the correct explanation of assertion.
If both assertion and reason are true but reason is not the correct explanation of assertion.
If assertion is true but reason is false.
If both assertion and reason are false.
A.
If both assertion and reason are true and reason is the correct explanation of assertion.
Resolving power of telescope
R =
R =
Here a is the diameter of the objective of the telescope.
Assertion: A secondary rainbow have inverted colours than the primary rainbow.
Reason: The secondary rainbow is formed by single total internal reflection.
If both assertion and reason are true and reason is the correct explanation of assertion.
If both assertion and reason are true but reason is not the correct explanation of assertion.
If assertion is true but reason is false.
If both assertion and reason are false.
C.
If assertion is true but reason is false.
In the formation of secondary rainbow, light suffers two total internal reflections instead of one. In this, ray undergoes two internal partial reflections having a maximum deviation of about 50° for red and 54° for violet. Also secondary rainbow will sometimes be formed with inverted colours. Secondary rainbow is fainter than the primary for two reason
(i) the light has undergone two internal reflections and has thereby becomes weakened,
(ii) there is greater angular dispersion in this rainbow than in the primary.
Sponsor Area
Sponsor Area
nucleus is taken to be RAl, then the radius of 
nucleus is nearly
and 
and R is the radius of curvature of the curved surface of the lenses, then the focal length of the combination is
emits one α - particle and two β- particles. The resulting nucleus is 
