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Ray Optics and Optical Instruments

More Topic from Physics

Question 1
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In an optics experiment, with the position of the object fixed, a student varies the position of a convex lens and for each position, the screen is adjusted to get a clear image of the object. A graph between the object distance u and the image distance v, from the lens, is plotted using the same scale for the two axes. A straight line passing through the origin and making an angle of 45o with the x-axis meets the experimental curve at P. The coordinates of P will be

  • (2f, 2f)

  • (f/2, f/2)

  • (f,f)

  • (4f, 4f)

Solution

A.

(2f, 2f)

It is possible when object kept at centre of curvature.
u = v
u = 2f,
v = 2f.

Question 2
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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

Solution

A.

1/15 m/s


The mirror formula
1 over straight u space plus 1 over straight v space equals space 1 over straight f 1 over straight v space plus space fraction numerator 1 over denominator negative 280 end fraction space equals space 1 over 20 1 over straight v space space equals space 1 over 20 plus 1 over 280 1 over straight v space plus fraction numerator 14 plus 1 over denominator 280 end fraction straight v space equals space 280 over 15 straight v subscript 1 space equals space minus space open parentheses straight v over straight u close parentheses squared. straight v therefore space straight v subscript 1 space equals space minus open parentheses fraction numerator 280 over denominator 15 space straight x space 280 end fraction close parentheses squared.15 space therefore space equals space straight v subscript 1 space equals space fraction numerator negative 15 over denominator 15 space straight x space 15 end fraction straight v subscript 1 space equals space minus 1 over 15 space straight m divided by straight s
Question 3
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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

Solution

A.

9.1 x 10-11 Wb


Let M12 be the coefficient of mutual induction between loops
straight capital phi subscript 1 space equals space straight M subscript 12 space straight i subscript 2 rightwards double arrow space fraction numerator straight mu subscript 0 straight i subscript 2 straight R squared over denominator 2 left parenthesis straight d squared plus straight R squared right parenthesis to the power of 3 divided by 2 end exponent end fraction πr squared space equals space straight M subscript 12 space straight i subscript 2 rightwards double arrow space straight M subscript 12 equals space fraction numerator straight mu subscript 0 straight R squared over denominator 2 left parenthesis straight d squared plus straight R squared right parenthesis to the power of 3 divided by 2 end exponent end fraction πr squared space straight capital phi space equals space straight M subscript 12 space straight i subscript 1 rightwards double arrow straight capital phi subscript 2 space equals space 9.1 space straight x space 10 to the power of negative 11 end exponent space weber
Question 4
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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

Solution

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
1 over straight v space minus fraction numerator 1 over denominator negative 40 end fraction space equals space 1 over 20
V = 40 cm from the converging lens.


Question 5
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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

Solution

D.

f

For open ends, fundamental frequency f in air
we have

straight lambda over 2 space equals space straight l straight lambda space equals space 2 straight l straight v space equals fλ straight f equals straight v over straight lambda space equals space fraction numerator straight v over denominator 2 straight l end fraction
When a pipe is dipped vertically in the water, so that half of it is in water, we have

straight lambda over 4 space equals space 1 half straight lambda space equals space 2 straight l rightwards double arrow space straight v equals space straight f apostrophe straight lambda straight f apostrophe space equals space straight v over straight lambda space equals fraction numerator straight v over denominator 2 straight l end fraction space equals space straight f
Thus, the fundamental frequency of the air column is now,
f=f'

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