Wave Optics

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Question
CBSEENPH12039877

A beam of light of straight lambda = 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

Solution

D.

2.4 mm


From the above figure,

tan space theta space equals space fraction numerator begin display style bevelled x over 2 end style over denominator 2 end fraction
For small straight theta and when straight theta is counted in rad, tan almost equal tostraight theta
So, 
space space space space space space straight theta almost equal to fraction numerator begin display style bevelled straight x over 2 end style over denominator 2 end fraction
rightwards double arrow space space straight lambda over 9 space almost equal to space straight x over 4
straight x space equals space fraction numerator 4 straight lambda over denominator straight a end fraction equals fraction numerator 4 space straight x space 600 space straight x space 10 to the power of negative 9 end exponent over denominator 10 to the power of negative 3 end exponent end fraction space straight m
space space space equals space 24 space straight x space 10 to the power of negative 4 end exponent space straight m
space space space equals space 2.4 space straight x space 10 to the power of negative 3 end exponent space straight m
space space space equals space 2.4 space mm

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Question
CBSEENPH12040057

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

Solution

A.

9.2 x 10-4 m

For the relation

straight I over straight O space equals space straight v over straight u
Here comma space straight O thin space equals space 1.39 space space straight x space 10 to the power of 9 comma space straight v equals space 0.1 space straight m comma
straight u space equals space 1.5 space straight x space 10 to the power of 11 space straight m
therefore comma
therefore space straight I space equals space fraction numerator 0.1 over denominator 1.5 space straight x space 10 to the power of 11 space end fraction space straight x space 1.39 space straight x space 10 to the power of 9
space
equals space 9.2 space straight x space 10 to the power of negative 4 end exponent space straight m space

Question
CBSEENPH12039893

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

Solution

B.

The angular width of the central maximum of the diffraction pattern will increase

Question
CBSEENPH12039787

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

Solution

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,
f' = open parentheses fraction numerator straight v over denominator straight v minus straight v subscript straight s end fraction close parentheses
where,
f = original frequency of source
v= velocity of the sound
vs = velocity of the source
So, straight f apostrophe space equals space open parentheses fraction numerator 330 over denominator 330 minus 15 end fraction close parentheses 800 space equals space 838 space Hz

Question
CBSEENPH12039813

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

Solution

A.

2Dλ/a

For the condition of maxima

sinθ space equals space straight lambda over straight a
from space the space geometry
sinθ space equals space straight theta space equals space straight lambda over straight D space space left parenthesis for space small space angle right parenthesis
So comma space straight Y over straight D space equals space straight lambda over straight a
straight Y space equals space λD over straight a
Hence comma space width space central space maxima space equals space 2 straight Y space equals space fraction numerator 2 λD space over denominator straight a end fraction

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