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 Multiple Choice QuestionsMultiple Choice Questions

1.

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:

  • straight a equals space straight lambda squared over straight L space and space straight b subscript min space equals space open parentheses fraction numerator 2 straight lambda squared over denominator straight L end fraction close parentheses
  • straight a space equals space square root of λL space and space straight b subscript min space equals open parentheses fraction numerator 2 straight lambda squared over denominator straight L end fraction close parentheses
  • straight a space equals space square root of λL space and space straight b subscript min space equals square root of space open parentheses 4 λL close parentheses end root
  • straight a space equals space square root of λL space and space straight b subscript min space equals square root of space open parentheses 4 λL close parentheses end root
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2.

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:

  • less than straight v space open parentheses 4 over 3 close parentheses to the power of 1 divided by 2 end exponent
  •  = v(4/3)1/2

  •  = v(4/3)1/2

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3.

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 spectrum of visible light whose frequency is more than that of green light will come out to the air medium.

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4.

Two beams, A and B, of plane polarised light with mutually perpendicular planes of polarisation are seen through a Polaroid. From the position when the beam A has maximum intensity (and beam B has zero intensity), a rotation of Polaroid through 300 makes the two beams appear equally bright. If the initial intensities of the two beams are IA and IB respectively, then IA/IB equals:

  • 3

  • 3/2

  • 1

  • 1

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5.

A beam of unpolarized light of intensity I0 is passed through a Polaroid A and then through another polaroid B which is oriented so that its principal plane makes an angle of 45° relative to that of A.The intensity of the emergent light is

  • I0

  • I0/2

  • I0/4

  • I0/4

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6.

Two coherent point sources S1 and S2 are separated by a small distance ‘d’ as shown. The fringes obtained on the screen will be:

  • point

  • straight lines

  • Semi-circle

  • Semi-circle

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7.

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

  • 3f/4

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8.

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 the correct explanation for statement 1

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9.

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

  • straight I subscript straight m over 9 left parenthesis 4 space plus 5 space cos space straight ϕ right parenthesis
  • straight I subscript straight m over 3 open parentheses 1 space plus space 2 space cos squared space straight ϕ over 2 close parentheses
  • straight I subscript straight m over 5 open parentheses 1 space plus 4 space cos squared straight ϕ over 2 close parentheses
  • straight I subscript straight m over 5 open parentheses 1 space plus 4 space cos squared straight ϕ over 2 close parentheses


D.

straight I subscript straight m over 5 open parentheses 1 space plus 4 space cos squared straight ϕ over 2 close parentheses

Let A1 = A0, A2 = 2A0
If amplitude of resultant wave is A then

straight A squared space equals space straight A subscript 1 superscript 2 space plus space straight A subscript 2 superscript 2 space plus 2 straight A subscript 1 straight A subscript 2 space cos space straight ϕ
 For maximum intensity

straight A subscript max superscript 2 space equals space straight A subscript 1 superscript 2 space plus space straight A subscript 2 superscript 2 space plus 2 straight A subscript 1 straight A subscript 2
therefore fraction numerator straight A squared over denominator straight A subscript max superscript 2 end fraction space equals space fraction numerator straight A subscript 1 superscript 2 space plus straight A subscript 2 superscript 2 space plus 2 straight A subscript 1 straight A subscript 2 space cos space straight ϕ over denominator straight A subscript 1 superscript 2 space plus straight A subscript 2 superscript 2 space plus space 2 straight A subscript 1 straight A subscript 2 end fraction
space equals space fraction numerator straight A subscript 0 superscript 2 space plus space 4 straight A subscript 0 superscript 2 space plus space 2 left parenthesis straight A subscript 0 right parenthesis left parenthesis 2 straight A subscript 0 right parenthesis cos space straight ϕ over denominator straight A subscript 0 superscript 2 space plus space 4 straight A subscript 0 superscript 2 space plus space 2 left parenthesis straight A subscript 0 right parenthesis left parenthesis 2 straight A subscript straight o right parenthesis end fraction
straight I over straight I subscript straight m space equals space fraction numerator 5 space plus 4 space cos space straight ϕ over denominator 9 end fraction space equals space fraction numerator 1 space plus space 8 space cos squared space left parenthesis straight ϕ divided by 2 right parenthesis over denominator 9 end fraction

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10.

The question has a paragraph followed by two statements, Statement 1 and Statement 2. Of the given four alternatives after the statements, choose the one that describes the statements. A thin air film is formed by putting the convex surface of a plane-convex lens over a plane glass plate. With monochromatic light, this film gives an interference pattern due to light reflected from the top (convex) surface and the bottom (glass plate) surface of the film.

Statement 1 :
When light reflects from the air-glass plate interface, the reflected wave suffers a phase change of π
Statement 2 :
The centre of the interference pattern is dark.

  • Statement 1 is true, statement 2 is false. 

  • Statement 1 is true, Statement 2 is true, Statement 2 is the correct explanation of Statement 1

  • Statement 1 is true, Statement 2 is true, Statement 2 is not the correct explanation of Statement 1

  • Statement 1 is true, Statement 2 is true, Statement 2 is not the correct explanation of Statement 1

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