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Indian Navy SSR Wave Optics — practice questions

Explore 6 free Wave Optics questions in Indian Navy SSR Physics. Open each question or exercise to read its content, reference answers and available explanation.

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Wave Optics questions and exercises with answers

Showing questions 1–6 of 6.

In Young's experiment, fringe width β equals:Read the answer and worked solution →Power of a convex lens of focal length 50 cm:Read the answer and worked solution →Two coherent sources have intensities I1 = 9I and I2 = I. Maximum and minimum intensities in interference pattern:Read the answer and worked solution →In single-slit diffraction, ratio of intensities of first and central maxima is approximately:Read the answer and worked solution →In YDSE, fringe width β =Read the answer and worked solution →In Young's double-slit experiment, the fringe width β is given by:Read the answer and worked solution →

Worked practice questions

In Young's experiment, fringe width β equals:

  1. λD/d
  2. λd/D
  3. Dd/λ
  4. λ/(Dd)
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Correct answer: A. λD/d

Fringe width β = λD/d, where λ is wavelength, D is slit-to-screen distance, d is slit separation. Wider screen distance or longer wavelength give wider fringes.

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Power of a convex lens of focal length 50 cm:

  1. 2 D
  2. 0.5 D
  3. -2 D
  4. 50 D
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Correct answer: A. 2 D

Power P = 1/f (in metres). f = 50 cm = 0.5 m. P = 1/0.5 = 2 dioptre (D). Positive sign means converging lens.

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Two coherent sources have intensities I1 = 9I and I2 = I. Maximum and minimum intensities in interference pattern:

  1. 10I, 8I
  2. 16I, 4I
  3. 9I, I
  4. 16I, 0
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Correct answer: B. 16I, 4I

Amplitudes A1 ∝ sqrt(9I) = 3 sqrt(I), A2 = sqrt(I). We have I_max = (A1 + A2)². So I_max = 16I. We have I_min = (A1 - A2)². So I_min = 4I. (Not 0 since intensities are unequal.)

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