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Indian Navy 10+2 B.Tech Cadet Wave Optics — practice questions

Explore 7 free Wave Optics questions in Indian Navy 10+2 B.Tech Cadet 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 1–7 of 7.

Snell's law states that the ratio sin(i)/sin(r) for two media is:Read the answer and worked solution →For maximum brightness in Newton's rings (reflected light, air film), what is the radius of n-th bright ring? (R = lens radius, λ = wavelength)Read the answer and worked solution →Two slits 0.5 mm apart with light of wavelength 5000 Å. The 5th bright fringe is at distance from central maximum on screen 1 m away:Read the answer and worked solution →Resolution limit of a telescope by Rayleigh criterion:Read the answer and worked solution →Wavelength of light in water (n = 4/3) if its vacuum wavelength is 600 nm:Read the answer and worked solution →Maximum number of bright fringes visible in YDSE with d/λ = 4:Read the answer and worked solution →HUYGENS' PRINCIPLE states that every point on a wavefront acts as a:Read the answer and worked solution →

Worked practice questions

Snell's law states that the ratio sin(i)/sin(r) for two media is:

  1. Variable
  2. Constant for given media
  3. Equal to 1
  4. Equal to 0
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Correct answer: B. Constant for given media

sin(i)/sin(r) = n21, the refractive index of medium 2 relative to medium 1. This ratio is constant for any pair of media.

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For maximum brightness in Newton's rings (reflected light, air film), what is the radius of n-th bright ring? (R = lens radius, λ = wavelength)

  1. sqrt(nλR)
  2. sqrt((2n+1)λR/2)
  3. sqrt(nλR/2)
  4. sqrt(nλ/R)
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Correct answer: B. sqrt((2n+1)λR/2)

In Newton's rings (reflected light), there is a phase change at the lower glass surface but not at the upper. **For bright ring:** 2t = (n + 1/2)λ where t = r²/(2R). **So r_n = sqrt((2n+1)λR/2).**

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Two slits 0.5 mm apart with light of wavelength 5000 Å. The 5th bright fringe is at distance from central maximum on screen 1 m away:

  1. 2.5 mm
  2. 5 mm
  3. 7.5 mm
  4. 10 mm
Show answer and solution

Correct answer: B. 5 mm

Position of n-th bright fringe: y_n = n λ D/d = 5 × (5000 × 10⁻¹⁰) × 1 / (0.5 × 10⁻³) = 5 × 10⁻³ m = 5 mm.

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