JEE Main 2023 · 6 April · morning · Single correct · +4 / −1

A monochromatic light wave with wavelength λ₁ and frequency v₁ in air enters another medium. If the angle of incidence and angle of refraction at the interface are 45° and 30° respectively, then the wavelength λ₂ and frequency v₂ of the refracted wave are:

  1. (a) λ₂=λ₁, v₂=1/√2 v₁
  2. (b) λ₂=λ₁, v₂=√2 v₁
  3. (c) λ₂=√2 λ₁, v₂=v₁
  4. (d) λ₂=1/√2 λ₁, v₂=v₁
Answer and solution

Answer (d) λ₂=1/√2 λ₁, v₂=v₁

Solution

Physics sir · 4 steps

Frequency is fixed by the source, so only the wavelength shrinks, by the factor n from Snell's law.

  1. Find the index. Snell's law with air on the incident side gives the refractive index of the second medium directly.

    n = (sin 45°)/(sin 30°) = (1/√2)/(1/2) = √2

  2. Frequency stays. Frequency is set by the source, and it does not change when light crosses a boundary. So v₂ = v₁.
  3. Wavelength shrinks. Speed drops to c/n while frequency stays the same. Since speed = frequency × wavelength, the wavelength drops by the same factor: λ₂ = λ₁/n = λ₁/√2.
  4. So. λ₂ = 1/√2λ₁ and v₂ = v₁, which is option (d).

The idea to keep. In refraction, frequency never changes. Speed and wavelength both divide by n when light enters a denser medium.

Answer from the NTA final answer key. Solution by PrepWiser.

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Asked about 200 times in JEE Main since 2020. This question is from 2023.