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JEE Advanced Physics Ray Optics 2022 JEE Advanced 2022 (Paper 2)

JEE Advanced Physics Question (2022) — Solution

Question

Consider a configuration of n  identical units, each consisting of three layers. The first layer is a column of air of height h = 1 3   cm , and the second and third layers are of equal thickness d = 3 - 1 2   cm , and refractive indices μ 1 = 3 2  and μ 2 = 3 , respectively. A light source O is placed on the top of the first unit, as shown in the figure. A ray of light from O is incident on the second layer of the first unit at an angle of θ = 60 °  to the normal. For a specific value of n , the ray of light emerges from the bottom of the configuration at a distance l = 8 3   cm , as shown in the figure. The value of  n is _______.

Options

  1. A. A
  2. B. B
  3. C. C
  4. D. D

Answer

A. A

Step-by-step solution

Considering a single unit with three layers. The figure below shows the refraction of light as it moves from one medium to others. For the first interface,  x 1 = 1 3 × tan 60 ° = 1 3   cm Using Snell's law, μ 1 sin θ 1 = μ 2 sin θ 2 ⇒ 1 × 3 2 = 3 2 × sin θ 2 ⇒ θ 2 = 45 ° For the second interface, ⇒ x 2 = d × tan 45 ° = d And again using Snell's law, 3 2 × sin 45 = 3 × sin θ 3 ⇒ 3 2 × 1 2 = 3 sin θ 3 ⇒ θ 3 = 30 ° ⇒ x 3 = d × tan 30 ° = d 3 ∴   x 1 + x 2 + x 3 = 1 3 + 3 - 1 2 1 + 1 3 = 2 3   cm ∴   n = 1 x 1 + x 2 + x 3 = 8 3 2 3 = 4 Thus, the value of  n = 4 .

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