JEE Advanced 2021 Paper 1 Q12 Physics Waves & Optics Ray Optics Hard

JEE Advanced 2021 Paper 1 · Q12 · Ray Optics

A wide slab consisting of two media of refractive indices n₁ and n₂ is placed in air as shown in the figure. A ray of light is incident from medium n₁ to n₂ at an angle θ, where sin θ is slightly larger than 1/n₁. Take refractive index of air as 1. Which of the following statement(s) is(are) correct?

[Figure: Horizontal slab with bottom layer of index n₁, top layer of index n₂, and air above. A ray enters at angle θ from the n₁ side at the n₁–n₂ interface.]

  1. A. The light ray enters air if n₂ = n₁
  2. B. The light ray is finally reflected back into the medium of refractive index n₁ if n₂ < n₁
  3. C. The light ray is finally reflected back into the medium of refractive index n₁ if n₂ > n₁
  4. D. The light ray is reflected back into the medium of refractive index n₁ if n₂ = 1
JEE Advanced multi-correct — pick every correct option, then check.
Reveal answer + step-by-step solution

Correct answer:B, C, D

Solution

Critical angle for the n₁–air interface satisfies sin θ_c = 1/n₁. Given sin θ > 1/n₁ at the n₁–n₂ interface. If n₂ = n₁ (option A), the ray passes straight into n₂ and then meets the n₂–air interface at the same θ, where sin θ > 1/n₁ = 1/n₂ ⇒ total internal reflection at n₂–air; ray does not enter air, so (A) is incorrect. For n₂ < n₁: ray totally internally reflects at the n₁–n₂ interface itself (since sin θ_c′ = n₂/n₁ < 1/n₁ when n₂ < 1, but in general the ray either TIRs there or — if it refracts into n₂ — TIRs at the n₂–air boundary), and finally returns to n₁ ⇒ (B) correct. For n₂ > n₁: ray refracts into n₂ at smaller angle, then meets n₂–air; by Snell at top boundary the effective sin in air = n₁ sin θ > 1, so TIR back; ray ultimately returns to n₁ ⇒ (C) correct. For n₂ = 1 (air), TIR happens at the n₁–air interface directly ⇒ (D) correct.

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