At the interface between two materials having refractive indices n₁ and n₂, the critical angle for reflection of an em wave is θ1C. The n₂ material is replaced by another material having refractive index n₃ such that the critical angle at the interface between n₁ and n₃ materials is θ2C. If n₃ > n₂ > n₁; (n₂)/(n₃) = 2/5 and sin θ2C - sin θ1C = 1/2, then θ1C is: (a) sin ⁻¹(1/6n₁) (b) sin ⁻¹(1/3n₁) (c) sin ⁻¹(5/6n₁) (d) sin ⁻¹(2/3n₁)

Answer:. Light goes from the denser medium ( n₂ or n₃) into n₁: sin θ1C = (n₁)/(n₂), sin θ2C = (n₁)/(n₃). sin θ2C = (n₁)/(n₂)·(n₂)/(n₃) = 2/5sin θ1C. The two values differ by 1/2: sin θ1C - 2/5sin θ1C = 3/5sin θ1C = 1/2. sin θ1C = 5/6, which matches option (c) sin ⁻¹(5/6n₁) when n₁ = 1. So the answer is (c).

At the interface between two materials having refractive indices n₁… | JEE Main 2025 (29 Jan, Shift 1)

Q10JEE Main 2025 · 29 JAN · SHIFT 1Medium

Total Internal Reflection

At the interface between two materials having refractive indices $n_1$ and $n_2$ , the critical angle for reflection of an em wave is $\theta_{1C}$ . The $n_2$ material is replaced by another material having refractive index $n_3$ such that the critical angle at the interface between $n_1$ and $n_3$ materials is $\theta_{2C}$ . If $n_3 > n_2 > n_1; \frac{n_2}{n_3} = \frac{2}{5}$ and $\sin\theta_{2C} - \sin\theta_{1C} = \frac{1}{2}$ , then $\theta_{1C}$ is:

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Answer:

Light goes from the denser medium ( $n_2$ or $n_3$ ) into $n_1$ : $\sin\theta_{1C} = \frac{n_1}{n_2}$ , $\sin\theta_{2C} = \frac{n_1}{n_3}$ .
$\sin\theta_{2C} = \frac{n_1}{n_2}\cdot\frac{n_2}{n_3} = \frac{2}{5}\sin\theta_{1C}$ .
The two values differ by $\frac{1}{2}$ : $\sin\theta_{1C} - \frac{2}{5}\sin\theta_{1C} = \frac{3}{5}\sin\theta_{1C} = \frac{1}{2}$ .
$\sin\theta_{1C} = \frac{5}{6}$ , which matches option (c) $\sin^{-1}\left(\frac{5}{6n_1}\right)$ when $n_1 = 1$ .
So the answer is (c).

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