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At the interface between two materials having refractive indices n_1 and n_2, the critical angle for reflection of an em wave is θ_1 C. The n_2…

JEE Main 2025 — Physics Optics

2025mcqhard

At the interface between two materials having refractive indices n1n_1n1​ and n2n_2n2​, the critical angle for reflection of an em wave is θ1C\theta_{1 C}θ1C​. The n2n_2n2​ material is replaced by another material having refractive index n3n_3n3​ such that the critical angle at the interface between n1n_1n1​ and n3n_3n3​ materials is θ2C\theta_{2 \mathrm{C}}θ2C​. If n3>n2>n1;n2n3=25\mathrm{n}_3\gt\mathrm{n}_2\gt\mathrm{n}_1 ; \frac{\mathrm{n}_2}{\mathrm{n}_3}=\frac{2}{5}n3​>n2​>n1​;n3​n2​​=52​ and sin⁡θ2C−sin⁡θ1C=12\sin \theta_{2 \mathrm{C}}-\sin \theta_{1 \mathrm{C}}=\frac{1}{2}sinθ2C​−sinθ1C​=21​, then θ1C\theta_{1 \mathrm{C}}θ1C​ is

Official previous-year question

Held on 29 Jan 2025 · Verified 6 Jul 2026.

Options

  1. A

    sin⁡−1(16)\sin ^{-1}\left(\frac{1}{6}\right)sin−1(61​)

  2. B

    sin⁡−1(13)\sin ^{-1}\left(\frac{1}{3}\right)sin−1(31​)

  3. C

    sin⁡−1(−56)\sin ^{-1}\left(\frac{-5}{6}\right)sin−1(6−5​)

  4. D

    sin⁡−1(23)\sin ^{-1}\left(\frac{2}{3}\right)sin−1(32​)

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