IIT-JEE 2008 · Paper 2 · Physics Q31

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PhysicsSingle correct+3 / -1Question 31 of 66

A light beam is traveling from Region I to Region IV (Refer Figure). The refractive index in Regions I, II, III and IV are $n_0$ , $\frac{n_0}2$ , $\frac{n_0}6$ and $\frac{n_0}8$ , respectively. The angle of incidence $\theta$ for which the beam just misses entering Region IV is

Four regions separated by vertical boundaries at x = 0, 0.2 m and 0.6 m: Region I (n₀), Region II (n₀/2), Region III (n₀/6) and Region IV (n₀/8). A ray in Region I strikes the first boundary at angle θ to the normal (dashed horizontal).

\sin^{-1}\left(\frac34\right)

\sin^{-1}\left(\frac18\right)

\sin^{-1}\left(\frac14\right)

\sin^{-1}\left(\frac13\right)

Answer and worked solution

Official answer: B

Parallel boundaries. $n\sin\theta$ is the same in every region: $n_0\sin\theta=n_2\sin\theta_2=\ldots$

Just missing Region IV means grazing incidence at the III-IV boundary:
$n_0\sin\theta=\frac{n_0}8\sin90°\implies\sin\theta=\frac18.$
(At this angle $n_0\sin\theta<\frac{n_0}6$ , so the beam does reach Region III.)

(B)

Topics:Total internal reflection Refraction at plane surfaces

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IIT-JEE 2008 · Paper 2, Physics question 31