Question

A light beam is travelling from region I to region IV (Refer figure). The refractive indices in regions I, II, III and IV are $\mu_{0}, \frac{\mu_{0}}{2}, \frac{\mu_{0}}{6}$ and $\frac{\mu_{0}}{8}$, respectively. The angle of incidence $\theta$ for which the beam just misses entering region IV is

• A. $\sin ^{-1}\left(\frac{3}{4}\right)$
• B. $\sin ^{-1}\left(\frac{1}{8}\right)$
• C. $\sin ^{-1}\left(\frac{1}{4}\right)$
• D. $\sin ^{-1}\left(\frac{1}{3}\right)$

Answer 1

Answer: (B)

As the beam just misses entering region IV, the angle of refraction in the region IV must be $90^{\circ}$.

Application of Snell's law successively at different interfaces gives
$\mu_{0} \sin \theta=\frac{\mu_{0}}{2} \sin \theta_{1}=\frac{\mu_{0}}{6} \sin \theta_{2}=\frac{\mu_{0}}{8} \sin 90^{\circ}$
$\Rightarrow \sin \theta=\frac{1}{8}$ or $\theta=\sin ^{-1}\left(\frac{1}{8}\right)$