A light beam is travelling from Region I to Region IV (refer...

A light beam is travelling from Region I to Region IV (refer figure). The refractive index in Region 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 $A light beam is travelling from Region I to Region IV (refer figure). The refractive index in Region I, II, III and IV$

Options

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 $\sin ^{-1}\left(\frac{1}{3}\right)$

Correct answer

B. $\sin ^{-1}\left(\frac{1}{8}\right)$

Step-by-step solution

Critical angle from region III to region IV \sin \theta_c=

\frac{n_0 / 8}{n_0 / 6}=\frac{3}{4}Now applying Snell's law in

region I and region III : n_0 \sin

\theta=\frac{n_0}{6} \sin \theta_

c or

\begin{aligned} \sin\theta& =\frac{1}{6} \sin \theta_c=\frac{1}{6}\left(\frac{3}{4}\right)=\frac{1}{8}\\ \theta& =\sin ^{-1}\left(\frac{1}{8}\right) \end{aligned}

\therefore correct option is (b).

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