A light beam is travelling from Region I to Region IV (refer figure). The refractive index in Region I, II,…
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
$\sin ^{-1}\left(\frac{3}{4}\right)$
$\sin ^{-1}\left(\frac{1}{8}\right)$
$\sin ^{-1}\left(\frac{1}{4}\right)$
$\sin ^{-1}\left(\frac{1}{3}\right)$
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)
