Optics 7 Question 13
13. A light beam is travelling 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
(2008, 3M)
(a) $\sin ^{-1} \frac{3}{4}$
(b) $\sin ^{-1} \frac{1}{8}$
(c) $\sin ^{-1} \frac{1}{4}$
(d) $\sin ^{-1} \frac{1}{3}$
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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
$$ \begin{aligned} n _0 \sin \theta & =\frac{n _0}{6} \sin \theta _C \\ \text { or } \quad \sin \theta & =\frac{1}{6} \sin \theta _C \\ & =\frac{1}{6} \frac{3}{4}=\frac{1}{8} \\ \therefore \quad \theta & =\sin ^{-1} \frac{1}{8} \end{aligned} $$
