Q.
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) $\sin^{- 1}\left( \frac{3}{4} \right)$
B) $\sin^{- 1}\left( \frac{1}{8} \right)$
C) $\sin^{- 1}\left( \frac{1}{4} \right)$
D) $\sin^{- 1}\left( \frac{1}{3} \right)$
Explanation
(b) 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 $\sin\theta = \frac{1}{6}\sin\theta_{C} = \frac{1}{6}\left( \frac{3}{4} \right) = \frac{1}{8}$
$\therefore\ \theta = \sin^{- 1}\left( \frac{1}{8} \right)$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.