Question

A light beam is traveling 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 :

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

Answer 1

The correct option is B $si{n}^{-1}\left(\frac{1}{8}\right)$

Since, the ray of light just misses entering the fourth medium, it will move at a grazing incidence or we can say that the beam will just suffers TIR at interface of region III and IV.

Applying Snell's law,

$n_{0}sin\theta =\frac{n_0}{2}sin{\theta }_1=\frac{n_0}{6}sin{\theta }_2=\frac{n_0}{8}sin{90}^0$

$n_{o}sin\theta =\frac{n_o}{8}sin90$

Thus, $sin\theta =\frac{1}{8}$

For $\theta ={sin}^{-1}\frac{1}{8}$, the ray just misses entering the 4th medium