Question

A light beam is travelling from Region $I$ to $IV$ (Refer Figure). The refractive index in 4 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. ${\sin }^{-1}\left(\frac{3}{4}\right)$
• B. ${\sin }^{-1}\left(\frac{1}{8}\right)$
• C. ${\sin }^{-1}\left(\frac{1}{4}\right)$
• D. ${\sin }^{-1}\left(\frac{1}{3}\right)$

Answer 1

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

The beam just misses entering region IV. It means it is striking at the interface at an angle of incidence $=$ critical angle.

For the interface of region III and IV -

$\Rightarrow {\theta }_c={\sin }^{-1}\left(\frac{3}{4}\right)$

For the interface of region II and III -

$\Rightarrow \frac{n_0}{2}\sin \left(r\right)=\frac{n_0}{6}\sin {\theta }_c$

$\sin \left(r\right)=\frac{1}{3}\left(\frac{3}{4}\right)=\frac{1}{4}$

For the interface region I and II

$\Rightarrow n_0\sin \theta =\frac{n_0}{2}\sin \left(r\right)$

$\Rightarrow \sin \theta =\frac{1}{2}\times \frac{1}{4}=\frac{1}{8}$

$\Rightarrow \theta ={\sin }^{-1}\left(\frac{1}{8}\right).$

Hence, the answer is ${\sin }^{-1}\left(\frac{1}{8}\right).$