Question

A light beam is travelling from Region $\text{I}$ to Region $\text{IV}$. The refractive index in Regions $\text{I}$ , $\text{II}$, $\text{III}$ and $\text{IV}$ are $n_o,\frac{n_o}{2},\frac{n_o}{6}$ and $\frac{n_o}{8}$ respectively. The angle of incidence $\theta$ for which the beam just could not enter region $\text{IV}$ is :-

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

Answer 1

The correct option is D

${\sin }^{-1}\left(\frac{1}{8}\right)$


Using Snell's law for consecutive refraction through multiple slabs or regions we get,

$n_o\sin \theta =\frac{n_o}{2}\sin {\theta }_1=\frac{n_o}{6}\sin {\theta }_2=\frac{n_o}{8}\sin {\theta }_3$

For just missing the region $\text{IV}$

${\theta }_2={\theta }_c$ & ${\theta }_3={90}^{\circ }$

$\Rightarrow n_o\sin \theta =\frac{n_o}{8}\sin {90}^{\circ }$

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

Hence, option (d) is the correct answer.

Why this question ? To understand the concept of total internal reflection & Snell's law.