Question

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

(IIT-JEE 2008)

• 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 $si{n}^{-1}\left(\frac{1}{8}\right)$

For first interface;
$\frac{sin\theta }{sinr}=\frac{n_0}{2\times n_0}=\frac{1}{2}\left(1\right)$

For second interface;
$\frac{sinr}{sin{r}^1}=\frac{n_0\times 2}{6\times n_0}=\frac{1}{3}\left(2\right)$

For the light to just miss region IV, the ray must graze through the third interface

So;
$\frac{sin{r}^1}{sin90}=\frac{n_0\times 6}{8\times n_0}=\frac{3}{4}\left(3\right)$

putting $sin{r}^1$ in (2)

$sinr=\frac{1}{3}\times \frac{3}{4}=\frac{1}{4}$

putting this value in (1)

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

$\therefore \theta =si{n}^{-1}\left(\frac{1}{8}\right)$