Question

A beam of light is traveling from Region I to Region IV (refer figure). The refractive index of 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. ${\text{sin}}^{-1}\left(\frac{3}{4}\right)$
• B. ${\text{sin}}^{-1}\left(\frac{1}{8}\right)$
• C. ${\text{sin}}^{-1}\left(\frac{1}{4}\right)$
• D. ${\text{sin}}^{-1}\left(\frac{1}{3}\right)$

Answer 1

The correct option is B

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

When a light ray enters from medium 1 to medium 2,
by snell's law we know that,
$\frac{sin\theta }{sinr}=\frac{absoluterefractiveindexmedium2}{absoluterefractiveindexmedium1}$
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 entering region IV
$\frac{sin{r}_1}{sin90}=\frac{n_0\times 6}{8\times n_0}=\frac{3}{4}\left(3\right)$
putting $sin{r}_1$ in (2)
$sinr=\frac{1}{3}\times \frac{3}{4}=\frac{1}{4}$
putting this value in (1)
$sin\theta =\frac{1}{2}\times \frac{1}{4}=\frac{1}{8}$
$\therefore \theta =si{n}^{-1}\left(\frac{1}{8}\right)$