Question

A ray of light passing through a prism having $\mu =\sqrt{2}$ suffer minimum deviation. It is found that angle of incidence is double the angle of refraction within the prism. Find angle of the prism.

Answer 1

if $\angle i=$ incidence angle

$\angle r=$ refracted angle

$\angle i=2\times \angle r$

$\frac{\sin i}{\sin r}=\mu$

$\frac{\sin 2r}{\sin r}=\frac{2\sin r\cos r}{\sin r}=2\cos r=\sqrt{2}$

$\cos r=\frac{1}{\sqrt{2}}$

$r={45}^{\circ }$

For minimum deviation, $r_1=r_2=r$

$\therefore a=2\times r$

$\angle a=2\times {45}^{\circ }$

$={90}^{\circ }$