Question

A ray of light passing through a prism $(\mu =\sqrt{3})$ suffers minimum deviation. It is found that the angle of incidence is double the angle of refraction within the prism. Then, the angle of prism is (in degrees)

Answer 1

At minimum deviation, $r_1=r_2=\frac{A}{2}$

Also given, $i=2r_1=A$

Now, using Snell's law,

${\mu }_{air}\sin i={\mu }_{prism}\sin r_1$

$\Rightarrow 1\sin A=\sqrt{3}\sin \frac{A}{2}$

$\Rightarrow 2\sin \frac{A}{2}\cos \frac{A}{2}=\sqrt{3}\sin \frac{A}{2}$

$\Rightarrow \cos \frac{A}{2}=\frac{\sqrt{3}}{2}\Rightarrow \frac{A}{2}={30}^{\circ }$

$\Rightarrow A={60}^{\circ }$

Correct answer:$60$