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

Answer: 60

At minimum deviation $r_1=r_2=\frac{A}{2}$
Also given $i=2r_1=A$
Now $1\cdot \sin i=\sqrt{3}\sin r_1$
$1\sin A=\sqrt{3}\sin \frac{A}{2}$
$\Rightarrow 2\sin \frac{A}{2}\cos \frac{A}{2}=\sqrt{3}\sin \frac{A}{2}$
$\Rightarrow \cos \frac{A}{2}=\frac{\sqrt{3}}{2}$
$\Rightarrow \frac{A}{2}=30^{\circ }$
$\Rightarrow A=60^{\circ }$