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 $i=2 r_{1}=A$
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}$
$\Rightarrow A=60^{\circ}$