Question

A ray $PQ$ incident on the refracting face $BA$ is refracted in the prism $BAC$ as shown in the figure and emerges from the other refracting face $AC$ as $RS$ such that $AQ = AR$. If the angle of prism $A = 60^{\circ}$ and the refractive index of the material of prism is $\sqrt{3}$ then the angle of deviation of the ray is

• A. $60^{\circ}$
• B. $35^{\circ}$
• C. $30^{\circ}$
• D. None of these

Answer 1

Answer: (A)

Given $A Q=A R$ and $\angle A=60^{\circ}$
$\therefore \angle A Q R=\angle A R Q=60^{\circ}$
$\therefore r_{1}=r_{2}=30^{\circ} $
Applying Snell's law on face $A B$.
$\sin i_{1}=\mu \sin r_{1}$
$\Rightarrow \sin i_{1}=\sqrt{3} \sin 30^{\circ}$
$=\sqrt{3} \times \frac{1}{2}=\frac{\sqrt{3}}{2}$
$\therefore i_{1}=60^{\circ}$
Similarly, $i_{2}=60^{\circ}$
In a prism, deviation
$\delta=i_{1}+i_{2}-A$
$=60^{\circ}+60^{\circ}-60^{\circ}=60^{\circ}$