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 $\mathrm{AC}$ as $\mathrm{RS}$ such that $\mathrm{AQ}=\mathrm{AR}$. If the angle of prism $\mathrm{A}=60^{\circ}$ and the refractive index of the material of prism is $\sqrt{3}$, then the angle of deviation of the ray is

• Correct answer
  A

  $60^{\circ}$
• B

  $45^{\circ}$
• C

  $30^{\circ}$
• D

  None of these

Solution

Given $A Q=A R$ and $\angle A=60^{\circ}$ $\therefore \quad \angle A Q R=\angle A R Q=60^{\circ}$ $\therefore \quad r_{1}=r_{2}=30^{\circ}$ Applying Snell's law on face $\mathrm{AB}$. $$ \begin{array}{l} \sin \mathrm{i}_{1}=\mu \sin \mathrm{r}_{1} \\ \Rightarrow \sin \mathrm{i}_{1}=\sqrt{3} \sin 30^{\circ}=\sqrt{3} \times \frac{1}{2}=\frac{\sqrt{3}}{2} \\ \therefore \quad \mathrm{i}_{1}=60^{\circ} \end{array} $$ Similarly, $\mathrm{i}_{2}=60^{\circ}$ In a prism, deviation $\delta=\mathrm{i}_{1}+\mathrm{i}_{2}-\mathrm{A}=60^{\circ}+60^{\circ}-60^{\circ}=60^{\circ}$

Hence, the correct answer is option: 'A'.