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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
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Verified Answer
The correct answer is:
$60^{\circ}$
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
$\delta=\mathrm{i}_{1}+\mathrm{i}_{2}-\mathrm{A}=60^{\circ}+60^{\circ}-60^{\circ}=60^{\circ}$
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