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(C) Given: Angle of incidence $i = 60^{\circ}$,Angle of prism $A = 30^{\circ}$,Angle of deviation $\delta = 30^{\circ}$.Using the formula for the angl

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$\sqrt{3}$

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(C) Given: Angle of incidence $i = 60^{\circ}$,Angle of prism $A = 30^{\circ}$,Angle of deviation $\delta = 30^{\circ}$.Using the formula for the angle of deviation: $\delta = i + e - A$.Substituting the values: $30^{\circ} = 60^{\circ} + e - 30^{\circ}$.Solving for the angle of emergence: $e = 0^{\circ}$.Since $e = 0^{\circ}$,the emergent ray is normal to the second surface,so the angle of refraction at the second surface $r_2 = 0^{\circ}$.Using the relation $A = r_1 + r_2$: $30^{\circ} = r_1 + 0^{\circ}$,which gives $r_1 = 30^{\circ}$.Using Snell's law at the first surface: $\mu = \frac{\sin i}{\sin r_1}$.$\mu = \frac{\sin 60^{\circ}}{\sin 30^{\circ}} = \frac{\sqrt{3}/2}{1/2} = \sqrt{3}$.

$A$ light ray is incident on a prism at an angle of $60^{\circ}$ and the refracting angle of the prism is $30^{\circ}$. The emergent ray makes an angle of $30^{\circ}$ with the incident ray. The refractive index of the prism is ...

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