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(D) Given: Refracting angle $A = 30^{\circ}$,angle of incidence $i_{1} = 45^{\circ}$.Since the ray retraces its path after reflection from the silvere

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$\sqrt{2}$

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(D) Given: Refracting angle $A = 30^{\circ}$,angle of incidence $i_{1} = 45^{\circ}$.Since the ray retraces its path after reflection from the silvered face,the ray must strike the silvered face normally. Therefore,the angle of refraction at the second face is $r_{2} = 0^{\circ}$.Using the relation $r_{1} + r_{2} = A$,we get $r_{1} + 0^{\circ} = 30^{\circ}$,so $r_{1} = 30^{\circ}$.Applying Snell's Law at the first face: $\mu = \frac{\sin i_{1}}{\sin r_{1}}$.Substituting the values: $\mu = \frac{\sin 45^{\circ}}{\sin 30^{\circ}} = \frac{1/\sqrt{2}}{1/2} = \frac{2}{\sqrt{2}} = \sqrt{2}$.Thus,the refractive index of the prism is $\sqrt{2}$.

One face of a prism with a refracting angle of $30^{\circ}$ is coated with silver. $A$ ray incident on the other face at an angle of $45^{\circ}$ is refracted,reflected from the silver-coated face,and retraces its path. The refractive index of the prism is:

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