The refractive index of a prism is sqrt{2} an | vedclass

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(C) Given,refractive index $\mu = \sqrt{2}$ and prism angle $A = 30^\circ$.Since the light ray retraces its path after reflection from the polished su

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$45$

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(C) Given,refractive index $\mu = \sqrt{2}$ and prism angle $A = 30^\circ$.Since the light ray retraces its path after reflection from the polished surface,it must strike the polished surface normally.Therefore,the angle of refraction at the second surface is $r_2 = 0^\circ$.From the relation $A = r_1 + r_2$,we have $30^\circ = r_1 + 0^\circ$,which gives $r_1 = 30^\circ$.Applying Snell's law at the first surface: $\mu = \frac{\sin i}{\sin r_1}$.$\sqrt{2} = \frac{\sin i}{\sin 30^\circ}$.$\sin i = \sqrt{2} 	imes \sin 30^\circ = \sqrt{2} 	imes \frac{1}{2} = \frac{1}{\sqrt{2}}$.Thus,$i = 45^\circ$.

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