The refractive index of the material of a pri | vedclass

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(B) For the light ray to retrace its path,it must strike the silvered surface normally (at an angle of $90^{\circ}$ to the surface).Let the angle of i

Vedclass · Accepted Answer

$45$

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(B) For the light ray to retrace its path,it must strike the silvered surface normally (at an angle of $90^{\circ}$ to the surface).Let the angle of incidence at the first surface be $i$ and the angle of refraction be $r_1$. The angle of the prism is $A = 30^{\circ}$.Since the ray strikes the second surface normally,the angle of refraction at the second surface is $r_2 = 0^{\circ}$.From the prism formula,$A = r_1 + r_2$,we have $30^{\circ} = r_1 + 0^{\circ}$,so $r_1 = 30^{\circ}$.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}}$Therefore,$i = 45^{\circ}$.

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