The angle of a prism is 30^o and its refracti | vedclass

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Question

(C) For a light ray to retrace its path after reflection from a silvered surface,it must strike the silvered surface normally (at an angle of $90^o$ t

Vedclass · Accepted Answer

$45^o$

Vedclass · Answer

(C) For a light ray to retrace its path after reflection from a silvered surface,it must strike the silvered surface normally (at an angle of $90^o$ to the surface,or $0^o$ to the normal).In a prism,the angle of the prism $A$ is given by $A = r_1 + r_2$. Here,the ray strikes the second surface normally,so $r_2 = 0^o$.Given $A = 30^o$,we have $30^o = r_1 + 0^o$,which implies $r_1 = 30^o$.Applying Snell's law at the first surface:$\mu = \frac{\sin i}{\sin r_1}$$\sqrt{2} = \frac{\sin i}{\sin 30^o}$$\sin i = \sqrt{2} 	imes \sin 30^o = \sqrt{2} 	imes \frac{1}{2} = \frac{1}{\sqrt{2}}$$i = \sin^{-1}\left(\frac{1}{\sqrt{2}}ight) = 45^o$.Thus,the angle of incidence is $45^o$.

The angle of a prism is $30^o$ and its refractive index is $\sqrt{2}$. One of its surfaces is silvered. At what angle of incidence should a ray be incident on the other surface so that after reflection from the silvered surface,it retraces its path?

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