A prism having refractive index sqrt{2} and r | vedclass

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(D) For the light to retrace its path,it must strike the silvered surface normally.Since the second surface is silvered,the angle of refraction at the

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$\sin ^{-1}\left(\frac{1}{\sqrt{2}}\right)$

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(D) For the light to retrace its path,it must strike the silvered surface normally.Since the second surface is silvered,the angle of refraction at the second surface $r_2$ must be $0^{\circ}$.Using the prism formula $A = r_1 + r_2$,where $A = 30^{\circ}$ is the refracting angle and $r_2 = 0^{\circ}$,we get $r_1 = 30^{\circ}$.Applying Snell's Law at the first surface: $n = \frac{\sin i}{\sin r_1}$.Given $n = \sqrt{2}$ and $r_1 = 30^{\circ}$,we have $\sqrt{2} = \frac{\sin i}{\sin 30^{\circ}}$.$\sin i = \sqrt{2} 	imes \sin 30^{\circ} = \sqrt{2} 	imes \frac{1}{2} = \frac{1}{\sqrt{2}}$.Therefore,$i = \sin^{-1}\left(\frac{1}{\sqrt{2}}ight)$.

$A$ prism having refractive index $\sqrt{2}$ and refracting angle $30^{\circ}$ has one of the refracting surfaces silvered. The beam of light incident on the other refracting surface will retrace its path,if the angle of incidence is $\left[\sin \frac{\pi}{6}=0.5\right]$

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