One of the refractive surfaces of a prism of | vedclass

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(B) When a ray of light retraces its path after reflection from a silvered surface,it must strike the silvered surface normally (at an angle of $90^{\

Vedclass · Accepted Answer

$\sqrt{3}$

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(B) When a ray of light retraces its path after reflection from a silvered surface,it must strike the silvered surface normally (at an angle of $90^{\circ}$).Let $r_1$ be the angle of refraction at the first surface and $r_2$ be the angle of incidence at the second (silvered) surface.Since the ray strikes the silvered surface normally,$r_2 = 0^{\circ}$.For a prism,the angle of the prism $A = r_1 + r_2$.Substituting the values,$30^{\circ} = r_1 + 0^{\circ}$,which gives $r_1 = 30^{\circ}$.The angle of incidence at the first surface is given as $i_1 = 60^{\circ}$.Using Snell's law at the first surface,$\mu = \frac{\sin i_1}{\sin r_1}$.Substituting the values,$\mu = \frac{\sin 60^{\circ}}{\sin 30^{\circ}} = \frac{\sqrt{3}/2}{1/2} = \sqrt{3}$.

One of the refractive surfaces of a prism of angle $30^{\circ}$ is silvered. $A$ ray of light incident at an angle of $60^{\circ}$ retraces its path. The refractive index of the material of the prism is:

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