A prism ABC of angle 30^circ has its face AC | vedclass

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(C) For the light ray to retrace its path,it must strike the silvered face $AC$ normally.Let $i = 45^\circ$ be the angle of incidence at face $AB$ and

Vedclass · Accepted Answer

$\sqrt{2}$

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(C) For the light ray to retrace its path,it must strike the silvered face $AC$ normally.Let $i = 45^\circ$ be the angle of incidence at face $AB$ and $r$ be the angle of refraction.From the geometry of the prism,the angle of the prism $A = 30^\circ$.Since the ray strikes face $AC$ normally,the angle of incidence at face $AC$ is $0^\circ$.In a prism,the angle $A = r_1 + r_2$. Here,$r_1 = r$ and $r_2 = 0^\circ$.Thus,$r = A = 30^\circ$.Using Snell's law at face $AB$: $\mu = \frac{\sin i}{\sin r} = \frac{\sin 45^\circ}{\sin 30^\circ}$.$\mu = \frac{1/\sqrt{2}}{1/2} = \frac{2}{\sqrt{2}} = \sqrt{2}$.Therefore,the refractive index of the material of the prism is $\sqrt{2}$.

$A$ prism $ABC$ of angle $30^\circ$ has its face $AC$ silvered. $A$ ray of light incident at an angle of $45^\circ$ at the face $AB$ retraces its path after refraction at face $AB$ and reflection at face $AC$. The refractive index of the material of the prism is

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