An isosceles prism of angle A=30^{circ} has o | vedclass

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

Vedclass · Accepted Answer

$1.732$

Vedclass · Answer

(C) For a light ray to retrace its path after reflection from the silvered surface,it must strike the silvered surface normally (at an angle of incidence of $0^{\circ}$).Let $i$ be the angle of incidence on the first surface and $r$ be the angle of refraction.Given,$i = 60^{\circ}$ and the prism angle $A = 30^{\circ}$.From the geometry of the prism,the angle of refraction $r$ at the first surface is related to the prism angle $A$ and the angle of incidence at the second surface $r'$ by $A = r + r'$.Since the ray strikes the second surface normally,$r' = 0^{\circ}$.Therefore,$r = A = 30^{\circ}$.Using Snell's Law at the first surface:$\mu = \frac{\sin i}{\sin r} = \frac{\sin 60^{\circ}}{\sin 30^{\circ}}$$\mu = \frac{\sqrt{3}/2}{1/2} = \sqrt{3} \approx 1.732$.Thus,the refractive index of the prism material is $1.732$.

An isosceles prism of angle $A=30^{\circ}$ has one of its surfaces silvered. Light rays falling at an angle of incidence $60^{\circ}$ on the other surface retrace their path after reflection from the silvered surface. The refractive index of the prism material is

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