A light ray is incident normally on one surfa | vedclass

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(D) Given,the angle of incidence $i = 0^{\circ}$.For an equilateral prism,the prism angle $A = 60^{\circ}$.At the first surface,since the ray is incid

Vedclass · Accepted Answer

$120$

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(D) Given,the angle of incidence $i = 0^{\circ}$.For an equilateral prism,the prism angle $A = 60^{\circ}$.At the first surface,since the ray is incident normally,the angle of refraction $r_1 = 0^{\circ}$.Using the relation $r_1 + r_2 = A$,we get $0^{\circ} + r_2 = 60^{\circ}$,so $r_2 = 60^{\circ}$.The critical angle $C$ for the prism is given by $\sin C = \frac{1}{\mu} = \frac{1}{\sqrt{2}}$,which means $C = 45^{\circ}$.Since the angle of incidence at the second surface $r_2 = 60^{\circ}$ is greater than the critical angle $C = 45^{\circ}$,the light ray undergoes total internal reflection at the second surface.After reflection,the ray strikes the third surface. The angle of incidence at the third surface is $r_3 = 180^{\circ} - (60^{\circ} + 60^{\circ}) = 60^{\circ}$ (from the geometry of the triangle). Since $60^{\circ} > 45^{\circ}$,it undergoes another total internal reflection.Finally,the ray emerges from the first surface normally. The total deviation $\delta$ is the angle between the incident ray and the emergent ray. Since the ray emerges parallel to the incident ray but in the opposite direction,the deviation is $180^{\circ} - 60^{\circ} = 120^{\circ}$.

$A$ light ray is incident normally on one surface of an equilateral prism. The angle of deviation of the light ray is (refractive index of the material of the prism $=\sqrt{2}$) (in $^{\circ}$)

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