A ray OP of monochromatic light is incident o | vedclass

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(A) $1$. At face $AB$: Using Snell's law,$1 \cdot \sin(60^{\circ}) = \sqrt{3} \cdot \sin(r_1)$.$\sin(r_1) = \frac{\sqrt{3}/2}{\sqrt{3}} = 1/2$,so $r_1

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$(A, B, C)$

Vedclass · Answer

(A) $1$. At face $AB$: Using Snell's law,$1 \cdot \sin(60^{\circ}) = \sqrt{3} \cdot \sin(r_1)$.$\sin(r_1) = \frac{\sqrt{3}/2}{\sqrt{3}} = 1/2$,so $r_1 = 30^{\circ}$.$2$. In the triangle formed inside the prism,the angle of incidence at face $CD$ is $r_2 = 180^{\circ} - 60^{\circ} - (90^{\circ} - 30^{\circ}) - 45^{\circ} = 45^{\circ}$.$3$. Critical angle $C = \sin^{-1}(1/\sqrt{3}) \approx 35.26^{\circ}$. Since $r_2 = 45^{\circ} > C$,the ray undergoes total internal reflection at face $CD$.$4$. After reflection,the ray strikes face $AD$ at an angle of incidence $i' = 30^{\circ}$.$5$. Using Snell's law at face $AD$: $\sqrt{3} \cdot \sin(30^{\circ}) = 1 \cdot \sin(e)$.$\sin(e) = \sqrt{3}/2$,so $e = 60^{\circ}$.$6$. The total deviation $\delta = 90^{\circ}$. Thus,$(A)$ and $(B)$ are correct,and the angle between incident and emergent ray is $90^{\circ}$,so $(C)$ is correct.

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