A ray of light is incident on the first face | vedclass

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(A) For an equilateral prism,the angle of the prism $A = 60^{\circ}$.Given that the emergent ray grazes the adjacent face,the angle of emergence $e =

Vedclass · Accepted Answer

$\sin^{-1}(\sqrt{2} \sin 15^{\circ})$

Vedclass · Answer

(A) For an equilateral prism,the angle of the prism $A = 60^{\circ}$.Given that the emergent ray grazes the adjacent face,the angle of emergence $e = 90^{\circ}$.From the condition of refraction at the second surface,$r_2 = C$,where $C$ is the critical angle.Using Snell's law at the second surface: $\mu \sin r_2 = 1 \sin e \implies \sqrt{2} \sin r_2 = \sin 90^{\circ} = 1$.Thus,$\sin r_2 = 1/\sqrt{2}$,which means $r_2 = 45^{\circ}$.Since $A = r_1 + r_2$,we have $60^{\circ} = r_1 + 45^{\circ}$,so $r_1 = 15^{\circ}$.Applying Snell's law at the first surface: $1 \sin i = \mu \sin r_1$.$\sin i = \sqrt{2} \sin 15^{\circ}$.Therefore,$i = \sin^{-1}(\sqrt{2} \sin 15^{\circ})$.

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