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(A) Let the prism be $PDC$ with base $DC$. The incident ray is parallel to the base $DC$. For grazing emergence at the second surface, the angle of re

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$60$

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(A) Let the prism be $PDC$ with base $DC$. The incident ray is parallel to the base $DC$. For grazing emergence at the second surface, the angle of refraction $r_2$ is equal to the critical angle $C$. Given $\mu = \sqrt{2}$, we have $\sin r_2 = \frac{1}{\mu} = \frac{1}{\sqrt{2}}$, which gives $r_2 = 45^{\circ}$. Since the incident ray is parallel to the base, the angle of incidence $i$ at the first surface is equal to the base angle $45^{\circ}$. Applying Snell's Law at the first surface: $1 	imes \sin(45^{\circ}) = \sqrt{2} 	imes \sin(r_1)$. $\frac{1}{\sqrt{2}} = \sqrt{2} \sin(r_1) \implies \sin(r_1) = \frac{1}{2} \implies r_1 = 30^{\circ}$. The angle of the prism $A$ at the top vertex is $A = r_1 + r_2 = 30^{\circ} + 45^{\circ} = 75^{\circ}$. In the triangle formed by the prism, the sum of angles is $180^{\circ}$. Thus, $45^{\circ} + 	heta + A = 180^{\circ}$. $45^{\circ} + 	heta + 75^{\circ} = 180^{\circ} \implies 	heta + 120^{\circ} = 180^{\circ} \implies 	heta = 60^{\circ}$.

As shown in the diagram, when the incident ray is parallel to the base of the prism, the emergent ray grazes along the second surface. If the refractive index of the material of the prism is $\sqrt{2}$, the angle $\theta$ of the prism is: (in $^{\circ}$)

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