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(A) $1$. From the figure, the incident ray strikes the first surface of the prism normally. Therefore, the angle of incidence $i = 0^\circ$, and the r

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$60$

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(A) $1$. From the figure, the incident ray strikes the first surface of the prism normally. Therefore, the angle of incidence $i = 0^\circ$, and the ray passes undeviated into the prism.$2$. The prism is a right-angled triangle with one angle $60^\circ$. The angle at the top vertex is $180^\circ - 90^\circ - 60^\circ = 30^\circ$.$3$. Inside the prism, the ray hits the second surface (the hypotenuse). The angle of incidence at this surface $(r_2)$ is the angle between the normal and the ray. Since the ray is perpendicular to the first surface, the angle of incidence at the second surface is $r_2 = 30^\circ$.$4$. Applying Snell's law at the second surface: $n_1 \sin(r_2) = n_2 \sin(e)$, where $n_1 = \sqrt{3}$, $n_2 = 1$, and $r_2 = 30^\circ$.$5$. $\sqrt{3} \sin(30^\circ) = 1 	imes \sin(e)$.$6$. $\sqrt{3} 	imes \frac{1}{2} = \sin(e) \implies \sin(e) = \frac{\sqrt{3}}{2}$.$7$. Therefore, $e = 60^\circ$.

Find the value of the angle of emergence for an emergent ray from the prism in the figure given below. (Refractive index of the material of the prism is $\sqrt{3}$ and the refractive index of air is $1$) (in $^\circ$)

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