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(D) Given: Prism angle $A = 30^\circ$,refractive index $\mu = \sqrt{2}$.Since the light ray is incident normally on one face,the angle of incidence $i

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$15$

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(D) Given: Prism angle $A = 30^\circ$,refractive index $\mu = \sqrt{2}$.Since the light ray is incident normally on one face,the angle of incidence $i = 0^\circ$,which implies the angle of refraction $r_1 = 0^\circ$.Using the relation $A = r_1 + r_2$,we get $30^\circ = 0^\circ + r_2$,so $r_2 = 30^\circ$.Applying Snell's law at the second surface (face $AC$): $\mu \sin r_2 = 1 \sin e$.Substituting the values: $\sqrt{2} \sin 30^\circ = \sin e$.$\sqrt{2} 	imes \frac{1}{2} = \sin e \Rightarrow \sin e = \frac{1}{\sqrt{2}}$.Thus,the angle of emergence $e = 45^\circ$.The angle of deviation $\delta$ is given by $\delta = e - A$ (since $i=0$ and $r_1=0$): $\delta = 45^\circ - 30^\circ = 15^\circ$.

$A$ ray of light is incident normally on one of the faces of a prism of angle $30^\circ$ and refractive index $\sqrt{2}$. The angle of deviation will be.....$^\circ$

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