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(C) Given: Apex angle $A = 30^{\circ}$,Refractive index $\mu = \sqrt{2}$.Since the ray is incident normally on one face,the angle of incidence $i_1 =

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$15$

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(C) Given: Apex angle $A = 30^{\circ}$,Refractive index $\mu = \sqrt{2}$.Since the ray is incident normally on one face,the angle of incidence $i_1 = 0^{\circ}$,which implies the angle of refraction at the first surface $r_1 = 0^{\circ}$.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: $\mu \sin r_2 = 1 \sin e$,where $e$ is the angle of emergence.$\sqrt{2} \sin 30^{\circ} = \sin e \Rightarrow \sqrt{2} 	imes \frac{1}{2} = \sin e \Rightarrow \sin e = \frac{1}{\sqrt{2}}$.Thus,$e = 45^{\circ}$.The angle of deviation $\delta$ is given by $\delta = (i_1 + e) - A$.$\delta = (0^{\circ} + 45^{\circ}) - 30^{\circ} = 15^{\circ}$.

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

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