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(D) For an equilateral prism,the angle of the prism is $A = 60^{\circ}$.When the light ray inside the prism moves parallel to the base,the ray undergo

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$60$

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(D) For an equilateral prism,the angle of the prism is $A = 60^{\circ}$.When the light ray inside the prism moves parallel to the base,the ray undergoes minimum deviation,which implies that the angle of incidence $i$ is equal to the angle of emergence $e$,and the angle of refraction $r_1$ is equal to $r_2$.Since $r_1 + r_2 = A$,we have $2r = 60^{\circ}$,which gives $r = 30^{\circ}$.Using Snell's Law at the first surface: $\mu = \frac{\sin i}{\sin r}$.Given $\mu = \sqrt{3}$ and $r = 30^{\circ}$,we have $\sqrt{3} = \frac{\sin i}{\sin 30^{\circ}}$.Since $\sin 30^{\circ} = 0.5$,we get $\sin i = \sqrt{3} 	imes 0.5 = \frac{\sqrt{3}}{2}$.Therefore,$i = 60^{\circ}$.

$A$ light ray is incident on an equilateral prism made of a material with a refractive index of $\sqrt{3}$. Inside the prism,if the light ray moves parallel to the base of the prism,then the angle of incidence of the light ray is: (in $^{\circ}$)

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