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(B) Let the point of incidence be $A$ and the point of reflection/refraction be $B$. The center of the sphere is $O$.By Snell's law at point $A$:$1

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$90$

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(B) Let the point of incidence be $A$ and the point of reflection/refraction be $B$. The center of the sphere is $O$.By Snell's law at point $A$:$1 	imes \sin 60^{\circ} = \sqrt{3} 	imes \sin 	heta$$\frac{\sqrt{3}}{2} = \sqrt{3} \sin 	heta$$\sin 	heta = \frac{1}{2} \Rightarrow 	heta = 30^{\circ}$.In the triangle $	riangle OAB$,$OA = OB = R$ (radius of the sphere). Thus,$	riangle OAB$ is an isosceles triangle.The angle of incidence at point $B$ is also $	heta = 30^{\circ}$.The angle of reflection at $B$ is $r' = 30^{\circ}$ (angle with the normal $OB$).By Snell's law at $B$ for refraction:$\sqrt{3} 	imes \sin 30^{\circ} = 1 	imes \sin r''$$\sqrt{3} 	imes \frac{1}{2} = \sin r'' \Rightarrow \sin r'' = \frac{\sqrt{3}}{2} \Rightarrow r'' = 60^{\circ}$.The angle between the normal $OB$ and the reflected ray is $30^{\circ}$.The angle between the normal $OB$ and the refracted ray is $60^{\circ}$.The angle between the reflected and refracted rays is the sum of these angles with the normal: $30^{\circ} + 60^{\circ} = 90^{\circ}$.

$A$ light ray enters a solid glass sphere of refractive index $\mu=\sqrt{3}$ at an angle of incidence $60^{\circ}$. The ray is both reflected and refracted at the farther surface of the sphere. The angle (in degrees) between the reflected and refracted rays at this surface is......

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