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(C) Let the point source be $A$,the top of the sphere be $B$,and the center of the sphere be $C$. The ray hits the interface at $E$ and refracts to be

Vedclass · Accepted Answer

$\frac{4}{\sqrt{5}}$

Vedclass · Answer

(C) Let the point source be $A$,the top of the sphere be $B$,and the center of the sphere be $C$. The ray hits the interface at $E$ and refracts to become tangent to the sphere at $D$.Given $AB = a/2$ and $BC = a$. The radius of the sphere is $a$.In $	riangle CED$,$CD = a$ (radius) and $ED$ is tangent,so $\angle CDE = 90^{\circ}$.The angle of refraction $r = \angle CED = 30^{\circ}$.In $	riangle CED$,$\sin r = \frac{CD}{CE} \Rightarrow \sin 30^{\circ} = \frac{a}{CE} \Rightarrow CE = 2a$.Using Pythagoras theorem in $	riangle CED$,$ED = \sqrt{CE^2 - CD^2} = \sqrt{(2a)^2 - a^2} = a\sqrt{3}$.Let $E$ be at a horizontal distance $x$ from the vertical axis. Then $BE = x$. In $	riangle ABE$,$	an i = \frac{BE}{AB} = \frac{x}{a/2} = \frac{2x}{a}$.From the geometry,$x = CE \sin(\angle ECD)$. Since $\angle CED = 30^{\circ}$ and $\angle CDE = 90^{\circ}$,$\angle ECD = 60^{\circ}$.Thus,$x = 2a \sin 60^{\circ} = 2a \frac{\sqrt{3}}{2} = a\sqrt{3}$.Then $	an i = \frac{2(a\sqrt{3})}{a} = 2\sqrt{3}$.Then $\sin i = \frac{2\sqrt{3}}{\sqrt{1 + (2\sqrt{3})^2}} = \frac{2\sqrt{3}}{\sqrt{13}}$.By Snell's Law: $1 \cdot \sin i = n \cdot \sin r \Rightarrow \frac{2\sqrt{3}}{\sqrt{13}} = n \cdot \frac{1}{2} \Rightarrow n = \frac{4\sqrt{3}}{\sqrt{13}}$.Wait,re-evaluating the geometry: The tangent point $D$ makes $\angle CDE = 90^{\circ}$. The angle of refraction $r$ is the angle with the normal at $E$. The normal at $E$ is vertical. Thus $\angle CED = r = 30^{\circ}$.$BE = CE \sin 30^{\circ} = 2a \cdot 0.5 = a$. $	an i = BE/AB = a/(a/2) = 2$. $\sin i = 2/\sqrt{5}$.$n = \sin i / \sin r = (2/\sqrt{5}) / (1/2) = 4/\sqrt{5}$.

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