Question

An opaque sphere of radius $a$ is just immersed in a transparent liquid as shown in figure. A point source is placed on the vertical diameter of the sphere at a distance $\frac{a}{2}$ from the top of the sphere. A ray from the point source after refraction from the air liquid interface forms tangent to the sphere. The angle of refraction for that particular ray is ${30}^{\circ }$. The refractive index of the liquid is

• A. $\frac{2}{\sqrt{3}}$
• B. $\frac{3}{\sqrt{5}}$
• C. $\frac{4}{\sqrt{5}}$
• D. $\frac{4}{\sqrt{7}}$

Answer 1

The correct option is D $\frac{4}{\sqrt{7}}$



Let $A$ be the projection of refracted ray on to the normal as shown.
From triangle $ACO$
$sin{30}^{\circ }=\frac{a}{\frac{3a}{2}+x}$
$\Rightarrow 2a=\frac{3a}{2}+x$
$\Rightarrow x=\frac{a}{2}$
From triangle $ABD$
$\tan {30}^{\circ }=\frac{y}{\frac{a}{2}+x}$
$\Rightarrow \frac{1}{\sqrt{3}}=\frac{y}{a}$
$\Rightarrow y=\frac{a}{\sqrt{3}}$
From triangle $PBD$
$\tan i=\frac{y}{a/2}=\frac{2}{\sqrt{3}}$
$\Rightarrow \sin i=\frac{2}{\sqrt{7}}$
Snell's law at incidence
$\sin i=n\sin {30}^{\circ }$
$\frac{2}{\sqrt{7}}=\frac{n}{2}$
$\Rightarrow n=\frac{4}{\sqrt{7}}$