Question

A sphere of radius $R$ is placed in air such that its centre is at origin. A ray of light traveling along $y=+\frac{\sqrt{3}}{2}R$ (in x-y plane) is incident on this sphere.

If ray travels through sphere as shown in figure, then:

• A. Refractive index of the sphere is 2
• B. Refractive index of the sphere $\sqrt{3}$
• C. Emergent ray travels along the line $\sqrt{3}x+y=\sqrt{3}R$
• D. Total deviation suffered by the ray is ${45}^o$

Answer 1

The correct option is B Refractive index of the sphere $\sqrt{3}$

$\begin{array}{c}\text{equation of sphere}{x}^2+y^2=R^2\\ \text{at point,}\begin{array}{cc}=R^2& y=+\frac{\sqrt{3}}{2}R\\ x^2+{\left(\frac{\sqrt{3R}}{2}\right)}^2& \\ x^2& =R^2-\frac{3R^2}{4}\\ x=\frac{R}{2}& \end{array}\end{array}$

considers $△PHO$

$\begin{array}{cc}\tan \theta =& \frac{\frac{\sqrt{3}R}{2}}{\frac{3R}{2}}\\ \tan \theta =\frac{1}{\sqrt{3}}& \\ \theta & ={30}^{\circ }\end{array}$

Consider $△PHc$

$\begin{array}{c}\tan \varphi =\frac{\frac{\sqrt{3P}}{2}}{\frac{R}{2}}\\ \tan \varphi =\sqrt{3}\\ \varphi ={60}^{\circ }\end{array}$

So, incident angle at $P={60}^{\circ }$

refracted angle at $p={30}^{\circ }$

${\mu }_1\sin i={\mu }_2\sin \gamma$ at point $p$

(1) $\sin 60=\mu \sin 30$

$\mu =\sqrt{3}]-1$

${\mu }_1\sin i={\mu }_2\sin \gamma$ at point 0

$\begin{array}{cc}\sqrt{3}\sin 30& =\left(1\right)\sin \alpha \\ \alpha & ={60}^{\circ }\end{array}$

slope of emergent ray $=\tan (90+\alpha )$

$=-\cot 60=-\frac{1}{\sqrt{3}}$

equation of emergent ray $(x-R)-\frac{1}{\sqrt{3}}=y$

By equation (1), option $B$ is correct, option $c$ is incorrect