Question

A transparent sphere of radius R has a cavity of radius R/2 as shown in Fig. If the refractive index of the sphere if a parallel beam of light falling on left surface focuses at point P is $\frac{3+\sqrt{5}}{X}$. Find out the value of X?

Answer 1

Given: A transparent sphere of radius R has a cavity of radius $\frac{R}{2}$. If the refractive index of the sphere if a parallel beam of light falling on left surface focuses at point P is $\frac{3+\sqrt{5}}{X}$

To find the value of X

Solution:

When parallel beam of light is falling in left surface, then obect distance will be infinity. Let$\mu$ be the refractive index of the sphere.

The equation for required situation becomes,

$\frac{1}{\infty }+\frac{\mu }{v_1}=\frac{\mu -1}{R}⟹v_1=\frac{\mu R}{\mu -1}$

This becomes object for the second boundary, hence object distance for second boundary be $u_2$

$⟹u_2=\frac{\mu R}{\mu -1}-R⟹u_2=\frac{R}{\mu -1}$

Hence the equation for the second boundary refraction is

$\frac{\mu }{-u_2}+\frac{1}{R}=\frac{1-\mu }{\frac{R}{2}}$

Substituting the values,we get

$\frac{\mu }{-\frac{R}{\mu -1}}+\frac{1}{R}=\frac{1-\mu }{\frac{R}{2}}⟹-\frac{\mu (\mu -1)}{R}+\frac{1}{R}=\frac{2(1-\mu )}{R}⟹\frac{1}{R}=\frac{\mu (\mu -1)}{R}+\frac{2(1-\mu )}{R}⟹1={\mu }^2-\mu +2-2\mu ⟹1={\mu }^2-3\mu +2⟹{\mu }^2-3\mu +1=0⟹\mu =\frac{3+\sqrt{5}}{2}$

is the refractive index of the sphere

Hencethe value of $X=2$