Question

A narrow parallel beam of light falls on a glass sphere of radius R and refractive index $\mu$ at normal incidence. The distance of the image from the outer edge is given by

• A. $\frac{R(2-\mu )}{2(\mu -1)}$
• B. $\frac{R(2+\mu )}{2(\mu -1)}$
• C. $\frac{R(2-\mu )}{2(\mu +1)}$
• D. $\frac{R(2+\mu )}{2(\mu +1)}$

Answer 1

The correct option is A $\frac{R(2-\mu )}{2(\mu -1)}$

Rays are coming from infinity so at first refraction surface the object distance is infinity and when image forms from these refracting surface at v1 then 2R-v1 will be the object distance for second refracting surface and v be the distance from second refracting surface where the final image is formed as shown in the figure
at the first refracting surface,

$\frac{\mu }{v_1}-\frac{1}{u}=\frac{(\mu -1)}{R}$

mu/v1-0=(mu-1)/R $\frac{\mu }{v1}-0=\frac{(\mu -1)}{R}$

v1=$\frac{\mu \times R}{(\mu -1)}$
object distance for second refracting surface is $v1-2R=\frac{R\times (2-\mu )}{(\mu -1)}$

$\frac{1}{v}-\frac{\mu \times (\mu -1)}{R\times (2-\mu )}=-\frac{(1-\mu )}{R}$

$\frac{1}{v}=\frac{\mu \times (\mu -1)}{R\times (2-\mu )}+\frac{(\mu -1)}{R}$

$\frac{1}{v}=\frac{(\mu -1)\times 2}{R\times (2-\mu )}$

$v=\frac{R\times (2-\mu )}{2\times (\mu -1)}$.