Question

A glass sphere having the refractive index (3/2) is having a small irregularity at its center. It is placed in a liquid of refractive index $\frac{4}{3}$ such that surface of the liquid is r high above sphere where r is a radius of a sphere. If the irregularity is viewed from above normally, the distance is 26/K cm from the center where the eye will observe the irregularity. Then find the value of K. (r = 20 cm)

Answer 1

Consider refraction at glass-water surface,

$\frac{{\mu }_2}{v}-\frac{{\mu }_1}{u}=\frac{{\mu }_2-{\mu }_1}{R}$

$⟹\frac{4}{3v}+\frac{3}{2r}=-\frac{1}{6r}$

$⟹v=-\frac{r}{5}$

Now consider refraction at water-air surface,

$u=-(r+\frac{4}{5}r)=-\frac{9r}{5}$

$\frac{{\mu }_2}{v}-\frac{{\mu }_1}{u}=\frac{{\mu }_2-{\mu }_1}{\infty }$

$⟹\frac{1}{v}=-\frac{20}{27r}$

$⟹v=-\frac{27r}{20}$

So the height above center=$2r-\frac{27r}{20}$

$=\frac{13}{20}r=13cm=\frac{26}{2}cm$

Therefore $K=2$