Question

A glass sphere having refractive index $\left(3/2\right)$ is having a small irregularity at its centre. It is placed in a liquid of refractive index $\frac{4}{3}$ such that the surface of the liquid is at a distance $r$ above the sphere, where $r=20cm$ is radius of the sphere. If the irregularity is viewed from above then what is it's distance (in $cm$ ) from the centre where eye will observe the irregularity?

Answer 1

Answer: 13

Consideration refraction at glass-water interface.
$\frac{{\mu }_2}{v}-\frac{{\mu }_1}{u}=\frac{{\mu }_2-{\mu }_1}{R}$
$\Rightarrow \frac{4}{3v}-\frac{3}{-2r}=\frac{\left(4/3\right)-\left(3/2\right)}{r}$

$\Rightarrow \frac{4}{3v}+\frac{3}{2r}=-\frac{1}{6r}$
$\therefore v=\frac{r}{5}$
Now refraction at water air surface
$u=-\left(r+\frac{4}{5}r\right)=-\frac{9r}{5}$
$\frac{{\mu }_2}{v}-\frac{{\mu }_1}{u}=\frac{{\mu }_2-{\mu }_1}{\infty }$
$\Rightarrow \frac{1}{v}+\frac{4\times 5}{3\times 9r}=\frac{1-\left(4/3\right)}{\infty }$
$\frac{1}{v}=-\frac{20}{27r}$
$v=-\frac{27r}{20}$
So height above center $=2r-\frac{27r}{20}$
$=\frac{40r-27r}{20}=\frac{13}{20}r=13cm$