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 $r=20 \mathrm{cm}$ is radius of the sphere. If the irregularity is viewed from above then what is it's distance (in $\mathrm{cm}$) from the centre where eye will observe the irregularity?
 

Solution

Consideration refraction at glass-water interface.

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

$\Rightarrow \frac{4}{3v}-\frac{3}{-2r}=\frac{\left(4/3\right)-\left(3/2\right)}{r}$



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

$\therefore \mathit{ }\mathit{ }\mathit{ }v=\frac{r}{5}$

Now refraction at water air surface

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

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

$\Rightarrow \frac{1}{v}+\frac{4\times 5}{3\times 9r}=\frac{1-\left(4/3\right)}{\infty }$

$\frac{1}{\mathrm{v}}=-\frac{20}{27r}$

$v=-\frac{27r}{20}$

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

$=\frac{40r-27r}{20}=\frac{13}{20}r=13\mathrm{cm}$

Hence, the correct answer is 13.