A glass sphere of radius $R=10 cm$ is kept inside water. A point object $O$ is placed at $20 cm$ from $A$ as shown in the figure. Find the position and nature of the image when seen from other side of the sphere. Also draw the ray diagram. Given ${\mu }_g=\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right. and {\mu }_w=\raisebox{1ex}{$4$}\!\left/ \!\raisebox{-1ex}{$3$}\right.$.

A ray of light starting from $O$ gets refracted twice. The ray of light is travelling in a direction of left to right. Hence the distances measured in this direction are taken positive.

Applying $\frac{{\mu }_2}{v}-\frac{{\mu }_1}{u}=\frac{{\mu }_2-{\mu }_1}{R}$, twice with proper signs. We have, 

$\frac{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}{A{I}_1}-\frac{{\displaystyle \raisebox{1ex}{$4$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}{-20}=\frac{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}-{\displaystyle \raisebox{1ex}{$4$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}{10} or A{I}_1=-30 cm$

Now, the first image $I_1$ acts as an object for the secon surface, where 

$B{I}_1=u=-(30+20)=-50 cm$

Therefore, $\frac{{\displaystyle \raisebox{1ex}{$4$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}{B{I}_2}-\frac{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}{-50}=\frac{{\displaystyle \raisebox{1ex}{$4$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}-{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}{-10} or B{I}_2=-100 cm$

That is, the final image $I_2$ is virtual and is formed at a distance of $100 cm$(towards left) from $B$.

The ray diagram is shown as :-