A glass sphere of radius $R=10 \mathrm{cm}$ having refractive index ${\mu }_g=\frac{3}{2}$ is kept inside water. A point object O is placed at $20 \mathrm{cm}$ from A as shown in figure. Find the position and nature of the image when seen from other side of the sphere. Also draw the ray diagram. Given refractive index of water is ${\mu }_w=\frac{4}{3}$ .

A ray of light starting from O gets refracted twice The ray of light is travelling in a direction from to right. Hence, the distances measured in this direction are taken positive. Applying $\frac{{\mu }_1}{-u}+\frac{{\mu }_2}{v}=\frac{{\mu }_2-\mu }{R}$ twice with appropriate signs at the two refractive surfaces, we get

$$\begin{gathered} \frac{\left(\frac{4}{3}\right)}{-(-20)}+\frac{\left(\frac{3}{2}\right)}{A{I}_1}=\frac{\left(\frac{3}{2}-\frac{4}{3}\right)}{10} \\ \Rightarrow A{I}_1=-30 \mathrm{cm} \end{gathered}$$

Now, the first image $I_1$, acts as an object for second surface, so, we have

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

Again applying $\frac{{\mu }_2}{-u}+\frac{{\mu }_1}{v}=\frac{{\mu }_1-{\mu }_2}{R}$, we get

$$\begin{gathered} \frac{\left(\frac{3}{2}\right)}{-(-50)}+\frac{\left(\frac{4}{3}\right)}{B{I}_2}=\frac{\frac{4}{3}-\frac{3}{2}}{-10} \\ \Rightarrow B{I}_2=-100 \mathrm{cm} \end{gathered}$$

i.e., the final image $I_2$ is virtual and is formed at a distance $100 \mathrm{cm}(towards left)$ from B. The ray diagram is as shown.

following points should be kept in mind while drawing the ray diagram.

(i) At P the ray travels from rarer to a denser medium. Hence, it will bend towards normal P C. At M, it travels from a denser to a rarer medium, hence, moves away from the normal M C.

(ii) The ray P M when extended backwards meets the principal axis at $I_1$ and the ray M N when extended meets the principal axis at $I_2$.