A parallel beam of light travelling in water refractive index$\frac{4}{3}$ is refracted by a sphere bubble of radius $2 \mathrm{mm}$ situated in water. A the light rays to be paraxial. Find the position of the image due to refraction of the first surface and the position of the final image.

To get the desired result(s), we shall be $\frac{{\mu }_2}{v}-\frac{{\mu }_1}{u}=\frac{{\mu }_2-{\mu }_1}{R}$, one by one on two surfaces.

For first refraction at A P_{1} B, we have

$$\begin{gathered} \frac{1}{v_1}-\frac{\frac{4}{3}}{\infty }=\frac{1-\frac{4}{3}}{+2} \\ \Rightarrow \frac{1}{v_1}=-\frac{1}{6} \\ \Rightarrow v_1=-6 \mathrm{mm} \end{gathered}$$

So, the first image will be formed So, the first image will be formed at 6mm towards left of $P_1$

For second refraction at $A{P}_{2}B$, the distance of first image $I_1$ from $P_2$ will be $6 \mathrm{mm}+4 \mathrm{mm}=10 \mathrm{mm}$ (towards left). So, we get

$$\begin{gathered} \frac{\frac{4}{3}}{v_2}-\frac{1}{-10}=\frac{\frac{4}{3}-1}{-2} \\ \Rightarrow \frac{4}{3v_2}=-\frac{1}{6}-\frac{1}{10}=-\frac{4}{15} \\ {v}_2=-5 \mathrm{mm} \end{gathered}$$