Question

A parallel beam of light travelling in water (refractive index$=\frac{4}{3}$) is refracted by a spherical air bubble of radius $2\text{mm}$ situated in water. Assuming the light rays to be paraxial. Find the distance ($\text{in mm}$) between position of the image due to refraction at the first surface and the position of the final image

Answer 1

The image of the object formed by the first refraction by the water-glass surface acts as the object for the second refraction at glass-water surface.
A parallel beam of light travelling in water is refracted by a spherical air bubble of $2\text{mm}$ situated in water.
(Position of image due to refraction at first surface and position of image due to refraction at second surface are determined by using
$\frac{{\mu }_2}{v}-\frac{{\mu }_1}{u}=\frac{{\mu }_2-{\mu }_1}{R}$
(Ray travels from ${\mu }_1$to ${\mu }_2$ for first surface,
$\frac{1}{v_1}-\frac{\frac{4}{3}}{\infty }=\frac{1-\frac{4}{3}}{2}$
This ray travels from water to air
$\therefore \frac{1}{v_1}=-\frac{1}{6}$
or $v_1=-6\text{mm}\to \left(1\right)$
First image $I_1$ will be formal a $6\text{mm}$ towards left of first surface
For second surface, the ray travels from air to water
$I_1$ acts as virtual object for second surface
Distance of $I_1$ from second surface$6+2+2)\text{mm}$
$\therefore u_2=-10\text{mm}$
Formula will now be $\frac{{\mu }_1}{v_2}-\frac{{\mu }_2}{u_2}=\frac{{\mu }_1-{\mu }_2}{R}$

$\frac{\frac{4}{3}}{v_2}-\frac{1}{-10}=\frac{\frac{4}{3}-1}{-2}$
$\therefore v_2=-\frac{30}{8}\times \frac{4}{3}=-5\text{mm}\to \left(2\right)$
Final image is formed at $I_2$, $5\text{mm}$ to the left of second surface
Distance between two images is $|u_2|-|v_2|=5\text{mm}$