Question

A parallel beam of light travelling in water (refractive index $=4/3$) is refracted by a spherical air bubble of radius 2 mm situated in water. Assuming the light rays to be paraxial, find the position of image due to refraction at first surface and position of the final image.

Answer 1

For refraction at spherical surface,
$\frac{{\mu }_2}{v}-\frac{{\mu }_1}{u}=\frac{{\mu }_2-{\mu }_1}{R_1}$ (i)
For refraction at the first surface,
${\mu }_2-1m{\mu }_1=4/3,u=\infty ,R=+2mm,v=v^{\prime }$ (say)
Position of image due to refraction at the first surface is given by
$\frac{1}{v^{\prime }}-\frac{4/3}{\infty }=\frac{1-\left(4/3\right)}{2}$
This given $v^{\prime }=-6mm$
That is the image is formed at a distance of 6 mm to the left of the first surface.
For refraction at the second surface,
$u^{\prime }=u=-(6+4)=-10mm,{\mu }_1=1,{\mu }_2=4/3$
$R_2=-2mm$
Substituting these values in Eq. (i), we get
$\frac{\left(4/3\right)}{v}-\frac{1}{(-10)}=\frac{\frac{4}{3}-1}{(-2)}$
$\Rightarrow \frac{\left(4/3\right)}{v}-\frac{1}{6}-\frac{1}{10}=\frac{-10-6}{60}$
$\Rightarrow v=-\frac{60\times \left(\frac{4}{3}\right)}{16}=-5mm$
The final image I is at a distance of $5mm$ to the left of the second surface.