Question

A parallel beam of light travelling in water (refrative index = $\frac{4}{3}$) is refracted by a spherical air bubble of radius 2 mm situated in water. Assuming the light rays to be paraxial, locate the position of the final image.

• A. -5 mm
• B. -6 mm
• C. 5 mm
• D. 6 mm

Answer 1

The correct option is A

-5 mm

The ray diagram is shown in figure. The equation for refraction at a spherical surface is.
$\frac{{\mu }_2}{2}-\frac{{\mu }_1}{2}=\frac{{\mu }_2-{\mu }_1}{R}$
For the first refraction (water to air); ${\mu }_1$ = 1.33, ${\mu }_2$ =1, u = $\infty$, R = +2 mm.
Thus, $\frac{1}{v}=\frac{1-1.33}{2mm}$
or, v = -6mm.
The negative sign shows that the image $I_1$ is virtual and forms at 6 mm from the surface of the bubble on the water side. The refracted rays ( which seems to come from $I_1$) are incident on the further surface of the bubble. For this refraction,
${\mu }_1=1,{\mu }_2=1.33,R=-2mm.$
The object distance is u = -(6 mm + 4 mm) = -10 mm.
Using equation (i),
$\frac{1.33}{v}+\frac{1}{10mm}=\frac{1.33-1}{-2mm}or,\frac{1.33}{v}=-\frac{0.33}{2mm}-\frac{1}{10mm}or,v=-5mm.$
The minus sign shows that the image is formed on the air side at 5 mm from the refracting surface.
Measuring from the centre of the bubble, the first image is formed at 8.0 mm from the centre and the second image is formed at 3.0 mm from the centre. Both images are formed on the side from which the incident rays are coming.