A parallel beam of light travelling in water (refractive index ` =4//3)` is refracted by a spherical air bublle of radius 2mm situated in water. Assuming the light rays to be paraxial,
a. Find the position of image due to refraction at first surface and position of the final image.
b. Draw a ray diagram showing the position of both images.

Explanation

a. For refraction at spherical surface,
`(mu_(2))/(v)-(mu_(1))/(u)=(mu_(2)-mu_(1))/(R_(1))` (i)
For refraction at the first surface,
`mu(2)=1, mu_(1)=4//3, u=oo, R_(1)=+2mm,v=v^(')` (say)
Position of image due to refraction at the first surface is give by
`(1)/(v^('))-(4//3)/(oo)=(1-(4//3))/(2)`
This give `v^(')=-6mm`
That is the image is formed at a distance of 6mm to the left of the first surface.
For refraction at the second surface,
`u^(')=u=-(6+4)=-10mm, mu_(1)=1, mu_(2)=4//3`
`R_(2)=-2mm`
Substituting these values in Eq. (i), we get
`((4//3))/(v)-(1)/((-10))=((4)/(3)-1)/((-2))`
`rArr ((4//3))/(v)-(1)/(6)-(1)/(10)=(-10-6)/(60)`
`rArr v=-(60xx((4)/(3)))/(16)=-5mm`
The final image I is at a distance of 5mm to the left of the second surface.
b. The ray diagram is shown in figure. `I^(')` is the virtual image formed by the first surface and the final image and is formed at I.