A parallel bean of light travelling in water (refractive index = 4/3) is refracted by a spherical bubble of radius 2 mm situation in water. Assuming the light rays to be paraxial. i. find the position of the image due to refraction at the first surface and the position of the final image, and ii. draw a ray diagram showing the positions of the 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.