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

The ray diagram for the given situation is as shown below . Note that the refractive index inside air bubble is less than that of the medium present outside it .

For the first refraction we can assume object is at infinity . First refraction is from water to air .
`(mu_(2))/(v) - (mu_(1))/(u) = (mu_(2) - mu_(1))/(R)`
Here `mu_(1) = 4//3` and `mu_(2) = 1` and `R = +0.2` cm
`(1)/(v) = (4//3)/(oo) = (1- 4//3)/(0.2)`
`implies " " (1)/(v) = - (1)/(0.6)`
`implies v = -0.6` cm
Hence , first image `I_(1)` is at a distance `0.6` cm from the surface of air bubble . Other surface of air bubble is further 0.4 cm (diameter of bubble) away from this point . Hence object distance for next refraction becomes `0.6 + 0.4 = 1` cm . Next refraction is from air to water .
`(mu_(2))/(v) - (mu_(1))/(u) = (mu_(2) - mu_(1))/(R)`
`mu_(1) = 1 , mu_(2) = 4//3 , R = -0.2` cm u = -1 cm
`(4//3)/(v) = (1)/(-1) = (4//3 -1)/(-0.2)`
`implies (4)/(3v) + 1 = - (1)/(0.6)`
`implies (4)/(3v) = - (5)/(3) - 1`
`implies (4)/(3v) = -(8)/(3)`
`implies v = -0.5` cm
Hence distance of the final image is 0.5 cm from the second surface.