Question

A jar of height h is filled with a transparent liquid of refractive index µ (Figure). At the centre of the jar on the bottom surface is a dot. Find the minimum diameter of a disc, such that when placed on the top surface symmetrically about the centre, the dot is invisible.

Answer 1

In the figure, ray 1 strikes the surface at an angle less than critical angle c and gets refracted in rear medium. Ray 2 strikes the surface making an angle greater than the critical angle and gets internally reflected. The locus of points where ray strikes at critical angle is a circle called circle of illuminance (C.O.I). All light rays striking inside the circle of illuminance get refracted in the rarer medium. If an observer is in the rear medium, he/she will see light coming out only from within the circle of illuminance. If a circular opaque plate covers the circle of illuminance, no light will get refracted in the rarer medium and then the object cannot be seen from the rarer medium.

In figure, O is a small dot at the bottom of the jar. The ray from the dot emerges out of a circular patch of water surface of diameter AB till the angle of incidence for the rays OA and OB exceeds the critical angle (i_C).

Rays of light incident at an angle greater than i_C, are totally reflected within the water and consequently cannot emerge out of the water surface.

As `sin i_C = 1/mu` ⇒ `tan i_C = 1/sqrt(mu^2 - 1)` 

Now, `(d/2)/h = tan i_c`

⇒ `d/2 = h tan i_c = h 1/sqrt(mu^2 - 1)`

⇒ `d = (2h)/sqrt(mu^2 - 1)`

This is the required expression of d.