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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.
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Solution
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 (iC).
Rays of light incident at an angle greater than iC, 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.
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