Question

Consider the situation in figure. The bottom of the pot is a reflecting plane mirror, S is a small fish and T is a human eye. Refractive index of water is μ. (a) At what distance(s) from itself will the fish see the image(s) of the eye? (b) At what distance(s) from itself will the eye see the image(s) of the fish.

Answer 1

Given,
Refractive index of water = μ.
Height of the pot = H
Let us take x as the distance of the image of the eye formed above the surface of the water as seen by the fish.

We can infer from from the diagram,
\[\frac{H}{x} = \frac{Real depth}{Apparent depth} = \frac{1}{\mu}\]
\[or x = \mu H\]
The distance of the direct image
\[= \frac{H}{2} + \mu H = H\left( \mu + \frac{1}{2} \right)\] 
\[\frac{H}{x} = \frac{Real depth}{Apparent depth} = \frac{1}{\mu}\]
\[or x = \mu H\]
Similarly, image through mirror = \[\frac{H}{2} + (H + x) = \frac{3H}{2} + \mu H = H\left( \frac{3}{2} + \mu \right)\]
b) We know that: 
\[\frac{H}{2} + (H + x) = \frac{3H}{2} + \mu H = H\left( \frac{3}{2} + \mu \right)\]
Where, y is the distance of the image of the fish below the surface as seen by the eye.
Direct image = \[H + y = H + \frac{H}{2\mu} = H\left( 1 + \frac{1}{2\mu} \right)\] 
Again another image of fish will be formed H/2 below the mirror.
Real depth for that image of fish becomes H + H/2 = 3H/2
So, apparent depth from the surface of water = 3H/2μ
So, distance of the image from the eye
\[= \frac{H}{2} + \frac{3H}{2\mu} = H\left( 1 + \frac{3}{2\mu} \right)\]