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$\mu$ . The distance (s) from itself will the fish see the image(s)  of the eye is                           

(a)Let x = distance of the image of the eye formed above the surface as seen by the fish

So, $\frac{H}{X}=\frac{\text{Real}\text{depth}}{\text{Apparent}\text{depth}}=\frac{1}{\mu }\left(or\right)X=\mu H$

So, distance of the direction image $=\frac{H}{2}+\mu H=\mu \left(\mu +\frac{1}{2}\right)$

Similarly, image through mirror$=\frac{H}{2}+\left(H+X\right)=\frac{3H}{2}+\mu H=H\left[\mu +\frac{3}{2}\right]$

(b)Here$\frac{H12}{Y}=\mu$ , so$Y=\frac{H}{2\mu }$

Where Y = distance of the image of fish below the surface as seen by eye

So, 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 below

The mirror

So, the real depth for that image of fish becomes$\text{H}+\text{H/2=}\frac{3H}{2}$

So, Apartment depth from the surface of water$=\frac{3H}{2\mu }$

So, distance of the image from the eye$=H+\frac{3H}{2\mu }=H\left(1+\frac{3}{2\mu }\right)$