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.
- At what distance(s) from itself will the fish see the image(s) of the eye?
- At what distance(s) from itself will the eye see the image(s) of the fish.
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- Let x = distance of the image of the eye formed above the surface as seen by the fish
So, distance of the direct image $=\frac{\text{H}}{2}+\mu\text{H}=\text{H}\Big(\mu+\frac{1}{2}\Big)$
Similarly, image through mirror $=\frac{\text{H}}{2}+(\text{H}\times\text{x})=\frac{3\text{H}}{2}+\mu\text{H}=\text{H}\Big(\mu+\frac{3}{2}\Big)$
- Here, $\frac{\frac{\text{H}}{2}}{\text{y}}=\mu,$ So, $\text{y}=\frac{\text{H}}{2\mu}$
So, Direct image $=\text{H}+\text{y}=\text{H}+\frac{\text{H}}{2\mu}=\text{H}\Big(1+\frac{1}{2\mu}\Big)$
Again another image of fish will be formed $\frac{\text{H}}{2}$ below the mirror.
So, the real depth for that image of fish becomes $\text{H}+\frac{\text{H}}{2}=\frac{3\text{H}}{2}$
So, Apparent depth from the surface of water $=\frac{3\text{H}}{2\mu}$
So, distance of the image from the eye $=\text{H}+\frac{3\text{H}}{2\mu}=\text{H}\Big(1+\frac{3}{2\mu}\Big).$
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