Question

A bird in air looks at a fish vertically below it and inside waterr $h_1$ is the height of the bird above the surface of water and $h_2$, the depth of the fish below the surface of water. If refractive index of the fish below the surface of water. If refractive index of water with respect to air be $\epsilon$, then the distance of the fish as observed by the bird is

• A. $h_1+h_2$
• B. $h_1+\frac{h_2}{\epsilon }$
• C. $\epsilon h_1+h_2$
• D. $\epsilon h_1+\epsilon h_2$

Answer 1

The correct option is B $h_1+\frac{h_2}{\epsilon }$

Given, $h_1$ is the height of the bird above surface and $h_2$ is the depth of the fish below surface.

And $\epsilon$ is the refractive index of the water.

Using, $\frac{1}{\epsilon }=\frac{apparentdepth}{realdepth}$

Real depth is $h_1+\frac{h_2}{\epsilon }$ as seen by the bird.