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A fish in water (refractive index $n$ ) looks at a bird vertically above in the air. If $y$ is the height of the bird and $x$ is the depth of the fish from the surface, then the distance of the bird as estimated by the fish is
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The correct answer is:
$x+n y$
When object is in rarer medium and observer is in denser medium
Normal shift, $\quad \mathrm{d}=(\mathrm{n}-1) \mathrm{h}$
where $h=$ real depth
Here, $\quad h=y$
Now, apparent depth or the apparent height of the bird from the surface of the water $=\mathrm{y}+(\mathrm{n}-1) \mathrm{y}=\mathrm{ny}$.
The total distance of the bird as estimated by fish is $x+n y$.
Normal shift, $\quad \mathrm{d}=(\mathrm{n}-1) \mathrm{h}$
where $h=$ real depth
Here, $\quad h=y$
Now, apparent depth or the apparent height of the bird from the surface of the water $=\mathrm{y}+(\mathrm{n}-1) \mathrm{y}=\mathrm{ny}$.
The total distance of the bird as estimated by fish is $x+n y$.
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