Question

A bird in air looks at a fish below it inside in a transparent liquid in a tank. If the distance of the fish as estimated by the bird is $h_1$ and that of the bird as estimated by the fish is $h_2$, then the refractive index of the liquid is

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

Answer 1

The correct option is A $\frac{h_2}{h_1}$


For fish, as an observer is in denser medium, the bird will appear at a farther distance.
Thus, apparent height,
$h_2=y+\mu x$ ..........$\left(i\right)$
However, for the bird, as an observer is in rarer medium, the fish will appear to it at a reduced distance.
Thus, apparent depth,
$h_1=x+\frac{y}{\mu }$
$\Rightarrow h_1=\frac{\mu x+y}{\mu }$
$\Rightarrow h_1=\frac{h_2}{\mu }$ $[$from $\left(i\right)]$
$\therefore \mu =\frac{h_2}{h_1}$

Why this question? It intends to challenge your visualization skill, when the observer is in denser and rarer medium.