To a bird in air, a fish in water appears to be at $30 \mathrm{~cm}$ from the surface. If refractive index…
- $60 \mathrm{~cm}$
- $30 \mathrm{~cm}$
- $40 \mathrm{~cm}$
- $50 \mathrm{~cm}$
Solution
Consider the labeled figure. The fish appears at an apparent depth $h^{\prime}$ while the real depth is $h$.
In triangle $\mathrm{OPF}: \tan (\mathrm{i})=\frac{\mathrm{P}}{\mathrm{h}} \simeq \sin (\mathrm{i})$
In triangle OPA: $\tan (\mathrm{r})=\frac{\mathrm{P}}{\mathrm{h}^{\prime}} \simeq \sin (\mathrm{r})$
For small angles $\theta, \tan \theta \simeq \sin \theta$ can be taken.
Using snell's law of refraction: $1 \times \sin (\mathrm{r})=\mu \times \sin (\mathrm{i})$
Inserting expressions for $\sin (\mathrm{i})$ and $\sin (\mathrm{r}) 1 \times \frac{\mathrm{P}}{\mathrm{h}^{\prime}}=\mu \times \frac{\mathrm{P}}{\mathrm{h}}$
$\mathrm{h}=\mu \mathrm{h}^{\prime}$
Given, $\mathrm{h}^{\prime}=30 \mathrm{~cm}$ and $\mu=\frac{4}{3}$, therefore real depth $\mathrm{h}=40 \mathrm{~cm}$.Asked in: MHT CET 2022 (05 Aug Shift 1)