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To a bird in air, a fish in water appears to be at $30 \mathrm{~cm}$ from the surface. If refractive index of water with respect to air is $\frac{4}{3}$, the real distance of bird from the surface is
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$40 \mathrm{~cm}$
The correct option is (C).
Consider the labeled figure. The fish appears at an apparent depth h๘ 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 $\theta, \operatorname{tam} \theta \simeq \sin \theta$ can be taken.
Using smell'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 $\mathrm{h}=40 \mathrm{~cm}$.
Consider the labeled figure. The fish appears at an apparent depth h๘ 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 $\theta, \operatorname{tam} \theta \simeq \sin \theta$ can be taken.
Using smell'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 $\mathrm{h}=40 \mathrm{~cm}$.
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