Question

The length of a vertical pole at the surface of a lake of water $(\mu =\frac{4}{3})$ is $24\text{cm}$. Then, for a fish just below the water surface, the upper tip of the pole appears to be at what distance?

Answer 1

We know that,
$\frac{{\mu }_2}{{\mu }_1}=\frac{\text{Real height}}{\text{Apparent height}}$
where ${\mu }_2$ is the refractive index of the medium in which the object is kept and ${\mu }_1$ is the refractive index of the medium in which the observer is present.
Here, ${\mu }_1$ is the refractive index of water and ${\mu }_2$ is the refractive index of air.
$\Rightarrow \frac{1}{4/3}=\frac{24}{\text{Apparent height}}$
$\Rightarrow \text{apparent height}=32\text{cm}$
so, for fish the upper tip of the pole will appear to be 32 cm above the water surface.