(N/A) As shown in the figure,the bottom of the tank is at a real depth $h_{2}$ in a denser medium (water) of refractive index $n_{2} = n$.
When the bottom is viewed normally,it is observed at $O^{\prime}$ instead of $O$ as shown in figure $(a)$. When it is viewed at some angle from the normal,it is observed at $O^{\prime}$ as shown in figure $(b)$. The real depth is $h_{2}$ and the apparent depth is $h_{1}$.
The relation between refractive indices and depths is given by:
$\frac{\text{Refractive index of air } (n_{1})}{\text{Refractive index of denser medium } (n_{2})} = \frac{\text{Apparent depth } (h_{1})}{\text{Real depth } (h_{2})}$
Since for air $n_{1} = 1$ and for the denser medium $n_{2} = n$:
$\frac{1}{n} = \frac{h_{1}}{h_{2}}$
Therefore,the relation is:
$n = \frac{h_{2}}{h_{1}} = \frac{\text{Real depth}}{\text{Apparent depth}}$
Thus,the apparent depth is:
$h_{1} = \frac{h_{2}}{n} = \frac{\text{Real depth}}{\text{Refractive index of denser medium}}$