$A$ tank is filled with water to a height of $12.5 \;cm$. The apparent depth of a needle lying at the bottom of the tank is measured by a microscope to be $9.4 \;cm$. What is the refractive index of water?
If water is replaced by a liquid of refractive index $1.63$ up to the same height,by what distance would the microscope have to be moved to focus on the needle again?

(A) Actual depth of the needle in water,$h_1 = 12.5 \;cm$.
Apparent depth of the needle in water,$h_2 = 9.4 \;cm$.
Refractive index of water,$\mu = \frac{\text{Actual depth}}{\text{Apparent depth}} = \frac{12.5}{9.4} \approx 1.33$.
Now,water is replaced by a liquid of refractive index $\mu' = 1.63$ up to the same height $h_1 = 12.5 \;cm$.
The new apparent depth $h_2'$ is given by:
$h_2' = \frac{h_1}{\mu'} = \frac{12.5}{1.63} \approx 7.67 \;cm$.
The microscope was initially at a position corresponding to an apparent depth of $9.4 \;cm$. To focus on the needle again,it must be moved by a distance $\Delta h = h_2 - h_2' = 9.4 - 7.67 = 1.73 \;cm$ upwards.