Question

A rectangular tank of depth $8\;{\text{m}}$ is full of water $\left( {\mu = \dfrac{4}{3}} \right)$, the bottom is seen at depth:
A. $6\;{\text{m}}$
B. $\dfrac{8}{3}\;{\text{m}}$
C. $8\;{\text{cm}}$
D. $10\;{\text{cm}}$

Answer 1

The real height and the apparent height while looking to the bottom of the tank will be different. This is due to the bending of the light rays. Therefore the refractive index has a major role.

Complete step-by-step solution :
Given the refractive index of the water with respect to air is $\mu = \dfrac{4}{3}$.
The rays of the light coming from the bottom bend away from the normal when they leave the water. As they reach the eye it comes from the virtual bottom of the tank and it seems to be above the actual bottom. Thus the depth of the virtual bottom will be less than the actual bottom. The change in direction of the light ray when it travels from one medium to another will cause the refraction. And it will result in the determination of the depth to be false when observed from the air.
Here the ratio of actual depth to the virtual depth can be taken as the refractive index of the water with respect to air.
Therefore,
$\mu = \dfrac{{{\text{actual}}\;{\text{depth}}}}{{{\text{virtual}}\;{\text{depth}}}}$
Actual depth of the tank is $8\;{\text{m}}$.
Substituting the values in the above expression,
$\ \dfrac{4}{3} = \dfrac{{8\;{\text{m}}}}{{{\text{virtual}}\;{\text{depth}}}} \\\ {\text{virtual}}\;{\text{depth}} = \dfrac{{8\;{\text{m}} \times 3}}{4} \\\ = 6\;{\text{m}} \\\$
Therefore the bottom is seen at the depth $6\;{\text{m}}$
The answer is option A.

Note: This refraction of light causes the swimming pools to be shallower than they really do. And when we put a stick into the water taken in a cup, then it seems to bend. The refractive index of the medium is a factor.