Question

A ball is dropped from a height of 20m above the surface of water in a lake. The refractive index of water is 4/3. A fish inside the lake, in the line of fall of the ball, looking at the ball. At an instant when the ball is 12.8m above the water surface, the fish sees the speed of ball as
$\text{A}\text{. }16m{{s}^{-1}}$
$\text{B}\text{. }12m{{s}^{-1}}$
$\text{C}\text{. 9}m{{s}^{-1}}$
$\text{D}\text{. 21}\text{.33}m{{s}^{-1}}$

Answer 1

Hint: The relation between apparent height (${{h}_{a}}$) and real height (${{h}_{r}}$) is given as $\dfrac{{{h}_{a}}}{{{h}_{r}}}=\dfrac{{{\mu }_{r}}}{{{\mu }_{i}}}$. Use this formula to relation the apparent speed and real speed of the ball. To calculate the speed of the ball when it is at a height of 12.8m, use one the kinematic equations.

Formula used:
$\dfrac{{{h}_{a}}}{{{h}_{r}}}=\dfrac{{{\mu }_{r}}}{{{\mu }_{i}}}$
$2as={{v}^{2}}-{{u}^{2}}$

Complete step-by-step answer:
Due to refraction of light at the interface of air and water, the fish will not see the ball at its real position. Due to refraction, the light rays coming from the ball appear to come from behind the real ball. Therefore, for the fish, the distance of the ball from the surface of water will be more than the distance of the real ball. This distance of the ball that the fish sees is called apparent height.

The relation between apparent height (${{h}_{a}}$) and real height (${{h}_{r}}$) is given as $\dfrac{{{h}_{a}}}{{{h}_{r}}}=\dfrac{{{\mu }_{r}}}{{{\mu }_{i}}}$
$\Rightarrow {{h}_{a}}={{h}_{r}}\dfrac{{{\mu }_{r}}}{{{\mu }_{i}}}$…….(i).
${{\mu }_{r}}$ and ${{\mu }_{i}}$ are the refractive indices of the mediums where the light is refracted and incident respectively.
Differentiate equation (i) with respect to time.
Therefore, $\dfrac{d{{h}_{a}}}{dt}=\dfrac{{{\mu }_{r}}}{{{\mu }_{i}}}\dfrac{d{{h}_{r}}}{dt}$.
We can write this as ${{v}_{a}}={{v}_{r}}\dfrac{{{\mu }_{r}}}{{{\mu }_{i}}}$.
${{v}_{r}}$ is the real speed of the ball and ${{v}_{a}}$ is the apparent speed of the ball.
Here, ${{\mu }_{r}}=\dfrac{4}{3}$ and ${{\mu }_{i}}=1$.
Therefore, ${{v}_{a}}=\dfrac{4}{3}{{v}_{r}}$……….(ii).
Let us now find the real speed of the balls when it is at a height of 12.8m from the surface of water.
For this, we will use the kinematic equation $2as={{v}^{2}}-{{u}^{2}}$.
Here, $a=g=-10m{{s}^{-2}}$, s = -20 - (-12.8) = 7.2m, $v={{v}_{r}}$ and u=0.
Hence, $2(-10)(-7.2)=v_{r}^{2}$.
$v_{r}^{2}=144\Rightarrow {{v}_{r}}=12m{{s}^{-1}}$.
Therefore, the real speed of the ball is $12m{{s}^{-1}}$ downwards.
Substitute the value of ${{v}_{r}}$ in equation (ii).
Therefore, ${{v}_{a}}=\dfrac{4}{3}.12=16m{{s}^{-1}}$.
Therefore, when the ball is at a height of 12.8m from the surface of water, the fish sees the ball coming at a speed of $16m{{s}^{-1}}$.
Hence, the correct option is A.

Note: One mistake that can be done in this solution is when we calculate the real speed of the ball by using the kinematic equation $2as={{v}^{2}}-{{u}^{2}}$. It is given that the ball is at a height of 12.8 metres, so students may substitute the value of s as 12.8 metres.