Question

A ball is dropped from a height of $20\text{m}$ above the surface of water in a lake. The refractive index of water is $\frac{4}{3}$. A fish inside the lake, in the line of fall of the ball, is looking at the ball. At an instant, when the ball is $12.8\text{m}$ above the water surface, the fish sees the speed of ball as

• A. $9{\text{ms}}^{-1}$
• B. $12{\text{ms}}^{-1}$
• C. $16{\text{ms}}^{-1}$
• D. $21.33{\text{ms}}^{-1}$

Answer 1

The correct option is C $16{\text{ms}}^{-1}$

Given that,
Refractive index of air, ${\mu }_1=1$
Refractive index of water, ${\mu }_2=\frac{4}{3}$
Height of drop of ball, $H=20\text{m}$

When the ball is $12.8\text{m}$ above the water surface, it has fallen through $(h=20-12.8=7.2\text{ms})$

Thus, velocity at this instant is
$v=\sqrt{2gh}$
$\Rightarrow v=\sqrt{2\times 10\times 7.2}=\sqrt{144}$
$\therefore v=12\text{m/s}\dots \left(i\right)$


The light ray from the ball is travelling from the rarer medium to denser medium.

Let the distance appears to fish is $x_{app}=v$ and real distance is $x_{real}=h=u$

Thus, $\frac{{\mu }_2}{{\mu }_1}=\frac{v}{u}=\frac{x_{app}}{x_{real}}$

$\Rightarrow \frac{{\mu }_2}{1}=\frac{x_{app}}{x_{real}}$
or $x_{app}={\mu }_2{x}_{real}$

Differentiate both sides.
$\frac{d{x}_{app}}{dt}={\mu }_2\frac{d{x}_{real}}{dt}$

By using equation $\left(i\right)$,
$\Rightarrow v_{app}={\mu }_2\times v$
or, $v_{app}=\frac{4}{3}\times 12=16\text{m/s}$