Question

A mango tree is at the bank of a river and one of the branches of the tree extends over the river. A tortoise lives in the river. A mango falls just above the tortoise. Find the acceleration of the mango falling from the tree as it appears to the tortoise (refractive index of water is $\dfrac{4}{3}$ and the tortoise is stationary).
a. $g$
b. $\dfrac{3g}{4}$
c. $\dfrac{4g}{3}$
d. None of these

Answer 1

Hint: When light passes from one medium to another medium having a different refractive index, its speed changes. Here the light from the mango passes from air to water as it reaches the tortoise. The falling mango is the object, water will act as a lens of a plane surface. The image of the mango, seen by the tortoise, would form at a lower height than the mango’s actual height from the surface of the water in the river.

Formula used:
The relation between the refractive indices of two mediums is given by, $\dfrac{{{n}_{2}}}{v}-\dfrac{{{n}_{1}}}{u}=\dfrac{{{n}_{2}}-{{n}_{1}}}{R}$ where ${{n}_{1}}$ and ${{n}_{2}}$ are the refractive indices of the first and second mediums, $v$ is the image distance, $u$ is the object distance and $R$ is the radius of curvature of the refracting surface.

Complete step by step answer:
Step 1:
List the parameters given about the problem at hand.
The refractive index of water is given to be ${{n}_{2}}=\dfrac{4}{3}$ .
The actual acceleration of the mango will be the acceleration due to gravity $g$ as it falls freely.
We have to determine the apparent acceleration $a$ of the mango as seen by the tortoise.
The refractive index of air is known to be ${{n}_{1}}=1$ .
The image distance $v={h}'$ will be the apparent height, as seen by the tortoise, from which the mango falls. The actual distance between the mango and water surface will be taken as the object distance $u=h$ .

Step 2:
Express the relation between the refractive indices of two mediums involved (air and water) to obtain the relation between the actual acceleration of the mango and its apparent acceleration.
The relation between the refractive indices of two mediums can be expressed as
$\dfrac{{{n}_{2}}}{{{h}'}}-\dfrac{{{n}_{1}}}{h}=\dfrac{{{n}_{2}}-{{n}_{1}}}{R}$ ---------- (1)
Since $R=\infty $ for the water surface, equation (1) becomes $\dfrac{{{n}_{2}}}{{{h}'}}-\dfrac{{{n}_{1}}}{h}=\dfrac{{{n}_{2}}-{{n}_{1}}}{\infty }=0$ .
$\Rightarrow \dfrac{{{n}_{2}}}{{{h}'}}=\dfrac{{{n}_{1}}}{h}\Rightarrow {h}'=\dfrac{{{n}_{2}}}{{{n}_{1}}}h$ --------- (2)
Differentiating equation (2) w.r.t time $t$ twice we get, $\dfrac{{{d}^{2}}{h}'}{d{{t}^{2}}}=\dfrac{{{n}_{2}}}{{{n}_{1}}}\dfrac{{{d}^{2}}h}{d{{t}^{2}}}$
$\Rightarrow a=\dfrac{{{n}_{2}}}{{{n}_{1}}}g$ ------------- (3)

Step 3:
Using equation (3) obtain the apparent acceleration of the mango.
Substituting ${{n}_{2}}=\dfrac{4}{3}$ and ${{n}_{1}}=1$ in equation (3) we get, $a=\dfrac{4}{3}g$
$\therefore $ the acceleration of the mango as seen by the tortoise is obtained to be $a=\dfrac{4}{3}g$ .

Hence, the correct answer is option (C).

Note: When light passes from a denser medium to a rarer medium, the speed of light increases and when it passes from a rarer medium to a denser medium its speed decreases. The radius of curvature of a plane surface is infinity and so we have the radius of curvature of the water surface as $R=\infty $ . The second derivative of distance w.r.t time gives the acceleration.