A mango tree is at the bank of a river and one of the branch of tree extends over the river. A tortoise lives in the river. A mango falls just ono the tortoise. The acceleration of the mango falling from tree as it appears to the tortoise is (refractive index of water is `4//3` and the tortoise is stationary)

Options

A. g
B. `3g//4`
C. `4g//3`
D. none of these

Explanation

To solve the problem, we need to determine the acceleration of a mango falling from a tree as it appears to a tortoise in the river, given the refractive index of water. Let's break this down step by step.

### Step 1: Understand the scenario
We have a mango tree on the bank of a river, and a mango falls from the tree onto a tortoise in the water. The refractive index of water is given as \( \mu = \frac{4}{3} \).

### Step 2: Identify the mediums
The mango is falling from air (first medium) into water (second medium). The refractive index of air is approximately 1.

### Step 3: Use the refraction formula
The formula for refraction at the interface between two mediums is given by:
\[ \frac{\mu_2}{v} - \frac{\mu_1}{u} = \frac{\mu_2 - \mu_1}{R} \]
where:
- \( \mu_1 = 1 \) (refractive index of air)
- \( \mu_2 = \frac{4}{3} \) (refractive index of water)
- \( R \) is the radius of curvature, which is infinite for a flat surface (water surface), so it becomes 0.

### Step 4: Simplify the equation
Since \( R \) is infinite, the equation simplifies to:
\[ \frac{\frac{4}{3}}{v} - \frac{1}{u} = 0 \]
This implies:
\[ \frac{4}{3}v = u \]
Thus, we can express the apparent distance \( v \) in terms of the actual distance \( u \):
\[ v = \frac{3u}{4} \]

### Step 5: Differentiate to find velocity
Now, we differentiate both sides with respect to time \( t \):
\[ \frac{dv}{dt} = \frac{3}{4} \frac{du}{dt} \]
Here, \( \frac{du}{dt} \) is the velocity of the mango falling (denoted as \( v_m \)), and \( \frac{dv}{dt} \) is the apparent velocity of the mango as seen by the tortoise.

### Step 6: Differentiate again to find acceleration
Next, we differentiate again with respect to time:
\[ \frac{d^2v}{dt^2} = \frac{3}{4} \frac{d^2u}{dt^2} \]
Here, \( \frac{d^2u}{dt^2} \) is the acceleration of the mango, which is equal to \( g \) (acceleration due to gravity).

### Step 7: Substitute to find the apparent acceleration
Substituting \( g \) into the equation gives:
\[ \frac{d^2v}{dt^2} = \frac{3}{4}g \]
Thus, the apparent acceleration of the mango as it appears to the tortoise is:
\[ \frac{d^2v}{dt^2} = \frac{4g}{3} \]

### Final Answer
The acceleration of the mango falling from the tree as it appears to the tortoise is \( \frac{4g}{3} \).

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