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 of determining the acceleration of the mango falling from the tree as it appears to the tortoise in the river, we can follow these steps:

### Step 1: Understand the scenario
We have a mango falling from a tree, and it falls directly onto a tortoise in the river. The tortoise is stationary, and we need to find out how the acceleration of the falling mango appears to the tortoise, taking into account the refractive index of the water.

### Step 2: Define the variables
Let:
- \( u \) = actual distance of the mango from the water surface (in meters).
- \( V \) = distance of the mango as observed by the tortoise (in meters).
- The refractive index of water, \( \mu = \frac{4}{3} \).

### Step 3: Use the formula for refraction
According to the principles of optics, the relationship between the actual distance \( u \) and the observed distance \( V \) through a medium is given by:
\[ \frac{\mu_2}{V} - \frac{\mu_1}{u} = \frac{\mu_2 - \mu_1}{R} \]
Here, \( \mu_1 \) is the refractive index of air (approximately 1), and \( R \) is the radius of curvature of the surface. Since the water surface is considered to be flat, \( R \) approaches infinity, making the right-hand side zero.

### Step 4: Simplify the equation
Given that \( R \) is infinite, we can simplify the equation:
\[ \frac{4/3}{V} - \frac{1}{u} = 0 \]
This leads to:
\[ \frac{4}{3V} = \frac{1}{u} \]
Rearranging gives:
\[ V = \frac{4}{3} u \]

### Step 5: Relate the accelerations
The acceleration of the mango in free fall is \( g \) (approximately \( 9.81 \, \text{m/s}^2 \)). We need to find the acceleration as observed by the tortoise. The relationship between the observed acceleration \( a \) and the actual acceleration \( g \) is derived from the second derivative of the distances:
\[ \frac{d^2 V}{dt^2} = \frac{4}{3} \frac{d^2 u}{dt^2} \]
Since \( \frac{d^2 u}{dt^2} = g \), we can substitute:
\[ \frac{d^2 V}{dt^2} = \frac{4}{3} g \]

### Step 6: Conclusion
Thus, the acceleration of the mango as it appears to the tortoise is:
\[ a = \frac{4}{3} g \]

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

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