A small bulb (assumed to be a point source) is placed at the bottom of a tank containing water to a depth of `80 cm`. Find out the area of the surface of water through which light from the bulb can emerge. Take the value of refractive index of water to be `4//3`.

Options

A. `2.6 m^(2)`
B. `3.6 m^(2)`
C. `4.2 m^(2)`
D. `5.8 m^(2)`

Explanation

To solve the problem of finding the area of the surface of water through which light from a bulb can emerge, we can follow these steps:

### Step 1: Understand the Given Data
- The depth of water (d) = 80 cm = 0.8 m
- The refractive index of water (μ) = 4/3

### Step 2: Calculate the Critical Angle
The critical angle (C) can be calculated using Snell's law, which states:
\[ \mu = \frac{\sin(90^\circ)}{\sin(C)} \]
This can be rearranged to find the critical angle:
\[ \sin(C) = \frac{1}{\mu} \]
Substituting the value of μ:
\[ \sin(C) = \frac{1}{4/3} = \frac{3}{4} \]
Now, calculate C:
\[ C = \sin^{-1}\left(\frac{3}{4}\right) \approx 48.75^\circ \]

### Step 3: Determine the Radius of the Circle of Light Emergence
The radius (r) of the circle through which light can emerge can be found using the tangent of the critical angle:
\[ r = d \cdot \tan(C) \]
Where \( d \) is the depth of the water.
Substituting the values:
\[ r = 0.8 \cdot \tan(48.75^\circ) \]
Calculating \( \tan(48.75^\circ) \):
\[ \tan(48.75^\circ) \approx 1.14 \]
Thus,
\[ r \approx 0.8 \cdot 1.14 \approx 0.912 \, \text{m} \]

### Step 4: Calculate the Area of the Circle
The area (A) of the circle through which light can emerge is given by the formula:
\[ A = \pi r^2 \]
Substituting the value of r:
\[ A \approx \pi \cdot (0.912)^2 \]
Calculating:
\[ A \approx \pi \cdot 0.831744 \approx 2.61 \, \text{m}^2 \]

### Conclusion
The area of the surface of water through which light from the bulb can emerge is approximately **2.61 m²**.

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