A point source of light is placed at a distance `h`below the surface of a large deep lake.What is the percentage of light energy that escapes directly form the water surface if `mu`of the water=`4//3`?(neglet partial reflection)

Options

A. `50%`
B. `21%`
C. `34%`
D. `17%`

Answer

To solve the problem, we need to determine the percentage of light energy that escapes directly from the water surface when a point source of light is placed at a distance \( h \) below the surface of a large deep lake, given that the refractive index \( \mu \) of the water is \( \frac{4}{3} \).

### Step-by-Step Solution:

1. **Determine the Critical Angle**:
The critical angle \( \theta_c \) can be calculated using the formula:
\[ \theta_c = \sin^{-1}\left(\frac{1}{\mu}\right) \]
Substituting \( \mu = \frac{4}{3} \):
\[ \theta_c = \sin^{-1}\left(\frac{3}{4}\right) \]
Calculating this gives:
\[ \theta_c \approx 48.59^\circ \]

2. **Calculate the Solid Angle**:
The solid angle \( \Omega \) subtended by the critical angle at the point source can be calculated using the formula:
\[ \Omega = 2\pi(1 - \cos(\theta_c)) \]
First, we need to find \( \cos(\theta_c) \):
\[ \cos(\theta_c) = \sqrt{1 - \sin^2(\theta_c)} = \sqrt{1 - \left(\frac{3}{4}\right)^2} = \sqrt{1 - \frac{9}{16}} = \sqrt{\frac{7}{16}} = \frac{\sqrt{7}}{4} \]
Now substituting into the solid angle formula:
\[ \Omega = 2\pi\left(1 - \frac{\sqrt{7}}{4}\right) \]

3. **Intensity Distribution**:
The intensity per unit solid angle \( I' \) is given by:
\[ I' = \frac{I}{4\pi} \]
where \( I \) is the total intensity of the light source.

4. **Intensity Escaping the Surface**:
The intensity escaping the surface can be calculated as:
\[ I_{escape} = I' \cdot \Omega = \frac{I}{4\pi} \cdot 2\pi\left(1 - \frac{\sqrt{7}}{4}\right) \]
Simplifying this gives:
\[ I_{escape} = \frac{I}{2}\left(1 - \frac{\sqrt{7}}{4}\right) \]

5. **Calculate the Percentage of Escaped Light**:
The percentage of light energy that escapes is given by:
\[ \text{Percentage} = \left(\frac{I_{escape}}{I}\right) \times 100 = \left(\frac{1}{2}\left(1 - \frac{\sqrt{7}}{4}\right)\right) \times 100 \]
Evaluating this expression will give us the final percentage.

6. **Final Calculation**:
Approximating \( \sqrt{7} \approx 2.64575 \):
\[ 1 - \frac{\sqrt{7}}{4} \approx 1 - 0.66144 \approx 0.33856 \]
Therefore:
\[ \text{Percentage} \approx \left(\frac{1}{2} \cdot 0.33856\right) \times 100 \approx 16.93\% \]
Rounding gives approximately \( 16.9\% \).

### Final Answer:
The percentage of light energy that escapes directly from the water surface is approximately **16.9%**.