If a point source is placed at a depth `h` in a liquid of refractive index `4/3` . Find percentage of energy of light that escapes from liquid. (assuming `100%` transmission of emerging light)

Options

A. `15%`
B. `17%`
C. `21%`
D. `34%`

Answer

To find the percentage of energy of light that escapes from a liquid when a point source is placed at a depth \( h \) in a liquid of refractive index \( \frac{4}{3} \), we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Geometry**:
- A point source is placed at a depth \( h \) in a liquid. The refractive index \( n \) of the liquid is given as \( \frac{4}{3} \).
- We need to find the angle \( \beta \) at which light rays emerge from the liquid surface.

2. **Using Snell's Law**:
- According to Snell's law, we have:
\[ n_1 \sin \theta_1 = n_2 \sin \theta_2 \]
- Here, \( n_1 = n \) (the refractive index of the liquid) and \( n_2 = 1 \) (the refractive index of air).
- Let \( \theta_1 = \beta \) and \( \theta_2 = 90^\circ \) (the angle of refraction when light exits into air). Thus, we can write:
\[ \frac{4}{3} \sin \beta = 1 \cdot \sin 90^\circ = 1 \]
- From this, we can solve for \( \sin \beta \):
\[ \sin \beta = \frac{3}{4} \]

3. **Finding \( \cos \beta \)**:
- Using the Pythagorean identity \( \sin^2 \beta + \cos^2 \beta = 1 \):
\[ \cos^2 \beta = 1 - \sin^2 \beta = 1 - \left(\frac{3}{4}\right)^2 = 1 - \frac{9}{16} = \frac{7}{16} \]
- Therefore, we find:
\[ \cos \beta = \sqrt{\frac{7}{16}} = \frac{\sqrt{7}}{4} \]

4. **Calculating the Solid Angle**:
- The solid angle \( \Omega \) for the light escaping is given by:
\[ \Omega = 2\pi (1 - \cos \beta) = 2\pi \left(1 - \frac{\sqrt{7}}{4}\right) \]

5. **Calculating the Total Solid Angle**:
- The total solid angle for a sphere is \( 4\pi \). Therefore, the fraction of the solid angle that escapes is:
\[ \text{Fraction escaping} = \frac{\Omega}{4\pi} = \frac{2\pi \left(1 - \frac{\sqrt{7}}{4}\right)}{4\pi} = \frac{1 - \frac{\sqrt{7}}{4}}{2} \]

6. **Finding the Percentage of Energy Escaping**:
- To find the percentage of energy that escapes, we multiply by 100:
\[ \text{Percentage} = \left(\frac{1 - \frac{\sqrt{7}}{4}}{2}\right) \times 100 \]
- Simplifying this gives:
\[ = \left(\frac{4 - \sqrt{7}}{8}\right) \times 100 = \frac{400 - 100\sqrt{7}}{8} \]

7. **Calculating the Numerical Value**:
- Approximating \( \sqrt{7} \approx 2.64575 \):
\[ \text{Percentage} \approx \frac{400 - 100 \times 2.64575}{8} \approx \frac{400 - 264.575}{8} \approx \frac{135.425}{8} \approx 16.928125\% \]
- Rounding this gives approximately \( 17\% \).

### Final Answer:
The percentage of energy of light that escapes from the liquid is approximately **17%**.