A point source of light is placed at a distance `h` below the surface of a large deep lake.
(a) Show that the fraction `f` of the light energy that escapes directly from the water surface is independent of `h` and is given by `f=(1)/(2)-(1)/(2n)sqrt(n^(2)-1)` where `n` is the index of refraction of water.
(Note: Absorption within the water and reflection at the surface, except where it is total, have been neglected)
(b) Evaluate this ratio for `n=4//3`.

Explanation

To solve the problem step by step, we will break it down into parts (a) and (b) as given in the question.

### Part (a): Show that the fraction of light energy that escapes is independent of `h`.

1. **Understanding the Setup**:
- We have a point source of light located at a distance `h` below the surface of water.
- The light rays emitted from the point source will travel in all directions, and some will hit the water surface.

2. **Critical Angle**:
- When light travels from a denser medium (water) to a less dense medium (air), it will refract away from the normal.
- The critical angle \( \theta_c \) is defined as the angle of incidence above which total internal reflection occurs. According to Snell's Law:
\[ n \sin(\theta_c) = 1 \cdot \sin(90^\circ) = 1 \]
- This gives us:
\[ \sin(\theta_c) = \frac{1}{n} \]
- Therefore, the critical angle can be expressed as:
\[ \theta_c = \sin^{-1}\left(\frac{1}{n}\right) \]

3. **Solid Angle Calculation**:
- The light escaping from the water surface can be represented as a solid angle.
- The solid angle \( \Omega \) for the entire sphere is \( 4\pi \).
- The solid angle for the circular area through which light escapes can be calculated using:
\[ \Omega = 2\pi(1 - \cos(\theta_c)) \]

4. **Using the Cosine Formula**:
- We know that \( \cos(\theta_c) = \sqrt{1 - \sin^2(\theta_c)} = \sqrt{1 - \left(\frac{1}{n}\right)^2} = \sqrt{\frac{n^2 - 1}{n^2}} \).
- Thus:
\[ 1 - \cos(\theta_c) = 1 - \sqrt{\frac{n^2 - 1}{n^2}} = \frac{n^2 - 1}{n^2} \]

5. **Fraction of Light Energy Escaping**:
- The fraction \( f \) of light escaping is given by the ratio of the solid angle of the circle to the total solid angle of the sphere:
\[ f = \frac{2\pi(1 - \cos(\theta_c))}{4\pi} = \frac{1}{2}(1 - \cos(\theta_c)) \]
- Substituting \( 1 - \cos(\theta_c) \):
\[ f = \frac{1}{2} \cdot \frac{n^2 - 1}{n^2} = \frac{1}{2} - \frac{1}{2n}\sqrt{n^2 - 1} \]
- Thus, we have shown that:
\[ f = \frac{1}{2} - \frac{1}{2n}\sqrt{n^2 - 1} \]
- This fraction is independent of `h`.

### Part (b): Evaluate this ratio for \( n = \frac{4}{3} \).

1. **Substituting the Value of n**:
- We substitute \( n = \frac{4}{3} \) into the expression for \( f \):
\[ f = \frac{1}{2} - \frac{1}{2 \cdot \frac{4}{3}} \sqrt{\left(\frac{4}{3}\right)^2 - 1} \]

2. **Calculating \( \sqrt{n^2 - 1} \)**:
- First, calculate \( n^2 \):
\[ n^2 = \left(\frac{4}{3}\right)^2 = \frac{16}{9} \]
- Now, calculate \( n^2 - 1 \):
\[ n^2 - 1 = \frac{16}{9} - 1 = \frac{16}{9} - \frac{9}{9} = \frac{7}{9} \]
- Therefore, \( \sqrt{n^2 - 1} = \sqrt{\frac{7}{9}} = \frac{\sqrt{7}}{3} \).

3. **Final Calculation**:
- Substitute back into the expression for \( f \):
\[ f = \frac{1}{2} - \frac{1}{\frac{8}{3}} \cdot \frac{\sqrt{7}}{3} \]
- Simplifying:
\[ f = \frac{1}{2} - \frac{3\sqrt{7}}{24} \]
- Converting \( \frac{1}{2} \) to a fraction with a denominator of 24:
\[ f = \frac{12}{24} - \frac{3\sqrt{7}}{24} = \frac{12 - 3\sqrt{7}}{24} \]

### Final Result:
The fraction of light energy that escapes is:
\[ f = \frac{12 - 3\sqrt{7}}{24} \]