A point source of light is placed at a distance h below the surface of a large and deep lake. Show that the fraction f of light that escape directly from water surface is independent of h and is given by
`f=[[1-sqrt(1-1//mu^2)])/2`

Explanation

To solve the problem of finding the fraction \( f \) of light that escapes directly from the water surface, we will follow these steps:

### Step 1: Understanding the Setup
We have a point source of light located at a depth \( h \) below the surface of a lake. The light emitted from the point source will travel in all directions, and we need to determine how much of this light can escape through the water surface.

### Step 2: Identify the Critical Angle
The critical angle \( \theta_C \) is the angle of incidence above which total internal reflection occurs. For light traveling from water (with refractive index \( \mu \)) to air (with refractive index approximately 1), the critical angle can be found using Snell's law:
\[ \sin \theta_C = \frac{1}{\mu} \]

### Step 3: Geometry of Light Escape
When light is emitted from the point source, the maximum angle at which light can escape the water surface is \( \theta_C \). The distance from the point source to the water surface is \( h \), and the light that escapes will form a cone with an apex at the point source.

### Step 4: Calculate the Radius of the Circle of Escape
The radius \( R \) of the circle on the water surface through which light can escape can be calculated using the relationship:
\[ R = h \tan \theta_C \]
However, we need to express this in terms of \( \cos \theta_C \) for further calculations.

### Step 5: Use Solid Angle Concept
The total solid angle for a point source is \( 4\pi \). The solid angle \( \Omega \) corresponding to the angle \( \theta_C \) can be calculated as:
\[ \Omega = 2\pi (1 - \cos \theta_C) \]
This represents the fraction of light that escapes.

### Step 6: Calculate the Fraction of Light Escaping
The fraction \( f \) of light that escapes can be expressed as:
\[ f = \frac{\Omega}{4\pi} = \frac{1 - \cos \theta_C}{2} \]

### Step 7: Substitute for \( \cos \theta_C \)
Using the relationship from Step 2, we can find \( \cos \theta_C \):
\[ \cos \theta_C = \sqrt{1 - \sin^2 \theta_C} = \sqrt{1 - \left(\frac{1}{\mu}\right)^2} = \sqrt{1 - \frac{1}{\mu^2}} \]
Substituting this into the expression for \( f \):
\[ f = \frac{1 - \sqrt{1 - \frac{1}{\mu^2}}}{2} \]

### Final Expression
Thus, the fraction \( f \) of light that escapes from the water surface is given by:
\[ f = \frac{1 - \sqrt{1 - \frac{1}{\mu^2}}}{2} \]
This shows that the fraction of light escaping is independent of the depth \( h \).

---