A glass prothole is made at the botton of a ahip for observing sea life. The hole diameter `D` is much larger than the thickness of the glass. Determine the area `S` of the field of vision at the sea botton for the porthole of the refractive index of water is `mu_(w)` and the sea depth is `h`.

Explanation

To solve the problem of determining the area \( S \) of the field of vision at the sea bottom for a porthole in a ship, we will follow these steps:

### Step 1: Understanding the Geometry
We start by visualizing the situation. We have a circular porthole at the bottom of a ship, with a diameter \( D \). The porthole is submerged in water at a depth \( h \). The refractive index of water is given as \( \mu_w \).

### Step 2: Critical Angle Calculation
The critical angle \( I_C \) can be determined using Snell's law. The critical angle is defined where light traveling from water to air is refracted at \( 90^\circ \):
\[ \sin I_C = \frac{1}{\mu_w} \]
From this, we can find \( I_C \):
\[ I_C = \arcsin\left(\frac{1}{\mu_w}\right) \]

### Step 3: Finding the Tangent of the Critical Angle
Using the relationship between sine and cosine, we can find \( \tan I_C \):
\[ \tan I_C = \frac{\sin I_C}{\cos I_C} \]
Where:
\[ \cos I_C = \sqrt{1 - \sin^2 I_C} = \sqrt{1 - \left(\frac{1}{\mu_w}\right)^2} = \frac{\sqrt{\mu_w^2 - 1}}{\mu_w} \]
Thus:
\[ \tan I_C = \frac{\frac{1}{\mu_w}}{\frac{\sqrt{\mu_w^2 - 1}}{\mu_w}} = \frac{1}{\sqrt{\mu_w^2 - 1}} \]

### Step 4: Relating Tangent to Geometry
From the geometry, we can relate the tangent of the critical angle to the dimensions of the porthole:
\[ \tan I_C = \frac{x}{h} \]
Where \( x \) is the horizontal distance from the edge of the porthole to the point directly below the porthole at the sea bottom.

### Step 5: Solving for \( x \)
Equating the two expressions for \( \tan I_C \):
\[ \frac{x}{h} = \frac{1}{\sqrt{\mu_w^2 - 1}} \implies x = \frac{h}{\sqrt{\mu_w^2 - 1}} \]

### Step 6: Calculating the Diameter of the Field of Vision
The total diameter \( D_0 \) of the field of vision at the sea bottom can be expressed as:
\[ D_0 = D + 2x = D + 2\left(\frac{h}{\sqrt{\mu_w^2 - 1}}\right) \]

### Step 7: Calculating the Area \( S \)
The area \( S \) of the field of vision is given by the area of a circle:
\[ S = \pi \left(\frac{D_0}{2}\right)^2 = \pi \left(\frac{D + 2\left(\frac{h}{\sqrt{\mu_w^2 - 1}}\right)}{2}\right)^2 \]
This simplifies to:
\[ S = \frac{\pi}{4} \left(D + \frac{2h}{\sqrt{\mu_w^2 - 1}}\right)^2 \]

### Final Result
Thus, the area \( S \) of the field of vision at the sea bottom for the porthole is:
\[ S = \frac{\pi}{4} \left(D + \frac{2h}{\sqrt{\mu_w^2 - 1}}\right)^2 \]