A point source of monochromatic light 'S' is kept at the centre of the bottom of a cylinder of radius `15.0 cm`. The cylinder contains water (refractive index 4//3) to a height of `7.0 cm`. Draw the ray diagram and calculate the area of water surface through which the light emerges in air.

Explanation

To solve the problem step by step, we will follow these procedures:

### Step 1: Understand the Setup
We have a point source of monochromatic light 'S' located at the center of the bottom of a cylinder filled with water. The cylinder has a radius of 15.0 cm and a height of 7.0 cm. The refractive index of water is given as \( \frac{4}{3} \).

### Step 2: Draw the Ray Diagram
1. Draw a vertical cylinder with a radius of 15 cm and a height of 7 cm.
2. Mark the center of the bottom of the cylinder as point 'S'.
3. Draw rays emanating from point 'S' towards the water surface at various angles.

### Step 3: Determine the Critical Angle
The light is transitioning from a denser medium (water) to a rarer medium (air). The critical angle \( C \) can be calculated using Snell’s law:
\[ \sin C = \frac{n_{\text{air}}}{n_{\text{water}}} \]
Given that \( n_{\text{air}} = 1 \) and \( n_{\text{water}} = \frac{4}{3} \):
\[ \sin C = \frac{1}{\frac{4}{3}} = \frac{3}{4} \]

### Step 4: Calculate the Critical Angle
To find the critical angle \( C \):
\[ C = \arcsin\left(\frac{3}{4}\right) \]

### Step 5: Use Geometry to Find the Radius of the Emerging Circle
Using the geometry of the situation, we can relate the radius \( R \) of the circle through which light emerges to the height of the water:
\[ \tan C = \frac{R}{h} \]
Where \( h = 7 \, \text{cm} \). From the triangle formed:
\[ \tan C = \frac{\sin C}{\cos C} \]
We already have \( \sin C = \frac{3}{4} \). To find \( \cos C \), we use:
\[ \cos^2 C + \sin^2 C = 1 \implies \cos^2 C = 1 - \left(\frac{3}{4}\right)^2 = 1 - \frac{9}{16} = \frac{7}{16} \implies \cos C = \frac{\sqrt{7}}{4} \]
Thus,
\[ \tan C = \frac{\frac{3}{4}}{\frac{\sqrt{7}}{4}} = \frac{3}{\sqrt{7}} \]
Now substituting back:
\[ \frac{R}{7} = \frac{3}{\sqrt{7}} \implies R = 7 \cdot \frac{3}{\sqrt{7}} = \frac{21}{\sqrt{7}} = 3\sqrt{7} \, \text{cm} \]

### Step 6: Calculate the Area of the Emerging Circle
The area \( A \) of the circle through which light emerges is given by:
\[ A = \pi R^2 \]
Substituting \( R = 3\sqrt{7} \):
\[ A = \pi (3\sqrt{7})^2 = \pi \cdot 9 \cdot 7 = 63\pi \, \text{cm}^2 \]

### Final Answer
The area of the water surface through which the light emerges in air is \( 63\pi \, \text{cm}^2 \).

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