A point source of light is placed 4 m below the surface of water of refractive index `(5)/(3).` The minimum diameter of a disc, which should be placed over the source, on the surface of water to cut off all light coming out of water is

Options

A. infinite
B. `6 m`
C. `4 m`
D. `3 m`

Answer

To solve the problem, we need to determine the minimum diameter of a disc that can be placed on the surface of the water to cut off all light coming from a point source located 4 meters below the water surface. The refractive index of water is given as \( \frac{5}{3} \).

### Step-by-Step Solution:

1. **Understanding the Problem:**
- We have a point source of light (S) located 4 meters below the surface of the water.
- We need to find the diameter of a disc that will block all light emerging from the water.

2. **Identifying the Critical Angle:**
- Light will undergo total internal reflection when it hits the water-air interface at an angle greater than the critical angle.
- The critical angle \( \theta_c \) can be calculated using Snell's law:
\[ n_1 \sin(\theta_c) = n_2 \sin(90^\circ) \]
Here, \( n_1 = \frac{5}{3} \) (water) and \( n_2 = 1 \) (air).
- Therefore, we have:
\[ \frac{5}{3} \sin(\theta_c) = 1 \]
\[ \sin(\theta_c) = \frac{3}{5} \]

3. **Calculating the Critical Angle:**
- To find \( \theta_c \), we can use the inverse sine function:
\[ \theta_c = \sin^{-1}\left(\frac{3}{5}\right) \]
- This gives us \( \theta_c \approx 36.87^\circ \).

4. **Finding the Distance from the Source to the Edge of the Disc:**
- The rays of light that are at the critical angle will define the boundary of the light that can escape.
- Using trigonometry, we can find the horizontal distance \( x \) from the point directly above the source (at the water surface) to the edge of the disc:
\[ \tan(\theta_c) = \frac{\text{opposite}}{\text{adjacent}} = \frac{h}{x} \]
Here, \( h = 4 \) m (the depth of the source).
- Therefore, we have:
\[ x = h \cdot \tan(\theta_c) \]

5. **Calculating \( \tan(\theta_c) \):**
- We know \( \tan(\theta_c) = \frac{\sin(\theta_c)}{\cos(\theta_c)} \).
- From the triangle, we can find \( \cos(\theta_c) \) using the Pythagorean identity:
\[ \sin^2(\theta_c) + \cos^2(\theta_c) = 1 \]
\[ \sin(\theta_c) = \frac{3}{5} \Rightarrow \cos(\theta_c) = \sqrt{1 - \left(\frac{3}{5}\right)^2} = \sqrt{\frac{16}{25}} = \frac{4}{5} \]
- Thus:
\[ \tan(\theta_c) = \frac{\frac{3}{5}}{\frac{4}{5}} = \frac{3}{4} \]

6. **Calculating \( x \):**
- Now substituting \( h = 4 \) m:
\[ x = 4 \cdot \tan(36.87^\circ) = 4 \cdot \frac{3}{4} = 3 \text{ m} \]

7. **Finding the Diameter of the Disc:**
- The diameter \( D \) of the disc is twice the distance \( x \):
\[ D = 2x = 2 \cdot 3 = 6 \text{ m} \]

### Final Answer:
The minimum diameter of the disc that should be placed over the source on the surface of the water to cut off all light coming out of the water is **6 meters**.