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. 6m
C. 4m
D. 3m

Explanation

To solve the problem of determining the minimum diameter of a disc that should be placed over a point source of light located 4 meters below the surface of water with a refractive index of \( \frac{5}{3} \), we can follow these steps:

### Step 1: Understand the Concept of Critical Angle
The critical angle (\( C \)) is the angle of incidence above which total internal reflection occurs. It can be calculated using the formula:
\[ \sin C = \frac{1}{\mu} \]
where \( \mu \) is the refractive index of the medium (water in this case).

### Step 2: Calculate the Critical Angle
Given that the refractive index \( \mu = \frac{5}{3} \):
\[ \sin C = \frac{3}{5} \]
Now, we can find the critical angle \( C \) using the inverse sine function:
\[ C = \arcsin\left(\frac{3}{5}\right) \]

### Step 3: Use Geometry to Relate the Radius to the Height
In the context of the problem, we can use the tangent of the critical angle to relate the radius (\( R \)) of the circle of light that can escape to the height of the water column (4 m):
\[ \tan C = \frac{R}{h} \]
where \( h = 4 \) m (the depth of the source).

### Step 4: Calculate the Tangent of the Critical Angle
Using the sine and cosine of the critical angle, we can find the tangent:
\[ \sin C = \frac{3}{5} \quad \text{and} \quad \cos C = \sqrt{1 - \sin^2 C} = \sqrt{1 - \left(\frac{3}{5}\right)^2} = \sqrt{\frac{16}{25}} = \frac{4}{5} \]
Thus,
\[ \tan C = \frac{\sin C}{\cos C} = \frac{\frac{3}{5}}{\frac{4}{5}} = \frac{3}{4} \]

### Step 5: Relate the Radius to the Height
Using the tangent value:
\[ \tan C = \frac{R}{4} \implies \frac{3}{4} = \frac{R}{4} \]
This gives:
\[ R = 3 \text{ m} \]

### Step 6: Calculate the Diameter
The diameter \( D \) of the disc is twice the radius:
\[ D = 2R = 2 \times 3 = 6 \text{ m} \]

### Final Answer
The minimum diameter of the disc that should be placed over the source to cut off all light coming out of the water is:
\[ \boxed{6 \text{ m}} \]