A small source of light is `4m` below the surface of a liquid of refractive index `5//3`.In order to cut off all the light comingout of liquid surface,minimum diameter of the disc placed on the surface of liquid is

Options

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

Explanation

To solve the problem step by step, we need to find the minimum diameter of the disc that can cut off all the light coming out of the liquid surface.

### Step 1: Understand the setup
We have a light source located 4 meters below the surface of a liquid with a refractive index of \( n = \frac{5}{3} \). We need to determine the minimum diameter of a disc that can block all the light rays emerging from the surface of the liquid.

### Step 2: Apply Snell's Law
Using Snell's law, we can relate the angles of incidence and refraction at the surface of the liquid. Snell's law states:
\[ n_1 \sin \theta_1 = n_2 \sin \theta_2 \]
Here, \( n_1 = \frac{5}{3} \) (refractive index of the liquid), \( n_2 = 1 \) (refractive index of air), \( \theta_1 \) is the angle of incidence (in the liquid), and \( \theta_2 \) is the angle of refraction (in air).

At the critical angle \( C \), the light will just graze the surface:
\[ \frac{5}{3} \sin C = 1 \cdot \sin 90^\circ \]
This simplifies to:
\[ \sin C = \frac{3}{5} \]

### Step 3: Calculate the distance of the light rays
Using the geometry of the situation, we can find the horizontal distance \( R \) from the point directly above the light source to the edge of the disc. From the right triangle formed, we have:
\[ \sin C = \frac{R}{\sqrt{R^2 + H^2}} \]
where \( H = 4 \, m \) (the depth of the light source).

Substituting the value of \( \sin C \):
\[ \frac{3}{5} = \frac{R}{\sqrt{R^2 + 4^2}} \]

### Step 4: Solve for \( R \)
Squaring both sides gives:
\[ \left(\frac{3}{5}\right)^2 = \frac{R^2}{R^2 + 16} \]
This results in:
\[ \frac{9}{25} = \frac{R^2}{R^2 + 16} \]
Cross-multiplying yields:
\[ 9(R^2 + 16) = 25R^2 \]
Expanding and rearranging gives:
\[ 9R^2 + 144 = 25R^2 \implies 16R^2 = 144 \implies R^2 = 9 \implies R = 3 \, m \]

### Step 5: Calculate the diameter of the disc
The minimum diameter \( D \) of the disc that can cut off all the light is twice the radius \( R \):
\[ D = 2R = 2 \times 3 = 6 \, m \]

### Final Answer
The minimum diameter of the disc placed on the surface of the liquid is \( 6 \, m \).

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