A disc is placed on a surface of pond which has refractive index `3/5` . A source of light is placed 4 m below the surface of liquid. The minimum radius of disc will be so light is not coming out

Options

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

Explanation

To solve the problem, we need to determine the minimum radius of the disc such that light from a source placed 4 meters below the surface of the pond does not escape. We will use Snell's law and the concept of total internal reflection.

### Step-by-Step Solution:

1. **Understanding the Setup**:
- We have a disc placed on the surface of a pond.
- The refractive index of the pond (n₁) is \( \frac{3}{5} \).
- The light source is located 4 meters below the surface of the pond.

2. **Identifying the Condition for Total Internal Reflection**:
- For light to not escape, it must undergo total internal reflection at the disc's edge.
- This occurs when the angle of incidence (i) is greater than the critical angle (θ_c).

3. **Finding the Critical Angle**:
- The critical angle can be found using Snell's law:
\[ n_1 \sin(\theta_c) = n_2 \sin(90^\circ) \]
Here, \( n_2 = 1 \) (refractive index of air).
- Substituting the values:
\[ \frac{3}{5} \sin(\theta_c) = 1 \cdot 1 \]
\[ \sin(\theta_c) = \frac{5}{3} \]
- Since this is not possible (as sine cannot exceed 1), we instead calculate the angle of incidence using the sine function.

4. **Using Geometry to Relate the Radius and Angle**:
- The light travels from the source to the edge of the disc.
- The vertical distance from the light source to the surface is 4 m, and the radius of the disc is r.
- The angle of incidence can be expressed using the triangle formed:
\[ \sin(i) = \frac{r}{\sqrt{16 + r^2}} \]
where 16 is the square of the vertical distance (4 m).

5. **Applying Snell's Law**:
- We know from Snell's law:
\[ \frac{3}{5} \sin(i) = 1 \cdot \sin(90^\circ) \]
\[ \frac{3}{5} \sin(i) = 1 \]
\[ \sin(i) = \frac{5}{3} \]

6. **Setting Up the Equation**:
- Substitute \( \sin(i) \) into the geometry relation:
\[ \frac{3}{5} \cdot \frac{r}{\sqrt{16 + r^2}} = 1 \]
- Rearranging gives:
\[ r = \frac{5}{3} \sqrt{16 + r^2} \]

7. **Squaring Both Sides**:
- Squaring both sides to eliminate the square root:
\[ r^2 = \frac{25}{9} (16 + r^2) \]
- Expanding gives:
\[ r^2 = \frac{400}{9} + \frac{25}{9} r^2 \]

8. **Rearranging the Equation**:
- Bringing all terms involving r² to one side:
\[ r^2 - \frac{25}{9} r^2 = \frac{400}{9} \]
\[ \left(1 - \frac{25}{9}\right) r^2 = \frac{400}{9} \]
\[ -\frac{16}{9} r^2 = \frac{400}{9} \]

9. **Solving for r²**:
- Multiplying through by -9/16:
\[ r^2 = 25 \]
- Therefore, \( r = 5 \) meters.

10. **Conclusion**:
- The minimum radius of the disc so that light does not come out is 5 meters.