A point source of light is placed at a depth of h below the surface of water of refractive index `mu`. A floating opaque disc is placed on the surface of water so that light from the source is not visible from the surface. The minimum diameter of the disc is

Options

A. `(2h)/((mu^(2) - 1)^(1//2))`
B. `2h(mu^(2)-1)^(1//2)`
C. `(h)/(2(mu^(2)-1)^(1//2))`
D. `h(mu^(2)-1)^(1//2)`

Explanation

To find the minimum diameter of the disc that prevents light from a point source located at a depth \( h \) below the surface of water (with refractive index \( \mu \)) from being visible from the surface, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Situation**:
- We have a point source of light at depth \( h \) in water. Light is emitted in all directions from this point source.
- An opaque disc is placed on the surface of the water to block the light from reaching the surface.

2. **Critical Angle Concept**:
- When light travels from a denser medium (water) to a rarer medium (air), it can undergo total internal reflection if the angle of incidence exceeds the critical angle \( C \).
- The critical angle \( C \) can be calculated using Snell's law:
\[ \sin C = \frac{1}{\mu} \]

3. **Using Geometry**:
- At the critical angle, the light will graze the surface of the water. We can visualize this with a right triangle where:
- The vertical side (height) is \( h \) (the depth of the source).
- The horizontal side (radius of the disc) is \( R \) (the distance from the point directly above the source to the edge of the disc).

4. **Relating Radius and Height**:
- In the right triangle formed, we can use the tangent of the critical angle:
\[ \tan C = \frac{R}{h} \]
- From the relationship of tangent and sine, we can express \( \tan C \) as:
\[ \tan C = \frac{\sin C}{\sqrt{1 - \sin^2 C}} = \frac{\frac{1}{\mu}}{\sqrt{1 - \left(\frac{1}{\mu}\right)^2}} = \frac{1}{\mu} \cdot \frac{\mu}{\sqrt{\mu^2 - 1}} = \frac{1}{\sqrt{\mu^2 - 1}} \]

5. **Equating the Two Expressions for \( \tan C \)**:
- Setting the two expressions for \( \tan C \) equal gives:
\[ \frac{R}{h} = \frac{1}{\sqrt{\mu^2 - 1}} \]
- Rearranging this equation gives:
\[ R = \frac{h}{\sqrt{\mu^2 - 1}} \]

6. **Finding the Diameter**:
- The diameter \( D \) of the disc is twice the radius:
\[ D = 2R = 2 \cdot \frac{h}{\sqrt{\mu^2 - 1}} = \frac{2h}{\sqrt{\mu^2 - 1}} \]

### Final Answer:
The minimum diameter of the disc is:
\[ D = \frac{2h}{\sqrt{\mu^2 - 1}} \]