An isotropic point source is placed at a depth h below the water surface. A floating opaque disc is placed on the surface of water so that the source is not visible from the surface. What is the minimum radius of the disc? Take refractive index of `"water"=mu`.

Explanation

To find the minimum radius of the opaque disc that will completely hide the isotropic point source from view at the water's surface, we can follow these steps:

### Step 1: Understand the Geometry
- The point source is located at a depth \( h \) below the water surface.
- An opaque disc is placed on the surface of the water to block the light emitted from the source.

### Step 2: Determine the Critical Angle
- The critical angle \( \theta_C \) is the angle of incidence above which total internal reflection occurs. It can be calculated using the formula:
\[ \theta_C = \sin^{-1}\left(\frac{1}{\mu}\right) \]
where \( \mu \) is the refractive index of water.

### Step 3: Relate the Radius of the Disc to the Critical Angle
- For the disc to completely block the light from the source, the light rays that emerge from the source must be at an angle greater than \( \theta_C \) when they reach the surface of the water.
- We can visualize this scenario as a right triangle where:
- The opposite side is the radius \( r \) of the disc.
- The adjacent side is the depth \( h \) of the source.
- Using trigonometric relations, we have:
\[ \tan(\theta_C) = \frac{r}{h} \]

### Step 4: Substitute the Critical Angle into the Equation
- We know that:
\[ \tan(\theta_C) = \frac{\sin(\theta_C)}{\cos(\theta_C)} \]
- Since \( \sin(\theta_C) = \frac{1}{\mu} \) and using the Pythagorean identity, we can find \( \cos(\theta_C) \):
\[ \cos(\theta_C) = \sqrt{1 - \sin^2(\theta_C)} = \sqrt{1 - \left(\frac{1}{\mu}\right)^2} = \sqrt{\frac{\mu^2 - 1}{\mu^2}} \]
- Therefore, we can express \( \tan(\theta_C) \) as:
\[ \tan(\theta_C) = \frac{\frac{1}{\mu}}{\sqrt{\frac{\mu^2 - 1}{\mu^2}}} = \frac{1}{\sqrt{\mu^2 - 1}} \]

### Step 5: Solve for the Radius \( r \)
- Now substituting this back into our equation:
\[ \frac{1}{\sqrt{\mu^2 - 1}} = \frac{r}{h} \]
- Rearranging gives:
\[ r = \frac{h}{\sqrt{\mu^2 - 1}} \]

### Conclusion
- The minimum radius \( r \) of the disc that will completely hide the light source is:
\[ r = \frac{h}{\sqrt{\mu^2 - 1}} \]