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 at the surface, we can follow these steps:

### Step 1: Understand the Geometry
We have a point source of light at a depth \( h \) below the water surface. When light travels from a denser medium (water) to a rarer medium (air), it bends away from the normal. The light rays that reach the surface will form a cone.

### Step 2: Define the Critical Angle
The critical angle \( \theta_c \) can be defined using Snell's law. At the critical angle, the angle of refraction is \( 90^\circ \):
\[ \sin \theta_c = \frac{1}{\mu} \]

### Step 3: Relate the Critical Angle to the Geometry
Using the definition of tangent in the context of our geometry:
\[ \tan \theta_c = \frac{\text{opposite}}{\text{adjacent}} = \frac{R}{h} \]
where \( R \) is the radius of the disc and \( h \) is the depth of the source.

### Step 4: Express Tangent in Terms of Sine
From the definition of the critical angle:
\[ \tan \theta_c = \frac{\sin \theta_c}{\sqrt{1 - \sin^2 \theta_c}} = \frac{\frac{1}{\mu}}{\sqrt{1 - \left(\frac{1}{\mu}\right)^2}} = \frac{1/\mu}{\sqrt{\frac{\mu^2 - 1}{\mu^2}}} = \frac{1}{\sqrt{\mu^2 - 1}} \]

### Step 5: Set Up the Equation
Now we can equate the two expressions for \( \tan \theta_c \):
\[ \frac{R}{h} = \frac{1}{\sqrt{\mu^2 - 1}} \]

### Step 6: Solve for \( R \)
Rearranging gives us:
\[ R = \frac{h}{\sqrt{\mu^2 - 1}} \]

### Step 7: Calculate 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
Thus, the minimum diameter of the disc is:
\[ D = \frac{2h}{\sqrt{\mu^2 - 1}} \]