A light bulb is at a depth of D below the surface of water. An opaque disc of radius R is placed on the surface of water just above the bulb. The bulb is not at all seen through the surface of water, then ( n = Refractive index of water)

Options

A. `R = (D)/(sqrt(n^(2) -1))`
B. `R gt (D)/(sqrt(n^(2) - 1))`
C. `R lt (D)/(sqrt(n^(2) - 1))`
D. `R = D sqrt(n^(2) - 1)`

Explanation

To solve the problem, we need to analyze the situation involving the light bulb, the opaque disc, and the refractive index of water. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the Setup
We have a light bulb located at a depth \( D \) below the surface of the water. Above the bulb, there is an opaque disc of radius \( R \) placed on the water's surface. The problem states that the bulb is not visible from above the water surface.

### Step 2: Critical Angle Concept
The fact that the bulb is not visible implies that light rays emanating from the bulb do not reach the observer above the water. This situation can be analyzed using the concept of the critical angle. The critical angle \( \theta_c \) is defined as the angle of incidence above which total internal reflection occurs.

### Step 3: Calculate the Critical Angle
The critical angle can be calculated using Snell's law:
\[ \sin(\theta_c) = \frac{n_2}{n_1} \]
where \( n_1 \) is the refractive index of water (which is \( n \)) and \( n_2 \) is the refractive index of air (which is approximately 1). Thus, we have:
\[ \sin(\theta_c) = \frac{1}{n} \]

### Step 4: Geometry of the Problem
Consider a right triangle formed by the light rays from the bulb to the edge of the disc. The height of the triangle is \( D \) (the depth of the bulb), and the base is the radius \( R \) of the disc.

### Step 5: Relate the Radius and Depth
Using the definition of sine in a right triangle:
\[ \sin(\theta_c) = \frac{R}{\sqrt{R^2 + D^2}} \]
Substituting the expression for \( \sin(\theta_c) \):
\[ \frac{1}{n} = \frac{R}{\sqrt{R^2 + D^2}} \]

### Step 6: Rearranging the Equation
Cross-multiplying gives:
\[ R = \frac{\sqrt{R^2 + D^2}}{n} \]
Squaring both sides results in:
\[ R^2 = \frac{R^2 + D^2}{n^2} \]
Rearranging this leads to:
\[ R^2 (n^2 - 1) = D^2 \]

### Step 7: Solve for \( R \)
Thus, we can express \( R \) as:
\[ R^2 = \frac{D^2}{n^2 - 1} \]
Taking the square root gives:
\[ R = \frac{D}{\sqrt{n^2 - 1}} \]

### Step 8: Conclusion
For the bulb to be completely obscured by the disc, the radius \( R \) must be greater than this value. Therefore, the condition is:
\[ R > \frac{D}{\sqrt{n^2 - 1}} \]

### Final Answer
From the analysis, we conclude that the radius \( R \) must be greater than \( \frac{D}{\sqrt{n^2 - 1}} \) for the bulb to not be seen.

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