A bulb is placed at a depth of `2sqrt7cm` in water and a floating opaque disc is placed over the bulb so that the bulb is not visible from the surface. What is the minimum diameter of the disc?

Options

A. 42 m
B. 6 m
C. `2sqrt7` m
D. 12 m

Answer

To find the minimum diameter of the disc that is placed over the bulb so that the bulb is not visible from the surface of the water, we can follow these steps:

### Step 1: Understand the Geometry
The bulb is placed at a depth of \(2\sqrt{7} \, \text{cm}\) in water. We need to determine the minimum diameter of the disc that will block the view of the bulb from the surface.

### Step 2: Use Snell's Law
According to Snell's law, the relationship between the angles of incidence and refraction is given by:
\[ \frac{\sin \theta_1}{\sin \theta_2} = \frac{n_2}{n_1} \]
In our case, \(n_1 = 1\) (air) and \(n_2 = \frac{4}{3}\) (water). Therefore, we can write:
\[ \sin \theta = \frac{1}{\mu} = \frac{3}{4} \]

### Step 3: Relate the Angles and Distances
Using the geometry of the situation, we can relate the radius \(R\) of the disc to the depth \(X\) of the bulb:
\[ \sin \theta = \frac{R}{X} \]
Substituting the value of \(\sin \theta\):
\[ \frac{R}{X} = \frac{3}{4} \]
This implies:
\[ R = \frac{3}{4}X \]

### Step 4: Calculate the Depth
The depth \(X\) is given as \(2\sqrt{7} \, \text{cm}\). Therefore, we can substitute this value into the equation:
\[ R = \frac{3}{4}(2\sqrt{7}) = \frac{3 \cdot 2\sqrt{7}}{4} = \frac{6\sqrt{7}}{4} = \frac{3\sqrt{7}}{2} \]

### Step 5: Find the Radius
Now, we need to square both sides to relate \(R\) to \(X\):
\[ X^2 = R^2 + (2\sqrt{7})^2 \]
Substituting \(R = \frac{3}{4}X\) into the equation:
\[ X^2 = \left(\frac{3}{4}X\right)^2 + 28 \]
Expanding and simplifying:
\[ X^2 = \frac{9}{16}X^2 + 28 \]
\[ 16X^2 - 9X^2 = 448 \]
\[ 7X^2 = 448 \]
\[ X^2 = \frac{448}{7} = 64 \]
\[ X = 8 \, \text{cm} \]

### Step 6: Calculate the Radius
Now substituting back to find \(R\):
\[ R = \frac{3}{4} \cdot 8 = 6 \, \text{cm} \]

### Step 7: Find the Diameter
The diameter \(D\) of the disc is given by:
\[ D = 2R = 2 \cdot 6 = 12 \, \text{cm} \]

### Final Answer
The minimum diameter of the disc is \(12 \, \text{cm}\).

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