There is an equilateral triangle ABC. Two monochromatic and coherent sources of light are placed at points A and B. Intensity of light from both the sources reaching the point C is same and equal to `2 W//m^2`. Resultant intensity of light at point C

Options

A. is `8 W//m^2`
B. is `4 W//m^2`
C. is zero
D. cannot be determined with the given information

Explanation

To solve the problem, we need to determine the resultant intensity of light at point C due to two coherent sources of light located at points A and B of an equilateral triangle ABC. The intensity of light from both sources at point C is given as 2 W/m².

### Step-by-Step Solution:

1. **Understanding the Setup**:
- We have an equilateral triangle ABC.
- Two coherent light sources are placed at points A and B.
- The intensity of light from both sources reaching point C is the same, given as \( I = 2 \, \text{W/m}^2 \).

2. **Using the Formula for Resultant Intensity**:
- When two coherent light sources interfere, the resultant intensity \( I_{net} \) at a point can be calculated using the formula:
\[ I_{net} = 4I \cos^2\left(\frac{\phi}{2}\right) \]
where \( I \) is the intensity from each source and \( \phi \) is the phase difference between the two waves at the point of observation.

3. **Determining the Phase Difference**:
- Since the triangle is equilateral, the distances from A to C and B to C are equal. Therefore, the path difference \( \Delta x \) is 0.
- The phase difference \( \phi \) can be calculated using:
\[ \phi = \frac{2\pi}{\lambda} \Delta x \]
- Since \( \Delta x = 0 \), we have:
\[ \phi = 0 \]

4. **Calculating the Resultant Intensity**:
- Substituting \( \phi = 0 \) into the intensity formula:
\[ I_{net} = 4I \cos^2\left(0\right) \]
- Since \( \cos(0) = 1 \):
\[ I_{net} = 4I \cdot 1^2 = 4I \]
- Now substituting \( I = 2 \, \text{W/m}^2 \):
\[ I_{net} = 4 \times 2 \, \text{W/m}^2 = 8 \, \text{W/m}^2 \]

5. **Final Result**:
- The resultant intensity of light at point C is:
\[ I_{net} = 8 \, \text{W/m}^2 \]