Two beam of light having intensities I and 4I interfere to produce a fringe pattern on a screen. The phase difference between the beams is `(pi)/(2)` at point A and `pi` at point B. Then the difference between resultant intensities at A and B is : `(2001 , 2M)`

Options

A. 2I
B. 4I
C. 5I
D. 7I

Explanation

To solve the problem, we need to calculate the resultant intensities at points A and B where two beams of light with intensities \(I\) and \(4I\) interfere, given their phase differences.

### Step-by-Step Solution:

1. **Understanding the Formula for Resultant Intensity**:
The resultant intensity \(I_r\) when two waves interfere is given by:
\[ I_r = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos \phi \]
where \(I_1\) and \(I_2\) are the intensities of the two beams, and \(\phi\) is the phase difference.

2. **Assigning Intensities**:
Let:
- \(I_1 = I\)
- \(I_2 = 4I\)

3. **Calculating Intensity at Point A**:
At point A, the phase difference \(\phi_A = \frac{\pi}{2}\).
Using the formula:
\[ I_A = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos \phi_A \]
Substitute the values:
\[ I_A = I + 4I + 2\sqrt{I \cdot 4I} \cos\left(\frac{\pi}{2}\right) \]
Since \(\cos\left(\frac{\pi}{2}\right) = 0\):
\[ I_A = I + 4I + 0 = 5I \]

4. **Calculating Intensity at Point B**:
At point B, the phase difference \(\phi_B = \pi\).
Using the formula:
\[ I_B = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos \phi_B \]
Substitute the values:
\[ I_B = I + 4I + 2\sqrt{I \cdot 4I} \cos(\pi) \]
Since \(\cos(\pi) = -1\):
\[ I_B = I + 4I - 2\sqrt{I \cdot 4I} \]
Calculate \(\sqrt{I \cdot 4I} = \sqrt{4I^2} = 2I\):
\[ I_B = 5I - 2(2I) = 5I - 4I = I \]

5. **Finding the Difference in Intensities**:
Now, we find the difference between the resultant intensities at points A and B:
\[ \Delta I = I_A - I_B = 5I - I = 4I \]

### Final Result:
The difference between the resultant intensities at points A and B is:
\[ \Delta I = 4I \]