The two light beams having intensities 9I and I interfere to produce a fringe pattern on a screen. The phase difference between the beams is `pi`/2 at point P and `pi`at point Q. Then the difference between the resultant intensities at P and Q will be :

Options

A. 2 I
B. 6 I
C. 5 I
D. 7 I

Explanation

To solve the problem, we need to determine the resultant intensities at points P and Q where two light beams with intensities \(9I\) and \(I\) interfere, and then find the difference between these intensities.

### Step-by-Step Solution:

1. **Identify the Intensities**:
- Let \(I_1 = 9I\) (intensity of the first beam)
- Let \(I_2 = I\) (intensity of the second beam)

2. **Resultant Intensity at Point P**:
- The phase difference at point P is \(\frac{\pi}{2}\).
- The formula for resultant intensity \(I_P\) when two beams interfere is:
\[ I_P = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos(\phi) \]
- Substituting the values:
\[ I_P = 9I + I + 2\sqrt{9I \cdot I} \cos\left(\frac{\pi}{2}\right) \]
- Since \(\cos\left(\frac{\pi}{2}\right) = 0\):
\[ I_P = 9I + I + 0 = 10I \]

3. **Resultant Intensity at Point Q**:
- The phase difference at point Q is \(\pi\).
- Using the same formula for resultant intensity \(I_Q\):
\[ I_Q = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos(\pi) \]
- Substituting the values:
\[ I_Q = 9I + I + 2\sqrt{9I \cdot I} \cos(\pi) \]
- Since \(\cos(\pi) = -1\):
\[ I_Q = 9I + I - 2\sqrt{9I^2} \]
- Simplifying further:
\[ I_Q = 10I - 2 \cdot 3I = 10I - 6I = 4I \]

4. **Calculate the Difference Between Intensities**:
- The difference between the resultant intensities at points P and Q is:
\[ I_P - I_Q = 10I - 4I = 6I \]

### Final Answer:
The difference between the resultant intensities at points P and Q is \(6I\).