Interference fringes are produced on a screen by using two light sources of intensities / and 9/. The phase difference between the beams `pi/2` is at point P and `pi`at point Q on the screen. The difference between the resultant intensities at point P and Q is

Options

A. (a)2I
B. (b)4I
C. (c)6I
D. (d)8I

Answer

To solve the problem of finding the difference between the resultant intensities at points P and Q, we can follow these steps:

### Step 1: Identify the given values
We have two light sources with intensities:
- \( I_1 = I \)
- \( I_2 = 9I \)

The phase differences at points P and Q are:
- At point P: \( \phi_P = \frac{\pi}{2} \)
- At point Q: \( \phi_Q = \pi \)

### Step 2: Calculate the resultant intensity at point P
The formula for the resultant intensity \( I_P \) when two waves interfere is given by:
\[ I_P = I_1 + I_2 + 2 \sqrt{I_1 I_2} \cos(\phi_P) \]

Substituting the values:
\[ I_P = I + 9I + 2 \sqrt{I \cdot 9I} \cos\left(\frac{\pi}{2}\right) \]

Since \( \cos\left(\frac{\pi}{2}\right) = 0 \):
\[ I_P = I + 9I + 0 = 10I \]

### Step 3: Calculate the resultant intensity at point Q
Using the same formula for point Q:
\[ I_Q = I_1 + I_2 + 2 \sqrt{I_1 I_2} \cos(\phi_Q) \]

Substituting the values:
\[ I_Q = I + 9I + 2 \sqrt{I \cdot 9I} \cos(\pi) \]

Since \( \cos(\pi) = -1 \):
\[ I_Q = I + 9I + 2 \sqrt{I \cdot 9I} (-1) \]
\[ = 10I - 2 \sqrt{9I^2} \]
\[ = 10I - 6I = 4I \]

### Step 4: Find the difference between the resultant intensities
Now, we calculate the difference between the intensities at points P and Q:
\[ \Delta I = I_P - I_Q = 10I - 4I = 6I \]

### Final Answer
The difference between the resultant intensities at point P and Q is:
\[ \Delta I = 6I \]