Two beams of ligth having intensities I and 4I interface 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 the resultant intensities at A and B is

Options

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

Explanation

To solve the problem of finding the difference between the resultant intensities at points A and B, we will follow these steps:

### Step 1: Identify the Intensities
We have two beams of light with intensities:
- \( I_1 = I \)
- \( I_2 = 4I \)

### Step 2: Calculate Resultant Intensity at Point A
The formula for resultant intensity \( I_A \) when two waves interfere is given by:
\[ I_A = I_1 + I_2 + 2 \sqrt{I_1 I_2} \cos(\phi_A) \]
Where \( \phi_A \) is the phase difference at point A, which is \( \frac{\pi}{2} \).

Substituting the values:
\[ I_A = I + 4I + 2 \sqrt{I \cdot 4I} \cos\left(\frac{\pi}{2}\right) \]

Now, calculate \( \cos\left(\frac{\pi}{2}\right) \):
\[ \cos\left(\frac{\pi}{2}\right) = 0 \]

Thus, the equation simplifies to:
\[ I_A = I + 4I + 0 = 5I \]

### Step 3: Calculate Resultant Intensity at Point B
Now, we calculate the resultant intensity \( I_B \) at point B where the phase difference \( \phi_B = \pi \):
\[ I_B = I_1 + I_2 + 2 \sqrt{I_1 I_2} \cos(\phi_B) \]
Substituting the values:
\[ I_B = I + 4I + 2 \sqrt{I \cdot 4I} \cos(\pi) \]

Calculate \( \cos(\pi) \):
\[ \cos(\pi) = -1 \]

Now substituting this into the equation:
\[ I_B = I + 4I + 2 \sqrt{I \cdot 4I} \cdot (-1) \]
Calculating \( \sqrt{I \cdot 4I} = \sqrt{4I^2} = 2I \):
\[ I_B = 5I - 2 \cdot 2I = 5I - 4I = I \]

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

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