Two beams 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 the resultant intensities at A and B is

Options

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

Explanation

To solve the problem of finding the difference between the resultant intensities at points A and B for two interfering beams of light, we can follow these steps:

### Step 1: Identify the Intensities and Phase Differences
We have two beams of light:
- Intensity of the first beam, \( I_1 = I \)
- Intensity of the second beam, \( I_2 = 4I \)

The phase differences at the points are:
- At point A, \( \phi_A = \frac{\pi}{2} \)
- At point B, \( \phi_B = \pi \)

### Step 2: Calculate Resultant Intensity at Point A
The formula for the resultant intensity \( I_R \) when two beams interfere is given by:
\[ I_R = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos(\phi) \]

For point A:
\[ I_{R_A} = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos\left(\frac{\pi}{2}\right) \]
Substituting the values:
\[ I_{R_A} = I + 4I + 2\sqrt{I \cdot 4I} \cdot \cos\left(\frac{\pi}{2}\right) \]
Since \( \cos\left(\frac{\pi}{2}\right) = 0 \):
\[ I_{R_A} = I + 4I + 0 = 5I \]

### Step 3: Calculate Resultant Intensity at Point B
Now, for point B:
\[ I_{R_B} = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos(\pi) \]
Substituting the values:
\[ I_{R_B} = I + 4I + 2\sqrt{I \cdot 4I} \cdot \cos(\pi) \]
Since \( \cos(\pi) = -1 \):
\[ I_{R_B} = I + 4I + 2\sqrt{I \cdot 4I} \cdot (-1) \]
Calculating \( \sqrt{I \cdot 4I} = \sqrt{4I^2} = 2I \):
\[ I_{R_B} = I + 4I - 2(2I) = I + 4I - 4I = I \]

### Step 4: Find the Difference in Resultant Intensities
Now, we find the difference between the resultant intensities at points A and B:
\[ \Delta I = I_{R_A} - I_{R_B} = 5I - I = 4I \]

### Final Answer
The difference between the resultant intensities at points A and B is \( 4I \).

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