Two coherent beams of wavelength `5000 Å` reaching point would individually produce in intensities 1.44 and 4.00 units . If they reach there together, the intensity is 10.24 units . Calculate the lowest phase difference with which the beams reach that point.

Options

A. zero
B. `pi/4`
C. `pi/2`
D. `pi`

Answer

To solve the problem, we need to find the lowest phase difference between two coherent beams of light that produce given intensities. Let's break down the solution step by step.

### Step 1: Understand the given data
We are given:
- Intensity of beam 1, \( I_1 = 1.44 \) units
- Intensity of beam 2, \( I_2 = 4.00 \) units
- Combined intensity when both beams are together, \( I = 10.24 \) units

### Step 2: Use the formula for combined intensity
The formula for the combined intensity of two coherent beams is given by:
\[ I = I_1 + I_2 + 2 \sqrt{I_1 I_2} \cos \phi \]
where \( \phi \) is the phase difference.

### Step 3: Substitute the known values into the formula
Substituting the values into the formula:
\[ 10.24 = 1.44 + 4.00 + 2 \sqrt{1.44 \times 4.00} \cos \phi \]

### Step 4: Calculate \( \sqrt{I_1 I_2} \)
First, we calculate \( \sqrt{1.44 \times 4.00} \):
\[ \sqrt{1.44 \times 4.00} = \sqrt{5.76} = 2.4 \]

### Step 5: Substitute back into the equation
Now, substitute \( \sqrt{1.44 \times 4.00} = 2.4 \) back into the equation:
\[ 10.24 = 1.44 + 4.00 + 2 \times 2.4 \cos \phi \]
This simplifies to:
\[ 10.24 = 5.44 + 4.8 \cos \phi \]

### Step 6: Rearrange to isolate \( \cos \phi \)
Rearranging gives:
\[ 10.24 - 5.44 = 4.8 \cos \phi \]
\[ 4.80 = 4.8 \cos \phi \]

### Step 7: Solve for \( \cos \phi \)
Dividing both sides by 4.8:
\[ \cos \phi = 1 \]

### Step 8: Find the phase difference \( \phi \)
The cosine of 1 corresponds to a phase difference of:
\[ \phi = \cos^{-1}(1) = 0 \text{ degrees} \]

### Conclusion
Thus, the lowest phase difference with which the beams reach that point is:
\[ \phi = 0 \text{ degrees} \]