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 will use the formula for the intensity of two coherent beams interfering with each other. The formula is given by:

\[ I = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos \phi \]

Where:
- \( I \) is the resultant intensity when both beams are present.
- \( I_1 \) and \( I_2 \) are the intensities of the individual beams.
- \( \phi \) is the phase difference between the two beams.

### Step-by-step Solution:

1. **Identify the given values:**
- \( I_1 = 1.44 \) units
- \( I_2 = 4.00 \) units
- \( I = 10.24 \) units

2. **Substitute the known values into the intensity formula:**
\[ 10.24 = 1.44 + 4.00 + 2\sqrt{1.44 \times 4.00} \cos \phi \]

3. **Calculate the sum of \( I_1 \) and \( I_2 \):**
\[ 1.44 + 4.00 = 5.44 \]

4. **Rearrange the equation:**
\[ 10.24 - 5.44 = 2\sqrt{1.44 \times 4.00} \cos \phi \]
\[ 4.80 = 2\sqrt{1.44 \times 4.00} \cos \phi \]

5. **Calculate \( \sqrt{1.44 \times 4.00} \):**
\[ \sqrt{1.44 \times 4.00} = \sqrt{5.76} = 2.4 \]

6. **Substitute back into the equation:**
\[ 4.80 = 2 \times 2.4 \cos \phi \]
\[ 4.80 = 4.8 \cos \phi \]

7. **Solve for \( \cos \phi \):**
\[ \cos \phi = \frac{4.80}{4.8} = 1 \]

8. **Determine the phase difference \( \phi \):**
Since \( \cos \phi = 1 \), the possible values for \( \phi \) are:
\[ \phi = 0, 2\pi, 4\pi, \ldots \]
The lowest phase difference is:
\[ \phi = 0 \]

### Final Answer:
The lowest phase difference with which the beams reach that point is \( \phi = 0 \).