When light from two sources (say slits `S_(1)` and `S_(2)`) interfere, they form alternate dark and bright fringes. Bright fringe is formed at all point where the path difference is an odd multiple of half wavelength. At the condition of equal amplitudes, `A_(1) = A_(2) = a`, the maximum intensity will be `4 a^(2)` and the visibility improves, The resultant intensity can also be indicated with phase factor as `I = 2 a^(2) cos^(2) (phi // 2)`.Using this passage, answer the following questions.
At point having a path difference of `(lambda)/(4)`, the intensity

Options

A. 0
B. `a^(2)`
C. `2 a^(2)`
D. `a^(2) // 2`

Explanation

To solve the problem, we need to find the intensity at a point where the path difference is \( \frac{\lambda}{4} \).

### Step-by-Step Solution:

1. **Understanding Path Difference and Phase Difference**:
The path difference \( \Delta x \) is related to the phase difference \( \Delta \phi \) by the formula:
\[ \Delta x = \frac{\Delta \phi}{2\pi} \lambda \]
Given that \( \Delta x = \frac{\lambda}{4} \), we can substitute this into the equation.

2. **Substituting Path Difference**:
Substituting \( \Delta x = \frac{\lambda}{4} \) into the equation:
\[ \frac{\lambda}{4} = \frac{\Delta \phi}{2\pi} \lambda \]
We can cancel \( \lambda \) from both sides (assuming \( \lambda \neq 0 \)):
\[ \frac{1}{4} = \frac{\Delta \phi}{2\pi} \]

3. **Solving for Phase Difference**:
Rearranging the equation to find \( \Delta \phi \):
\[ \Delta \phi = \frac{2\pi}{4} = \frac{\pi}{2} \]

4. **Using the Resultant Intensity Formula**:
The resultant intensity \( I \) can be expressed as:
\[ I = 2a^2 \cos^2\left(\frac{\Delta \phi}{2}\right) \]
Now, substituting \( \Delta \phi = \frac{\pi}{2} \):
\[ I = 2a^2 \cos^2\left(\frac{\pi}{2}/2\right) = 2a^2 \cos^2\left(\frac{\pi}{4}\right) \]

5. **Calculating \( \cos^2\left(\frac{\pi}{4}\right) \)**:
We know that:
\[ \cos\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} \]
Therefore:
\[ \cos^2\left(\frac{\pi}{4}\right) = \left(\frac{1}{\sqrt{2}}\right)^2 = \frac{1}{2} \]

6. **Final Calculation of Intensity**:
Now substituting back into the intensity formula:
\[ I = 2a^2 \cdot \frac{1}{2} = a^2 \]

### Conclusion:
The intensity at the point having a path difference of \( \frac{\lambda}{4} \) is:
\[ \boxed{a^2} \]