In Young's double slit experiment, the constant phase difference between two source is `(pi)/(2)` . The intensity at a point equidistant from theslits in terms of maximum intensity `I_(0)` is

Options

A. `I_(0)`
B. `I_(0)//2`
C. `3I_(0)//4`
D. `3I_(0)`

Explanation

To solve the problem, we need to determine the intensity at a point equidistant from the two slits in Young's double slit experiment, given that the constant phase difference between the two sources is \(\frac{\pi}{2}\).

### Step-by-Step Solution:

1. **Understanding the Setup**:
In Young's double slit experiment, two coherent light sources (slits) emit light waves that interfere with each other. The intensity of the resultant wave at any point depends on the phase difference between the waves arriving at that point.

2. **Identify the Phase Difference**:
We are given that the constant phase difference between the two sources is \(\theta = \frac{\pi}{2}\).

3. **Determine Path Difference**:
Since we are considering a point that is equidistant from both slits, the path difference (\(\Delta x\)) at this point is zero. Thus, the phase difference due to the path difference is also zero.

4. **Total Phase Difference**:
The total phase difference at the point of interest is the sum of the phase difference due to path difference and the constant phase difference:
\[ \text{Total Phase Difference} = \text{Phase Difference due to Path Difference} + \text{Constant Phase Difference} = 0 + \frac{\pi}{2} = \frac{\pi}{2} \]

5. **Intensity Formula**:
The intensity \(I\) at any point in terms of the maximum intensity \(I_0\) is given by:
\[ I = I_0 \cos^2\left(\frac{\theta}{2}\right) \]
where \(\theta\) is the total phase difference.

6. **Substituting the Phase Difference**:
Now, substituting \(\theta = \frac{\pi}{2}\) into the intensity formula:
\[ I = I_0 \cos^2\left(\frac{\frac{\pi}{2}}{2}\right) = I_0 \cos^2\left(\frac{\pi}{4}\right) \]

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

8. **Final Intensity Calculation**:
Substituting this back into the intensity equation:
\[ I = I_0 \cdot \frac{1}{2} = \frac{I_0}{2} \]

### Conclusion:
The intensity at the point equidistant from the slits, given a constant phase difference of \(\frac{\pi}{2}\), is:
\[ \boxed{\frac{I_0}{2}} \]