In Young's double slit experiment two disturbances arriving at a point P have phase difference fo `(pi)/(3)`. The intensity of this point expressed as a fraction of maximum intensity `I_(0)` is

Options

A. `(3)/(2)I_(0)`
B. `(1)/(2)I_(0)`
C. `(4)/(3)I_(0)`
D. `(3)/(4)I_(0)`

Explanation

To solve the problem of finding the intensity at point P in Young's double slit experiment with a phase difference of \( \frac{\pi}{3} \), we can follow these steps:

### Step-by-Step Solution:

1. **Understand the relationship between intensity and phase difference**:
The intensity \( I \) at a point in Young's double slit experiment can be expressed in terms of the maximum intensity \( I_0 \) and the phase difference \( \phi \) between the two waves arriving at that point. The formula is given by:
\[ I = I_0 \cos^2\left(\frac{\phi}{2}\right) \]

2. **Identify the given phase difference**:
We are given that the phase difference \( \phi \) is \( \frac{\pi}{3} \).

3. **Substitute the phase difference into the formula**:
Substitute \( \phi = \frac{\pi}{3} \) into the intensity formula:
\[ I = I_0 \cos^2\left(\frac{\frac{\pi}{3}}{2}\right) = I_0 \cos^2\left(\frac{\pi}{6}\right) \]

4. **Calculate \( \cos\left(\frac{\pi}{6}\right) \)**:
The value of \( \cos\left(\frac{\pi}{6}\right) \) is known to be \( \frac{\sqrt{3}}{2} \).

5. **Substitute this value into the intensity formula**:
Now we substitute \( \cos\left(\frac{\pi}{6}\right) \) back into the equation:
\[ I = I_0 \left(\frac{\sqrt{3}}{2}\right)^2 \]

6. **Calculate the square of \( \frac{\sqrt{3}}{2} \)**:
\[ \left(\frac{\sqrt{3}}{2}\right)^2 = \frac{3}{4} \]

7. **Final expression for intensity**:
Therefore, substituting this back gives:
\[ I = I_0 \cdot \frac{3}{4} \]

8. **Express the intensity as a fraction of maximum intensity**:
The intensity at point P expressed as a fraction of maximum intensity \( I_0 \) is:
\[ \frac{I}{I_0} = \frac{3}{4} \]

### Final Answer:
The intensity at point P expressed as a fraction of maximum intensity \( I_0 \) is \( \frac{3}{4} \).