In Young's double slit experiment two disturbances arriving at a point P have phase difference of `(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, we need to find the intensity of the light at point P in Young's double slit experiment when the phase difference between the two disturbances is \( \frac{\pi}{3} \).

### Step-by-Step Solution:

1. **Understand the Formula for Intensity**:
The intensity \( I \) at a point where two waves interfere can be expressed in terms of the maximum intensity \( I_0 \) and the phase difference \( \phi \) between the two waves. The formula is given by:
\[ I = I_0 \cos^2\left(\frac{\phi}{2}\right) \]

2. **Identify the Phase Difference**:
In this problem, the phase difference \( \phi \) is given as \( \frac{\pi}{3} \).

3. **Substitute the Phase Difference into the Formula**:
We substitute \( \phi = \frac{\pi}{3} \) into the intensity formula:
\[ I = I_0 \cos^2\left(\frac{\frac{\pi}{3}}{2}\right) \]

4. **Calculate \( \frac{\phi}{2} \)**:
Calculate \( \frac{\pi}{3} \div 2 = \frac{\pi}{6} \).

5. **Find the Cosine Value**:
Now, we need to find \( \cos\left(\frac{\pi}{6}\right) \):
\[ \cos\left(\frac{\pi}{6}\right) = \cos(30^\circ) = \frac{\sqrt{3}}{2} \]

6. **Substitute the Cosine Value Back into the Intensity Formula**:
Substitute \( \cos\left(\frac{\pi}{6}\right) \) into the intensity formula:
\[ I = I_0 \left(\frac{\sqrt{3}}{2}\right)^2 \]

7. **Calculate the Square**:
Now calculate the square:
\[ I = I_0 \cdot \frac{3}{4} \]

8. **Express as a Fraction of Maximum Intensity**:
The intensity \( I \) expressed as a fraction of the 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 \( \frac{3}{4} \).

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