In a Young's double slit experiment, `I_0` is the intensity at the central maximum and `beta` is the fringe width. The intensity at a point P distant x from the centre will be

Options

A. `I_(0) cos (pix)/beta`
B. `4I_(0)cos^(2)(pix)/beta`
C. `I_(0)cos^(2)(pix)/beta`
D. `I_(0)/4 cos^(2)(pix)/beta`

Answer

To find the intensity at a point P distant \( x \) from the center in a Young's double slit experiment, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Setup**: In a Young's double slit experiment, the intensity at the central maximum is denoted as \( I_0 \). The fringe width is denoted as \( \beta \).

2. **Relate Position to Fringe Width**: The distance \( y \) from the central maximum to a point P can be expressed in terms of the fringe width \( \beta \) and the phase difference \( \delta \). The relationship can be given as:
\[ y = \frac{d}{D} \cdot \delta x \]
where \( d \) is the distance between the slits, \( D \) is the distance from the slits to the screen, and \( \delta x \) is the path difference.

3. **Path Difference and Phase Difference**: The path difference \( \delta \) is related to the phase difference \( \Delta \) by the equation:
\[ \delta = \frac{2\pi}{\lambda} \cdot \text{(path difference)} \]
where \( \lambda \) is the wavelength of the light used.

4. **Substituting for \( \delta \)**: Since the path difference at point P can be expressed in terms of \( x \) (the distance from the center), we can write:
\[ \delta = \frac{2\pi x}{\beta} \]

5. **Intensity at Point P**: The intensity \( I \) at point P can be expressed in terms of the phase difference as:
\[ I = I_0 \cos^2\left(\frac{\delta}{2}\right) \]
Substituting for \( \delta \):
\[ I = I_0 \cos^2\left(\frac{2\pi x}{2\beta}\right) = I_0 \cos^2\left(\frac{\pi x}{\beta}\right) \]

6. **Final Expression**: Therefore, the intensity at point P distant \( x \) from the center is given by:
\[ I = I_0 \cos^2\left(\frac{\pi x}{\beta}\right) \]

### Final Answer:
\[ I = I_0 \cos^2\left(\frac{\pi x}{\beta}\right) \]