In YDSE, having slits of equal width, let `beta` be the fringe width and `I_(0)` be the maximum intensity. At a distance x from the central brigth fringe, the intensity will be

Options

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

Explanation

To find the intensity at a distance \( x \) from the central bright fringe in Young's Double Slit Experiment (YDSE), we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Setup**: In YDSE, we have two slits (S1 and S2) that produce interference patterns on a screen. The distance from the central bright fringe (the point of maximum intensity) to a point \( P \) on the screen is \( x \).

2. **Fringe Width and Maximum Intensity**: Let \( \beta \) be the fringe width and \( I_0 \) be the maximum intensity at the central bright fringe.

3. **Phase Difference Calculation**: The intensity at point \( P \) depends on the phase difference \( \phi \) between the two waves arriving from the slits S1 and S2. The phase difference can be calculated from the path difference \( \Delta x \).

4. **Finding Path Difference**: The path difference \( \Delta x \) can be expressed in terms of the angle \( \theta \):
\[ \Delta x = d \sin \theta \]
For small angles, \( \sin \theta \approx \tan \theta \approx \theta \), where \( d \) is the distance between the slits.

5. **Using Small Angle Approximation**: From the geometry of the setup, we have:
\[ \tan \theta = \frac{x}{D} \]
where \( D \) is the distance from the slits to the screen. Therefore, \( \theta \approx \frac{x}{D} \).

6. **Substituting for Path Difference**: Now substituting for \( \sin \theta \):
\[ \Delta x = d \cdot \frac{x}{D} \]

7. **Finding Phase Difference**: The phase difference \( \phi \) is given by:
\[ \phi = \frac{2\pi}{\lambda} \Delta x \]
Substituting for \( \Delta x \):
\[ \phi = \frac{2\pi}{\lambda} \cdot \frac{d \cdot x}{D} \]

8. **Relating to Fringe Width**: The fringe width \( \beta \) is defined as:
\[ \beta = \frac{d \lambda}{D} \]
Hence, we can express \( \frac{d}{D} \) as \( \frac{\beta}{\lambda} \). Substituting this back into the phase difference:
\[ \phi = \frac{2\pi x}{\beta} \]

9. **Intensity Formula**: The intensity at point \( P \) can be calculated using the formula:
\[ I = I_0 \cos^2\left(\frac{\phi}{2}\right) \]
Substituting for \( \phi \):
\[ I = I_0 \cos^2\left(\frac{2\pi x}{2\beta}\right) = I_0 \cos^2\left(\frac{\pi x}{\beta}\right) \]

10. **Final Result**: Therefore, the intensity at a distance \( x \) from the central bright fringe is:
\[ I = I_0 \cos^2\left(\frac{\pi x}{\beta}\right) \]