In YDSE of equal width slits, if intensity at the center of screen is `I_(0)`, then intensity at a distance of `beta // 4` from the central maxima is

Options

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

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the given information
In Young's double slit experiment (YDSE), the intensity at the center of the screen is given as \( I_0 \). We need to find the intensity at a distance of \( \frac{\beta}{4} \) from the central maximum.

### Step 2: Define the parameters
In YDSE, \( \beta \) is the fringe width, which is given by the formula:
\[ \beta = \frac{\lambda D}{d} \]
where:
- \( \lambda \) is the wavelength of light used,
- \( D \) is the distance from the slits to the screen,
- \( d \) is the distance between the slits.

### Step 3: Calculate the phase difference at \( y = \frac{\beta}{4} \)
At a distance \( y = \frac{\beta}{4} \), we can express the path difference \( \Delta x \) as:
\[ \Delta x = d \sin \theta \approx d \tan \theta \approx \frac{d}{D} y \]
For \( y = \frac{\beta}{4} \):
\[ \Delta x = \frac{d}{D} \cdot \frac{\beta}{4} \]
Substituting the expression for \( \beta \):
\[ \Delta x = \frac{d}{D} \cdot \frac{\lambda D}{4d} = \frac{\lambda}{4} \]

### Step 4: Calculate the phase difference
The phase difference \( \Delta \phi \) corresponding to the path difference \( \Delta x \) is given by:
\[ \Delta \phi = \frac{2\pi}{\lambda} \Delta x = \frac{2\pi}{\lambda} \cdot \frac{\lambda}{4} = \frac{\pi}{2} \]

### Step 5: Calculate the intensity at \( y = \frac{\beta}{4} \)
The intensity \( I \) at a point in YDSE can be calculated using the formula:
\[ I = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos(\Delta \phi) \]
Given that the intensity at the center \( I_0 \) is the sum of the intensities from both slits:
\[ I_0 = I_1 + I_2 = 2I \quad \text{(where } I = \frac{I_0}{2} \text{ for each slit)} \]
At \( y = \frac{\beta}{4} \), we substitute \( \Delta \phi = \frac{\pi}{2} \):
\[ I = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos\left(\frac{\pi}{2}\right) \]
Since \( \cos\left(\frac{\pi}{2}\right) = 0 \):
\[ I = I_1 + I_2 = 2I = I_0 \]
But we need to consider the contributions from each slit:
\[ I_1 = I_2 = \frac{I_0}{4} \]
Thus:
\[ I = \frac{I_0}{4} + \frac{I_0}{4} + 0 = \frac{I_0}{2} \]

### Final Answer
The intensity at a distance of \( \frac{\beta}{4} \) from the central maximum is:
\[ \boxed{\frac{I_0}{2}} \]