In a Young's double slit experiment with light of wavelength `lamda` the separation of slits is d and distance of screen is D such that `D gt gt d gt gt lamda`. If the fringe width is `bea`, the distance from point of maximum intensity to the point where intensity falls to half of maximum intensity on either side is

Options

A. `beta/4`
B. `beta/3`
C. `beta/3`
D. `beta/2`

Answer

To solve the problem regarding the Young's double slit experiment, we need to determine the distance from the point of maximum intensity to the point where the intensity falls to half of the maximum intensity on either side. Let's break this down step by step.

### Step 1: Understanding Fringe Width
In a Young's double slit experiment, the fringe width (β) is given by the formula:
\[ \beta = \frac{\lambda D}{d} \]
where:
- \( \lambda \) is the wavelength of light,
- \( D \) is the distance from the slits to the screen,
- \( d \) is the separation between the slits.

### Step 2: Maximum Intensity
The points of maximum intensity (bright fringes) occur at positions given by:
\[ y_m = \frac{m \lambda D}{d} \]
where \( m \) is the order of the maximum (0, 1, 2,...).

### Step 3: Intensity Formula
The intensity \( I \) at a point on the screen can be expressed as:
\[ I = I_0 \cos^2\left(\frac{\pi d y}{\lambda D}\right) \]
where \( I_0 \) is the maximum intensity.

### Step 4: Finding Half Maximum Intensity
To find the position where the intensity falls to half of the maximum intensity, we set:
\[ I = \frac{I_0}{2} \]
Thus, we have:
\[ \frac{I_0}{2} = I_0 \cos^2\left(\frac{\pi d y}{\lambda D}\right) \]
This simplifies to:
\[ \cos^2\left(\frac{\pi d y}{\lambda D}\right) = \frac{1}{2} \]
Taking the square root gives:
\[ \cos\left(\frac{\pi d y}{\lambda D}\right) = \frac{1}{\sqrt{2}} \]

### Step 5: Finding the Angle
The angle corresponding to \( \cos^{-1}\left(\frac{1}{\sqrt{2}}\right) \) is:
\[ \frac{\pi d y}{\lambda D} = \frac{\pi}{4} \]
Solving for \( y \):
\[ y = \frac{\lambda D}{d} \cdot \frac{1}{4} = \frac{\beta}{4} \]

### Step 6: Distance from Maximum to Half Maximum
Since this distance \( y \) is on either side of the maximum, the total distance from the point of maximum intensity to the point where the intensity falls to half of maximum intensity on either side is:
\[ \text{Total distance} = 2y = 2 \cdot \frac{\beta}{4} = \frac{\beta}{2} \]

### Final Answer
Thus, the distance from the point of maximum intensity to the point where intensity falls to half of maximum intensity on either side is:
\[ \frac{\beta}{2} \]