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)/(3)`
B. `(beta)/(6)`
C. `(beta)/(2)`
D. `(beta)/(4)`

Explanation

To solve the problem, we will use the principles of interference in the Young's double slit experiment. Here are the steps to find the distance from the point of maximum intensity to the point where the intensity falls to half of the maximum intensity on either side.

### Step 1: Understand the fringe width
The fringe width (β) in a Young's double slit experiment 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 and intensity formula
The intensity \(I\) at a point on the screen in a double slit experiment can be expressed as:
\[ I = I_0 \cos^2\left(\frac{\pi d \sin \theta}{\lambda}\right) \]
where \(I_0\) is the maximum intensity, and \(\theta\) is the angle of the point from the central maximum.

### Step 3: Condition for half maximum intensity
To find the points where the intensity falls to half of the maximum intensity, we set:
\[ I = \frac{I_0}{2} \]
Thus,
\[ \frac{I_0}{2} = I_0 \cos^2\left(\frac{\pi d \sin \theta}{\lambda}\right) \]
This simplifies to:
\[ \cos^2\left(\frac{\pi d \sin \theta}{\lambda}\right) = \frac{1}{2} \]
Taking the square root gives:
\[ \cos\left(\frac{\pi d \sin \theta}{\lambda}\right) = \frac{1}{\sqrt{2}} \]

### Step 4: Finding the angle θ
The angle corresponding to this condition is:
\[ \frac{\pi d \sin \theta}{\lambda} = \frac{\pi}{4} + n\pi \quad (n \in \mathbb{Z}) \]
For the first order (n=0), we have:
\[ \sin \theta = \frac{\lambda}{4d} \]

### Step 5: Relating angle to distance on the screen
For small angles, \(\sin \theta \approx \tan \theta\), and the distance \(y\) from the central maximum to the point where the intensity falls to half is given by:
\[ y = D \tan \theta \approx D \sin \theta = D \cdot \frac{\lambda}{4d} \]

### Step 6: Final distance calculation
Thus, the distance from the point of maximum intensity to the point where the intensity falls to half of the maximum intensity on either side is:
\[ y = \frac{D \lambda}{4d} \]

### Summary
The distance from the point of maximum intensity to the point where the intensity falls to half of maximum intensity on either side is:
\[ \frac{D \lambda}{4d} \]