Two identical coherent sources produce a zero order bright fringe on a screen. If `beta` is the band width, the minimum distance between two points on either side of the bright where the intensity is half that of maximum intensity is half of maximum intensity is

Options

A. `beta//2`
B. `beta//4`
C. `beta//3`
D. `beta//6`

Explanation

To solve the problem step by step, we will analyze the situation of two coherent sources producing interference patterns and find the minimum distance between two points on either side of the central maximum where the intensity is half of the maximum intensity.

### Step 1: Understand the setup
We have two coherent sources producing a central maximum on a screen. The intensity at the central maximum is denoted as \( I_0 \).

**Hint:** Identify the central maximum and the conditions for half intensity.

### Step 2: Write the intensity formula
The intensity \( I \) at any point in the interference pattern can be expressed as:
\[ I = I_0 \cos^2\left(\frac{\phi}{2}\right) \]
where \( \phi \) is the phase difference.

**Hint:** Recall the relationship between intensity and phase difference in interference patterns.

### Step 3: Set up the equation for half intensity
We need to find the points where the intensity is half of the maximum intensity:
\[ \frac{I_0}{2} = I_0 \cos^2\left(\frac{\phi}{2}\right) \]
Dividing both sides by \( I_0 \) (assuming \( I_0 \neq 0 \)):
\[ \frac{1}{2} = \cos^2\left(\frac{\phi}{2}\right) \]

**Hint:** Simplify the equation to find the cosine value.

### Step 4: Solve for the phase difference
Taking the square root of both sides gives:
\[ \cos\left(\frac{\phi}{2}\right) = \frac{1}{\sqrt{2}} \]
This implies:
\[ \frac{\phi}{2} = \frac{\pi}{4} \quad \Rightarrow \quad \phi = \frac{\pi}{2} \]

**Hint:** Understand the relationship between phase difference and the angle in the interference pattern.

### Step 5: Relate phase difference to path difference
The phase difference \( \phi \) is related to the path difference \( \Delta x \) by the formula:
\[ \phi = \frac{2\pi}{\lambda} \Delta x \]
Substituting \( \phi = \frac{\pi}{2} \):
\[ \frac{\pi}{2} = \frac{2\pi}{\lambda} \Delta x \]
From this, we can solve for the path difference:
\[ \Delta x = \frac{\lambda}{4} \]

**Hint:** Remember the relationship between phase difference and path difference.

### Step 6: Find the distance between the two points
The distance from the central maximum to the point where the intensity is half is given by:
\[ y_n = n \cdot \frac{\lambda D}{d} \]
where \( n \) is the order of the fringe. For our case, we have:
\[ n = \frac{1}{4} \]
Thus:
\[ y_n = \frac{1}{4} \cdot \frac{\lambda D}{d} \]
We know that the fringe width \( \beta \) is given by:
\[ \beta = \frac{\lambda D}{d} \]
Substituting this into our equation gives:
\[ y_n = \frac{\beta}{4} \]

**Hint:** Relate the fringe width to the distance from the central maximum.

### Step 7: Calculate the total distance between the two points
Since we have two points on either side of the central maximum, the total distance \( D \) between these two points is:
\[ D = 2y_n = 2 \cdot \frac{\beta}{4} = \frac{\beta}{2} \]

**Hint:** Remember to consider both sides of the central maximum for the total distance.

### Final Answer
The minimum distance between two points on either side of the central maximum where the intensity is half of the maximum intensity is:
\[ \frac{\beta}{2} \]