In a double-slit experiment, the slits are separated by a distance d and the screen is at a distance D from the slits. If a maximum is formed just opposite to each slit, then what is the order or the fringe so formed?

Options

A. `(d^(2))/(2 lambda D)`
B. `(2 d^(2))/( lambda D)`
C. `(d^(2))/( lambda D)`
D. `(d^(2))/(4 lambda D)`

Explanation

To solve the problem, we need to find the order of the fringe formed in a double-slit experiment when a maximum is formed just opposite to each slit. Here’s a step-by-step solution:

### Step 1: Understand the setup
In a double-slit experiment, two slits are separated by a distance \( d \), and a screen is placed at a distance \( D \) from the slits. The light waves from each slit interfere to produce a pattern of bright and dark fringes on the screen.

### Step 2: Identify the position of the maxima
The position of the maxima (bright fringes) on the screen can be described by the formula:
\[ x = \frac{n \lambda D}{d} \]
where:
- \( x \) is the distance from the central maximum to the nth maximum,
- \( n \) is the order of the fringe,
- \( \lambda \) is the wavelength of the light used,
- \( D \) is the distance from the slits to the screen,
- \( d \) is the distance between the slits.

### Step 3: Determine the position of the maxima in this scenario
Since a maximum is formed just opposite to each slit, the distance \( x \) from the central maximum to the position of the maximum directly in front of one slit is:
\[ x = \frac{d}{2} \]

### Step 4: Set up the equation
Now, we can set the two expressions for \( x \) equal to each other:
\[ \frac{d}{2} = \frac{n \lambda D}{d} \]

### Step 5: Solve for \( n \)
To find \( n \), we can rearrange the equation:
\[ n = \frac{d^2}{2 \lambda D} \]

### Step 6: Conclusion
Thus, the order of the fringe formed when a maximum is located just opposite to each slit is given by:
\[ n = \frac{d^2}{2 \lambda D} \]

### Final Answer
The order of the fringe is \( n = \frac{d^2}{2 \lambda D} \).

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