In a double slit experiment , the coherent sources are spaced 2d apart and the screen is placed a distance D from the slits. If `n^("th")` bright fringe is formed on the screen exactly opposite to a slit , the value of n must be -

Options

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

Explanation

To solve the problem, we need to analyze the double slit experiment setup and derive the condition for the nth bright fringe to be formed directly opposite to one of the slits.

### Step-by-Step Solution:

1. **Understanding the Setup**:
- We have two coherent sources (slits) separated by a distance of \(2d\).
- The screen is placed at a distance \(D\) from the slits.
- We need to find the value of \(n\) such that the nth bright fringe appears directly opposite to one of the slits (let's say slit S1).

2. **Condition for Bright Fringes**:
- The condition for bright fringes in a double slit experiment is given by:
\[ y_n = \frac{n \lambda D}{d} \]
- Here, \(y_n\) is the position of the nth bright fringe on the screen, \(\lambda\) is the wavelength of the light used, and \(d\) is the distance between the two slits.

3. **Position of the nth Bright Fringe**:
- Since the distance between the slits is \(2d\), we can modify our equation for the nth bright fringe:
\[ y_n = \frac{n \lambda D}{2d} \]

4. **Finding the Position of the nth Bright Fringe**:
- The nth bright fringe is formed directly opposite to slit S1, which means that the path difference between the light coming from S1 and S2 at this point must be an integer multiple of the wavelength:
\[ \text{Path difference} = n \lambda \]

5. **Calculating the Path Difference**:
- The path difference can also be expressed in terms of the geometry of the setup. Since the distance to the screen is \(D\) and the distance between the slits is \(2d\), the path difference at the position of the nth fringe can be approximated as:
\[ \text{Path difference} = \frac{2d \cdot y_n}{D} \]
- Setting this equal to \(n \lambda\):
\[ \frac{2d \cdot y_n}{D} = n \lambda \]

6. **Substituting for \(y_n\)**:
- From our earlier equation for \(y_n\):
\[ \frac{2d \cdot \left(\frac{n \lambda D}{2d}\right)}{D} = n \lambda \]
- This confirms the condition for the nth bright fringe.

7. **Finding the Value of \(n\)**:
- Rearranging gives:
\[ n = \frac{2D}{\lambda} \]
- Thus, the value of \(n\) must be an integer.

### Final Answer:
The value of \(n\) must be \(n = \frac{2D}{\lambda}\).