The distance of `n^(th)` bright fringe to the `(n+1)^(th)` dark fringe in Young's experiment is equal to:

Options

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

Explanation

To solve the problem, we need to find the distance between the \( n^{th} \) bright fringe and the \( (n+1)^{th} \) dark fringe in Young's double-slit experiment.

### Step-by-Step Solution:

1. **Understanding Bright and Dark Fringes**:
- The position of the \( n^{th} \) bright fringe is given by the formula:
\[ y_b = \frac{n \lambda D}{d} \]
where \( y_b \) is the position of the \( n^{th} \) bright fringe, \( \lambda \) is the wavelength of light, \( D \) is the distance from the slits to the screen, and \( d \) is the distance between the slits.

2. **Position of Dark Fringes**:
- The position of the \( n^{th} \) dark fringe is given by:
\[ y_d = \frac{(2n - 1) \lambda D}{2d} \]
- For the \( (n+1)^{th} \) dark fringe, the position is:
\[ y_{d(n+1)} = \frac{(2(n+1) - 1) \lambda D}{2d} = \frac{(2n + 1) \lambda D}{2d} \]

3. **Distance Between Bright and Dark Fringe**:
- Now, we need to find the distance between the \( n^{th} \) bright fringe and the \( (n+1)^{th} \) dark fringe:
\[ \Delta y = y_{d(n+1)} - y_b \]
- Substituting the expressions for \( y_{d(n+1)} \) and \( y_b \):
\[ \Delta y = \frac{(2n + 1) \lambda D}{2d} - \frac{n \lambda D}{d} \]

4. **Simplifying the Expression**:
- To simplify, we can find a common denominator:
\[ \Delta y = \frac{(2n + 1) \lambda D}{2d} - \frac{2n \lambda D}{2d} \]
- This simplifies to:
\[ \Delta y = \frac{(2n + 1 - 2n) \lambda D}{2d} = \frac{\lambda D}{2d} \]

5. **Final Result**:
- Therefore, the distance between the \( n^{th} \) bright fringe and the \( (n+1)^{th} \) dark fringe is:
\[ \Delta y = \frac{\lambda D}{2d} \]

### Final Answer:
The distance of the \( n^{th} \) bright fringe to the \( (n+1)^{th} \) dark fringe in Young's experiment is equal to \( \frac{\lambda D}{2d} \).