A beam of light consists of two wavelengths, `6300Å and 5600Å`. This beam of light is used to obtain an interference pattern in YDSE. If `4^("th")` bright fringe of `6300Å` coincides with the `n^("th")` dark fringe of `5600Å` from the central line, then find the value of n.

Answer

To solve the problem, we need to find the value of \( n \) such that the 4th bright fringe of the wavelength \( 6300 \, \text{Å} \) coincides with the \( n \)-th dark fringe of the wavelength \( 5600 \, \text{Å} \) in a Young's Double Slit Experiment (YDSE).

### Step-by-Step Solution:

1. **Understanding Fringe Positions**:
- The position of the \( m \)-th bright fringe for a wavelength \( \lambda \) is given by:
\[ y_b = \frac{m \lambda D}{d} \]
- The position of the \( n \)-th dark fringe for a wavelength \( \lambda \) is given by:
\[ y_d = \frac{(n + \frac{1}{2}) \lambda D}{d} \]

2. **Setting Up the Equation**:
- According to the problem, the 4th bright fringe of \( 6300 \, \text{Å} \) coincides with the \( n \)-th dark fringe of \( 5600 \, \text{Å} \).
- Therefore, we can set the two equations equal to each other:
\[ \frac{4 \cdot 6300 \cdot D}{d} = \frac{(n + \frac{1}{2}) \cdot 5600 \cdot D}{d} \]

3. **Canceling Common Terms**:
- Since \( D \) and \( d \) are common in both terms, we can cancel them out:
\[ 4 \cdot 6300 = (n + \frac{1}{2}) \cdot 5600 \]

4. **Calculating the Left Side**:
- Calculate \( 4 \cdot 6300 \):
\[ 4 \cdot 6300 = 25200 \]

5. **Setting Up the Equation**:
- Now we have:
\[ 25200 = (n + \frac{1}{2}) \cdot 5600 \]

6. **Solving for \( n \)**:
- Divide both sides by \( 5600 \):
\[ \frac{25200}{5600} = n + \frac{1}{2} \]
- Calculate \( \frac{25200}{5600} \):
\[ \frac{25200}{5600} = 4.5 \]
- Therefore:
\[ 4.5 = n + \frac{1}{2} \]
- Subtract \( \frac{1}{2} \) from both sides:
\[ n = 4.5 - 0.5 = 4 \]

### Final Answer:
Thus, the value of \( n \) is \( 4 \).