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 analyze the interference pattern created by two wavelengths of light in Young's Double Slit Experiment (YDSE). We are given two wavelengths:

- \( \lambda_1 = 6300 \, \text{Å} \) (or \( 6300 \times 10^{-10} \, \text{m} \))
- \( \lambda_2 = 5600 \, \text{Å} \) (or \( 5600 \times 10^{-10} \, \text{m} \))

We know that the \( m^{th} \) bright fringe position for a wavelength \( \lambda \) is given by:
\[ y_m = \frac{m \lambda D}{d} \]
where \( D \) is the distance from the slits to the screen and \( d \) is the distance between the slits.

For dark fringes, the position is given by:
\[ y_n = \frac{(n + \frac{1}{2}) \lambda D}{d} \]
where \( n \) is the order of the dark fringe.

According to the problem, the \( 4^{th} \) bright fringe of \( 6300 \, \text{Å} \) coincides with the \( n^{th} \) dark fringe of \( 5600 \, \text{Å} \).

### Step 1: Write the equation for the \( 4^{th} \) bright fringe of \( 6300 \, \text{Å} \)
Using the formula for bright fringes:
\[ y_4 = \frac{4 \cdot 6300 \times 10^{-10} \cdot D}{d} \]

### Step 2: Write the equation for the \( n^{th} \) dark fringe of \( 5600 \, \text{Å} \)
Using the formula for dark fringes:
\[ y_n = \frac{(n + \frac{1}{2}) \cdot 5600 \times 10^{-10} \cdot D}{d} \]

### Step 3: Set the two equations equal to each other
Since the two positions coincide:
\[ \frac{4 \cdot 6300 \times 10^{-10} \cdot D}{d} = \frac{(n + \frac{1}{2}) \cdot 5600 \times 10^{-10} \cdot D}{d} \]

### Step 4: Cancel \( D \) and \( d \) from both sides
\[ 4 \cdot 6300 \times 10^{-10} = (n + \frac{1}{2}) \cdot 5600 \times 10^{-10} \]

### Step 5: Simplify the equation
Cancelling \( 10^{-10} \) from both sides:
\[ 4 \cdot 6300 = (n + \frac{1}{2}) \cdot 5600 \]

### Step 6: Calculate \( 4 \cdot 6300 \)
\[ 25200 = (n + \frac{1}{2}) \cdot 5600 \]

### Step 7: Divide both sides by \( 5600 \)
\[ \frac{25200}{5600} = n + \frac{1}{2} \]
\[ 4.5 = n + \frac{1}{2} \]

### Step 8: Solve for \( n \)
\[ n = 4.5 - 0.5 = 4 \]

### Conclusion
The value of \( n \) is \( 4 \). Therefore, the \( 4^{th} \) dark fringe of \( 5600 \, \text{Å} \) coincides with the \( 4^{th} \) bright fringe of \( 6300 \, \text{Å} \).