In a Young's double slit experiment using red and blue lights of wavelengths 600 nm and 480 nm respectively, the value of n for which the nth red fringe coincides with `(n+1)` th blue fringe is

Options

A. 5
B. 4
C. 3
D. 2

Explanation

To solve the problem of finding the value of \( n \) for which the \( n \)th red fringe coincides with the \( (n+1) \)th blue fringe in a Young's double slit experiment, we can follow these steps:

### Step 1: Understand the fringe formation
In a Young's double slit experiment, the position of the \( n \)th fringe for light of wavelength \( \lambda \) is given by the formula:
\[ y_n = \frac{n \lambda D}{d} \]
where:
- \( y_n \) is the position of the \( n \)th fringe,
- \( D \) is the distance from the slits to the screen,
- \( d \) is the distance between the slits,
- \( n \) is the fringe order.

### Step 2: Set up the equations for red and blue fringes
Let:
- \( \lambda_r = 600 \, \text{nm} \) (wavelength of red light),
- \( \lambda_b = 480 \, \text{nm} \) (wavelength of blue light).

The position of the \( n \)th red fringe is:
\[ y_n^{(r)} = \frac{n \lambda_r D}{d} = \frac{n \cdot 600 \, \text{nm} \cdot D}{d} \]

The position of the \( (n+1) \)th blue fringe is:
\[ y_{n+1}^{(b)} = \frac{(n+1) \lambda_b D}{d} = \frac{(n+1) \cdot 480 \, \text{nm} \cdot D}{d} \]

### Step 3: Set the equations equal to each other
Since we want the \( n \)th red fringe to coincide with the \( (n+1) \)th blue fringe, we can set the two equations equal:
\[ \frac{n \cdot 600 \, \text{nm} \cdot D}{d} = \frac{(n+1) \cdot 480 \, \text{nm} \cdot D}{d} \]

### Step 4: Simplify the equation
We can cancel \( \frac{D}{d} \) from both sides (as long as \( D \) and \( d \) are not zero):
\[ n \cdot 600 = (n + 1) \cdot 480 \]

### Step 5: Expand and rearrange the equation
Expanding the right side gives:
\[ 600n = 480n + 480 \]
Now, rearranging the equation:
\[ 600n - 480n = 480 \]
\[ 120n = 480 \]

### Step 6: Solve for \( n \)
Dividing both sides by 120:
\[ n = \frac{480}{120} = 4 \]

### Conclusion
Thus, the value of \( n \) for which the \( n \)th red fringe coincides with the \( (n+1) \)th blue fringe is:
\[ \boxed{4} \]