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, we need to find 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.

### Step 1: Understand the fringe width formula
The fringe width (\( \beta \)) in a Young's double slit experiment is given by the formula:
\[ \beta = \frac{\lambda D}{d} \]
where:
- \( \lambda \) is the wavelength of light,
- \( D \) is the distance from the slits to the screen,
- \( d \) is the distance between the slits.

### Step 2: Write the equations for red and blue light
For red light with wavelength \( \lambda_R = 600 \, \text{nm} \):
\[ \beta_R = n \lambda_R \]
For blue light with wavelength \( \lambda_B = 480 \, \text{nm} \):
\[ \beta_B = (n + 1) \lambda_B \]

### Step 3: Set the fringe widths equal
Since the \( n \)th red fringe coincides with the \( (n + 1) \)th blue fringe, we can equate the two fringe width equations:
\[ n \lambda_R = (n + 1) \lambda_B \]

### Step 4: Substitute the wavelengths
Substituting the values of \( \lambda_R \) and \( \lambda_B \):
\[ n \times 600 \, \text{nm} = (n + 1) \times 480 \, \text{nm} \]

### Step 5: Simplify the equation
We can cancel out the units of nm:
\[ 600n = 480(n + 1) \]

### Step 6: Expand and rearrange the equation
Expanding the right side:
\[ 600n = 480n + 480 \]
Now, rearranging gives:
\[ 600n - 480n = 480 \]
\[ 120n = 480 \]

### Step 7: Solve for \( n \)
Dividing both sides by 120:
\[ n = \frac{480}{120} = 4 \]

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