In Young's double slit experiment, two wavelength `lamda_(1)=780nm` and `lamda_(2)=520` nm are used to obtain interference fringes. If the nth bright band due to `lamda_(1)` coincides with `(n+1)^(th)` bright band due to `lamda_(2)` then the value of n is

Options

A. 4
B. 3
C. 2
D. 6

Answer

To solve the problem, we need to find the value of \( n \) such that the \( n^{th} \) bright fringe due to the wavelength \( \lambda_1 = 780 \, \text{nm} \) coincides with the \( (n+1)^{th} \) bright fringe due to the wavelength \( \lambda_2 = 520 \, \text{nm} \).

### Step-by-Step Solution:

1. **Understanding Fringe Width**:
The fringe width \( \beta \) in Young's double slit experiment is given by the formula:
\[ \beta = \frac{\lambda D}{d} \]
where \( \lambda \) is the wavelength, \( D \) is the distance from the slits to the screen, and \( d \) is the distance between the slits.

2. **Fringe Width for Each Wavelength**:
For the first wavelength \( \lambda_1 = 780 \, \text{nm} \):
\[ \beta_1 = \frac{\lambda_1 D}{d} = \frac{780 D}{d} \]
For the second wavelength \( \lambda_2 = 520 \, \text{nm} \):
\[ \beta_2 = \frac{\lambda_2 D}{d} = \frac{520 D}{d} \]

3. **Setting Up the Equation**:
According to the problem, the \( n^{th} \) bright fringe of \( \lambda_1 \) coincides with the \( (n+1)^{th} \) bright fringe of \( \lambda_2 \). The positions of the bright fringes can be expressed as:
\[ n \beta_1 = (n + 1) \beta_2 \]

4. **Substituting the Fringe Widths**:
Substitute \( \beta_1 \) and \( \beta_2 \):
\[ n \left(\frac{780 D}{d}\right) = (n + 1) \left(\frac{520 D}{d}\right) \]

5. **Canceling Common Terms**:
Since \( D \) and \( d \) are common in both terms, we can cancel them out:
\[ n \cdot 780 = (n + 1) \cdot 520 \]

6. **Expanding the Equation**:
Expand the right side:
\[ 780n = 520n + 520 \]

7. **Rearranging the Equation**:
Rearranging gives:
\[ 780n - 520n = 520 \]
\[ 260n = 520 \]

8. **Solving for \( n \)**:
Divide both sides by 260:
\[ n = \frac{520}{260} = 2 \]

### Final Answer:
The value of \( n \) is \( 2 \).