In Young's double slit experiment, the wavelength of red light is `7800 Å`. The value of n for which nth bright band due to red light coincides with `(n+1)th` bright band due to blue light`(lambda=5200Angstrom)`, is

Options

A. 1
B. 2
C. 3
D. 4

Answer

To solve the problem, we need to find the value of \( n \) for which the \( n \)th bright band due to red light coincides with the \( (n+1) \)th bright band due to blue light in Young's double slit experiment.

### Step-by-Step Solution:

1. **Identify the Wavelengths**:
- Wavelength of red light, \( \lambda_r = 7800 \, \text{Å} \)
- Wavelength of blue light, \( \lambda_b = 5200 \, \text{Å} \)

2. **Write the Condition for Bright Bands**:
- The position of the \( n \)th bright band for red light is given by:
\[ y_r = \frac{n \lambda_r D}{d} \]
- The position of the \( (n+1) \)th bright band for blue light is given by:
\[ y_b = \frac{(n+1) \lambda_b D}{d} \]

3. **Set the Positions Equal**:
- According to the problem, these positions coincide:
\[ y_r = y_b \]
- Therefore, we can equate the two expressions:
\[ \frac{n \lambda_r D}{d} = \frac{(n+1) \lambda_b D}{d} \]

4. **Cancel Common Terms**:
- Since \( D \) and \( d \) are common in both terms, they can be canceled out:
\[ n \lambda_r = (n+1) \lambda_b \]

5. **Substitute the Wavelengths**:
- Substitute the values of \( \lambda_r \) and \( \lambda_b \):
\[ n \cdot 7800 = (n + 1) \cdot 5200 \]

6. **Expand and Rearrange**:
- Expanding the equation gives:
\[ 7800n = 5200n + 5200 \]
- Rearranging this leads to:
\[ 7800n - 5200n = 5200 \]
\[ 2600n = 5200 \]

7. **Solve for \( n \)**:
- Dividing both sides by 2600:
\[ n = \frac{5200}{2600} = 2 \]

### Conclusion:
The value of \( n \) for which the \( n \)th bright band due to red light coincides with the \( (n+1) \)th bright band due to blue light is:
\[ \boxed{2} \]