A monochromatice light beam of wavelength `5896 A^(0)` is used in double slit experiment to get interference pattern on a screen 9th bright fringe is seen at a particular position on the screen. At the same point on the screen, if 11th bright fringe is to be seen, teh wavelength of the light that is needed is (nearly)

Options

A. `7014 A^(0)`
B. `3525 A^(0)`
C. `6780 A^(0)`
D. `4824 A^(0)`

Answer

To solve the problem, we need to find the wavelength of light required to see the 11th bright fringe at the same position on the screen where the 9th bright fringe is seen with a wavelength of \( 5896 \, \text{Å} \).

### Step-by-Step Solution:

1. **Understanding the Fringe Position Formula**:
The position of the bright fringes in a double-slit experiment is given by the formula:
\[ y_n = \frac{n \lambda D}{d} \]
where:
- \( y_n \) = position of the nth bright fringe,
- \( n \) = order of the fringe (9 for the 9th fringe, 11 for the 11th fringe),
- \( \lambda \) = wavelength of the light,
- \( D \) = distance from the slits to the screen,
- \( d \) = distance between the slits.

2. **Setting Up the Equations**:
For the 9th bright fringe with wavelength \( \lambda_1 = 5896 \, \text{Å} \):
\[ y_9 = \frac{9 \lambda_1 D}{d} \]
For the 11th bright fringe with the unknown wavelength \( \lambda_2 \):
\[ y_{11} = \frac{11 \lambda_2 D}{d} \]

3. **Equating the Positions**:
Since both fringes are at the same position \( y \), we can set the two equations equal to each other:
\[ \frac{9 \lambda_1 D}{d} = \frac{11 \lambda_2 D}{d} \]
The \( D \) and \( d \) terms cancel out, leading to:
\[ 9 \lambda_1 = 11 \lambda_2 \]

4. **Solving for \( \lambda_2 \)**:
Rearranging the equation gives:
\[ \lambda_2 = \frac{9}{11} \lambda_1 \]
Substituting \( \lambda_1 = 5896 \, \text{Å} \):
\[ \lambda_2 = \frac{9}{11} \times 5896 \, \text{Å} \]

5. **Calculating \( \lambda_2 \)**:
Performing the multiplication:
\[ \lambda_2 = \frac{9 \times 5896}{11} = \frac{53064}{11} \approx 4824 \, \text{Å} \]

### Final Answer:
The wavelength of light needed to see the 11th bright fringe at the same position is approximately \( 4824 \, \text{Å} \).