Young's experiment is performed with light of wavelength 6000 Ã… wherein 16 fringes occupy a certain region on the screen. If 24 frings occupy the same region with another light of wavelength `lambda`, then `lambda` is

Options

A. 6000 Ã…
B. 4500 Ã…
C. 5000 Ã…
D. 4000 Ã…

Explanation

To solve the problem, we will use the relationship between the number of fringes, the wavelength of light, and the fringe width in Young's double-slit experiment.

### Step-by-Step Solution:

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

2. **Set up the equations for the two scenarios**:
- For the first light with wavelength \( \lambda_1 = 6000 \, \text{Å} \) (or \( 6000 \times 10^{-10} \, \text{m} \)), and 16 fringes, the total width occupied by the fringes is:
\[ W_1 = n_1 \cdot \beta = 16 \cdot \beta \]
- For the second light with wavelength \( \lambda_2 = \lambda \), and 24 fringes, the total width occupied by the fringes is:
\[ W_2 = n_2 \cdot \beta' = 24 \cdot \beta' \]

3. **Relate the two scenarios**: Since the width of the region on the screen is the same for both cases, we can equate \( W_1 \) and \( W_2 \):
\[ 16 \cdot \beta = 24 \cdot \beta' \]

4. **Express fringe widths**: The fringe widths for both cases can be expressed as:
\[ \beta = \frac{\lambda_1 D}{d} \quad \text{and} \quad \beta' = \frac{\lambda_2 D}{d} \]
Substituting these into the equation gives:
\[ 16 \cdot \frac{\lambda_1 D}{d} = 24 \cdot \frac{\lambda_2 D}{d} \]

5. **Cancel out common terms**: Since \( D \) and \( d \) are the same for both cases, they can be canceled out:
\[ 16 \lambda_1 = 24 \lambda_2 \]

6. **Solve for \( \lambda_2 \)**: Rearranging the equation gives:
\[ \lambda_2 = \frac{16}{24} \lambda_1 \]
Simplifying the fraction:
\[ \lambda_2 = \frac{2}{3} \lambda_1 \]

7. **Substitute the value of \( \lambda_1 \)**: Now substituting \( \lambda_1 = 6000 \, \text{Å} \):
\[ \lambda_2 = \frac{2}{3} \times 6000 \, \text{Å} = 4000 \, \text{Å} \]

### Final Answer:
Thus, the wavelength \( \lambda \) is:
\[ \lambda = 4000 \, \text{Å} \]