In Young's double slit experiment with a mono - chromatic light of wavelength `4000 A^(@)`, the fringe width is found to be 0.4 mm. When the slits are now illuminated with a light of wavelength `5000A^(@)` the fringe width will the

Options

A. 0.32 mm
B. 0.5 mm
C. 0.6 mm
D. 0.8 mm

Answer

To solve the problem, we need to find the fringe width when the wavelength of light changes from 4000 Å to 5000 Å in Young's double slit experiment.

### Step-by-Step Solution:

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

2. **Identify Given Values**:
- For the first case:
- Wavelength \( \lambda_1 = 4000 \, \text{Å} = 4000 \times 10^{-10} \, \text{m} \)
- Fringe width \( \beta_1 = 0.4 \, \text{mm} = 0.4 \times 10^{-3} \, \text{m} \)
- For the second case:
- Wavelength \( \lambda_2 = 5000 \, \text{Å} = 5000 \times 10^{-10} \, \text{m} \)

3. **Relate the Fringe Widths**:
Since the distance \( D \) and the slit separation \( d \) remain constant, we can set up a ratio of the fringe widths:
\[ \frac{\beta_1}{\beta_2} = \frac{\lambda_1}{\lambda_2} \]

4. **Substitute the Known Values**:
Substituting the known values into the ratio:
\[ \frac{0.4 \times 10^{-3}}{\beta_2} = \frac{4000 \times 10^{-10}}{5000 \times 10^{-10}} \]

5. **Simplify the Right Side**:
Simplifying the right side gives:
\[ \frac{4000}{5000} = \frac{4}{5} \]

6. **Solve for \( \beta_2 \)**:
Rearranging the equation to find \( \beta_2 \):
\[ \beta_2 = \beta_1 \cdot \frac{5}{4} \]
Substituting \( \beta_1 = 0.4 \times 10^{-3} \):
\[ \beta_2 = 0.4 \times 10^{-3} \cdot \frac{5}{4} = 0.5 \times 10^{-3} \, \text{m} = 0.5 \, \text{mm} \]

7. **Final Answer**:
The fringe width when the wavelength is 5000 Å is:
\[ \beta_2 = 0.5 \, \text{mm} \]