In Young's double-slit experiment, the separation between the slits is halved and the distance between the slits and the screen in doubled. The fringe width is

Options

A. unchanged
B. halved
C. doubled
D. quadrupled

Explanation

To solve the problem, we need to understand how the fringe width in Young's double-slit experiment is affected by changes in the slit separation and the distance from the slits to the screen.

### Given:
1. Original slit separation = \( d \)
2. New slit separation = \( \frac{d}{2} \) (halved)
3. Original distance from the slits to the screen = \( D \)
4. New distance from the slits to the screen = \( 2D \) (doubled)

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

### Step-by-Step Solution:

1. **Calculate the Original Fringe Width**:
\[ \beta = \frac{\lambda D}{d} \]

2. **Calculate the New Fringe Width**:
The new fringe width \( \beta' \) after the changes can be calculated using the new values of \( d \) and \( D \):
- New slit separation \( d' = \frac{d}{2} \)
- New distance to the screen \( D' = 2D \)

Substitute these values into the fringe width formula:
\[ \beta' = \frac{\lambda D'}{d'} = \frac{\lambda (2D)}{\frac{d}{2}} \]

3. **Simplify the Expression**:
\[ \beta' = \frac{\lambda \cdot 2D}{\frac{d}{2}} = \frac{\lambda \cdot 2D \cdot 2}{d} = \frac{4\lambda D}{d} \]

4. **Relate New Fringe Width to Original Fringe Width**:
Notice that:
\[ \beta' = 4 \cdot \frac{\lambda D}{d} = 4\beta \]

### Conclusion:
The new fringe width \( \beta' \) is four times the original fringe width \( \beta \):
\[ \beta' = 4\beta \]

### Final Answer:
The fringe width is quadrupled.

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