In Young's double sit experiment two light sources when placed at a distance d apart, then Interference pattem having fringe width w. If the distance between the sources is reduced to d/3, then fringe width would be

Options

A. `w/3`
B. 3w
C. w
D. `w/2`

Answer

To solve the problem, we need to understand the relationship between the fringe width in Young's double slit experiment and the distance between the slits.

### Step-by-Step Solution:

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

2. **Initial Conditions**:
We are given that the initial distance between the slits is \( d \) and the corresponding fringe width is \( w \).

3. **Changing the Distance Between Slits**:
Now, the distance between the slits is reduced to \( \frac{d}{3} \). We need to find the new fringe width \( w' \) when the distance between the slits is changed.

4. **Applying the Fringe Width Formula Again**:
Substituting \( \frac{d}{3} \) into the fringe width formula, we get:
\[ w' = \frac{\lambda D}{\frac{d}{3}} = \frac{3\lambda D}{d} \]

5. **Relating New Fringe Width to Initial Fringe Width**:
We can relate \( w' \) to the initial fringe width \( w \):
\[ w' = 3 \left( \frac{\lambda D}{d} \right) = 3w \]

6. **Conclusion**:
Therefore, the new fringe width \( w' \) when the distance between the slits is reduced to \( \frac{d}{3} \) is:
\[ w' = 3w \]

### Final Answer:
The new fringe width would be \( 3w \).

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