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 will use the formula for fringe width in Young's double slit experiment. The fringe width \( w \) is given by the formula:

\[ w = \frac{\lambda D}{d} \]

where:
- \( \lambda \) is the wavelength of the light used,
- \( D \) is the distance from the slits to the screen,
- \( d \) is the distance between the slits.

### Step 1: Write down the initial fringe width
The initial fringe width \( w \) is given by:

\[ w = \frac{\lambda D}{d} \]

### Step 2: Change the distance between the slits
Now, if the distance between the slits is reduced to \( \frac{d}{3} \), we need to find the new fringe width \( w' \).

### Step 3: Substitute the new distance into the fringe width formula
The new fringe width \( w' \) can be calculated as follows:

\[ w' = \frac{\lambda D}{d'} \]

where \( d' = \frac{d}{3} \). Substituting this into the equation gives:

\[ w' = \frac{\lambda D}{\frac{d}{3}} = \frac{\lambda D \cdot 3}{d} = 3 \cdot \frac{\lambda D}{d} \]

### Step 4: Relate the new fringe width to the initial fringe width
Since \( \frac{\lambda D}{d} = w \), we can substitute this into the equation:

\[ w' = 3w \]

### Conclusion
Thus, the new fringe width \( w' \) when the distance between the sources is reduced to \( \frac{d}{3} \) is:

\[ w' = 3w \]

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

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