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

Options

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

Explanation

To solve the problem, we will use the formula for fringe width in Young's double slit experiment, which is given by:

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

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

### Step 1: Write the initial fringe width
Let the initial separation between the slits be \(d\) and the initial distance from the slits to the screen be \(D\). The initial fringe width (\(\beta\)) can be expressed as:

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

### Step 2: Modify the parameters according to the problem
According to the problem:
- The separation between the slits is halved, so the new separation \(d' = \frac{d}{2}\).
- The distance from the slits to the screen is doubled, so the new distance \(D' = 2D\).

### Step 3: Calculate the new fringe width
Now, we can calculate the new fringe width (\(\beta'\)) using the modified values:

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

Substituting the new values:

\[ \beta' = \frac{\lambda (2D)}{\frac{d}{2}} \]

### Step 4: Simplify the expression
Now simplify the expression:

\[ \beta' = \frac{\lambda \cdot 2D \cdot 2}{d} = \frac{4\lambda D}{d} \]

### Step 5: Relate the new fringe width to the initial fringe width
Now we can relate the new fringe width to the initial fringe width:

\[ \beta' = 4 \cdot \frac{\lambda D}{d} = 4\beta \]

### Conclusion
Thus, the new fringe width is four times the original fringe width. Therefore, the final answer is:

\[ \text{New fringe width} = 4\beta \]