In young's double slit experiment, if the distance between the slits is halved and the distance between the slits and the screen is doubled, the fringe width becomes

Options

A. Half
B. Double
C. Four times
D. Eigth times

Answer

To solve the problem regarding the change in fringe width in Young's double slit experiment, let's follow these steps:

### Step 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:
- \( \lambda \) = wavelength of light used
- \( D \) = distance from the slits to the screen
- \( d \) = distance between the slits

### Step 2: Define the initial conditions
Let:
- Initial distance between slits, \( d_i = d \)
- Initial distance from slits to screen, \( D_i = D \)

Thus, the initial fringe width \( \beta_i \) is:
\[ \beta_i = \frac{\lambda D}{d} \]

### Step 3: Define the new conditions
According to the problem:
- The distance between the slits is halved: \( d_f = \frac{d}{2} \)
- The distance from the slits to the screen is doubled: \( D_f = 2D \)

### Step 4: Calculate the final fringe width
Using the new values in the fringe width formula, we get the final fringe width \( \beta_f \):
\[ \beta_f = \frac{\lambda D_f}{d_f} = \frac{\lambda (2D)}{\frac{d}{2}} = \frac{\lambda \cdot 2D \cdot 2}{d} = \frac{4\lambda D}{d} \]

### Step 5: Compare the final and initial fringe widths
Now, we can compare the final fringe width \( \beta_f \) with the initial fringe width \( \beta_i \):
\[ \beta_f = 4 \cdot \frac{\lambda D}{d} = 4 \beta_i \]

### Conclusion
Thus, the fringe width becomes 4 times the initial fringe width.

### Final Answer
The fringe width becomes **4 times**.

---