In a YDSE setup, the distance between the slits varies as `d = d_0 + A sin(wt)` What is the difference between maximum and minimum fringe width?

Options

A. `(2 lambda A D)/((d_0)^2)`
B. `(2 lambda A D)/((d_0)^2- A^2)`
C. `(2 lambda A D)/((d_0)^2+ A^2)`
D. `(2 lambda A d_0)/((d_0)^2- A^2)`

Answer

To solve the problem, we need to determine the difference between the maximum and minimum fringe widths in a Young's Double Slit Experiment (YDSE) setup where the distance between the slits varies as \( d = d_0 + A \sin(\omega t) \).

### Step-by-Step Solution:

1. **Understanding Fringe Width**:
The fringe width \( \beta \) in YDSE is given by the formula:
\[ \beta = \frac{\lambda D}{d} \]
where \( \lambda \) is the wavelength of light, \( D \) is the distance from the slits to the screen, and \( d \) is the distance between the slits.

2. **Substituting the Variable Slit Distance**:
Given that the distance between the slits varies with time as:
\[ d = d_0 + A \sin(\omega t) \]
we can substitute this into the fringe width formula:
\[ \beta = \frac{\lambda D}{d_0 + A \sin(\omega t)} \]

3. **Finding Maximum and Minimum Fringe Width**:
To find the maximum and minimum fringe widths, we need to evaluate \( \beta \) at the maximum and minimum values of \( \sin(\omega t) \):
- **Maximum Fringe Width** (\( \beta_{\text{max}} \)):
When \( \sin(\omega t) = 1 \):
\[ \beta_{\text{max}} = \frac{\lambda D}{d_0 + A} \]
- **Minimum Fringe Width** (\( \beta_{\text{min}} \)):
When \( \sin(\omega t) = -1 \):
\[ \beta_{\text{min}} = \frac{\lambda D}{d_0 - A} \]

4. **Calculating the Difference**:
Now, we find the difference between the maximum and minimum fringe widths:
\[ \Delta \beta = \beta_{\text{max}} - \beta_{\text{min}} = \frac{\lambda D}{d_0 + A} - \frac{\lambda D}{d_0 - A} \]

5. **Finding a Common Denominator**:
To simplify the expression, we can find a common denominator:
\[ \Delta \beta = \lambda D \left( \frac{(d_0 - A) - (d_0 + A)}{(d_0 + A)(d_0 - A)} \right) \]
Simplifying the numerator:
\[ (d_0 - A) - (d_0 + A) = -2A \]
Thus, we have:
\[ \Delta \beta = \lambda D \left( \frac{-2A}{(d_0 + A)(d_0 - A)} \right) = \frac{-2\lambda D A}{d_0^2 - A^2} \]

6. **Final Result**:
The absolute value of the difference in fringe widths is:
\[ \Delta \beta = \frac{2\lambda D A}{d_0^2 - A^2} \]

### Final Answer:
The difference between the maximum and minimum fringe width is:
\[ \Delta \beta = \frac{2\lambda D A}{d_0^2 - A^2} \]