Question

In the Young's double slit experiment, the distance between the slits varies in time as $d(t) = d_0 + a_0 \sin \omega t$; where $d_0, \omega$ and $a_0$ are constants. The difference between the largest fringe width and the smallest fringe width obtained over time is given as :

• A. $\frac{\lambda D}{d_0^2} a_0$
• B. $\frac{2\lambda D (d_0)}{(d_0^2 - a_0^2)}$
• C. $\frac{\lambda D}{d_0 + a_0}$
• D. $\frac{2\lambda D a_0}{(d_0^2 - a_0^2)}$

Hint:

Fringe width \(\beta\) and slit distance \(d\) are inversely related.
Maximum \(\beta\) occurs at minimum \(d\), and minimum \(\beta\) occurs at maximum \(d\).

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
The fringe width (\(\beta\)) in YDSE is given by \(\beta = \frac{\lambda D}{d}\), where \(D\) is the screen distance and \(d\) is the slit separation.
Since \(d\) varies with time, the fringe width \(\beta\) also varies.
Step 2: Key Formula or Approach:
The distance between slits is \(d(t) = d_0 + a_0 \sin \omega t\).
The range of \(\sin \omega t\) is \([-1, 1]\).
Thus, the maximum value of \(d\) is \(d_{\text{max}} = d_0 + a_0\).
The minimum value of \(d\) is \(d_{\text{min}} = d_0 - a_0\).
Step 3: Detailed Explanation:
The fringe width is inversely proportional to the slit distance.
Largest fringe width occurs when \(d\) is minimum:
\[ \beta_{\text{max}} = \frac{\lambda D}{d_{\text{min}}} = \frac{\lambda D}{d_0 - a_0} \]
Smallest fringe width occurs when \(d\) is maximum:
\[ \beta_{\text{min}} = \frac{\lambda D}{d_{\text{max}}} = \frac{\lambda D}{d_0 + a_0} \]
The difference is:
\[ \Delta \beta = \beta_{\text{max}} - \beta_{\text{min}} = \lambda D \left[ \frac{1}{d_0 - a_0} - \frac{1}{d_0 + a_0} \right] \]
\[ \Delta \beta = \lambda D \left[ \frac{(d_0 + a_0) - (d_0 - a_0)}{(d_0 - a_0)(d_0 + a_0)} \right] \]
\[ \Delta \beta = \lambda D \left[ \frac{2 a_0}{d_0^2 - a_0^2} \right] = \frac{2\lambda D a_0}{d_0^2 - a_0^2} \]
Step 4: Final Answer:
The difference is \(\frac{2\lambda D a_0}{d_0^2 - a_0^2}\).