Animated Solution for Physics - Optics: 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

Imagine a classic Young's Double Slit Experiment (YDSE), but with a fascinating twist! Instead of the slits being rigidly fixed in place, they are oscillating. This means the distance between the two slits, denoted as $d$, is no longer a constant but a function of time.

The problem states that this distance varies sinusoidally as:

Here, $d_0$ is the mean separation between the slits, $a_0$ is the amplitude of their oscillation, and $\omega$ is the angular frequency. Because the slit separation is constantly changing, the interference pattern on the screen will also be dynamic. The fringes will appear to "breathe," expanding and contracting over time.

To understand this breathing effect, we need to recall the fundamental formula for the fringe width $\beta$ in a YDSE setup:

Since our slit distance $d$ is a function of time, the fringe width $\beta$ also becomes a function of time:

We are interested in the extreme cases: the largest and the smallest fringe widths.

1. Maximum Fringe Width (${\beta }_{\max }$): The fringe width is inversely proportional to the slit separation. Therefore, to get the maximum fringe width, the denominator must be as small as possible. The sine function, $\sin \omega t$, reaches its minimum value at $-1$. Substituting this in, we get:

2. Minimum Fringe Width (${\beta }_{\min }$): Conversely, to get the smallest fringe width, the denominator must be at its maximum. The sine function reaches its maximum value at $+1$. Substituting this in, we get:

The question asks for the difference between these two extreme fringe widths. Let's set up the subtraction:

To simplify this, we can factor out the common term $\lambda D$:

Now, we take the common denominator for the terms inside the bracket. The common denominator is $(d_0-a_0)(d_0+a_0)$, which simplifies to $d_0^2-a_0^2$ using the difference of squares formula.

In the numerator, the $d_0$ terms cancel each other out ($d_0-d_0=0$), and the $a_0$ terms add up ($a_0-(-a_0)=2a_0$). This leaves us with our final, elegant expression:

This perfectly matches option (b). The dynamic nature of the slits translates into a beautifully predictable oscillation of the interference pattern!