In the Young's double slit experiment, the distance between the slits varies in time as $$dt = d_0 + a_0 \sin\omega t$$; 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:

In Young’s double-slit experiment the fringe width is governed by the well-known relation $$\beta=\dfrac{\lambda D}{d}$$ where $$\lambda$$ is the wavelength of the light, $$D$$ is the distance from the slits to the screen and $$d$$ is the separation between the two slits.

In this problem the slit separation is not constant; it varies with time according to $$d(t)=d_0+a_0\sin\omega t,$$ where the symbols $$d_0,\;a_0,\;\omega$$ are fixed constants and the condition $$a_0<d_0$$ is implied so that the distance between the slits never becomes negative.

Because $$\beta=\dfrac{\lambda D}{d},$$ the fringe width will attain its maximum value when $$d(t)$$ is minimum and will attain its minimum value when $$d(t)$$ is maximum.

The minimum value of $$d(t)$$ occurs when $$\sin\omega t=-1$$, giving

$$d_{\text{min}}=d_0-a_0.$$

Correspondingly, the maximum value of $$d(t)$$ occurs when $$\sin\omega t=+1$$, giving

$$d_{\text{max}}=d_0+a_0.$$

Now the largest possible fringe width (call it $$\beta_{\text{large}}$$) is obtained by substituting $$d_{\text{min}}$$ into the formula for $$\beta$$:

$$\beta_{\text{large}}=\dfrac{\lambda D}{d_{\text{min}}}=\dfrac{\lambda D}{d_0-a_0}.$$

Similarly, the smallest possible fringe width (call it $$\beta_{\text{small}}$$) arises when $$d=d_{\text{max}}$$:

$$\beta_{\text{small}}=\dfrac{\lambda D}{d_{\text{max}}}=\dfrac{\lambda D}{d_0+a_0}.$$

The problem asks for the difference between these two extreme values:

$$\Delta\beta=\beta_{\text{large}}-\beta_{\text{small}}=\dfrac{\lambda D}{d_0-a_0}-\dfrac{\lambda D}{d_0+a_0}.$$

We combine the two fractions by taking a common denominator:

$$\Delta\beta=\lambda D\left[\dfrac{1}{d_0-a_0}-\dfrac{1}{d_0+a_0}\right]$$

$$=\lambda D\left[\dfrac{d_0+a_0-(d_0-a_0)}{(d_0-a_0)(d_0+a_0)}\right]$$

$$=\lambda D\left[\dfrac{d_0+a_0-d_0+a_0}{d_0^2-a_0^2}\right]$$

$$=\lambda D\left[\dfrac{2a_0}{d_0^2-a_0^2}\right].$$

Simplifying, we obtain

$$\boxed{\Delta\beta=\dfrac{2\lambda D a_0}{\,d_0^2-a_0^2\,}}.$$

This expression matches Option B in the list provided.

Hence, the correct answer is Option 2.