In a Young's double slit experiment, the intensity at a point is $$\left(\frac{1}{4}\right)^{th}$$ of the maximum intensity, the minimum distance of the point from the central maximum is _______ $$\mu$$m. (Given : $$\lambda = 600$$ nm, $$d = 1.0$$ mm, $$D = 1.0$$ m)

In Young’s double slit experiment, intensity at a point is given by:

$$I = I_{\max}\cos^2\left(\frac{\phi}{2}\right)$$

When $$I = \frac{I_{\max}}{4}$$, we have

$$\cos^2\left(\frac{\phi}{2}\right) = \frac{1}{4} \implies \cos\left(\frac{\phi}{2}\right) = \frac{1}{2} \implies \frac{\phi}{2} = \frac{\pi}{3} \implies \phi = \frac{2\pi}{3}$$

The phase difference is related to the path difference by $$\phi = \frac{2\pi}{\lambda}\Delta$$, where $$\Delta = \frac{yd}{D}$$.

Thus,

$$\frac{2\pi}{3} = \frac{2\pi}{\lambda} \cdot \frac{yd}{D}$$

Rearranging to solve for $$y$$ gives

$$y = \frac{\lambda D}{3d} = \frac{600 \times 10^{-9} \times 1}{3 \times 10^{-3}} = \frac{600 \times 10^{-9}}{3 \times 10^{-3}} = 200 \times 10^{-6} \text{ m} = 200 \text{ }\mu\text{m}$$

Therefore, the answer is $$\boxed{200}$$ μm.