In a Young's double slit experiment, the intensity at a point can be expressed as a function of the phase difference between the light arriving from the two slits. The intensity at any point on the screen is given by:

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

Where:

• $$I_{\text{max}}$$ is the maximum intensity.
• $$\delta$$ is the phase difference between the light waves from the two slits.

Given the intensity at a point is $$\left(\frac{1}{4}\right)^{\text {th }}$$ of the maximum intensity, we can write:

$$\frac{I}{I_{\text{max}}} = \frac{1}{4}$$

Substituting this into the intensity equation:

$$\frac{1}{4} = \cos^2 \left( \frac{\delta}{2} \right)$$

Taking the square root of both sides, we get:

$$\cos \left( \frac{\delta}{2} \right) = \frac{1}{2}$$

The possible solutions for $$\delta$$ are:

$$\frac{\delta}{2} = \frac{\pi}{3}$$ or $$\frac{\delta}{2} = \left(\pi - \frac{\pi}{3}\right)$$ which gives $$\delta = \frac{2\pi}{3}$$ or $$\delta = \frac{4\pi}{3}$$.

Considering the smallest phase difference, $$\delta = \frac{2\pi}{3}$$, we use the relation for the phase difference due to path difference:

$$\delta = \frac{2 \pi}{\lambda} \cdot \Delta x$$

Thus, substituting the value of $$\delta$$, we have:

$$\frac{2\pi}{\lambda} \cdot \Delta x = \frac{2\pi}{3}$$

Solving for $$\Delta x$$, we get:

$$\Delta x = \frac{\lambda}{3} = \frac{600 \, \mathrm{nm}}{3} = 200 \, \mathrm{nm} = 0.2 \, \mu \mathrm{m}$$

The minimum distance of the point from the central maximum on the screen can be found using the interference equation:

$$y = \frac{\Delta x \cdot D}{d}$$

Substituting the known values:

$$y = \frac{0.2 \, \mu \mathrm{m} \cdot 1.0 \, \mathrm{m}}{1.0 \, \mathrm{mm}} = \frac{0.2 \cdot 10^{-6} \, \mathrm{m} \cdot 1.0 \, \mathrm{m}}{1.0 \cdot 10^{-3} \, \mathrm{m}}$$

Simplifying this expression:

$$y = 0.2 \, \mathrm{mm} = 200 \, \mu \mathrm{m}$$

Hence, the minimum distance of the point from the central maximum is:

$$200 \, \mu \mathrm{m}$$