Intensity distribution in YDSE

For two identical coherent sources, the intensity at a point is $I = I_{\max} \cos^2\left(\frac{\phi}{2}\right)$ where $I_{\max}$ is the maximum intensity.

Given: $I = \frac{1}{4} I_{\max}$

So, $\cos^2\left(\frac{\phi}{2}\right)=\frac{1}{4}$

Find the least phase difference

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

For the minimum distance from central maximum, we need the smallest non-zero phase difference.

Hence, $\frac{\phi}{2}=\frac{\pi}{3} \quad \Rightarrow \quad \phi=\frac{2\pi}{3}$

Relate phase difference to path difference

$\phi = \frac{2\pi}{\lambda}\,\Delta$

Therefore, $\frac{2\pi}{\lambda}\,\Delta = \frac{2\pi}{3}$ $\Delta = \frac{\lambda}{3}$

Relate path difference to position on screen

In Young's double slit experiment, $\Delta = \frac{dy}{D}$

So, $\frac{dy}{D} = \frac{\lambda}{3}$ $y = \frac{D\lambda}{3d}$

Substitute the given values

$\lambda = 600\text{ nm} = 600\times 10^{-9}\text{ m}$ $d = 1.0\text{ mm} = 10^{-3}\text{ m}$ $D = 1.0\text{ m}$

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

Final answer

The minimum distance from the central maximum is $\boxed{200\,\mu\text{m}}$