In Young's double-slit experiment, the intensity \( I \) at a point is described by \( I = I_0 \cos^2(\frac{\pi d \sin \theta}{\lambda}) \), where \( I_0 \) is the maximum intensity, \( d \) is the slit separation, \( \theta \) is the angle of the point from the center, and \( \lambda \) is the wavelength.
Given that \( I = \left(\frac{1}{4}\right) I_0 \), we find that:
\[ \cos^2\left(\frac{\pi d \sin \theta}{\lambda}\right) = \frac{1}{4} \]
This simplifies to:
\[ \cos\left(\frac{\pi d \sin \theta}{\lambda}\right) = \frac{1}{2} \]
Consequently:
\[ \frac{\pi d \sin \theta}{\lambda} = \frac{\pi}{3} \text{ or } \frac{2\pi}{3} \]
Solving for \( \sin \theta \) yields:
\[ \sin \theta = \frac{\lambda}{3d} \text{ (for } \frac{\pi}{3}) \text{ or } \sin \theta = -\frac{\lambda}{3d} \text{ (for } \frac{2\pi}{3}) \]
Applying the small angle approximation (\( \sin \theta \approx \tan \theta \approx \frac{x}{D} \), where \( x \) is the distance from the central maximum), we get:
\[ \frac{x}{D} = \frac{\lambda}{3d} \]
Therefore, the distance \( x \) is given by:
\[ x = \frac{\lambda D}{3d} \]
Substituting the given values \( \lambda = 600 \, \text{nm} = 600 \times 10^{-9} \, \text{m} \), \( d = 1.0 \, \text{mm} = 1.0 \times 10^{-3} \, \text{m} \), and \( D = 1.0 \, \text{m} \):
\[ x = \frac{600 \times 10^{-9} \times 1.0}{3 \times 1.0 \times 10^{-3}} \]
This results in:
\[ x = 200 \times 10^{-6} \, \text{m} = 200 \, \mu\text{m} \]
The minimum distance calculated is \( 200 \, \mu\text{m} \), which falls within the specified range (200,200).