To solve this problem, we will use the formula for the position of dark fringes in a single-slit diffraction pattern.$\\$The condition for the first dark fringe in single-slit diffraction is given by:$\\ a \sin \theta = m \lambda \\$where:$\\$• $a$ is the width of the slit (1 mm = $1 \times 10^{-3}$ m),$\\$• $\lambda$ is the wavelength of light (600 nm = $600 \times 10^{-9}$ m),$\\$• $m$ is the order of the dark fringe (m = 1 for the first dark fringe),$\\$• $\theta$ is the angle of diffraction.$\\$For small angles, $\sin \theta \approx \tan \theta \approx \frac{y}{D}$, where $y$ is the distance from the central maximum to the dark fringe on the screen,$\\$and $D$ is the distance from the slit to the screen (2 m).$\\$Substituting these into the equation, we get:$\\ a \frac{y}{D} = m \lambda \\$Rearranging for $y: \\ y = \frac{m \lambda D}{a} \\$Substitute the known values:$\\ y = \frac{1 \times 600 \times 10^{-9} \times 2}{1 \times 10^{-3}} \\ y = \frac{1200 \times 10^{-9}}{1 \times 10^{-3}} \\ y = 1200 \times 10^{-6}$ m$\\ y = 1.2$ mm$\\$The distance between the first dark fringes on either side of the central bright fringe is $2y: \\ 2y = 2 \times 1.2$ mm$\\ 2y = 2.4$ mm$\\$Therefore, the distance between the first dark fringes is $2.4$ mm, which corresponds to Option 4.