Given: Wavelength of light, $\lambda = 600\text{ nm} = 600 \times 10^{-9}\text{ m}$ Slit width, $a = 1\text{ mm} = 10^{-3}\text{ m}$ Distance of the screen, $D = 2\text{ m}$

The distance between the first dark fringes on either side of the central bright fringe is equal to the linear width of the central maximum. Width of central maximum $= \frac{2\lambda D}{a}$ $= \frac{2 \times 600 \times 10^{-9} \times 2}{10^{-3}}$ $= \frac{2400 \times 10^{-9}}{10^{-3}}$ $= 2400 \times 10^{-6}\text{ m}$ $= 2.4 \times 10^{-3}\text{ m} = 2.4\text{ mm}$