A narrow slit of width 1 mm is illuminated by monochromatic light of wavelength 600 nm. The distance between the first minima on either side of a screen at a distance of 2 m is

$$\begin{gathered} For first dark fringe on either side, \\ d\sin \theta =\lambda \\ \Rightarrow \frac{dy}{D}=\lambda [\sin \theta \simeq \tan \theta ] \\ \therefore y=\frac{\lambda D}{d} \\ Therefore, dis\tan ce between two dark fringes on either side = 2y=\frac{2\lambda D}{d} \\ Substituting the values, we have dis\tan ce = \frac{2\left(600\times {10}^{-6}mm\right)\left(2\times {10}^{3}mm\right)}{\left(1.0mm\right)}=2.4mm \end{gathered}$$