A beam of light of wave length 600 nm from a distant source falls on a single slit 1 mm wide and the resulting diffraction pattern is observed on a screen 2 m away. The distance between the first dark fringes on either side of the central bright is

2.4 mm

For first minimum putting n = 1 in the condition for minima given as $d \sin θ = nλ$

we obtain $d \sin θ = λ$ …(1)

where $\sin θ ≅ θ ≅ \frac{2/x}{D}$

$⇒ sin θ = \frac{x}{2D}$ …(2)

using (1) and (2) we obtain,

$\frac{λ}{d}=\frac{x}{2D}$

$x=\frac{2λD}{d}=\frac{2(600×10^{-9})(2)}{10^{-3}}$

$= 2.4 × 10^{-3} m$

= 2.4 mm