Light of wavelength 6000 12 × 10^–6 m is incident on a single slit. First minimum is obtained at a distance of 0.4 cm from the centre. If width of the slit is 0.3 mm, then distance between slit and screen will be

2.0 m

Using $a\, \sin θλ = nλ$

or $a.\frac{x}{D}= nλ$ or $a =nλ.\frac{D}{x}$

or $D=\frac{a.x}{nλ}$

$\frac{3×10^{-4}×4×10^{-3}}{1×6000×10^{-10}}m=2.0m$