Explanation

(b) Here, $a = 2mm = 2 \times 10^{- 3}m$
$\lambda = 500nm = 500 \times 10^{- 9}m = 5 \times 10^{- 7}m$
$D = 1m$
The distance between the first minima on either side on a screen is
$= \frac{2\lambda D}{a} = \frac{2 \times 5 \times 10^{- 7} \times 1}{2 \times 10^{- 3}}$
$= 5 \times 10^{- 4}m = 0.5 \times 10^{- 3}m = 0.5mm$