Two slits 0.5 mm apart are illuminated with light of 600 nm. The fringes are observed on a screen 1 m away. The fringe width is:

We have $w = \lambda D/d$. That is $w = (600 \times 10^{-9})(1)/(0.5 \times 10^{-3})$. Then $w = 1.2 \times 10^{-3}$. So m = 1.2 mm.