d
Here,  \(\lambda=600 \mathrm{nm}=600 \times 10^{-9} \mathrm{m}\)

\(a = 1\,{\text{mm}} = {10^{ - 3}}{\text{m}},\) \(D = 2\,{\text{m}}\)

Distance between the first dark fringes on either side of the central bright fringe is also the width of central maximum.

Width of central maximum \(=\frac{2 \lambda D}{a}\)

\( = \frac{{2 \times 600 \times {{10}^{ - 9}}{\text{m}} \times 2{\text{m}}}}{{{{10}^{ - 3}}{\text{m}}}}\)

\( = 24 \times {10^{ - 4}}\,{\text{m}}\) \( = 2.4 \times {10^{ - 3}}{\text{m}} = 2.4\,{\text{mm}}\)