Given data,
\(d=1\;\mathrm{mm}=10^{-3}\;\mathrm{m}\)
\(D=1\;\mathrm{m}\)
\(\lambda =500\;\mathrm{nm}\)
We know, fringe separation,
\(W=\frac{\mathit{\lambda D}} d\)
\(=\frac{500\times 10^{-9}\times 1}{10^{-3}}\)
\(=5\times 10^{-4}\;\mathrm{m}\)
\(=0{\cdot}5\;\mathrm{mm}\)