(d) Distance of $n$th dark fringe from central fringe
$x_{n} = \frac{(2n - 1)\lambda D}{2d}$
$\therefore x_{2} = \frac{(2 \times 2 - 1)\lambda D}{2d} = \frac{3\lambda D}{2d}$
$\Rightarrow 1 \times 10^{- 3} = \frac{3\lambda \times 1}{2 \times 0.9 \times 10^{- 3}}$
$\Rightarrow \lambda = 6 \times 10^{- 5}cm$