a b c d

b

$\text{The position of dark fringe in the interference is given by} \theta = (2n -1) \frac{\lambda}{2d}$ $\text{Angular position of first dark fringe is } \theta_1 = \frac{\lambda}{2d}=\frac{5460\times 10^{-10}}{2\times 10^{-4}} = 2730\times 10^{-6} rad = 2730\times 10^{-6} \times \frac{180}{\pi} = 0.16^0$