When a single slit is used, the interference pattern is due to the diffraction phenomenon. Since, corresponding wavelets from two halves of a single slit have a path difference of $\;\frac{λ}{2}$., the overlapping of their contributions in the first scenario results in a minimum. Because these wavefronts have a path difference of $λ$ in the second example, the overlapping of the wavefronts from the two slits results in the first maximum.