 Solution

Fresnel diffraction retains quadratic aperture-dependent propagation phases and generally depends on both aperture and screen distance. Fraunhofer diffraction neglects those phases and gives the Fourier transform of the aperture in angle space, either sufficiently far away or in a suitable lens's focal plane.

The calculations below use the Fraunhofer, paraxial approximation. For an overall aperture scale $D\sim 1000\lambda$, free propagation requires $L\gg D^2/\lambda \sim 10^6\lambda$. The printed $L\gg 1000\lambda$ alone only makes the separation small relative to distance; it is not by itself the full far-field condition.