by static intensity distributions do you mean that an observer at that point would just be receiving a single sine wave? Or that they would just be receiving a value 1 (or say always in the interval [1-e,1+e] for very small e? What you mean? Can we think of this as sound or no, light gives something new to interpret here?

As a sidenote: This might be trivial to most of you, but this visualization made me realize what "a single frequency" really means. If you look closely at the wave fields, especially the intense superposition in the region between source and slit, the snapshot fields really are "tiled/patterned" by a single spatial frequency (or in the case of slightly detuned emitters, a very narrow distribution) corresponding to the wavelength. => Harmonic functions of one frequency always add up ...

... to another harmonic function, no matter the spatial, phase, or amplitude distribution of sources! (Not sure if there exist any other class of functions with this characteristic property.)

didn't really answer my question

obviously there are ways to do it even with 1 dimension by using a spectrum of emitters, you see how right?

@Snared To start from the last question: This problem applies to anything that can be described in terms of waves, i.e. sines, cosines, and arbitrary superpositions thereof. Which is why this is such a fundamental topic. and stationary intensity distribution means that, on average, you measure a constant intensity at every point in space (most sensors do not measure field amplitudes $\Psi(x,t)$, but intensities $I(x.t) \propto |\Psi(x,t)|^2$). ...

... If you could/would measure the oscillating field itself, then yes, you would indeed measure a sine wave at every point in space (except where you have 100% destructive interference).

So it's not that there exists a point for which the intensities are approximately constant at that point? You need it to be for the whole observer space?

(Sorry, I was just typing the "sidenote" when you posted your comment...)

@Snared No, for a perfectly coherent wave field, the intensities will be constant at every point in space! Such that no matter where you'd place a screen or image sensor, you'll always see some stationary pattern.

seems impossible with only finitely many emitters, certainly none of your photos are even close to that being the case

@Snared Sure, cases I and II are exactly that! The entire field is stationary w.r.t. to intensities. Field amplitudes will, obviously, oscillate everywhere... I would strongly recommend to download the simulation program (or any similar 2D wave simulator) and see for yourself. The linked program is an EXE doesn't require installation.

@GerryMyerson Your point being...? The edit history is public and therefore 100% transparent. No significant changes to the question were made, just added simulations/calculations/small corrections in the text.

Point 1: it repeatedly bumps your question to the front page, thereby bumping off some more recent question. Point 2: it doesn't inspire confidence in anyone tempted to write an answer, as the question could change at any minute. Point 3: an unsympathetic observer could interpret all the edits to mean that OP has no idea what he/she really wants to ask, and said observer could, again, be deterred from trying to help. Please, pick a version and stick to it, and stop all these attention-grabbing edits.

@GerryMyerson Interesting, I wasn't aware of this "bump" functionality triggered by edits...sure you're not making this up? Considering the rest - I think these problems exist mainly inside your head. A more positive outlook would help - and some gratitude/appreciation: SE exists solely because people spend their free time here. Without our contributions, this platform would be an empty shell.