Is manufacturing roughness really the only reason we don't see optical interference in thick dielectrics like windows? I had always kind of wondered why we didn't see interference in things like windows -- we were taught that the condition is that the thickness of the film/slab/medium just has to be an integral number of wavelengths, $d=m\lambda$. My logic was, if visible light has wavelengths ~400-780nm, any window pane has got to be an integral number of wavelengths (or close to integral) for some of the wavelengths in the visible.
For example, if a window is, $6mm = 6\times10^6nm$ thick, then you should see interference for wavelengths (and some range around them) 200nm, 250nm, 300nm, 400nm, 500nm, 600nm... Because 6mm is an integral number of each of those.
Obviously that would be really annoying and isn't the case. I found this (warning: PDF) source that seems to explain it and I was wondering if what they're saying is sound. On the 10th page (p171 for them), it says that the reason we don't see this interference is that no window is that smooth, it has a variance in the thickness of at least several wavelengths, so the interference effect is effectively average out among all those thicknesses, which usually gets rid of it. They do the math and show that.
Is that the reason?
If so, that leads me to two questions: Could we, if we were really careful and used something like MBE, make a glass window with that sort of accuracy, and then see interference effects in the optical range? I know that quartz crystal monitors can easily give you nanometer accuracy for evaporators and sputterers, but they also have pretty limited growth rates.
My second question is, then, do we see interference effects in glass windows with electromagnetic radiation of a comparable wavelength, like microwaves?
Thanks!
 A: Very simply, when a plate is quite thick, the fringe patterns will be very close together - because a tiny change in angle will result in an additional wavelength's worth of path difference. Different colors will have a different repeat distance (because of different wavelengths); and light will typically arrive at the eye from more than one direction (spatially, the light source is not a single point).
An example of how white light "washes out" when wavelengths are different is given by the following image (from http://www.doitpoms.ac.uk/tlplib/optical-microscopy/images/wedge.jpg):

which shows what an optical (quartz) wedge will do in terms of interference: different path lengths (from almost nothing on the left, so about 0.08 mm on the right) result in different degrees of interference between the components of white light (which each have their own spatial frequency across this plate). A few more wavelengths and the fringes will be washed out.
Taking all those things together, you don't expect to be able to resolve any fringe patterns. If you used a single, coherent light source (say a laser that was properly collimated and expanded) you might see fringes even on a thick plate of "common" glass at just the right angle - but typically they would be too close together (due to angle of incidence and/or flatness spec) to be visible.
A: In addition to @Floris response:
You have missed a lot of wavelengths in your list of wavelengths that would experience interference.  Take your example of a $6,000,000 \text{ nanometer}$ pane of glass, and consider that 15,000 waves of $400 \text{ nanometer}$ wavelength light exactly fills this space.  So, indeed, this light will experience some sort of interference effect, depending on the nature of the medium on the far side of the glass.
But the next wavelengths to consider as subject to interference effects are only very slightly different: ones where $15,001$ wavelengths or  $14,999$ wavelengths fit into the thickness of the glass.  This corresponds to $400.02666$ nm and $399.9733$ nm 
So, what you basically have is a continuous spectrum, with a huge number of very sharp interference lines.  These lines are so sharp and so numerous that the effect is merely a slight reduction in overall light intensity...
