# 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!

• To see interference one needs reflection AND a coherent source. Most light sources that we use are not coherent enough to observe strong interference patterns on thick glass plates. A skilled experimentalist should have no problems demonstrating interference with a laser on a reasonably thick window glass plate of good optical quality. Commented Oct 15, 2014 at 20:34
• Did you see youtube.com/watch?v=nqdi3R64lHE? I can't guarantee that they didn't fake the interference patterns and I don't know it the glass plate qualifies as "thick enough" for your question, but it's a simple attempt at a demonstration of interference, anyway. That the glass industry cares about the phenomenon can be seen in this article: "CHARACTERISTIC INTERFERENCE PATTERNS OF PLATE AND WINDOW GLASS*", F. W. Adams andP. W. French, Journal of the American Ceramic Society, 2006, Volume 24, Issue 11 Pages 341–381 Commented Oct 15, 2014 at 20:54
– user4552
Commented Oct 15, 2014 at 22:59
• @BenCrowell: Honestly, I should back up my claim about the experiment. I have a laser and a camera and I know halfway what I am doing... if I can manage to take a useful photo, I'll post it. Commented Oct 16, 2014 at 4:53

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.

• Nice(+1), I wanted to add that unintentional etalon fringes in the windows of optical instruments is a known hazard, and you'll sometimes see windows with a slight wedge. Commented Oct 15, 2014 at 23:55
• No, this is wrong. IYou can have interference without fringes, e.g., the entire plate could be black due to destructive interference. This is all about coherence length, as described in CuriousOne's comment.
– user4552
Commented Oct 16, 2014 at 3:05
• @BenCrowell - The entire plate cannot be black since it is not possible that all the different colors have an integer number of wavelengths that fit. Colored light == short coherence length of light. I made this point quite explicitly - for both the spatial coherence (point source vs extended leading to a blurring of fringes) and length coherence (color). I don't believe that my answer contradicts the comment that CuriousOne made, it expands on it. It is possible to make fringes appear; but I believe that when OP said "see interference effects in glass windows" he talked about everyday life Commented Oct 16, 2014 at 5:38
• @Floris: I was referring to the monochromatic case. Colored light == short coherence length of light. Huh? For example, red laser light is colored, and has a long coherence length.
– user4552
Commented Oct 17, 2014 at 0:10
• @BenCrowell meant polychromatic light ; multi colored. I thought OP was talking about the fact that you can see interference effects in very thin reflectors (oil film, soap bubble) but not in window panes. That was the angle from which I wrote my answer. Commented Oct 17, 2014 at 2:41

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...