How do anti-reflective coatings work?

This has bugged me for a long time. Do anti-reflective coatings really increase the amount of light going through a lens? The simple answer is, “if less is being reflected, then there must be more going through”. But my understanding (which could be faulty) of how anti-reflective coatings work is that the coatings are made to create two reflections that are half a wave out of phase so that they interfere destructively. So the reflections are still there, you just can’t see them.

Certainly less internal reflection in an eyepiece will increase contrast, but my question remains – if the reflections are really still there, then how can the throughput be increased? I suppose that there’s a quantum-mechanical related answer that I may be unaware of. Anyone know?

>>>This has bugged me for a long time. Do anti-reflective coatings really increase the amount of light going through a lens? The simple answer is, “if less is being reflected, then there must be more going through”. But my understanding (which could be faulty) of how anti-reflective coatings work is that the coatings are made to create two reflections that are half a wave out of phase so that they interfere destructively. So the reflections are still there, you just can’t see them.
----------
I am far from an expert on this matter but my understanding the Anti-reflection coatings work by partially matching the index of refraction of the glass with that of air...

Your phase concept only works at 1 specific frequency (actually maybe a series of frequencies) and probably depends on the angle of incidence as well.

jon

Destructive interference is part of it, but it is much simpler to look at it as a reflectance "game".
Reflectance at a boundary between two different mediums of refractive indici n0 and n1 is given by
[(n1-n0)/(n1+n0)]^2. So if we have reflectve coating with n=1.35, the amount of light reflected from its front surface will be 1.5%. The amount of light reflected from the coating/glass (n=1.5) surface is only 0.3%, of which 1.5% would be reflected back from the coating/air (n=1) boundary, and so on, and so on.

If the coating layer thickness T=lambda/4n (with "n" being the refractive index of the layer) the 0.3% secondary reflection will destructively interfere with the 1.5% primary reflection, resulting in a total reflection of 1.2%.
Total transsmission is, therefore, 98.2% vs. 96% w/o coating.

This is very simplified picture, but it's also much easier to grasp than going into propagation of electromagnetic waves.