I don’t understand how though; the light that destructively interferes was on it’s way back toward the source (being reflected); how does it end up being transmitted?
I don’t think the “it turns to heat” explanation was correct. CalMeacham’s explanation seems to suggest that the energy always turns up as a wave somewhere else.
I think it probably does turn to heat in the case of sound waves, beacuse there is a medium that can absorb them; cancellation of sound waves could be thought of as absorption on the molecular scale (I’m making this up as I go along, so somebody please correct me if I’m wildly wrong) - sound waves consist of moving particles; if force A tries to move a particle in one direction and equal force B tries to move it in exactly the opposite direction, the particle absorbs the energy and warms up a little.
Or am I just completely wrong on that one?
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Well, first, you gotta think about the light as waves, or at least as particles with wave-like features. And you have to consider the entire system.
I can see your confusion if you view photons like baseballs. You throw it at a coated lens – let’s say it’s a simple quarter-wave AR coating . Your photon bounces off the first surface it hits. Now it’s on its way back to you. Light has been reflected, case closed.
But it hasn’t it’s a wave, and only part of it gets reflected back at you. Part goes through, and strikes the coating-glass interface. It gets reflected as well. Since the coating thickness is small you’re still within the coherence length of the light, so the amplitudes add. Because the thickness and index of the coating were properly chosen, the two reflected waves cancel each other out, and there’s no net reflection. Actually, you’re going to get a lot of rattling back and forth inside that thin film, so you’ll keep getting coherent waves that add to the amplitude until you run out of coherence length. And if the coating is properly designed, they’ll cancel out perfectly. All of the other light goes into the lens, and re-inforces instead of cancelling, so you have virtually perfect transmission.
If you view the photons as baseballs, you have to consider that a.) there are a lot of them, so you use statistics to get around having part of the photon return; and b.) these are quantum baseballs that can cancel each other out when they meet; and c.) the light is going on for some time, so that the time delay between reflected baseballs isn’t an issue – there’s always a baseball reflected from the inside interface at just the right time to cancel one reflected from the air/coating interface.
Actually, just thinking about the noise-cancelling headphones more, it seems to me that the air in the ear canal will be cooler than it would without the noise cancellation, simply because the pressure waves won’t be knocking all the molecules around.
I understand the deal with light and photons, but at a gut level, there seems to be something different about sound waves. I’m having trouble articulating why I believe there’s a difference, however.
You can try to work out a scheme. I’ve never been able to find a way to completely combine beams without something leaking away.
Incidentally, rainbows are caused by the combination of beams with slightly different path lengths that have been refracted twice and reflected once. They told you in school that rainbows were due to refraction of light, but it’s really an interference effect.
I’m going to pick a small nit, because it might confuse someone. This is true for telescopes, microscopes, cameras, binoculars, and such, but for eyeglasses, I think the main point is to reduce reflections, so people don’t see them when they look at your eyes.
Cal
To check my understanding, in Mangetout’s diagram, what would happen if you made D a mirror? Am I correct in thinking that you would get complete cancellation on paths 2,3,4 and 5 (and no path 6, of course) but have a full-strength beam going “vertically” from A?
If you made D a mirror you don’t get any canellation at all, I’m afraid. In order to get cancellation, you have to combine two out-of-phase beams, and making D a mirror prevents that from happening on path 6. Paths 1-5 each only have the contribution from one input, so there’s no cancellation on those arms, either. And (assuming a two-sided mirror) the path going up from mirror D will only have light from path 5, unmixed with light from path 4, so there’s no cancellation there, either.
I think you mean electromagnetic waves as described by Maxwell’s equations. Sorry for the nitpick.
A possibly easier way to think about this is that Maxwell’s equations as applied to this system simply prohibits any reflection and therefore all the incident energy is transmitted.
On a related note Let’s say you are able to beat two laser beams of slightly different frequencies, you’ll then get nodes where they reinforce, and nodes where they cancel. What happens when you put a piece of matter at the point where they cancel? In this case it can be shown that the lasers have emitted no energy. (however this can’t actually be done)
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Can you actually do this experiment? How do you generate two nearly equal frequencies and “beat” them together?
But, assuming you can, why do you say the laser hasn’t emitted energy? What if you put the paper where the waves add, rather than cancelling? Would the paper catch fire?
There are points inside a laser resonator where there is no net field – there have to be, because the laser is a resonator, with standing nodes. Just because nodes exist at one point in a system doesn’t mean that the power density can’t be nonzero elsewhere. Halfway between two nodes the power density is maximum, and if I put something there it feels it.
Ok, I think I see that. If you make D into a mirror, it is impossible to get the waves out-of phase.
What I was trying to get at is what happens when the waves are out-of-phase but travelling in opposite directions. Let’s suppose that instead of our interferometer, we have two lasers tuned to the same frequency. We set them up facing each other with the beam paths aligned and turn them on. Can we get the beams to cancel?
Oh, that’s easy – I give it in my post just above. Counter-propagating waves of the same frequency = standing waves. You get those in a laser. The node regions, with zero amplitude, gives you regions with no energy. They alternate with regions of high power density.
You can work this backwards – you can build a laser with no mirrors by modulating your gain region high-low-high-low. It’s called a distribued feedback laser (DFB).
Another interesting fact connected with this is that the nodal regions in a laser cavity are underutilized, since the amplitude never gets high enough to pull the energy out of the population inversion there. One reason they came up with the Unstable Resonator design was to “sweep out” those “dead” regions and extract maximum power from the laser.
Interesting stuff, optics.
As I said, no you can’t actually do this. The ony way to get complete cancellation across the whole beam is to send the two beams out of the same source, with same intensity and opposite phase. This is just another way of saying “turn the laser off”.
Yes, and in this case there is an energy flow.
If you place an opaque absorber at any node you will completely disrupt the emitting media within the lasers.
CalMeacham just to be absolutely clear it is evident that you know far more about lasers than me, I just happened to have been in on a discussion on this subject in another forum.
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Indeed it is!
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Hmmm. Is it possible to “tune” our two-laser system so that rather than forming a standing wave, the beams “cancel?” In other words, is it possible to have “no energy” along the entire beam path?
Actually, it is not correct that there is no energy in the node regions. You can easily demonstrate this by calculating the poynting vector of two counter-propogating plane waves. You will find that the the poynting vector oscillates back and forth with the same frequency of the light IIRC (it may be 2[symbol]w[/symbol]). Most people forget this as there are some very handy formulas for the time averaged poyting vectror (ie, the intensity) which is indeed zero.
Actually I would think you could do this experiment using third order optical effects like stimulated brillouin scattering, but it would be very, very tricky. Stimulated Raman scattering could also work, however the frequency shift between the incident light and the raman scattered light can be pretty large. With stimulated Brillouin scattering, the incident light creates (and destroys) acoustic phonons in the material. When creating a phonon, the frequency of the incident light is lowered (in accordance with conservation of energy) by the frequncy of the accoustic wave. This can give you two waves, the incident wave, and the scattered wave which have frequencies separated by anywhere from 10 megahertz to several hundred gigahertz. Since the incident field is coherent, the scattered field is coherent also. Should be doable.
There are several problems of course. Since this is a third order effect, you will need to crank up the intensity of the incident light to the order of 1 GW- 10 TW/cm^2 depending on the material. Also the scattered wave will have a very low intensity.
Alas, I still don’t understand the lens coating thin, Cal, are we saying that the lens with the coating actually transmits more light than a similar uncoated lens?
Forgive me for asking the same question over and over, but I can’t grasp how it is that the rays that that are being reflected back away from the lens and then desctructively interfereing with each other can end up contributing to the light which is passing through the lens in the opposite direction.
Alas, I still don’t understand the lens coating thing, Cal, are we saying that the lens with the coating actually transmits more light than a similar uncoated lens?
Forgive me for asking the same question over and over, but I can’t grasp how it is that the rays that that are being reflected back away from the lens and then desctructively interfereing with each other can end up contributing to the light which is passing through the lens in the opposite direction.
Actually I spoke too quickly, this is actually done now. This is the foundation of optical heterodyne detection.
Perhaps the easiest way to understand this is if you check out the Fabry-Perot Etalon (resonator/interferometer) as this is basically what the coating is. Take a look at this site also.