Lets assume we are talking about Newtonian waves, to avoid all that odd photon behaviours stuff.
When two waves interfere destructively, where does their energy disappear to? In the diffraction grating experiment (I know that’s photons and we were going to try to avoid them), the dark spots from destructive interference are balanced out by doubly bright spots from constructive interference, but is that necessarily always the case? - is it possible to have entirely destructive interference?
In general, when waves cancel out, their coherent energy is transformed into incoherent energy, otherwise known as heat, often mistaken for a late-night dorm beer bash.
Mangetout, yes, it is entirely possible to have entirely destructive interference. This does, however, require you to have a perfectly spatially and temporally coherent source of light, i.e. a laser.
What I mean by spatially coherent is that a light ray originating from one part of the light source has the same frequency and amplitude as a light ray originating from any other part of the ligh source. This doesn’t happen with normal light.
By temporally coherent, I mean that a light ray given off at time t0 will have the same frequency and amplitude as a light ray given out at a later time t1.
Then, if you put a filter, consisting of two very thin slits in front of the laser aperture, and place a screen about a metre away from the laser, you’ll see a series of bright spots with dark gaps between them. The dark gaps are regions of destructive interference between the two light rays.
OK, now where has the energy gone from the two waves that interfered destructively?
Sound waves, I could understand, as the destructive interference could simply result in the air being heated up a little, but light does not require a medium for transmission…
(I have a feeling that we’re straying in the direction I didn’t want to go)
Not true. Perfectly ordinary sunlight is coherent, although over a much shorter range (the coherence length). Keep in mind that Dennis Gabor invented the hologram – which is the recording of coherent wave interference – in 1948. The laser wasn’t invented until 1960, and wasn’t applied to holograms until 1962. So people were making holograms without lasers for 14 years.
Where does the energy from destructive interference go? nto the simultaneously-created constructive interference. When Thomas Young did his famous two-slit experiment circa 1800 (complete destructive interference 160 years before the laser ! And Grimaldi was doing it even earlier.) He noted that one slit produced a broad swath of light, but two slits created a series of alternating light and dark bands. Well, the dark bands were obviously darker than the pettern with only one slit, but the light bands were also brighter than the corresponding points in the single-slit pattern. Energy isn’t destroyed, it’s just re-arranged.
Years ago, at the Institute of Optics in Rochester, a similar problem came up during a coffee hour. Let’s say you build a two-arm interferometer, they said – you send a laser onto a 50/50 beamsplitter at 45 degrees. It sends half the light one way (as idf it were a window) and half the other way (as it the beam splitter were acting as a mirror). You take those two beams and bounce them off mirrors so they travel along the sides of a rectangle and recombine them at another beamsplitter (at the corner of the rectangle diagonally across from the first one). Now you have two beams coming off that second beamsplitter. By arranging the lengths of the sides of the rectangle you can get one of those emerging beams to perfectly cancel out. Where does the energy go?
This one had even the professors arguing about it. “Do you mean to say that if I did this with a megawatt laser the energy would be completely cancelled out, and I could put my hand in there?” “Yes, but heaven help you if the path length shifts while your hand is in there.”
So where did the energy go?
Into the other arm! Even though you’ve caused complete destructive interference in one arm of the interferometer, the same conditions cause complete constructive interference in the other arm. All you’ve done is re-arrange the energy, not destroyed it.
Your explanation is clearly a good one Cal, I just don’t quite understand the interferometer bit; can you explain it in more depth? (I’m trying to picture it).
Me too…
Is it called a two-arm interferometer because the two beams of the laser follow two different paths? If so, when the destructive interference takes place, the two beams have re-combined and so there is no “other arm” for the energy to appear in… or have I misunderstood?
Grim
From another messageboard tackling the same question
Not sure whether that makes things better or worse…
Grim
I’m guilty of not using the proper terms. Forgive me.
I’m thinking of something like a Mach-Zender interferometer. Imagine a rectangle. Inject the laser beam at the upper left corner. You’ve got a beam-splitter here that lets one beam pass through to become the upper line on the rectangle, but which acts like a mirror to the other half, directing it downward. There’s a mirror set at 45 degrees at the upper right corner of the rectangle, directing the beam downwards, and another at the lower left corner, directing that beam horizontally (so it becomes the lower line on the rectangle). These two beams meet at the lower right corner of the rectangle, where there’s anotherr beamplitter. This a.) transmits half of the horizontal beam and reflects half of the incoming vertical beam, sending the resulting combined beam out horizontally; and b.) transmits half of the incoming vertical beam and reflects half of the incoming horizontal beam, sending the combined beam out vertically (downwards on my diagram). You can completely eliminate one of the two combined beams emerging from this beamsplitter beam by adjusting the lengths of the “arms” (the sides of the rectangles), but you’ll just reinforce the other emerging beam.
As far as I can tell, you’re describing something like this.
If that is correct, please could you describe the effect again, relating what happens at each labelled stage?
(If it isn’t correct, please advise and I will amend the diagram first…)
Somewhat – but you’re recombining the beams at the upper right corner, and I suggested lower right.
No matter. The point is that your diagram is incomplete because you left out the other beam that must emergy from your upper right corner. Not only is there a beam emerging horizoontally and moving to the right (your beam 6), there must also be a beam emerging vertically (upward on your diagram). After all, your beamsplitter will not only reflect part of beam 4, it must transmit part as well. And it can’t just transmit beam 5, it has to reflect part as well.
Of course, people tend to forget these beams – they’re not a useful part of the interferometer, so they don’t draw them in the diagram. In real life you put a Beam Dump there, or you turn that face of a cube beamsplitter matte, or paint it black. People just forget about that “unimportant” arm, which is why even professors were arguing about this issue. It’s the optical equivalent of “The Butler Did It.”
This came up during those discussions.
“THe laser beam isn’t perfectly monochromatic.”
“If you left it on for all eternity it would be.”
“Not in my lab.”
But it’s a dead end. The effect of a laser beam that turns on and off is a broadening of its spectral line. That will cause your destructive interference to get out of p[hase and not be perfect, but that can’t counteract the massive observable effect you see. You really could set up a powerful laser and destructively interfere in one arm, so that you could put a piece of paper in there and not have it burn all the way along the path (out to the wall of your lab). That’s a real effect, and it has observable consequences. The fact that the paper remains unburned all along that onme arm of the interferometer, even though it’s a real laser that we turn on and off over a pretty short time period can’t be explained away on the basis of the eventual dephasing of beams thousands of miles away due to spectral differences. There’s a helluva lot od energy going someplace. It’s in the other arm.
In other setups it may seem that there is no “other arm”. Look closer – some of that energy might be retroreflected back along its original path. That’s another easily-neglected arm, too.
I see, so it isn’t possible to fully recombine the split beams?
Wait a second. If beams 4 and 5 meet at mirror D at opposite phase, wouldn’t that minimize the intensity of both of the exit beams?
Hmmmm. It’s been some ti8me since I looked over this. IIRC, it has to do with the phase changes upon reflection. If you count up the number of reflections in each path you find that they’re not the same for both sets of beams, so one ends up with an excess of them and the other a surfeit, so in one case the two beams are exactly in phase and in the other they’re out of phase. The situation is not symmetric, because you inject your laser beam on one side of beamsplitter A and not the other, so you can’t say both arms are equivalent “by symmetry” – the injection direction splits that symmetry.
Just a thought… is it possible to fully recombine the beams refractively?
Actually, the whole question can probably be simplified by reverting to the idea that made me start this thread in the first place; non-reflective coatings on camera lenses; suppose there is a thin-film coating on a lens that is tuned to just the right thickness that light of a certain wavelength (say green 540nm) reflecting off the top and bottom surfaces of the coating interferes destructively… where where does the energy go?
If (as I suspect may be the case) more of it is transmitted, isn’t that counterintuitive, as the destructive interference occurred to rays that were reflecting back toward the source?
In your case the AR coating is designed to transmit virtually all of the light. The destructive interference keeps it from reflecting, so (aside from the tiny portion absorbed or scattered), it’s all transmitted.
This is really the point of AntiReflection (AR) coatings – they’re not there to reduce the annoying light reflected from your lens, they’re there to maximize the light transmitted. If you don’t have a coating, your typical n = 1.5 glass will lose 4% per surface to reflection. That doesn’t sound like a lot, but it adds up fast. For n surfaces the transmitted light is (1-0.04)^n. For 10 surfaces (not unusual in a camera lens), assuming they’re all free, you lose 1/3 of your light. For 16 surfaces you lose 1/2 of it. AR coatings let you use larger f/stops and shorter exposure times. Photographers are always trying to get more light.
That’s what I was thinking, but if you take a vertically polarized light (with the interferometer laid out on a horizontal surface), reflection won’t change the phase, would it? If you imagine the two vertically polarized rays coming together at the second beam splitter, the E field of the two will cancel each other out.
I’m not saying you are wrong, I’m just confused and am trying to sort it out. Sorry for messing up the thread but it might help someone who’s as confused as I am.
Hang on a second… If I may, let’s go back to Mangetout’s OP and the first reply. Does a conversion of energy into heat mean that, for example, the air between someone’s eardrum and a noise-cancelling headphone would heat up? How much (in relation to decibels of input, for example)?
Or, are there constructive avenues for the energy to follow, such as radiating out the other side of the headphone earpiece?
