Pop Quiz: Michelson Interferometer

Inspired by one of yesterday's easy questions, a pop quiz for you. The figure below shows a Michelson Interferometer:


A laser falls on a beamsplitter, which allows half of the light to pass straight through, and reflects the other half downward. Each of those beams then hits a mirror that reflects it directly back where it come from. The beams are recombined at the beamsplitter, and then fall on the viewing screen at the top of the figure.

When we add together the light from the two paths, we find that if the lengths of the two arms (that is, the distance from beamsplitter to mirror) are exactly the same, then we see a bright spot on the viewing screen that is as bright as the original laser (this is called "constructive interference"). if we move the right-hand mirror back by one half of the wavelength of the laser, we see no light at all on the viewing screen ("destructive interference").

Here's the question: When there's no light on the viewing screen, we have light going into the interferometer, but no light coming out. What happened to the light?

Feel free to elaborate on your answer in the comments, or suggest additional answers that I may have left off.

(Interferometer picture grabbed from Stony Brook.)

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Goes out the way it came in, but isn't visible due to destructive interference. Light (or any kind of energy) can't be "destroyed".

Have to admit, this is one of those things I never thought to ask when in high school, and now I'm puzzled by it. As Waterdog says, the energy can't just disappear, so nothing is destroyed by destructive interference. But where does it go? My best guess is "heat" (that seems to be the correct answer to 90% of all questions like this), but that doesn't really answer anything.

I thought the Copenhagen interpretation and instrumental interpretations of QM were not so dead that they could be left out. There is nothing between the preparation apparatus and the measurement apparatus of which we can speak, perhaps? One can speak of what different measurement results one obtains when the beam-splitter is or is not present in the experimental apparatus, but from the different rates and statistics of the thermodynamic transitions of the "detectors" one cannot suppose either that there are particles or that there is a field between the preparation and measurement apparatuses. Old-fashioned in these days of realism about the quantum state, of course, and it's just one of many interpretations. Another choice, the antipode of "I was told there would be no math", The experimental results of this and similar experiments can be modeled and engineered effectively using quantum mechanics.

I can't guess, but I can start a very wild guess (I dispatch trucks for a living). Phase 2 photon can be interpreted as the anti-particle of phase 1 photon, so the destructive interference can be interpreted as a particle-anti-particle annihilation. I have no idea if or how such a photon annihilation produces other energetic particles that are not detectable to the viewing screen, but if I did, then maybe I would guess that they would conserve the energy of the light, while not showing up on the viewing screen.

The problem comes in thinking of light as a thing, a particle. If you think about water waves, then obviously if you have and up wave and a down wave and they have exactly the same amplitude and energy etc they will meet and cancel, and you never ask "where did the water go?" or "where did the wave go?". The water is still there, just the interactions cancelled out... Same with light, right? The electrons are still there, you've just stopped them from interacting in such a way that visible light is emitted at the end...


or LIGO's two installations, or the interferometer watching an Eotvos rotor. When the output is dark the energy recirculates. When non-zero output appears that is leakage. Look at the energy contained in a running multiple pass LIGO interferometer! One wonders whey its mirrors are not deflected by light pressure.

There is a fun part, as with total internal reflection. What happens if a lossy interface is within evanescent field distance *on the output side* of the beamsplitter? Frustrated total internal reflectance does spectroscopy IR to visible.

The light can't go back where it came from, either, because the phase shift between the two beams on that path must be the same as the phase shift on the path leading to the detector. But the energy from the incident laser beam has to go somewhere, and in the idealized version of the problem the beamsplitter is the only place it can go.

By Eric Lund (not verified) on 14 Jul 2009 #permalink

Light is just an electric field and a magnetic field propagating through space in a configuration in which all three are mutually perpendicular. So when two beams of light destructively interfere, it's just their EM fields "canceling each other out".

So whatever is happening to the light is the same thing that's happening to any electric or magnetic field which is being destructively superimposed upon. Which probably doesn't answer the question as simply as intended. I blame J.D. Jackson.

Also, there can't be no light on the viewing screen - there will be rings of constructive and destructive interference.

The constructive interference has "more light" (energy). In addition, the light going out the input port should be opposite in interference pattern (due to the phase shifts of the differing numbers of reflections).

In the net, no energy is lost or gained...

Joe beat me to it. It's called an interference pattern for a reason. Integrated over the whole surface, the energy in the bright and dark fringes is that of the incoming beams.

Glad my question was so inspirational. Those are some of my "stumpers" that I like asking to scientists who you would think would know the answer but haven't really thought about it before. I was happy to see the 'pyrolysis' answer, and happy to see the example of its common manifestation in cooking.

RE: the interference:
Born and Wolf (7th extended) appendix XI has a general proof of energy conservation, although it's not very approachable, like anything in B&W.


I'm think it's supposed to go back the way it came (minus a small amount of absorbed energy). The beam-splitting process is, after all, time-reversible.

As several commenters said, if you add together all the light and dark fringes, the energy should remain constant. The same occurs here, except there is really only one "dark fringe" (the viewing screen) and one "light fringe" (returning towards the laser). Add these two fringes together, and the result will remain constant regardless of the position of the mirrors.

If you move the mirror back by one half wavelength, doesn't that create a total path difference of one full wavelength and thus leave the interference pattern unchanged?

By Anonymous (not verified) on 14 Jul 2009 #permalink

If you move the mirror back by one half wavelength, doesn't that create a total path difference of one full wavelength and thus leave the interference pattern unchanged?

By Anonymous (not verified) on 14 Jul 2009 #permalink

If you move the mirror back by one half wavelength, doesn't that create a total path difference of one full wavelength and thus leave the interference pattern unchanged?

By Anonymous (not verified) on 14 Jul 2009 #permalink

Of course, this is a modern Michelson interferometer. Michelson didn't have lasers for the original.

The provided answers are a little oversimplified so here's my best guess: From a wave perspective the light arrives out of phase and therefore the two beams cancel each other out. From a particle perspective the photons are still there. They're just not visible due to their out-of-phase wavelengths. The energy of one beam is essentially being used to make the energy of the other beam invisible but the sum of the energies of the two beams has not changed. Thus the mass/energy of the photons is conserved but they are now invisible.

By Anonymous (not verified) on 14 Jul 2009 #permalink

To clarify (or maybe confuse) things a bit: this is a hypothetical idealized Michelson interferometer in which the beams have zero divergence. Thus, you don't get a ring pattern, just a single spot of light.

One thing maybe people didn't make fully clear: the interference can reallocate the energy but not get rid of it or add to it. I always used to wonder, what happens in a symmetrical beam splitter (they exist!) such as a fused glass cube with metal film along a diagonal. In such a case, relative amplitude changes must the same regardless of which way the light enters. Hence I long ago puzzled: if we send two beams into the SBS, in phase, and their combined split waves exit in phase, then the net output is double the entering intensity! (Do the math.) I couldn't figure out, what "gave" in this case. In a Usenet discussion, someone claimed there had to be a 90° phase difference between transmission and reflection at each face, which would indeed fix things for conservation. (REM an SBS doesnât require equal phases for transmission & reflection, only equal handling regardless of which entering face.) But I never found out, if thatâs what they really do â anyone here?

Another weird thing, is that putting a grey filter in one leg of say a Mach-Zehnder interferometer should actually attenuate the amplitude of the transmitted wave when thereâs no absorption. (Well, a partially silvered mirror will, so ....) Hence the split beams arenât balanced anymore, and so the output pattern (post-recombiner) is different! Thatâs weird, because you wonder how passing through something that didnât absorb the photon anyway, can make that portion of the wave weaker. (After all, it isnât a stream of photons, some being absorbed! Itâs just the one photon ....)

Another issue with interferometers and beam splitters is time reversal. AFAIK, we can take the paths of light in an interferometer (or really, just around one BS) and run the path backward. The interference (such as destructive) will produce the same stuff coming out, as was formerly going in. So far so good AFAIK. But consider a polarizing beam splitter. An input âfrom the leftâ of mixed |R> and |L> bases will be split, with e.g. the |R> going through and the |L> reflecting âupâ. Well, try to run that back in time â it doesnât work! The reversed |R> and |L> beams canât interfere such as to produce no beam going down, etc. They are independent bases. Anyone ever think of this before?

That reminds me, that the term âinterferenceâ is rather loose and non-rigourous anyway. The fundamental phenomenon is âsuperpositionâ, and the careless expression âinterferenceâ tends to mean, when squared amplitudes of superpositions produce noticeable patterns. Really, light at 90° phase difference does âinterfereâ in the strict sense, it just doesnât produce a distinct âpattern.â I think that bears on the âdecoherenceâ explanation, but I won't start a whole thrash about that again just yet ;-)

Here's a tip for those considering the "it's still there" solution:

Power is going in. Potentially a lot, but for physics that's not important. Therefore, for energy to be conserved, something is getting hotter. Pick one.

Or, alternately, consider the possibility that your solution is arguing for nonconservation. Which would be a wonderful thing to demonstrate and doubtless make you eternally famous ...

By D. C. Sessions (not verified) on 14 Jul 2009 #permalink

I voted for the destructive interference since the question was asking about where the "light" went. But I believe that if it's a question of where the energy emitted by the laser goes, the laser would see the bright spot configuration as one output impedance, and the other configuration as a different output impedance, similarly to how different antennas manifest themselves to a radio transmitter. In the second configuration, the laser would be unable to get rid of its input energy and heat up, thus explaining what happens to the energy.

Consider the half of the light that reflects off the beamsplitter and hits the fixed mirror. It bounces back up and hits the beamsplitter again, and gets split again. So half of the half, one quarter, passes through the beamsplitter and hits the viewing screen, while one quarter reflects back to the laser.

Similarly, of the half of the original light that passes through the beamsplitter, half of the half will pass through again and proceed to the laser, and half of the half will reflect to the screen.

What's the difference between the two quarter-beams that combine on the screen, and the two quarter-beams that return to the laser? Well, the two quarter-beams that hit the screen have both reflected off the beamsplitter once, and passed through it once. But one of the quarter-beams that returns to the laser has reflected twice and passed no times, while the other has reflected no times and passed twice.

By Bob Hawkins (not verified) on 14 Jul 2009 #permalink

It goes out at the source. For the light at the detector to be equal to the inbound light, no light must be leaving elsewhere, the source is the only other place it can leave if it isn't absorbed (which is ruled out by the "ideal" materials), and the only way we get no light coming back out at the source is if the phases work out such that they cancel heading back to the source. Phase shift shifts occur both from traveling a distance not evenly divisible by the wavelength and being reflected.

The light getting to the detector via M1 has phase shifts from the emitter to the BS, being reflected in the BS, the BS to M1, being reflected by M1, M1 to the BS, and the BS to the detector. The light getting to the detector via M2 has phase shifts from the emitter to the BS, the BS to M2, being reflected by M2, M2 to the BS, being reflected by the BS, and the BS to the detector. Cancelling out the common elements, the net phase shift should be the 2 times differences in phase shift on the paths from the mirrors to the BS. So when the paths are the same, no phase shift and full constructive interference. When the mirror is moved so that the difference in the phase shift is 180 degrees, full destructive inteference.

For the light heading back to the source, the light taking the path involving M1 gets the the emitter to BS, reflected in the BS, BS to M1, reflected in M1, M1 to BS, reflected in the BS, BS to emitter phase shifts and the light taking the path involving M2 gets the emitter to BS, BS to M2, reflected in M2, M2 to BS, and BS to emitter phase shifts. So the difference in phase shift is 2 times difference in the path lengths from the BS to the mirrors plus 2 times the phase shift from being reflected in the BS.

When n2<

Continued due to forgetting not to use less-than signs in blog comments...

When n2 is much less than n1 the phase shift for total internal reflection with an angle of incidence of 45 degrees is 270 degrees. Two times that is 540 degrees, which is equivalent to 180, which gives us differences betweeen the phase shifts within for the light returning at the emitter and the detector being 180 degrees from each other. So it seems like a BS that works via fustrated total internal reflection will exhibit the phase shifts necessary.

What happened to the light?

It reflected back into the laser, messing with the wavelength of the output light and (if it's a diode, at least) trying to ruin the laser. :)

This is freshman physics

By Spankmaster S (not verified) on 14 Jul 2009 #permalink

I heard on another thread, that I misconstrued the functioning of polarizing beam splitters: one base state will be transmitted regardless of which way it comes in, and the other will be reflected. I thought it depended on which way it went (well, I suppose it can't.) So that ends up making sense. I don't know where or how I got a wrong take on a PBS.

So, when do you post the answer?

Tomorrow, around 10:30 am.
I just scheduled it.

I heard on another thread, that I misconstrued the functioning of polarizing beam splitters: one base state will be transmitted regardless of which way it comes in, and the other will be reflected. I thought it depended on which way it went (well, I suppose it can't.) So that ends up making sense. I don't know where or how I got a wrong take on a PBS.

The beamsplitter in this example is not a polarizing beamsplitter. If it were, the two beams would have opposite polarizations, and then would not be able to interfere with each other.

The light goes back out where it came from.

RE: LIGO, where I did an internship last year. Yes the beam intensities (~20kW in the arms) do cause a force on the mirror, however this is dynamically corrected for, I don't remember the magnitude of the force however.

Part of the reason the power is so high is that the light is recycled in two places. LIGO has a power recycling mirror (PRM) between the laser and the rest or the interferometer that reflects the light from the symmetric port returning to the laser back into the interferometer. This is in addition the light held resonant in the arms cavities.

Re: alkali's comment about messing with the laser try looking up Faraday rotators which can solve this problem.


By Philip Roberts (not verified) on 14 Jul 2009 #permalink

The beamsplitter in this example is not a polarizing beamsplitter. If it were, the two beams would have opposite polarizations, and then would not be able to interfere with each other.
Right, I got on a tangent about time-reversal (which applies to all wave phenomena, true?)

The issue with gray filters is really weird - how a gray filter can attenuate the amplitude of a single photons' WF even though no absorption took place - and even long after we find out it didn't (say, by checking for increased energy in the filter.) As someone noted elsewhere, that blends into the issue of an opaque filter, which spoils the interference pattern even before it ever gets hit (quantum "seeing in the dark.")

I don't think you can posit an ideal non-diverging beam when discussing a quantum effect. You can try, but there will be escapes sideways, so you will get interference rings, so the energy just manifests off-axis, but it is still there.

Of course, my freshman physics was in 1965. My career as a logic designer never needed this part.

And shame on you, Spankmaster @26, for your glib dismissal. If it is so easy, say what the answer is.

By Gray Gaffer (not verified) on 14 Jul 2009 #permalink

In the case of reflecting back into the laser, suppose you replace the two totally reflective mirrors with 2 lasers and eliminate the original laser and screen.

Adjust so that one beam after the 50% reflective mirror is canceled out. I'm not seeing why the other one should not be canceled out too. Which seems like it would be a most bizarre situation to observe.

By Ret Jekel (not verified) on 14 Jul 2009 #permalink

Rohde & Plaut argue that connectionist networks intrinsically extract more basic covariations in training data before extracting more.

The light goes back into the laser. The beams reflected by the mirrors are actually split again by the beam splitter.

When doing light interference experiments, the total amount of energy is conserved. I.e., if you don't see the light at your detector, it must go somewhere else.

This doesn't work for sound interference: sound waves can be converted to heat.

It's been 15 years since I had my last optics course, but we did cover this case.

The interferometer needs its geometry changed; as it's drawn the other output appears where the laser is.

By MadScientist (not verified) on 15 Jul 2009 #permalink

It goes back to the source. It's sorta like when you put interference layers on a glass plate to reduce reflection you also increase transmission.

And the correct answer is: The light isn't going anywhere. In the geometry where no light hits the screen you have a standing wave pattern, so it can be argued that the light is just sitting there. This is especially true if you think of the medium you're lasing as lossless - the geometry of a basic laser consists of the lasing medium sandwiched between a partially silvered mirror and a fully silvered one combined with some way to excite the laser into action. In the real world there will be losses in each mirror and the lasing medium.

You can see a similar effect with a simpler geometry by looking at the reflection and transmission of a Fabry-Perot interferometer - two partially silvered mirrors can be 100% transparent given the right separation.

By BlackGriffen (not verified) on 15 Jul 2009 #permalink