Yesterday’s Michelson Interferometer quiz was surprisingly popular– as of 8:30 pm Tuesday (when I’m writing this), just under 1500 people have voted in the poll, three and a half times as many as in the next most popular poll I’ve done. Who says there’s no audience for physics?

The correct answer finished a distant second (as of 8:30 it has 21% of the votes, to 52% for the leading wrong answer). The answer is that the light goes back where it came from. Bob Hawkins and MattXIV have the right explanation: on the return trip, half of each beam goes to the screen, and half of each beam goes back to the laser. You get interference on each of those paths, and they complement one another– whatever fraction of the initial light goes to the screen, the remainder of the light goes back to the laser.

Another way to see this is by comparing the Michelson interferometer to everybody’s second favorite interferometer, the Mach-Zehnder. Here’s the Michelson again:

If you take the two mirrors, and turn them each by 45%, then insert a second beamsplitter, you get a Mach-Zehnder interferometer, which looks like this:

The Michelson is basically a Mach-Zehnder interferometer “folded” back on itself. If you look at the Mach-Zehnder interferometer, you can clearly see that there are two ways for the light to leave. When folded into the Michelson configuration, one of those paths leads to the detector screen, and the other goes back to the laser.

The other answers turned out to be even more distracting than I thought they would be, but they’re wrong. The light can’t simply be destroyed, because light carries energy, and energy can neither be created nor destroyed. All of the energy that flows into the interferometer has to end up going somewhere– it can’t just disappear.

It also can’t be absorbed by the beamsplitter, because we said it was an ideal 50/50 beamsplitter. If it were absorbing significant amounts of light, it wouldn’t be sending 50% into each arm of the interferometer. And the energy can’t be disappearing into parallel universes because this isn’t Star Trek, and the parallel “universes” of Many-Worlds aren’t real universes, but rather branches of a larger universal wavefunction.

Anyway, many thanks to “h” for proposing this question. I had no idea it would get such a huge response. It’s nice to see that interferometry has lots of fans.

1. #1 rob
July 15, 2009

wrong. it’s magic.

:)

i got the poll question correct. then again i use an optical measurement system to monitor growth of thin films.

2. #2 Prem
July 15, 2009

I got the quiz wrong, but I think it’s because in your diagram you don’t show that the light can return to its source.
As I’m not familiar with the properties of a beam splitter I did not assume the light could return to it, but as displayed in the diagram thought it only went in certain directions (but don’t ask to me to explain why it would have this strange property)
:-P

3. #3 Ben Lillie
July 15, 2009

Excellent poll, Chad. There was a question on my qualifying exam to describe all the workings of a Michelson interferometer. I don’t think I’ve thought about them since, but the answer was still burned into my brain, so maybe the test was good for something after all.

I wonder if some of the confusion came from the diagram you used. There are double arrows on the paths to the mirrors, but not the laser. That could easily throw someone off.

4. #4 rachel
July 15, 2009

Shouldn’t you move M2 back by 1/4 (not 1/2) the wavelength? The light has to travel the extra distance twice. If M2 is moved back by 1/2 wavelength, the return beam from M2 will have traveled 1 full wavelength farther than the beam from M1 and the two will be back in phase.

5. #5 Prem
July 15, 2009

I got the quiz wrong, but I think it’s because in your diagram you don’t show that the light can return to its source.
As I’m not familiar with the properties of a beam splitter I did not assume the light could return to it, but as displayed in the diagram thought it only went in certain directions (but don’t ask to me to explain why it would have this strange property)
:-P

July 15, 2009

Shouldn’t you move M2 back by 1/4 (not 1/2) the wavelength?

Yes, that’s right.
The mis-statement in the original post was an editing problem– I started with “increase the path length difference by \lambda/2” and tried to change it to “move the mirror back by \lambda/4” but didn’t make it all the way there.

7. #7 James
July 15, 2009

As a physics educator – do you find it depressing that 52% of the (presumably scientifically biased sample (given they check out your physics orientated blog)) went for the wrong answer?

8. #8 Andy B
July 15, 2009

So I’m guessing you don’t want to use a Michelson interferometer with a super-powerful laser? Seems like pumping that power back to the source could potentially cause some damage.

9. #9 lee
July 15, 2009

Question – For the Michelson interferometer, the path length delta for the beams reflected back to the laser from the 2 mirrors is the same as the delta for the 2 routes to the screen (twice the difference of the beam splitter to mirrors distance). So isn’t it the case that whenever there is destructive interference at the screen, there will also be destructive interference of the beams reflected back to the source? Am I missing something?

July 15, 2009

Question – For the Michelson interferometer, the path length delta for the beams reflected back to the laser from the 2 mirrors is the same as the delta for the 2 routes to the screen (twice the difference of the beam splitter to mirrors distance). So isn’t it the case that whenever there is destructive interference at the screen, there will also be destructive interference of the beams reflected back to the source? Am I missing something?

There’s a shift in the wave that is reflected off the beamsplitter relative to the wave that is transmitted through the beamsplitter. The resulting difference in phase is the same as if one had traveled a quarter of a wavelength farther than the other.

If you look at the paths, you’ll see that the path going to the screen involves two beams that have each been reflected once and transmitted once, so the net effect is the same for both. The path going back to the screen, though, adds together a beam that has been reflected twice to a beam that has been transmitted twice, giving a net difference between the two that’s the same as if one beam had gone an extra half-wavelength farther than the other. Meaning that when you have constructive interference on one path, you get destructive interference on the other.

11. #11 lee
July 15, 2009

Chad – thanks for the response.

Is this true of all types of beamsplitters? That is, is there something fundamental about the beam-splitting mechanism that guarantees a 1/4 wavelength phase difference between the transmitted and reflected beam regardless of the physical implementation?

12. #12 Neil B ♪
July 16, 2009

Lee, look at it this way: if you have a symmetrical BS (like metal film inside a glass cube, or thin gap in a cube for FTIR), then it has to have a 1/4 lambda phase difference to conserve energy and be consistent. IOW, if it was all zero, then incoming waves would constructively interfere out of both exits and you’d have “free energy” etc. But the PD doesn’t have to specifically be 1/4. With a BS/recombiner, imagine rightward input w1 and up input w2. Then we could have rightward A-output = w1(same) + w2(same) and upward B-output = w1(1/2 phase) + w2(same). That way there’s constructive interference in A and destructive in B: “it’s all relative” since we can adjust the overall relation between w1 and w2 or the total A v. B.

BTW, I point out that “interference” is a sloppy term anyway. It is used to mean noticeable patterns, but waves are supposed to superpose regardless of whether they make “interference patterns.” This is not just amplitude per se, but full vector addition because the light vector has orientation (polarization). That means, combine |x> + |y> in phase and get 45 linear pol., etc. I understand saying of perpendicular polarizations (ie, orthogonal bases) that “they don’t interfere” based on historical tradition about intensity variations (what we can see just looking), but it’s poor usage. A rigorous description wouldn’t even include the word “interference” but would say “superposition” directly. If the squared amplitudes thereby make distinct patterns, call it “interference” if you want but that’s sloppy talk. I think this relates to the issue of “the states don’t interfere anymore” in decoherence, as follows.

I propose something, perhaps original, that is likely surprising and relates to precise versus careless talk of “interference”, and to decoherence (but no tirade, just the idea speaking for itself): I argue we can experimentally *show* that the combined wave trains of single photons are still distributed into two paths, despite complete prior randomization (incoherence) of their relative phases. IOW, we can show they are distinguishable from single-path photons coming out of either channel A or B but “not both.” What I mean: split a photon at BS1 in a Mach-Zehnder. Have something get in the way of one or both legs (air densities, etc.) to randomize phase as they approach recombiner BS2. The output as such, is randomized and “shows no interference.” But suppose we tag the input legs using polarization. We can start with say |x> LP original input. We put an optical rotator to convert the wave in leg II to |y> LP. When they recombine in BS2, transmitted |x> and reflected |y> superpose in channel A, and transmitted |y> and reflected |x> superpose in channel B. We can arrange for the separated |x> output components to share the same relative phase. We can arrange for the |y> components to be 180° out of phase (both, same-to-same polarization and not within a single channel.)

By itself, that doesn’t do anything. But we can recombine these beams a second time in BS3. Now, the |x> components can constructively interfere only out of the second channel A, “A2.” The |y> components can constructively interfere only out of the second channel B, “B2.” Hence, only |x> LP will come out of A2 and only |y> LP comes out of B2. It doesn’t matter what the phase between |x> and |y> is, because we can show another way that two wave trains existed in transit between BS2 and BS3. If isolated (one-path) photons had been approaching and/or exiting BS2, each could be either reflected or transmitted there or at BS3. When the hit BS3, they again could go either way. So after the photons were polarization tagged, we could get either |x> or |y> in either A2 or B2. But, we don’t because of phase relations – “superposition” of vectors despite not being “interference” in the sloppy, traditional sense. There’s the difference.

Readers may need to play around with diagrams etc. to get this, let me know what you think. This may relate to arguments about randomized phases being “decoherence” and the supposed implications thereof.

13. #13 Neil B ♪
July 17, 2009

Actually, I have a better point, since I don’t think the above really proves the incoming wave wasn’t restricted to a single “leg” at a time. I apparently had a misfire on the randomization issue. This point isn’t about randomization – it’s about the which-way tagging. Tagging each leg as |x>, |y> respectively should (intuitively) provide “which way information” and spoil the chance to get “interference” later. But if we appreciate detecting “superposition” even if it’s not classic “interference”, then there’s a way to show that both paths are still active despite the apparent WWI. For example, in-phase inputs will go out of channel A as a 45° diagonal, which we can measure and distinguish from a mix of random |x> and |y>. Note that we could use actual LP filters, which have a chance of absorbing the photon. One might intuitively” expect that passage of a split-wave through one or both such filters would surely “collapse” it into a localized wave of the corresponding polarization, but it doesn’t – both waves continue, but changed into that polarization.

14. #14 joezairosh
August 23, 2009

thank you very much for the learnings cuz’ i passed the test