Hi all,

this may be an old hat for you guys, but it's a question I have pondered many a times, but never came up with a good solution:

Say you split a laser beam and re-merge it later, but (by diverting one of them a bit) so that they completely annihilate each other because of the phase difference.
Now after the merging there isn't any beam anymore, so where did the energy go?

Thanks in advance,
rumborak

Hi all,

this may be an old hat for you guys, but it's a question I have pondered many a times, but never came up with a good solution:

Say you split a laser beam and re-merge it later, but (by diverting one of them a bit) so that they completely annihilate each other because of the phase difference.
Now after the merging there isn't any beam anymore, so where did the energy go?

Thanks in advance,
rumborak

Excellent issue. I am interested in seeing what resonses you get.

Hi rumborak,
I think that it is not possible to have complete destructive interference... that you can always find areas of constructive interference as well.

I could be wrong, of course.

Pete is correct. You can never have only destructive interference – there will always be constructive interference in some places and destructive interference in others. The energy goes to zero at the points of total destructive interference, but increases in the areas of constructive interference.

If you do manage to set up a system that has total destructive interference, than your waves will never exist in the first place, and your 'lost' energy will be the energy dissipated by the two competing wave generators.

Here's a tricky exercise:

<img src="/attachment.php?attachmentid=3166&stc=1"> (http://www.sciforums.com/attachment.php?attachmentid=3166&stc=1)

The attached image shows a laser, two half-silvered mirrors (1 and 4, silvering is on the side nearest the laser), two full mirrors (2 and 3), and two detectors.
The mirrors split the laser then recombine it.
The difference in the two paths is an integer-and-a-half wavelengths

Exercise: What do the detectors see, and why?

I have to admit that I'm not sure of the result for this one - there are some tricky things involved with the reflections.

PS - all you geniuses (genii??) please let rumborak have a go first!

Heh, well, it's not as if I had a solution for it, otherwise I wouldn't have asked! :D

The "destructive interference always has a compensating constructive interference" rule I know.
But that would mean there HAS to be a beam, somewhere else, because we're eliminating the one (or the two rather, in Pete's experiment) after the merging mirror completely.

So, the only thing I can imagine is that instead of being merged, the two partial beams would get reflected at that mirror. After all reflection is caused by the electromagnetic field response of the electron gas in the metal of the mirror, instigated by the field of the incoming light wave.
So, maybe exactly that phase relationship of the two beams creates a different response of the mirror?

I'm so clueless...

rumborak

The response of the mirror is not dependent on the nature of the incident light.

Forget the "destructive interference always has a compensating constructive interference" rule for the moment - if it really is a valid rule, then we should be able to independently confirm it for particular scenarios, such as this one. If the rule is correct, then an analysis will show that the sum of energy detected at A and B is the same as emitted by the laser (or that there is another outgoing beam that I haven't considered). On the other hand, if a correct analysis shows that energy out does not equal energy in, then the rule must be incorrect. This is a standard falsification test.

So, on analysis, what do you think A and B should detect?

Well, obviously I think A and B won't receive any signal because of the interference.

Sorry Pete, I'm not too much in the mood for "hide and seek". Do you know the answer?

rumborak

I'm not sure.
I do know that it is important that light reflected from a medium with a higher index of refraction undergoes a 180° phase shift, as shown in this diagram found at U.Penn:
http://dept.physics.upenn.edu/courses/gladney/phys151/lectures/images/wave_reflection_different_media.gif (http://dept.physics.upenn.edu/courses/gladney/phys151/lectures/lecture_apr_18_2003.shtml)

I'm not completely sure of the implications, because I'm not sure about the silvering and half-silvering on the mirrors... is the silvering a medium, or do we only need to consider the air and glass mediums?

I'm assuming that silvering automatically does the above. In that case, B would see nothing and A would see the original laser.

How do you measure the index of refraction for a metal? Or one that's used in silvering?

In the opening scenario, I'd guess the energy goes into the act of cancellation. It takes 1/2 the energy to cancel the other half, so you get zero output. Hmm, is that a perfectly elastic collision?

It seems to me it almost becomes a case of statics (assuming ideal conditions). Half the force of the beam is just enough to cancel the other half. At the point of collision, it's two equal forces pushing against one another. The force that propagates the wave is met by an equal and opposite force; yielding equilibrium.

Something tells me I'm not quite thinking correctly about that.

In Pete's diagram I'd think the detectors would detect the 1/2 the beam's energy minus the energy absorbed by the mirrors exactly 180 degree phase offset, since the beams don't interact when normal to one another. That's just a guess though, as I was mostly lost during that part of physics and don't remember much about it, so I'm not sure if the beams would interfere or not when normal to one another.

Here's what I think happens in the illustrated scenario.
The blue bit is the kicker.

Long path to Detector A:
Mirror 1 - beam reflects off the front of the mirror. 180° phase shift.
Mirror 2 - beam reflects off the front of the mirror. 180° phase shift.
Mirror 3 - beam reflects off the front of the mirror. 180° phase shift.
Mirror 4 - beam passes through. No phase shift.
Total phase shift due to reflections: 180°

Short path to Detector A:
Mirror 1 - beam passes through. No phase shift.
Mirror 4 - beam reflects off the front of the mirror. 180° phase shift.
Total phase shift due to reflections: 180°

Detector A summary:
Phase difference due to reflections = 0°.
Phase difference due to different path length = 180°.
Total phase difference = 180°. Destructive interference, no signal detected


Long path to Detector B:
Mirror 1 - beam reflects off the front of the mirror. 180° phase shift.
Mirror 2 - beam reflects off the front of the mirror. 180° phase shift.
Mirror 3 - beam reflects off the front of the mirror. 180° phase shift.
Mirror 4 - beam reflects off the back of the half-silvered mirror surface. No phase shift.
Total phase shift due to reflections: 180°

Short path to Detector B:
Mirror 1 - beam passes through. No phase shift.
Mirror 4 - beam passes through. No phase shift.
Total phase shift due to reflections: 0°

Detector B summary:
Phase difference due to reflections = 180°.
Phase difference due to different path length = 180°.
Total phase difference = 0°. Constructive interference, full recombined signal detected

If that is the case, what do you think would happen if the phase shift due to the path difference alone was an integer multiple of the wavelength instead of a half-integer multiple?

Interactive toy for interference:

http://www.colorado.edu/physics/2000/applets/fourier.html

The easiest method of creating the null is to watch the two balls in the lower right quadrant and to set on on top of the other after dragging one of the balls into a negative (below the line) while the other is positive (above the line).

Also:

http://www.kettering.edu/~drussell/Demos/superposition/superposition.html

The question at hand however is interesting in theory. Travelling waves do not remain in a constructive or destructive pattern and result in the null and dual intensity energy densities, wherein on the average the energy is conserved.

However, consider an Op Amp with two perfectly matched singals one + and one -. The input energy of the + vs the output is calcuable, likewise for the - input.

However if both inputs are driven simultaneously there is no output. Energy hasn't disappeared. The Op Amp still draws power but the efficiency of transfer to i.e. acustical in case of a speaker, has dropped to zero.

If in theory, however, one could generate signals where there is an energy output and they could be perfectly matched and aligned 180 degrees out of phase with the same vector, energy conservation appears to be violated.

Light is a funny beast and I don't think it resolves the issue. Perhaps an acustical laser (there are such things BTW) where the beams are merged in a simular fashion to mirrors and light would be interesting.

If that is the case, what do you think would happen if the phase shift due to the path difference alone was an integer multiple of the wavelength instead of a half-integer multiple?

Hi James,
Are you talking to me?

Actually, I've stuffed up... the beam is being split twice, right?
Now I figure that Detector A sees nothing (destructive interference of two 1/4 strength signals), while Detector B sees a 1/2 strength signal (constructive interference of two 1/4 strength signals)... I'm confused?

Hi Mac,

Doesn't an ideal OpAmp draw no power from its inputs, and all output power comes through the voltage rails?

If so, then how could energy conservation appear to be violated?

Hi Mac,

Doesn't an ideal OpAmp draw no power from its inputs, and all output power comes through the voltage rails?

If so, then how could energy conservation appear to be violated?

You misconstrued my post. I stated that the Op amp input/output efficiency would go to zero but that isn't the same as loss of conservation.

IF two acustical signals could be produced and perfectly matched out of phase and aligned in a comoving beam, then that would raise a question of energy conservation.

And yes Op Amp circuits can be high impedance and are used to design buffer circuits but pragmatically I don't believe any are considered perfect insulators.

Gotcha. It's a big IF, of course.

Gotcha. It's a big IF, of course.

Gotcha, if you choose to see it that way but it is obvious the question is hypothetical since nothing can ever be perfectly synchronized and aligned impertuitedly and thus there will be harmonic nulls and constructive envelopes.

Except in the case of the universe which seems perfectly harmonized and may well be a simple case of existance in a constructive interference vs "Nothingness" which would be "Destructive" interference.

Bifurcated +/- s. :D

Gotcha.

:o Bad communication - sorry!

My intended meaning was "Ah! Now I get what you mean!"

Pete

:o Bad communication - sorry!

My intended meaning was "Ah! Now I get what you mean!"

Pete

Not a problem. ;)

Say you split a laser beam and re-merge it later, but (by diverting one of them a bit) so that they completely annihilate each other because of the phase difference.
Now after the merging there isn't any beam anymore, so where did the energy go?
At the point of merging their phase difference is always 0, or at least in Pete's setup. Can you give any setup where the merging beams are out of phase with each other so that they completely annihilate each other?

As for the interference pattern, it is rather energy distribution pattern than destruction & construction of energy.

At the point of merging their phase difference is always 0, or at least in Pete's setup.
Can't you adjust the mirrors so that the difference in path length gives destructive interfence?

In your setup, the number of reflections only matters irrespective of path length.

In the longer path, only the fractions of the wavelength count, not the integer multiples of wavelength, am i right ? Then, the fractions of wavelengths, before and after each reflection, would add up to yield 180 degree phase difference between the incident and reflected beams no matter what is the fraction before a reflection.

Since there are 4 reflections before merging the total phase difference is 4*180 degrees (0 phase difference) for any path length.

No... sorry!
Change in path length directly produces a phase shift.

No... sorry!
Change in path length directly produces a phase shift.
That might be the case as in a long overlapping ray of light travelling around just above the event horizon of a black hole. But in your case there are reflections involved.

Lets see your fig :

<img src="/attachment.php?attachmentid=3166&stc=1"> (http://www.sciforums.com/attachment.php?attachmentid=3166&stc=1)


The ray 1-2 is out of phase with ray Laser-1 by 180.

The ray 2-3 is out of phase with ray 1-2 by 180 and out of phase with ray Laser-1 (or ray 1-4) by 0. That is the horizontal rays 1-4 and 2-3 are in phase.

The ray 3-4 is out of phase again with both rays 1-4 and 2-3 by 180.

The finally merged ray 4-detectorB has both rays are in phase.

The same is the case (in phase) for 4-detectorA too. Both the detectors A&B would detect the laser rays with lesser intensity than the orginal ray from the source.

everneo, the path length difference affects the phase shift as well as the reflections. For example, a difference of n+0.25 wavelengths in path length would produce a 90 degree phase shift, in addition to the phase shifts produced by the reflections.

Nope, the fractional difference due to path length difference are made to result in integer multiples of wavelength after every even number of reflections, for any path. Both the rays will be in phase after even number of reflections irrespective of the location of the mirrors.

That's crap.

Look up thin film interference (http://users.erols.com/renau/thinfilm.html)

yeah, i have been carrying on your holy crap for too long even after you discarded it. :p James R was right.

here is another holy crap for you :

Keep the mirrors 2 & 3 at a distance n wavelengths from mirrors 1 & 4 respectively & You are free to move mirrors 1 & 2 or 3 & 4 together horizontally without the merging beams falling out of phase.

Wouldn’t it simply result in the energy becoming heat? That seems logical to me, then again this is my second post so I could be wrong. don’t want to step on any toes =]

It seems to me that you could adjust the distance between mirrors 1-2 so that it is 1/4 wavelength (+ some integer) and do the same with 3-4 (by mounting 2 and 3 on the same platform, and moving it with a caliper). Thus, the path 1-2-3-4 would be 1/2 wavelength longer (+n) than the direct path 1-4, and so 180 degrees out of phase.

Since the beam taking the path 1-2-3-4-A will have 4 reflections (which results in no net phase change), it should destructively interfere with the beam taking the path 1-4-A. Similarly, the beam taking the path 1-2-3-4-B will have 3 reflections, and the beam taking the path 1-4-B will have 1, so it too should destructively interfere.

If you did mount the mirrors 2 and 3 on a little platform with a caliper to adjust the distance, you should be able to note the intensitites at A&B, and find the spot where the detectors at A&B show a null intensity, so you wouldn't actually need to measure the tiny wavelength distances.

Experiments with Michelson interferometers appear to bear this out. See, for example, http://electron9.phys.utk.edu/optics421/modules/m5/Interferometers.htm (or google complete destructive interference).

This gets really interesting if you do something similar with particle beams, in which case the destructive interference of the wave function of the particles should cause them to stop existing! (this is what I was thinking about when I found this interesting discussion). Someone seems to be working on this (http://wwwuser.gwdg.de/~mpisfto/www.e/interfermtr_e.html).

Hi Bruce,
The question is whether you would get destructive interference at A and B at the same time. I think that if there was destructive interference at A, there would be constructive interference at B, and vice-versa.

Yeah, I just read an article on the Michelson intererometers at http://www.eng.warwick.ac.uk/~espbc/courses/undergrad/lec3/michelson.pdf

It said "We get complete destructive interference such that no light hits the screen (and thus all of the light reflects back to the source) when [there is]..destuctive interference. I guess the key is to note that, as Pete said, there is no phase shift when the beam is reflected off the back of the mirror, so there will always be an assymetry in the phase changes. This is why "all the light reflects back to the source" in the Michelson stuff, or goes to the other detector in the current setup.

It sure is tempting to try and figure out some other way of getting the beams to annihilate each other, though -- the phase difference is just sitting there, after all -- , but my guess is that it's impossible for some profound underlying reason.

Bruce

Regarding the original question about where does the energy go if there is destructive interference.

Lets assume a single photon or whatever particle we want to use, is fired through a diffraction grating at a time. The single photon will of course, 'interfere' with itself. But it can't simply vanish from the universe and not show up on the other side. "Destructive interference" simply prevents the photon from being found in certain locations. If you watched the photodetector behind the grating, you'd see a flash of light every time you fired a photon and the photon made it through the slits. And after a while you would notice that there are locations where the photon never lands, but it always lands somewhere. The places where the photon never lands we say it can't land due to 'destructive interference' but that doesn't mean it disappears. It simply goes somewhere else.

Of course if you fire a single photon at a diffraction grating, there is a VERY high likelihood that it will simply impact the grating itself and not make it through. The probability of that I believe is the area of the grating divided by the entire area it could possibly travel. In other words, if the photon is fired in a conical region, and 80% of that area is the grating itself, and the remaining 20% is the slit area through which the photon can pass, there is an 80% chance it will actually hit the grating and not go through the slits. The photon exists, and can always be detected somewhere.