-3
$\begingroup$

I have always wondered what happens when two opposite EM waves which completely destructs each other hits a wall? Does it reflects back or does it go through it. Would be good if anyone could give a(quantum) description of what would happen.

$\endgroup$
  • $\begingroup$ How do you mean they are hitting a wall? On opposite sides of the wall? Then there will be some fresnel reflections, and if the conditions are still fulfilled, destructive interference where they meet. If they are destructing before they hit the wall, or AT THE LOCATION where they hit they wall, then there will be no reflection as they will have destroyed themselves. $\endgroup$ – DakkVader Sep 1 '19 at 9:13
  • $\begingroup$ The waves are completely destructed. I guess it doesn't matter where they destruct? They are always destructed along the wave path. $\endgroup$ – einstein Sep 1 '19 at 9:23
  • 2
    $\begingroup$ You'll have to clarify your set up. What do you mean by "opposite EM waves"? Do you mean that they are propagating in opposite directions? $\endgroup$ – garyp Sep 2 '19 at 10:54
  • $\begingroup$ I mention "opposite EM waves, which completely destructs each other". Isn't that clear enough? $\endgroup$ – einstein Sep 2 '19 at 17:32
1
$\begingroup$

"Where does energy go in destructive interference" is a frequently asked question. (For example: What happens to the energy when waves perfectly cancel each other?)

However, it is very counter-intuitive, and at some point there has even been a debate about it in serious physics journals, (probably Phys Rev, I do not recall at the moment.)

Of course, destructive interference does happen in practice, and the light does indeed vanish completely. This happens all the time in interferometers. (You can see a demonstration here: https://youtu.be/RRi4dv9KgCg?t=158)

But the essence of the paradox of seemingly vanishing energy has nothing to do with quantum mechanics -- the same effect can be demonstrated with waves in rubber bands and it is just as confusing!

(A very good discussion and additional references can be found here: http://www.physics.princeton.edu/~mcdonald/examples/destructive.pdf)

The correct short answer is: this question is subtly ill-posed. To answer it properly, you have to include into consideration how the waves were generated and combined. If you do that, it will turn out that no matter how you generate the two waves which cancel each other, you will also end up generating some side effects which ensure conservation of energy in the whole system. The energy does not disappear, it simply gets re-distributed. (For example, "idler beams" somewhere else in the system may interfere constructively, with the power in their superposition seemingly increasing beyond the sum of the power of the source beams. Or, in another situation, a standing wave sets in the system and the source stops supplying any power at all, if we assume no losses, etc.)

Figuring out how exactly the power gets redistributed can be quite involved -- if not done carefully, it easily leads to wrong answers -- that's why there has been such a debate about this situation.

But this redistribution happens even in the systems with classical mechanical waves -- the paradox of "seemingly disappearing in destructive interference power" it is not a quantum phenomenon at its core!

(Of course, providing a detailed description of the mechanism for an optical system would depend on treating the quantum behavior of light properly -- including treating photons not as particles but excitation states of fields in the system. This is not a trivial exercise, and the additional complexity only makes the essence of the seeming paradox more difficult to grasp.)

$\endgroup$
0
$\begingroup$

Photons (light rays) and radio waves are the most common examples of EM waves. If 2 photons of opposite phase hit a wall at the same point the electron there will not sense them or absorb them but it would likely scatter them a tiny tiny bit so they are slightly out of phase. When they hit the next atom or electron they would get scattered again, the more they are scattered the more they are out of phase, eventually they are absorbed or reflected. Photons never ever cancel (destroy) themselves, it is a violation of conservation of energy. Photon "interference" is an historical term but is still in use today, modern thinking is to say certain paths are preferred (like in thin film interference).

For radio waves many photons that are say 180 degrees all together can meet with many photons that are 0 degrees in phase. The antenna electrons will not get a strong signal and the radio waves keep on going.

QM says that probability rules the interactions, it is possible that 2 photons that are in phase meet a perfectly positioned electron that manages to ignore one photon and absorb the other.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.