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When I began learning about optical interference, I came to know about destructive interference in which light waves cancel each other.

How the energy is still conserved ?

I found that the conservation is not disturbed and the photon emission is stopped instantaneously. How can the electron know what to do, whether to return to ground state giving back energy in the form of photons or to undergo other phenomenon like electron scattering...

Consider two laser sources emitting laser undergo perfect destructive interference after a distance d by means of a mirror. Then the light from both the laser goes uninterfered up to distance d. How can the energy be conserved?

ryanafrish7
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  • In interference pattern you can see that, light gets cancelled in some places, but add up in some places. Energy is conserved. – Sensebe Nov 15 '14 at 09:17
  • It do happen with a hemispherical wave front but what with a plane wave front or a single non divergent beam of ray – ryanafrish7 Nov 15 '14 at 09:19
  • "Consider two laser sources emitting laser undergo perfect destructive interference after a distance d by means of a mirror."-Please add a GIF or a diagram or add a link of description of your experiment, it is not understandable by the way the question stands now. – Sensebe Nov 15 '14 at 09:29
  • Sorry, I'm accessing SE through android. When I get access to a computer, I will surely address your request. – ryanafrish7 Nov 15 '14 at 10:08
  • See also Where does energy go in destructive interference? – John Rennie Nov 15 '14 at 11:39
  • How can the interference of two waves result in the absorbance of the energy by one of the sources? – ryanafrish7 Nov 15 '14 at 11:48
  • I have answered this and posted it on the duplicate http://physics.stackexchange.com/questions/23930/what-happens-to-the-energy-when-waves-perfectly-cancel-each-other/146951#146951 – anna v Nov 15 '14 at 16:53
  • Moderators .The question to which is duplicate is not addressing electromagnetic waves per se. There exist other questions that do, but the one chosen is not the best. I deleted my answer there and added one relevant to light , which is this question. – anna v Nov 15 '14 at 17:02

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Generally, there are only two things that can happen in the hypothetical experiment you describe. Either photons are emitted and absorbed in some place (detector, mirror, anything), or they are not emitted (and of course not absorbed). Both cases conserve energy.

The latter possibility is the one that is somewhat counter-intuitive but it goes to the heart of radiative processes, radioactive decay and in general all processes, and means that if your example interaction really sets up an interaction-scenario with completely destructive probability amplitudes at all places and at all times, then that interaction simply does not happen (more or less that is the definition of probability in QM of course).

You need to treat the problem in the example as a whole, and not look at individual photons and what happens to them "along their track", as that surely leads to some confusion.

BjornW
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