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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?

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marked as duplicate by John Rennie, Prahar, Rob Jeffries, ACuriousMind, Danu Nov 15 '14 at 15:56

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  • $\begingroup$ In interference pattern you can see that, light gets cancelled in some places, but add up in some places. Energy is conserved. $\endgroup$ – Immortal Player Nov 15 '14 at 9:17
  • $\begingroup$ It do happen with a hemispherical wave front but what with a plane wave front or a single non divergent beam of ray $\endgroup$ – ryanafrish7 Nov 15 '14 at 9:19
  • $\begingroup$ "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. $\endgroup$ – Immortal Player Nov 15 '14 at 9:29
  • $\begingroup$ Sorry, I'm accessing SE through android. When I get access to a computer, I will surely address your request. $\endgroup$ – ryanafrish7 Nov 15 '14 at 10:08
  • $\begingroup$ See also Where does energy go in destructive interference? $\endgroup$ – John Rennie Nov 15 '14 at 11:39
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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.

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