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I was recently told by my physics teacher that when two waves interfere then the energy lost in the destructive region is transferred to the constructive region and hence no violation of conservation of energy.

Now the speed of EM waves is the speed limit of the universe. Keeping this in mind , we can conclude that when two EM waves interfere , for energy to be transferred from destructive region to the constructive region, the energy must go faster than its initial traveling speed (the speed of light) so as to cover up the extra distance but this can't be true since not even light can cross its speed (right ?).

Where am I making a mistake ? How does the energy actually gets transferred and at what speed (for EM waves) ?

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  • $\begingroup$ In the DSE there are no photons (i.e. NO energy) in the dark areas. All the photons (i.e. all the energy) goes to the bright areas. In high school you are taught the simplest theory .... but if you think about it you can see the flaws. Richard Feynman and Dirac stated every photon determines its own path, Feynman discovered the "path integral theory" to explain why light chooses a path. $\endgroup$ Jan 23 at 0:29

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for energy to be transferred from destructive region to the constructive region, the energy must go faster than its initial traveling speed (the speed of light) so as to cover up the extra distance

This is incorrect. The key problem is the idea of “extra distance”. There is no extra distance. A region of interference only exists where EM waves from both sources have already arrived. This is never ahead of either of the two sources wavefronts and is often well behind one of them. There is no risk at any time of superluminal energy transfer to regions of destructive interference because all regions of interference (both constructive and destructive) are always within the past light cone of both sources.

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    $\begingroup$ It's like the laser-dot-on-the-face-of-the-moon thing. Sure, the dot might move faster than light, but it's not an actual thing, only the interaction between two things - the laser beam and the moon. Likewise, the interference pattern is not a thing, only the interaction between two waves. In both cases, the lightspeed limit does not apply. $\endgroup$
    – No Name
    Jan 22 at 1:11
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    $\begingroup$ @NoName : Not really. If you wave the detector towards and away from the slits, you find that the interference pattern is really a pattern in space. The waves don't travel to the detector and then suddenly interfere -- they've been interfering as far back (in time and space) as they have been interacting. So the energy was redistributed long before arriving at the detector. What you are describing is different -- the apparent dot is not moving superluminally on virtual surfaces at all distances, this is an artifact of surfaces at large distance. $\endgroup$ Jan 22 at 2:26
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In classical electromagnetic waves that are described by solutions of the Maxwell equations, the energy carried by the wave is given by the Poynting vector, as a rate of energy transport

poynting

and is in term of averages per unit area, and cannot be used for determining any direction in energy,afaik.

To answer your question, one has to go to the quantization frame where a light beam is composed out of zillions of photons. Photons also go with velocity c.

The simplest demonstration of interference patterns and how single photons build up the double slit interference experiment may help:

enter image description here

Single-photon camera recording of photons from a double slit illuminated by very weak laser light. Left to right: single frame, superposition of 200, 1’000, and 500’000 frames

Note that the little dot footprint each photon leaves is at an angle with the center of the slits, and yes, the photon travels a larger distance than the straight line of a classical light ray. Photons still travel at speed c, so individual photons take longer to hit the screen, BUT, photons are not classical light, in a similar way that bricks build bridge but the bridge is not a brick. They build up quantum mechanically the classical electromagnetic radiation in a complicated quantum field theory , QED, way.

This is discussed in this link How classical fields, particles emerge from quantum theory is discussed.

Interference of classical light and quantum phenomena have unexpected consequences as can be seen in this MIT video.

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  • $\begingroup$ And yet, in single photon interference, all the energy of the electromagnetic wave packet associated with the photon, which can be spread over many centimeters is absorbed by a single electron. How long does that take? $\endgroup$
    – R.W. Bird
    Jan 22 at 17:09
  • $\begingroup$ @R.W.Bird One can get from the energy-time heisenberg uncertainty an envelope of the time it takes for the transition of the atom, The photon is a point particle. it is the probability of being absorbed that has a space distribution, the wave packet ( an extra distribution convoluted with the probability distribution ) is an unnecessary complication . $\endgroup$
    – anna v
    Jan 22 at 18:26
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I was recently told by my physics teacher that when two waves interfere then the energy lost in the destructive region is transferred to the constructive region and hence no violation of conservation of energy.

This is exactly how constructive and destructive interference work together. For mechanical waves. In areas of destructive interference, the up and down movement of particles in opposite directions causes an increase in density and the particles are pushed away to the side in the direction of the reduced density in constructive interference.

What does the picture look like with electromagnetic radiation?

...we can conclude that when two EM waves interfere, ...energy to be transferred from destructive region to the constructive region,...

For EM radiation, too, there is a density, the radiation density. However, a displacement of photons from areas of higher density into such lower density does not take place. Simply because in the energy range, in which these experiments take place, an interaction between photons is a more than rare event.

The concept of interference is too much of a simplification for EM radiation to hold up in the realm of physics. And I also think it is not correct to present this in school lessons. (But of course students have the duty to reproduce correctly the interpretation taught in this way. This is how the inquiry of a learning process works).

The intensity distribution of the EM radiation behind edges is the result of the deflection of this radiation at these very edges. Here an interaction takes place. Without edge no intensity distribution on an observation screen.

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  • $\begingroup$ Yes the em field of an aperture or edge interacts with the em field of the photon, the photon path is "predetermined" likely by virtual em forces which also interact with the edges. The edges reduce the number of allowable paths and dark areas and bright areas become more visible. $\endgroup$ Jan 24 at 19:58
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In addition to the other answers, I'd like to point out something important:

Your beam of light with an interference pattern does not travel at the speed of light!

The speed of light $c$ is not the speed of a beam of light of any size and profile, it is the speed that an infinite plane wave of light that's existed throughout time would move.

As discussed in @flippiefanus's answer to Wouldn't any structured beam of light be expected to travel slower than a plane wave? once you have transverse modulation like an interference pattern and/or standing waves, the longitudinal propagation is slower than the speed of light.

For further details I refer the reader to the linked answer.

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In the DSE there are no photons (i.e. NO energy) in the dark areas. All the photons (i.e. all the energy) goes to the bright areas. In high school you are taught the simplest theory .... but if you think about it you can see the flaws. Richard Feynman and Dirac stated every photon determines its own path, Feynman discovered the "path integral theory" to explain why light chooses a path.

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