"Total" destructive interference of plane waves

[angrystudent]

Hello.

Let's suppose that we have a Michelson interferometer to study interference patterns of light. This time we use plane waves.

If we set the whole thing up so that the two separated beams have a phase difference of π when they superpose, destructive interference ensues. Since we're talking about plane waves propagating in the same direction, no interference pattern can be observed: the only output is darkness.

Where does the energy go in such a case?. It looks like electromagnetic energy disappears in the process, but I see no way it could be absorbed or deflected, instead of simply being "destroyed".

 

[Dale]

You cannot have a beam that is a plane wave. A beam is limited to a small region, a plane wave is not. They are mutually contradictory descriptions. Which one do you want to discuss?

 

[angrystudent]

Both of those cases sound interesting to me, I didn't even realize that such a distinction could be relevant here.

 

[sophiecentaur]

The only way you could contemplate having total destructive interference from two sources in all directions is to have the emitting sources in exactly the same location, dissipating their output power between themselves. At least that answers the question about where the wasted Power would actually be going. IMO, that's better than just saying it can't be done.
I'd bet that the equivalent to this has been observed in RF transmitter systems when virtually nothing has been radiated from the antenna but two transmitters have been getting very hot.

 

[DaveE]

A perfect plane wave will propagate as a plane wave throughout the entire universe, i.e. infinitely wide. 2 perfect plane waves out of phase (and going in the same direction, of course) will be out of phase everywhere and destructively interfere everywhere. So, IMO there will be no radiation at all, anywhere in this case.
Also, BTW, there is no such thing as a perfect plane wave, it is a useful abstraction used to simplify problems.
In your example, it's really ok that the waves destructively interfere when the interferometer arms are certain lengths. The point of the interferometer is to look at how the beam interference changes when the lengths change. If you zoom in on a point on the detector (i.e. plane wave approximation), sometimes it's dark, sometimes it's light. You don't need (or even want) fringes, they are the consequence of the difficulty of building a perfect interferometer with a perfect light source.

 

[CWatters]

This includes plane electromagnetic waves but the maths is too hot for me..http://www.physics.princeton.edu/~mcdonald/examples/destructive.pdf

Hope that link works. It's to a paper/article "
Does Destructive Interference Destroy Energy" by Kirk T. McDonald

 

[Dale]

Both of those cases sound interesting to me, I didn't even realize that such a distinction could be relevant here.

So for a beam you will have regions of destructive interference and regions of constructive interference. So it is fairly easy to see that the overall energy is conserved, it is just moved around a bit.

For a plane wave to get perfectly destructive interference everywhere means that you have no energy anywhere. So zero everywhere means no energy which is clearly conserved.