Where does light energy go in destructive interference?

[mbe]

Let us arrange a coherent monochromatic light beam, e.g. a laser beam, to
be split and directed along two paths. With repect to the light's
wavelength, the paths are suitably arranged such that a half-cycle net
time delay exists between them. These beams are then recombined
downstream, so as to destructively interfere there. Question: where
does the light energy "go"?

 

[mbe]

any comments will be appreciated...

 

[jdavel]

mbe,

As close as the laser beam is to having no spreading it's not perfectly so. Eventually the beam spreads enough to get constructive and destructive interference and conserve energy.

 

[shyboy]

the simplest way of the beam splitting is the double slit experiment. We know what will happen- the energy will be spatially redistributed. That means, once you mentioned the word "phase", you should deal with difraction.
So, if you have more complicated beam splitting mechanism, that means that difraction will be more complicated, but energy will be conserved. So, as jdavel hinted, the energy will be spatially redistributed.

 

[mbe]

I am not entirely sure what you mean. Are you saying that unless we can find a laser with a beam whose thickness is the size of one photon the light would always spread giving you fringes of dark and bright spots? And that if we did have a laser with thickness of one photon we would never be able to split it via diffraction?

Well if that is what you mean then let's take two identical lasers with a beam thickness of one photon and shine it at one point on a surface. Where would the energy go then?

 

[shyboy]

one photon does not have any phase, only waves have phase. If you "shine" one pont you exclude any interference, because you are working with particles. If you want to include the interference, you cannot "shine" one point.

 

[mbe]

Well but i thought that all things could act as waves even photons as said by debroghlie (not sure that's how you spell his name). He was the guy who said that all things can exhibit wave like properties. So that means that a single photon can ineterfere.

 

[shyboy]

Yes, you may say that single photon can interfere, but , as I said, that also means no "shininig in a certain point". That why it is called wave-particle duality.

 

[mbe]

ok i don't mean to be a thorn in ur side but its just bothering me. It just doesn't seem to be much of an explanation to say the chances of it happening are so low that we won't have to worry about it. What if it does happen what if two photons arrive in the same spot by accident being out of phase by lambda by two. What would be the result then?

 

[shyboy]

photon does not have phase, it has a probability which behaves like waves. So when we speak about the wave properties of a single particle, we mean that the probability to find the particle at some point has wave properties. The same is true for a probability of two photons- it may look like the interference but it does not mean that a single photon have any phase. Actually, the interference means that there is no such thing as separate photons.

 

[Doc Al]

mbe: In the future, do not start multiple threads on the same topic!

Followups here: https://www.physicsforums.com/showthread.php?t=73919