Can Coherent Light Waves Interfere with Each Other?

[Dipto]

If two beams of light, in a vacuum, are projected towards each other will the beams be reflected ? If not what will happen ?

 

[Delta2]

No there will be no reflection , there will be wave interference instead. In short there will be a new beam that will be brighter than any of the two but what i just said is probably an oversimplification , check http://en.wikipedia.org/wiki/Interference_(wave_propagation ) especially the section about optical interference.

 

[ZapperZ]

If two beams of light, in a vacuum, are projected towards each other will the beams be reflected ? If not what will happen ?

So, let me first start by saying that this is a good opportunity for you to learn a little bit about the correct "phrase" to use, so that your question will come out the way you intended to ask, or it'll come out more accurately based on what we know about physics.

When two things collide or interact, one seldom use the term "reflected", because reflection is a well-defined observation. The more accurate term to be used here is "scatter". So it is my guess that you're asking of two light beam scatter off each other. This is a more well-defined question.

The answer is, no, at least at the everyday level. In classical wave picture, when two light beam overlaps, they produce a superposition effect (the same one that produces the interference effect). So they simply "add" according to the phase of the light wave.

The situation when you include quantum electrodynamics (QED) isn't that easy anymore. One can have photon-photon collisions, but this is extremely rare, often requires high energy photos (gammas), and it doesn't always just result in a scattering of these photons, but also the creation of other particles. This is not something you normally encounter everyday.

So if you want just one take-home answer, it would be "No."

Zz.

 

[Dipto]

So, let me first start by saying that this is a good opportunity for you to learn a little bit about the correct "phrase" to use, so that your question will come out the way you intended to ask, or it'll come out more accurately based on what we know about physics.

When two things collide or interact, one seldom use the term "reflected", because reflection is a well-defined observation. The more accurate term to be used here is "scatter". So it is my guess that you're asking of two light beam scatter off each other. This is a more well-defined question.

The answer is, no, at least at the everyday level. In classical wave picture, when two light beam overlaps, they produce a superposition effect (the same one that produces the interference effect). So they simply "add" according to the phase of the light wave.

The situation when you include quantum electrodynamics (QED) isn't that easy anymore. One can have photon-photon collisions, but this is extremely rare, often requires high energy photos (gammas), and it doesn't always just result in a scattering of these photons, but also the creation of other particles. This is not something you normally encounter everyday.

So if you want just one take-home answer, it would be "No."

Zz.

thanks

 

[echaniot]

As an addition to ZapperZ's answer, this QED process is called Delbruck scattering . Photon's spontaneously create virtual electron-positron pairs which interact.
This is actually one of the most prominent mechanisms of gamma ray attenuation in high energy astrophysics.

 

[sophiecentaur]

No there will be no reflection , there will be wave interference instead. In short there will be a new beam that will be brighter than any of the two but what i just said is probably an oversimplification , check http://en.wikipedia.org/wiki/Interference_(wave_propagation ) especially the section about optical interference.

Not true, in general. The interfering beam needs to be coherent with the beam that you want to affect. That more or less demands that both beams come from the same source (or from two synchronised lasers). The resulting interference pattern will not consist of 'a beam' because you cannot get the two beams, of finite width, to coincide so that the path differences are the same across a whole plane. You will produce a set of fringes or a standing wave pattern and this even happens if the two beams are launched along the ends of an optical fibre.
(NB. The total power is not altered so input power is 'redistributed' but not reduced or enhanced)