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Basic question (Laser, wave interaction)

Basic question (Laser, wave interaction) - Physics Forum

Basic question (Laser, wave interaction) - Physics Forum. Discuss and ask physics questions, kinematics and other physics problems.


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  #1  
Old 08-06-2005, 09:25 AM
Sebastien Auclair
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Default Basic question (Laser, wave interaction)



Greetings !



I need to know if the following assumption is correct...



Assuming a type of laser beam defined as A and a different one called B. (
"Type" meaning wavelength, amplitude, etc.)

In other words, lets say that A is green and B is red.



Two laser beams of type A crossing each other at a right angle (i.e. 90)
will produce, at the intersection point only, a new wave that could
correspond to the wave characteristics of the type B.



Thanks !


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  #2  
Old 08-06-2005, 10:23 AM
Clemens W
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Default Basic question (Laser, wave interaction)

Sebastien Auclair wrote:

Which means they emit photons of different frequency (i.e. wavelength)


Since both laser beams have the same frequency, the resulting
interference will also have the same frequency, only the amplitude will
be different.

In other words: Two reds don't make a green.

See also:
[Only registered users see links. ]

Good luck,

A. Friend

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  #3  
Old 08-06-2005, 10:36 AM
Jan Panteltje
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Default Basic question (Laser, wave interaction)

On a sunny day (Sat, 6 Aug 2005 05:25:20 -0400) it happened "Sebastien
Auclair" <[Only registered users see links. ]> wrote in
<Nm%Ie.25080$[Only registered users see links. ]>:

if the beams cross in a non-linear medium multiplication may occur.

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  #4  
Old 08-06-2005, 12:07 PM
Sam Goldwasser
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Default Basic question (Laser, wave interaction)

"Sebastien Auclair" <[Only registered users see links. ]> writes:



Beams of light do not interact in a linear medium (e.g., vacuum, air).
So, even at the intersection, only the original frequencies/wavelengths
are present unless the power density is so high that the medium behaves
non-linearly.

However, if they are both incident on a photodiode with sufficient
freuquency response, the difference frequency between them will be seen.
(The sub frequency is also produced but no know photodiode will respond
that fast.)

In a non-linear medium, there can be new frequencies generated including
the sum, difference, doubling of either of the original frequencies, etc.

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Repair | Main Table of Contents: [Only registered users see links. ]
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| Mirror Sites: [Only registered users see links. ]

Note: These links are hopefully temporary until we can sort out the excessive
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  #5  
Old 08-06-2005, 12:16 PM
redbelly
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Default Basic question (Laser, wave interaction)


Jan Panteltje wrote:


The OP asked if this could happen "at the intersection point ONLY".
With non-linear media, beams are emitted that are not confined to this
area. I'd have to say the answer is simply "no".

Mark

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  #6  
Old 08-06-2005, 02:23 PM
tadchem
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Default Basic question (Laser, wave interaction)


"Sebastien Auclair" <[Only registered users see links. ]> wrote in message
news:Nm%Ie.25080$[Only registered users see links. ]...

<snip>


A laser beam is a large collections of photons, disciplined such that the
photons are all in phase and of the same frequency.

Look at the photons.

When two photons interact with each other in a vacuum, the interaction
depends on the relative phase and the directions.

In other words, two photons of the same wavelength moving in exactly the
same direction will either reinforce (add) each other or interfere
(subtract) from each other, with the exact magnitude of the effect depending
on the phase difference.

Two photons two photons of the same wavelength moving in different
directions will either reinforce (add) each other or interfere (subtract)
from each other, with the exact magnitude of the effect depending on the
phase difference *at the exact location where they are being observed.*
Shift your location a fraction of a wavelength in either direction and you
will see a different magnitude of interaction. No 'new waves' and no new
photons of a different wavelength will be created.

Two photons of different wavelengths meeting in a vacuum will not interact
with each other at all. They may both interact with matter which is placed
at the intersection point. Under the right circumstances, a photon of one
wavelength may be absorbed by an atom, exciting it into a state in which the
second photon *also* interacts, permitting phenomena which may not be
possible with the same material and only a single laser beam.


Tom Davidson
Richmond, VA


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  #7  
Old 08-06-2005, 03:14 PM
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Default Basic question (Laser, wave interaction)

On 8/6/05 2:25 AM, in article Nm%Ie.25080$[Only registered users see links. ],
"Sebastien Auclair" <[Only registered users see links. ]> wrote:

NOT TRUE!

While there may be something funny happening at high intensity in a
nonlinear medium, such as mixing and harmonic generation, the intensity for
that to happen in vacuum is orders of magnitude greater.

Bill

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  #8  
Old 08-07-2005, 04:55 AM
the softrat
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Default Basic question (Laser, wave interaction)

On Sat, 6 Aug 2005 05:25:20 -0400, "Sebastien Auclair"
<[Only registered users see links. ]> wrote:

Nop.e

A) If there is nothing at the intersection point, the two laser beams
will not interact at all.

B) If there is something (real -- not imaginary Cavourite, please) at
the interaction point, interactions with it may result in virtually
any frequency light being emitted. The most probable outcome would be
light of frequency A and light of frequency B in ratio as their input
intensities.


the softrat
Sometimes I get so tired of the taste of my own toes.
mailto:[Only registered users see links. ]
--
"I get to go to lots of overseas places, like Canada." --
Britney Spears
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  #9  
Old 08-07-2005, 09:00 AM
mmeron@cars3.uchicago.edu
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Default Basic question (Laser, wave interaction)

In article <[Only registered users see links. ]>, "tadchem" <[Only registered users see links. ]> writes:
With the exception of higher order processes (involving the
interaction of virtual electron-positron pairs) photons *do not*
interact with each other in vacuum. The so called "interference" is
a result of the fact that they *do not interfere*.

Mati Meron | "When you argue with a fool,
[Only registered users see links. ] | chances are he is doing just the same"
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  #10  
Old 08-08-2005, 05:16 PM
Andy Resnick
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Default Basic question (Laser, wave interaction)

Sam Goldwasser wrote:
<snip>
<snip>

Let's say I engineered the material 'unobtanium' with an extremely high
suseptibility (linear and or nonlinear) in the kHz-MHz range. I suppose
that the output could be used as part of a stabilization circuit.

Does such a material exist? What about using resonant cavities?

It's kind of interesting because on one hand, there's the timescale set
by the beam frequencies (THz- EHz) and on the other is the timescale set
by the difference frequency- many orders of magnitude lower. In
nonequilibrium situations, for example the glass transition, there are
many timescales present all at once, all equally affecting the dynamics
of the system.

--
Andrew Resnick, Ph.D.
Department of Physiology and Biophysics
Case Western Reserve University
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