| Topic: |
Science > Physics |
| User: |
"Sebastien Auclair" |
| Date: |
06 Aug 2005 04:25:20 AM |
| Object: |
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 !
.
|
|
| User: "Sam Goldwasser" |
|
| Title: Re: Basic question (Laser, wave interaction) |
06 Aug 2005 07:07:04 AM |
|
|
"Sebastien Auclair" <sebastien_auclair@videotron.ca> writes:
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.
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.
--- sam | Sci.Electronics.Repair FAQ Mirror: http://repairfaq.ece.drexel.edu/
Repair | Main Table of Contents: http://repairfaq.ece.drexel.edu/REPAIR/
+Lasers | Sam's Laser FAQ: http://repairfaq.ece.drexel.edu/sam/lasersam.htm
| Mirror Sites: http://repairfaq.ece.drexel.edu/REPAIR/F_mirror.html
Note: These links are hopefully temporary until we can sort out the excessive
traffic on Repairfaq.org.
Important: Anything sent to the email address in the message header above is
ignored unless my full name is included in the subject line. Or, you can
contact me via the Feedback Form in the FAQs.
.
|
|
|
| User: "Andy Resnick" |
|
| Title: Re: Basic question (Laser, wave interaction) |
08 Aug 2005 12:16:25 PM |
|
|
Sam Goldwasser wrote:
<snip>
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.)
<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
.
|
|
|
| User: "Phil Hobbs" |
|
| Title: Re: Basic question (Laser, wave interaction) |
08 Aug 2005 01:20:39 PM |
|
|
Andy Resnick wrote:
Sam Goldwasser wrote:
<snip>
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.)
<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.
You need some nonlinearity to cause the waves to couple to the
low-frequency resonance. Otherwise there's no forcing at that
frequency, and therefore no response (linear systems are exponential-in,
exponential out). The case you seem to have in mind, namely an
electrical output and an electronic control system, is not too different
in character from an ordinary phase locked loop--apart from the noise
behaviour, it doesn't matter too much where in the circuit the frequency
selectivity comes in.
Having it happen in an all-optical setup, e.g. an externally-pumped
optical cavity locker, might be quite an interesting device. Something
like a strong Kerr medium with a strongly polar molecule might fit the
bill--you might even be able to do it with carbon disulphide.
Cheers,
Phil Hobbs
.
|
|
|
|
|
|
| User: "tadchem" |
|
| Title: Re: Basic question (Laser, wave interaction) |
06 Aug 2005 09:23:46 AM |
|
|
"Sebastien Auclair" <sebastien_auclair@videotron.ca> wrote in message
news:Nm%Ie.25080$De.947467@wagner.videotron.net...
<snip>
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.
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
.
|
|
|
| User: "" |
|
| Title: Re: Basic question (Laser, wave interaction) |
07 Aug 2005 04:00:35 AM |
|
|
In article <o7WdnZ2dnZ0HJRfqnZ2dncNVad-dnZ2dRVn-yZ2dnZ0@comcast.com>, "tadchem" <tadchemNOSPAM@comcast.net> writes:
"Sebastien Auclair" <sebastien_auclair@videotron.ca> wrote in message
news:Nm%Ie.25080$De.947467@wagner.videotron.net...
<snip>
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.
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.
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,
meron@cars.uchicago.edu | chances are he is doing just the same"
.
|
|
|
|
|
| User: "" |
|
| Title: Re: Basic question (Laser, wave interaction) |
06 Aug 2005 10:14:47 AM |
|
|
On 8/6/05 2:25 AM, in article Nm%Ie.25080$De.947467@wagner.videotron.net,
"Sebastien Auclair" <sebastien_auclair@videotron.ca> wrote:
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 !
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
.
|
|
|
|
| User: "the softrat" |
|
| Title: Re: Basic question (Laser, wave interaction) |
06 Aug 2005 11:55:42 PM |
|
|
On Sat, 6 Aug 2005 05:25:20 -0400, "Sebastien Auclair"
<sebastien_auclair@videotron.ca> wrote:
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 !
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:softrat@pobox.com
--
"I get to go to lots of overseas places, like Canada." --
Britney Spears
.
|
|
|
|
| User: "Jan Panteltje" |
|
| Title: Re: Basic question (Laser, wave interaction) |
06 Aug 2005 05:36:37 AM |
|
|
On a sunny day (Sat, 6 Aug 2005 05:25:20 -0400) it happened "Sebastien
Auclair" <sebastien_auclair@videotron.ca> wrote in
<Nm%Ie.25080$De.947467@wagner.videotron.net>:
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 !
if the beams cross in a non-linear medium multiplication may occur.
.
|
|
|
| User: "redbelly" |
|
| Title: Re: Basic question (Laser, wave interaction) |
06 Aug 2005 07:16:19 AM |
|
|
Jan Panteltje wrote:
if the beams cross in a non-linear medium multiplication may occur.
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
.
|
|
|
|
|
| User: "Clemens W" |
|
| Title: Re: Basic question (Laser, wave interaction) |
06 Aug 2005 05:23:22 AM |
|
|
Sebastien Auclair wrote:
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.
Which means they emit photons of different frequency (i.e. wavelength)
Two laser beams of type A crossing each other at a right angle (i.e. 90=
=B0)
will produce, at the intersection point only, a new wave that could
correspond to the wave characteristics of the type B.
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:
http://en.wikipedia.org/wiki/Interference
Good luck,
A=2E Friend
.
|
|
|
|
|
Related Articles |
|
|
