what happens when you shoot 2 lasers at the same spot?
science_geek
Posts: 247
lets just say that 2 lasers are·mount on a turret, one 5 mw laser cant burn a hole in a material, but a 7-10 mw laser will burn a hole. If i were to shine 2 5mw lasers at the same spot, would it burn a hole or do the lasers not combine.
i know this doesnt sound good, but i dont know how else to put it...
i know this doesnt sound good, but i dont know how else to put it...
Comments
Two lasers aimed at the same spot theoretically might cancel out or they might add together or anywhere in-between. In practice, the kinds of lasers you would use (cheap) with the optics you would have (fair to poor) would mostly add their output together. You might not have exactly the sum of the two amounts of energy, but it would be close.
Look at some of the Wikipedia's articles on lasers and optics including interference for further information.
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This does bring up an interesting thought experiment, though. Suppose it were possible (e.g. with a half-silvered mirror) to create colinear beams that exactly cancel. Now each laser emits light having a certain amount of energy. Due to conservation of energy, the energy in the light beams cannot simply disappear. So where does it go when the beams cancel?
-Phil
My head hearts.
I guess I'm not fully wrapping my aching head around it, but the two waves are not just "canceling" each other out, they are "interfering" - they would be absorbing and emitting each others energy in balance. Sorta. I guess.
This is almost as bad as:
- Work = Mass X Height Moved
- Hold a copy of the "Handbook of Chemistry and Physics" at arms length.
- Don't raise or lower the book.
- You haven't done any work (you have not created any additional potential energy).
- Why is your arm getting tired?
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Click here to see my Nomad Build Log
Into the Burmda triangle, reverbertating and oskalliting off the Mohorovičić discontinuity, and back up beneath the shopping mall in Boulder, Colorado - where all the UFO's are built.
But you knew that already, Phil, so why are you asking?
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As for 7 to 10 mW burning anything - don't expect much! One of my active hobbies is making VERY high power lasers, and I like to make handheld units with arrays of diode lasers. Here's a video of a single red diode laser I ran up a few years ago putting out somewhere around 150 to 200mW, igniting a match:
Ignore the rest of the circuitry running on the board - that light show was another project for a friend's vehicle and has nothing to do with the laser, which is just a current regulator, a resistor and a capacitor.
There are now single diode lasers that put out around 500mW of red. The diodes for BluRay DVD Burners put out a near UV beam that is around 405nm and now are getting up around 150 to 200mW, and so are capable of popping small holes in things with good optics.
If you are VERY CAREFUL and take ALL the precautions inherent in these items, you can do a lot with these items. An IR Laser Diode can put out 50 WATTS of power and cut thick materials rapidly (downside to IR: you can't see the beam, and you can lose your vision in a second with stray reflections. USE LASER SAFETY GOGGLES!!!). One of my "to do" personal projects is to use a BS2 to control an X-Y plotter style device that has an IR laser diode instead of a pen, that I can program to cut out parts for other projects.
But I digress. If you want to combine beams, you want to use a polarized beam combiner cube. Laser diodes put out polarized radiation. If you put one polarization into one face of the cube, and rotate another laser diode 90 degrees and put its beam into the adjacent face of the cube, the resultant output beam will be a randomly polarized beam that is roughly equal to the input power of the first laser plus the input power of the second, minus a very small percentage of losses through the cube and whatever collimating optics you use.
You may also use these little cubes to combine lasers of different wavelengths. I tend to go for combining the highest power devices I have got, but I have also combined my 632nm reds and my 405nm UVs to get some really pretty colors!
One last thing - once you have combined two beams, you cannot use another cube to add in beams - as only ONE polarity will pass through the next cube face, so you'll lose 1/2 your beam power in the attempted second recombine.
If you want to learn a LOT about high power diode lasers, there are several forums you can go to, the best two I have found are laserpointerforums.com and laserenthusiast.com. Be forewarned that those boards are no where NEAR the level of professionalism you will find here. There's a lot of "drama" (which is why I tend to use them in a "read only" fashion now). But there is no doubt that theere is a wealth of highly valuable and useful information to be had there, and there are a handful of helpful and intelligent folks on those boards who contribute regularly.
I hope this info has been useful to you.
Lasers are fun. Microcontrollers are fun. Looking forward to combining the two!
Dave X
The experiment goes something like this. You get two colinear beams that are just SLIGHTLY different wavelengths - say the frequency is just 1Hz difference. Shine this colinear beam on a wall, with a second wall behind it. The first wall would need to be SUPER thin, but still opaque. As the waves come in and out of phase, the net EM field is zero, so at some points, the light SHOULD pass through the first wall, go out of phase, and hit the wall behind it. Basically, it's like shining a flashlight on an opaque wall, but having some of the light "magically" pass through and hit whatever is on the otherside.
Spooky. I'm going to talk to some physics professors once my normal semester starts back up this fall at my university. I'm sure there's a simple explaination.
Phil Pilgrim you are correct the 2 waves do not cancel out exactly like sine waves do unless they were some how traveling along exactly the same tragectory 180 deg out of phase. would become an interesting output because the angle between them would make the phase difference chancge from point to point along the paper.
As for what happens to the energy. If we had to lasers of different frequency shining through mirrors to some how line up so that there beams ran in the same trajectory. as the beams went in and out of phase with each other it would brighten and dim. When it hits the paper we are trying to light up it will either reflect of pass through. Cool thing is if we get it so it perfectly cancels when it hits the paper the paper would not light up but we could see the beam on the other side(dimmed because some will have reflected but none will have been absorbed)
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Don't worry. Be happy
I think my point in my previous post was that there is no "negative light" in a laser beam, so, unlike a "regular" electromagnetic wave, there is only a wave of increasing energy density or decreasing energy density, but at its lowest energy point over time, there is no energy present to "cancel" the wave of the opposite phase.
Dave
This is an argument I had with a teacher or two in high school. First define what you mean by light and crash (sound). If you define light and sound as something seen and heard respectively then by that definition the answer is no. If you define light as energy emitted at specific wavelengths and sound as vibrations produced in a medium then the answer is yes. Nothing personal John, but consider this to be one of the dumbest pseudo philosophical questions of all time, and I spent several hours in detention for my opinion.
As for the " * Why is your arm getting tired? " question, that is a much more interesting one. Your muscles require energy to maintain contraction against any applied force. Think of a rope wrapped around a motor shaft with a weight on the end of it. To keep the weight at any but the fully extended position requires power to the motor even though you are not adding energy to the weight.
You are correct. the double slit experiment shows us that light indeed adds together and cancels itself ot like we would expect waves to do. the really interesting thing with that experiment is not only does light cancel itself out but if you do the experient 1 photon at a time where there is no other photon to interfear as long as you are going to send another it will still interfear. the reason because the photon do not interfear with each other but in fact interfere with the probability of each other.
This experiment was the first proof that quantum mechanics was real.
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Excuse me while I adjust my chaise. Mozo! Una otra margarita, por favor!
-Phil
if you send one photon at a time you have to use film so you can see the patern after firing a few million photons.
going back to the first experiment there is no slits but two beams overlapping but rotating in there phase to one another. you should see a sine wave like intensity over the length of the beam. if the rotation was small enough say 3nHz it could be easily seen with naked eye.
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Post Edited (mctrivia) : 7/19/2009 4:53:25 AM GMT
http://www.dailytech.com/Northrop+Grumman+Pumps+Lasers+Power+to+WarReady+100+kW/article14647.htm
Mike
Is that not the case?
If so, I've seen some really nice 50 - 75 WATT (yeow!) laser diodes I'd considered using in an "intruder dissuading" project.
Smiles,
DJ
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Instead of:
"Those who can, do.· Those who can't, teach." (Shaw)
I prefer:
"Those who know, do.· Those who understand, teach." (Aristotle)
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That said, I've seen an 80 Watt IR cut a fried chicken leg in half in about 6 seconds.
Dave
Good book on this subject is
Fabric of the cosmos : Space time and the texture of reality by Brian Greene
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Post Edited (mctrivia) : 7/20/2009 5:43:56 PM GMT
I don't question Brian Green's credentials and have not read those two books, but i do know that both books are in the realm of theoretical physics, not proven science.
In normal white light there is so many different frequencies that no noticeable difference will be seen. However in the case of lasers that have 1 frequency and all the light is lined up in phase the effect can be very noticeable.
The 2 slit experiment would not give an output like the diagram shown above if light did not interfere with each other. The output seen by the experiment is what would be expected from 2 sine waves adding and subtracting from each other.
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If two beams of light did not interfere with each other at the surface of a substrate, holography would not be possible. But holography uses coherent light from a laser. When you cite the example of the sun reflecting from mirrors, you're talking about incoherent light. In such a case, the lightwaves will still interact at the surface, but any interference effects will be so transient and disassociated as to be undetectable.
Also, it should be noted that beams from two separate lasers will not be mutually coherent, having emanated from independent sources. So they will not form a static, detectable interference pattern at the target. Multiple lasers aimed at a target containing deuterium or tritium are being tried as a means of exciting nuclear fusion in the sample to produce power.
-Phil
> I was taught that light beams do not interfere with one another.
That was quantum I and maybe II - you need to get deeper into QED (quantum electro-dynamics) to look at Phil's rather devilish - and perhaps - trick, question. :-P
Dave (xantos)> (although a photon is its own antiparticle, I believe, like neutrinos).
AFAIR, Dirac noted first in "Principles of Quantum Mechanics" that a photon can only interfer with itself. But that was really before QED kicked into high gear. Since a photon is a _boson_ they can interfere with other photons. A neutrino is a lepton, not a boson.
Keep in mind that a photon's probablity state ("probability amplitude") of *where* it is, differs from where it is observed, (where it is annihilated).
The question really is, can two photons from different sources be superpositioned?
- Howard
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If Schr
Ordinarily the light beams will not generate new frequencies when they pass through the same region of space, but their electric and magnetic fields will still add or subtract where the beams cross.
In cases where the beams are extremely closely matched, such as when recombining beams from a single light source as in an interferometer, the addition and subtraction can result in large areas of low signal, or darkness, alternating with areas of strong signal, so you see a pattern like a bulls-eye or a picket fence series of lines.
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Phil did mention the question immediately after, and in the context of, interference 'across' and 'energy imparted to a target'. To which your post with the interference amplitude waveforms was a nice explainer when the wave function is solved, when the photons are observed and annihilated. However, that is *not* what Phil's thought experiment is, the actual question was much more subtle and I requote:
" Suppose it were possible (e.g. with a half-silvered mirror) to create colinear beams that exactly cancel. Now each laser emits light having a certain amount of energy. Due to conservation of energy, the energy in the light beams cannot simply disappear. So where does it go when the beams cancel? "
Phil can correct me if I'm wrong, but I read this to be asking what is happening to the light beams *in transit, before* any targets are hit. This is a *very* different question than what you address (and also different from what the OP asked about).
He should have put TWO to FIVE devils at the end of his question - it's been a thorn in optical quantum mechanic's side for decades. In fact, it's a thorn in the side of not only Q.M. but also touches on the entire "standard model" and the unresolved "grand theory" bridges needed between QM and relativistic theories. It goes to the very foundations of modern science. (That's why I was so flippant in my first answer, because from what I've read in Phil's other posts, he *usually* ;-P knows what he's talking about --- I suspect he knew where this was coming from too.)
If, to this question, you give an answer that physicists accept, this forum will become very hot and famous - because you will certainly win a Noble.
cheers
- H
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My guess is it would become the perfect laser never losing any energy no mater how much it reflects. Though its energy would not be in a visible form but in a probabilistic form only
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