If lasers are collimated, what causes them to decollimate? Their production system seems to suggest a completely linear, collimated light source, but they do spread out over large distances. The same holds for synchrotrons. Why does this happen?
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There is a fundamental limit to the collimation of a laser due to diffraction. Assuming the laser beam profile is a uniform disk it will be diffracted to an Airy disk at large distances, and the angular spread is approximately given by:
$$ \theta \approx 1.22 \frac{\lambda}{d} $$
where $d$ is the beam diameter. Assuming a diameter of 1 mm, which seems a reasonable estimate for most lasers I've seen, you get an angular divergence of about 0.6 milliradians for 500nm light.
I know next to nothing about the design of lasers, but Wikipedia reports the divergence as commonly less than 1 milliradian, which fits with the above estimate.
John Rennie
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If the beam is a gaussian, the constant will be $\frac{1}{\pi}$ rather than the Airy constant – lurscher Jul 22 '13 at 16:46
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Following John Rennie's answer I read the 1st paragraph of Collimated light at Wikipedia .
Last 3 lines stated that "Collimated light is sometimes said to be focused at infinity.Thus as the distance from a point source increases, the spherical wavefronts become flatter and closer to plane waves, which are perfectly collimated."
Does it mean that for a light source with enough power we will have a perfectly collimated light after a certain distance far away from source when it has already passed its divergent phase because of spherical wavefronts ? So after a certain distance we have a perfectly parallel plane wavefronts and beam won't be divergent any more!
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Lasers are not perfectly collimated.
In fact, according to an analogue of the Heisenberg Uncertainty Principle, it's fundamentally impossible to create a perfectly-collimated beam of light from a finite-sized source.
Why? If the light is traveling in the z direction, and it's perfectly collimated, then the photon momentum in the x and y direction is exactly zero. Which means the photon position in the x and y direction is completely unknown. Again, only a wave with an infinitely-wide cross-section can be exactly collimated.
A laser beam is never very large, and certainly not infinitely large. The size of the laser beam depends on the laser active area (the area producing laser light), and also depends on the lenses you put after the laser. For example, a laser pointer might have its light in a 1mm circular spot (before it spreads out). As a rule of thumb, you can divide the wavelength of the laser light by 1mm. You get a very small number, but it's not zero. That number is more-or-less the smallest possible divergence angle (in radians) that this beam can have.
Although it depends on the application, people who make lasers usually try to have them output something close to a gaussian beam, which is more-or-less the most collimated possible given the finite size of the beam. If you read about gaussian beams, you'll see the tradeoff very directly between the cross-sectional area of the beam and its divergence angle.
Well, it's impossible even to put 100% of the energy into a perfect gaussian beam (once again because the laser has a finite size), but you can get close to 100%. See M2 and beam parameter product for how these things are characterized.
Steve Byrnes
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This is the particle point of view, photos are better described as waves for this case, and most others. A laser can be perfectly collimated at a single z-position. And it is not "impossible even to put 100% of the energy into a perfect Gaussian beam" . – Luke Burgess Dec 01 '13 at 16:42
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@LukeBurgess -- You have a funny definition of "collimated" if a wave can be "collimated at a single z-position". The whole point of collimating a beam is that it is supposed to stay collimated as it travels, until it hits the next lens. A beam of finite size can never be perfectly collimated. You can prove it by thinking about photons, or by working through Maxwell's equations, or by thinking about Huygens' principle, or by thinking about diffraction, or by thinking about fourier transforms and plane-wave decompositions. (Actually, these all amount to the same thing.) – Steve Byrnes Dec 01 '13 at 18:37
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@LukeBurgess - As you go away from the center of a gaussian beam, the intensity reduces exponentially, but never reaches zero in a finite distance. Therefore a finite-sized source or cavity CANNOT be 100% mode-matched to a gaussian beam. Maybe you can find a source that puts 99.99% of the energy into a perfect gaussian beam, but not 100%. – Steve Byrnes Dec 01 '13 at 18:38
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If any portion of the beam is not Gaussian, it can be said that such a portion is no part of the beam. This portion is due to interference with other wave-forms. Read http://arxiv.org/pdf/1308.1326.pdf – Luke Burgess Dec 01 '13 at 19:02
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I wrote "it's impossible to put 100% of [a laser's optical] energy into a perfect gaussian beam". This is obviously true because lasers have finite lateral size while gaussian beams do not. Are you really arguing that this statement is not true? Your comment above ("...not part of the beam") sounds like "100% of the energy goes into the gaussian beam, if you ignore the energy that does not go into the gaussian beam". Are you really arguing something so silly? If not, I don't understand you. The paper you linked to references a ~99.3%-efficient coupling, which they rounded to 100%. – Steve Byrnes Dec 01 '13 at 21:48
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If you were talking about the device and not the beam in the answer, then what are you trying to say? Yes, "Lasers" the devices are not 100% efficient, and the beams do scatter due to air particles. But you said "it's fundamentally impossible to create a perfectly-collimated beam of light". If you add qualifiers like "maintain", "keep", "over all space", or "indefinitely", then I am OK. But as it stands the statement is misleading. BTW: I hope you are aware that a Gaussian beam is not a plane wave, except at a single position along its axis of travel. – Luke Burgess Dec 03 '13 at 18:45
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I hope you are aware that a Gaussian beam is not a plane wave anywhere, full stop. It is not a plane wave at the beam waist, it is not a plane wave anywhere else. Nor is it "collimated" at the beam waist, or anywhere else. I think you're confused about the definition of "plane wave", and "collimated". – Steve Byrnes Dec 03 '13 at 22:20
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A plane wave front, or collimated wave front is any wave front with a radius of curvature at infinity. – Luke Burgess Dec 04 '13 at 17:06
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"At the waist, the wavefront is a plane." http://www.optique-ingenieur.org/en/courses/OPI_ang_M01_C03/co/Contenu_08.html – Luke Burgess Dec 04 '13 at 17:29
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"Plane wave" is a technical term with a standard definition that you can (and should) look up in any textbook. You'll see that the definition of "plane wave" is not what you think it is. (Sure, if you look around enough, you can find someone who has misused the term "plane wave". But I promise, the term "plane wave" is completely standard and unambiguous.) I'm glad you seemingly understand the physics of light propagation, but you do NOT know the terminology for talking about it, e.g. the standard definitions of "plane wave" and "collimated". – Steve Byrnes Dec 05 '13 at 02:11
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What is this more "unambiguous" definition than "a wavefront with a radius of curvature at infinity"? "A collimated beam has a radius of curvature approaching infinity" from http://laser.physics.sunysb.edu/~ariana/tm/ – Luke Burgess Dec 05 '13 at 05:02
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is this better:"As the source is at infinity the waves reaching the obstacle are plane waves. The curvature of wavefront is negligible." PRINCIPLES OF PHYSICS - By P.V. NAIK – Luke Burgess Dec 05 '13 at 05:05
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You need the correct definition, not just an unambiguous definition. For example, if you say "'neutron' is a term for any neutral particle", it's an unambiguous definition, but it's not the correct definition. :-P You can't call something a "plane wave" unless it has planar wavefronts (note the plural; one planar wavefront is not enough) AND (related to that) it has constant intensity in the lateral directions. The Naik quote is consistent with that. Again, you can look it up in any textbook or ask any expert what "plane wave" means. – Steve Byrnes Dec 05 '13 at 15:54
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how about replacing "perfectly collimated" with "a perfect plane wave". I can agree to that. The one time I used the words "plane wave" I ended it with "front", to make it clear I was not saying in all space time. (Except the quoted material). If this must be turned into a question... – Luke Burgess Dec 05 '13 at 16:36
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