# Laser Light Dispersion **URL:** \ **Category:** Factual Questions\ **Created:** [February 16, 2000, 7:08pm UTC](https://boards.straightdope.com/t/laser-light-dispersion/7588 "2000-02-16T19:08:00Z")\ **Posts on this page:** 20\ **Page:** 1
**Author:** ![pluto](https://avatars.discourse-cdn.com/v4/letter/p/f08c70/32.png) [@pluto](https://boards.straightdope.com/u/pluto)\ **Post date:** [February 17, 2000, 6:12am UTC](https://boards.straightdope.com/t/laser-light-dispersion/7588/1 "2000-02-17T06:12:00Z")
In an earlier thread that discussed beam weapons someone stated that one of the problems with laser weapons was that the beam was subject to an inverse-square law. A beam that was at a nominal strength of 1 at a certain distance was only 1/4 as strong at twice the distance, etc. The statement caught my eye at the time but I didn’t really give it much thought. Now I’m convinced it is wrong. The inverse-square law is for a spherical wave. If a light source gives off a certain number of photons to cover a spherical area then the number of photons per area does drop off with distance because the area increases as the square of the distance. But for a focused beam, even a flashlight beam, the intensity falls off more slowly since the area increases more slowly. Ordinary light, even if initially focused with a parabolic mirror, disperses over distance. But coherent light, from a laser, does not. (I know there are minor effects, the beam is not perfect.) So a laser beam should not diminish in intensity over distance. That’s why we can fire a laser at the moon and get a detectable reflection. There will be some dispersion and some degradation because of passing through the atmosphere, etc., so if you are talking about a beam passing through several miles of atmosphere there will be a significant loss. But not an inverse-square loss. ## Right? well well said the royal desiccation my political opponents back home always maintained that i would wind up in hell and it seems they had the right dope Don Marquis _archy interviews a pharaoh_ ---
**Author:** ![Padeye](https://avatars.discourse-cdn.com/v4/letter/p/a9a28c/32.png) [@Padeye](https://boards.straightdope.com/u/Padeye)\ **Post date:** [February 17, 2000, 6:34am UTC](https://boards.straightdope.com/t/laser-light-dispersion/7588/2 "2000-02-17T06:34:00Z")
That’s a myth about laser light, that there is no beam dispertion. Coherence does not mean that the waves somehow “stick” together but that all waves are in phase with each other rather than being random. Laser light must be focused as with any other light source but it’s much easier to focus accurately with a refrective lense than ordinary light. Still no focusing device is perfect so there will always be some dispersion and the inverse squares law applies. IIRC the laser used to bounce off the Apollo 11 planted mirror was many miles across by the time it reached the moom. ---
**Author:** ![Patty\_O\_Furniture](https://avatars.discourse-cdn.com/v4/letter/p/96bed5/32.png) [@Patty\_O\_Furniture](https://boards.straightdope.com/u/Patty_O_Furniture)\ **Post date:** [February 17, 2000, 6:35am UTC](https://boards.straightdope.com/t/laser-light-dispersion/7588/3 "2000-02-17T06:35:00Z")
Right! * * * Recent polls revealed that some people have never been polled, until recently. ---
**Author:** ![Padeye](https://avatars.discourse-cdn.com/v4/letter/p/a9a28c/32.png) [@Padeye](https://boards.straightdope.com/u/Padeye)\ **Post date:** [February 17, 2000, 6:52am UTC](https://boards.straightdope.com/t/laser-light-dispersion/7588/4 "2000-02-17T06:52:00Z")
> [@](#): > > Originally posted by pluto: > \*\*Ordinary light, even if initially focused with a parabolic mirror, disperses over distance. \*\* The problem with a parabolic reflector is not the kind of light but that the source would have to be a single point to make a non-dispersing beam. With a theoretical point light source exactly at the focus of a **perfect** parabolic mirror you could make a non-dispersing beam. ---
**Author:** ![andros](https://avatars.discourse-cdn.com/v4/letter/a/e480ec/32.png) [@andros](https://boards.straightdope.com/u/andros)\ **Post date:** [February 17, 2000, 6:56am UTC](https://boards.straightdope.com/t/laser-light-dispersion/7588/5 "2000-02-17T06:56:00Z")
from a nuclear engineer friend of mine: > [@](#): > > I actually had a laser physics class, back in the dark ages. I’ll look it up tonight and get you a definitive answer. > > But my initial thought is that your friend is partly right. There is a rule for the intensity of a laser beam as a function of distance. A laser beam will not remain perfectly collimated, mostly because is dispersion through photon collisions with matter. In a perfect vacuum, there is no matter for the little photons to bounce off of, but as you well know, there is no such thing as a perfect vacuum. That is the real reason that we can’t bounce a laser beam off of the surface of Alpha Centauri and get a return reflection that is detectable. (A perfectly collimated laser beam is a fine, highly intense point of light. At a distance, the beam begins to spread and becomes a bigger point of light with a lower intensity per area than at the source. So less signal gets back to our detectors on Mother Earth.) So a higher power (and therefore, higher intensity) laser will be more visible at a distance. > > I don’t think that the cross-sectional area of a laser beam increases at the inverse square of distance, though. But I will crack open the dusty tomes tonight and find you a real answer. -andros- ---
**Author:** ![frolix8](https://avatars.discourse-cdn.com/v4/letter/f/f04885/32.png) [@frolix8](https://boards.straightdope.com/u/frolix8)\ **Post date:** [February 16, 2000, 7:08pm UTC](https://boards.straightdope.com/t/laser-light-dispersion/7588/6 "2000-02-16T19:08:00Z")
Here’s a quote from Britannica: The beam from such a laser typically diverges by less than one part in a thousand, approaching the theoretical limit. The beam’s divergence can be reduced by passing it backward through a telescope, although fluctuations in the atmosphere then limit the sharpness of a beam over a long path. The company I work for makes a laser that can destroy a missile 100 miles away. What is the lasers power 1 foot from the device? 278 billion times more powerful than it is at 100 miles. I don’t think so. There may be an inverse square in the equation, but it is reduced very significantly. If you don’t believe this, take a common laser pointer and compare the bright spot from 1 foot with the bright spot from 10 feet. 1/100 of the power? The light from a bulb drops off with the square, try lighting an object with a fractional watt laser pointer and a 100 watt bulb from a foot away and from 10 feet away. Compare the relative brightness. ---
**Author:** ![InutilisVisEst](https://avatars.discourse-cdn.com/v4/letter/i/d2c977/32.png) [@InutilisVisEst](https://boards.straightdope.com/u/InutilisVisEst)\ **Post date:** [February 16, 2000, 7:35pm UTC](https://boards.straightdope.com/t/laser-light-dispersion/7588/7 "2000-02-16T19:35:00Z")
I believe it’s an inverse-square relationship for the dispersion, not the intensity. In other words, if your cheap laser pointerbeam has spread by 100% (width) at 10m, you should expect it to be 200% wider at 20m. Of course, ignoring atmospheric scattering, the beam is actually the same strength no matter what the distance, it’s just more diffuse. If you can get the entire beam on a target, you might still heat it to destruction. Fine focus to pierce a target’s skin is a different thing. * * * Sure, I’m all for moderation – as long as it’s not excessive. ---
**Author:** ![Stephen](https://avatars.discourse-cdn.com/v4/letter/s/96bed5/32.png) [@Stephen](https://boards.straightdope.com/u/Stephen)\ **Post date:** [February 16, 2000, 8:18pm UTC](https://boards.straightdope.com/t/laser-light-dispersion/7588/8 "2000-02-16T20:18:00Z")
Typical beam divergence from a laser with good mode is about 1.1 mrad. If you know your target distance, you can collimate, spatial filter, focus, etc. * * * Stephen [Stephen’s Website](http://stephen.fathom.org) Satellite Hunting 2.0.1 (Y2K compliant!) visible satellite pass prediction shareware available for download at [Satellite Hunting](http://stephen.fathom.org/sathunt.html) ---
**Author:** ![frolix8](https://avatars.discourse-cdn.com/v4/letter/f/f04885/32.png) [@frolix8](https://boards.straightdope.com/u/frolix8)\ **Post date:** [February 16, 2000, 8:29pm UTC](https://boards.straightdope.com/t/laser-light-dispersion/7588/9 "2000-02-16T20:29:00Z")
Here’s the formula: ``` Divergence Angle (in radians) = Theta/2 = (Wavelength * 4 )/ pi * Beam Diameter ``` I hope that looks OK . Here’s a source with an amazing amount of information about lasers: [http://plop.phys.cwru.edu/repairfaq/sam/laserioi.htm#ioilpm6](http://plop.phys.cwru.edu/repairfaq/sam/laserioi.htm#ioilpm6) ---
**Author:** ![douglips](https://avatars.discourse-cdn.com/v4/letter/d/e68b1a/32.png) [@douglips](https://boards.straightdope.com/u/douglips)\ **Post date:** [February 16, 2000, 9:43pm UTC](https://boards.straightdope.com/t/laser-light-dispersion/7588/10 "2000-02-16T21:43:00Z")
For a theoretical laser beam with a perfect point source and a non-zero beam width at a distance _d_, you do get an inverse square law for power per unit area - as you double the distance, the cross section of the beam will quadruple in area. However, if your target is big enough to cover the entire cross section, then all the energy is absorbed by the target no matter how far away, so there is only loss due to interactions with the atmosphere or other media. For a laser with a finite width at the source and a slow dispersion, the fall off of power per unit area will not be inverse square of the distance, but something less than that. However, it is only per unit area that the dispersion matters for - the entire beam will have the same energy minus the losses described above. ---
**Author:** ![ZenBeam](https://avatars.discourse-cdn.com/v4/letter/z/3ab097/32.png) [@ZenBeam](https://boards.straightdope.com/u/ZenBeam)\ **Post date:** [February 16, 2000, 10:41pm UTC](https://boards.straightdope.com/t/laser-light-dispersion/7588/11 "2000-02-16T22:41:00Z")
> [@](#): > > Divergence Angle (in radians) = Theta/2 = (Wavelength \* 4 )/ pi \* Beam Diameter Valid when you are far from the source. There’s no hard cut-off for how far _far_ is, but the formula I have for the spot diameter is D(z) = D0 \* sqrt(1 + z^2/z0^2) where z0 = pi \* D^2 / (4 \* wavelength) and D0 is the diameter at the laser. So, roughly, farther than D^2 / wavelength the spot follows the inverse-square law. For a laser with D = 2 mm (I’m guessing here. Is this typical for those pointers?), and for wavelength = 600 nm = 6E-7 meters, this is 6.7 meters = 22 feet. At this distance, you’ve still less than doubled the spot size. Just for kicks, to get z0 = 100 miles, I get D = 31 cm = 1 foot. Am I close, Frolix? * * * _It is too clear, and so it is hard to see._ ---
**Author:** ![frolix8](https://avatars.discourse-cdn.com/v4/letter/f/f04885/32.png) [@frolix8](https://boards.straightdope.com/u/frolix8)\ **Post date:** [February 16, 2000, 11:43pm UTC](https://boards.straightdope.com/t/laser-light-dispersion/7588/12 "2000-02-16T23:43:00Z")
ZenBeam, you could be on target. Which is almost a MIRACL. 'Nuff said? ---
**Author:** ![Geezer](https://avatars.discourse-cdn.com/v4/letter/g/49beb7/32.png) [@Geezer](https://boards.straightdope.com/u/Geezer)\ **Post date:** [February 18, 2000, 6:42am UTC](https://boards.straightdope.com/t/laser-light-dispersion/7588/13 "2000-02-18T06:42:00Z")
Lasers most certainly do obey the inverse square law. The rays of a laser beam do diverge (not “disperse”; that’s a different phenomenon) and as a result, the area of the beam increases as the square of the distance along the beam. A laser beam _is_ a spherical wave–it’s just that the center of curvature of the sphere is not at the laser, but some large distance _behind_ the instrument itself. A plane wave is simply a spherical wave of infinite radius of curvature. Forget lasers for a moment. Suppose you have a point light source–a pinhole with a light bulb behind it, say. In front of the pinhole at a distance r0, we place a mask with a circular aperture in it. Light emerges from the aperture into a cone with its vertex at the source. Now how do you apply the inverse square law? The flux falling onto a patch of area A at a distance r from the pinhle is some amount F. At a distance 2r, the same flux falls onto an area 4A, so that the illuminance is quartered. Now this analysis depends on measuring _from the pinhole_, that is the center of curvature of the waves emerging from the aperture. If you measure r and 2r from the aperture (i.e. the front of the instrument), you won’t get “double the distance, quadruple the area” behavior. Now suppose you fit a lens into the aperture such that the focal length of the lens matches the distance to the pinhole. The result is a collimated beam of light emerging from the lens (i.e. the aperture). Now if you perform the “double the distance” experiment, you won’t get “quadruple the area” results, even if you measure from the pinhole. The reason is that the lens has had the effect of optically moving the pinhole to a “virtual position” an infinite distance _behind_ the actual pinhole. That is, the vertex of the cone from which the rays diverge is a _long_ way away, so doubling the distance from the pinhole doesn’t double the distance from the “source.” A laser beam is very well collimated–but not perfectly. That is, it is nearly a plane wave–a spherical wave with very large (but not infinite) radius of curvature. It obeys the inverse square law, but you have to measure distances from way back behind the laser itself. Divergence in lasers arise from two souces: intrinsic divergence from the “beam waist” inside the laser cavity, and diffraction at the exit aperture. Further errors may be introduced by additional beam-shaping optics. ---
**Author:** ![frolix8](https://avatars.discourse-cdn.com/v4/letter/f/f04885/32.png) [@frolix8](https://boards.straightdope.com/u/frolix8)\ **Post date:** [February 17, 2000, 8:00pm UTC](https://boards.straightdope.com/t/laser-light-dispersion/7588/14 "2000-02-17T20:00:00Z")
OK, let’s see if we can agree on this: The inverse square law, from my Physics text, is stated: “The intensity of radiation from a point source decreases as the square of the distance” My view is that a laser doesn’t qualify as a point source, so the law doesn’t apply. Others argue that the law _does_ apply if we consider a virtual point source to exist far “behind” the laser. Either way, the same equations are used to calculate the lux at distance D from the laser. In any case, lasers are way cool. ---
**Author:** ![ZenBeam](https://avatars.discourse-cdn.com/v4/letter/z/3ab097/32.png) [@ZenBeam](https://boards.straightdope.com/u/ZenBeam)\ **Post date:** [February 17, 2000, 10:22pm UTC](https://boards.straightdope.com/t/laser-light-dispersion/7588/15 "2000-02-17T22:22:00Z")
Geezer wrote: > [@](#): > > Lasers most certainly do obey the inverse square law. > … > It obeys the inverse square law, but you have to measure distances from way back behind the laser itself. This is wrong. The power density (for a Gaussian beam) falls off as 1 / (1 + z^2 / z0^2). This is **not** inverse square. Geezer, I challenge you to tell me what distance L behind the laser I should use. That is, what is the value of L such that L^2 / (z + L)^2 = 1 / (1 + z^2 / z0^2) (I included L^2 on top of the left side so the equation would be correct at z = 0) * * * _It is too clear, and so it is hard to see._ ---
**Author:** ![Zor](https://avatars.discourse-cdn.com/v4/letter/z/848f3c/32.png) [@Zor](https://boards.straightdope.com/u/Zor)\ **Post date:** [February 18, 2000, 5:01am UTC](https://boards.straightdope.com/t/laser-light-dispersion/7588/16 "2000-02-18T05:01:00Z")
Lasers do obey the inverse square law in certain situations, and don’t in some others. We certainly don’t have to split hairs to see why however 🙂 Simply put, if the distance traveled by the laser beam is significantly larger than the original diameter of the beam itself, then the laser can be considered to be a point source. On the other hand, if the distance traveled is significantly smaller than the original diameter of the beam, then you’ve got yourself a planar source, and similar arguments hold for line sources. The problem we have here is that there’s a gray area in between the point and planar/line sources. Usually you just consider your light source to be composed of a plane/line of point sources in this area. So, the real question we should be asking ourselves is how large our original beam is, and what distances of travel we’re talking about. Since the original post was referring to laser weapons, it’s probably a good idea to use a point source for simplicity of calculation. ---
**Author:** ![ZenBeam](https://avatars.discourse-cdn.com/v4/letter/z/3ab097/32.png) [@ZenBeam](https://boards.straightdope.com/u/ZenBeam)\ **Post date:** [February 19, 2000, 6:08am UTC](https://boards.straightdope.com/t/laser-light-dispersion/7588/17 "2000-02-19T06:08:00Z")
> [@](#): > > Simply put, if the distance traveled by the laser beam is significantly larger than the original diameter of the beam itself, then the laser can be considered to be a point source. ## _Too_ simply put. By this criteria, if I were, say, twenty feet from a one foot diameter beam of light, I’d expect to be able to treat it as a point source. In fact, I need to be more like 100 _miles_ away. _It is too clear, and so it is hard to see._ ---
**Author:** ![ZenBeam](https://avatars.discourse-cdn.com/v4/letter/z/3ab097/32.png) [@ZenBeam](https://boards.straightdope.com/u/ZenBeam)\ **Post date:** [February 18, 2000, 7:33pm UTC](https://boards.straightdope.com/t/laser-light-dispersion/7588/18 "2000-02-18T19:33:00Z")
> [@](#): > > Simply put, if the distance traveled by the laser beam is significantly larger than the original diameter of the beam itself, then the laser can be considered to be a point source. ## _Too_ simply put. By this criteria, if I were, say, twenty feet from a one foot diameter beam of light, I’d expect to be able to treat it as a point source. In fact, I need to be more like 100 _miles_ away. _It is too clear, and so it is hard to see._ ---
**Author:** ![frolix8](https://avatars.discourse-cdn.com/v4/letter/f/f04885/32.png) [@frolix8](https://boards.straightdope.com/u/frolix8)\ **Post date:** [February 18, 2000, 8:57pm UTC](https://boards.straightdope.com/t/laser-light-dispersion/7588/19 "2000-02-18T20:57:00Z")
As an aside, the exit port for the MIRACL laser is about 4.5 feet in diameter. ---
**Author:** ![Zor](https://avatars.discourse-cdn.com/v4/letter/z/848f3c/32.png) [@Zor](https://boards.straightdope.com/u/Zor)\ **Post date:** [February 18, 2000, 11:20pm UTC](https://boards.straightdope.com/t/laser-light-dispersion/7588/20 "2000-02-18T23:20:00Z")
It doesn’t matter if the MIRACL’s exit port is 4.5 feet or 4.5 miles in diameter; the beam itself is only 21 cm high and 3 cm wide. It is also used to shoot at targets miles high in the sky. I still remember a couple years back when they pointed that thing at a satellite in orbit. So, for similar purposes of blowing things out of the sky, a point source is a very good approximation. And ZenBeam, \* significantly larger \* is generally denoted as \*\* \>\> \*\*in mathematical equations. A 20:1 distance to diameter ratio is not even remotely close to this criterion. I suppose it’s my fault for not clarifying the terms however. On the other hand, as for having to be 100 miles away to treat said beam as a point source, that is surely a gross exaggeration; a mere 100 feet is enough for all practical purposes. I should also add that I have never heard of a laser that generates a 1 foot diameter beam. Most laser beams are barely the size of your palm, if not much smaller. There are engineering difficulties in producing larger beams, not to mention there is no purpose in manufacturing one anyway; a simple lens system could magnify the beam to whatever sizes you need. [Next page](https://boards.straightdope.com/t/laser-light-dispersion/7588.md?page=2)