# Why is laser light a cone?

What about the production of laser light disallows it to be perfectly straight as opposed to a cone? I feel like it should be a plane wave, not a very tight cone.

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Can anything man-made be "perfectly" anything? –  tpg2114 Jul 16 '13 at 2:53
See also my description (physics.stackexchange.com/a/70394/26076) of diffraction. A non diverging plane wave has infinite sideways extent. You only have a finite output aperture for a laser, although you can make it very big. –  WetSavannaAnimal aka Rod Vance Jul 16 '13 at 3:19
Why can't you just have the source be some flat surface? (I don't really know how lasers work.) I mean, what if you just have some small flat surface that emits light-would that then be a plane wave-even though it isn't an infinite plane source? –  Anthony Jul 16 '13 at 5:37
Because Christian Huygens wanted it to be a cone :=) –  Georg Jul 16 '13 at 11:24
You can make the source a big flat surface as you say. But the bigger you make it (and it has to be optically perfect to well within $\lambda /10$) the harder and costlier it is to do. The beam width is also set by the gain medium - if this isn't a gas, then this is hard to make wide too. It comes down to technology limitations, and you'll never get the aperture to be infinite, so there'll always be some divergence. –  WetSavannaAnimal aka Rod Vance Jul 16 '13 at 11:45

Plane waves are an idealization and can never be achieved in the real world because they must have an infinite spatial extent (and thus carry infinite energy) to work. A truncated plane wave is not a solution to the wave equation. The feature that real lasers have that forces them to have nonzero angular divergence is a finite spatial extent. Like all finite-size wave sources, their output will diffract.

This spatial extent will usually be a few millimeters, which is about 10,000 times the wavelength (of about a few tenths of a micron). This means that the beams can have angular divergences as low as ~1 milliradian, which they are very close to: over 10 m, the beam has to diffract to at least 1 cm, and it usually isn't bigger than that. Thus even cheap laser pointers are usually pretty close to 'as good as it gets' as regards angular divergence.

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So, holding wavelength constant, the wider the aperture of the laser cavity the lower the angular divergence of the beam? –  feetwet Jul 25 at 22:21
That is sort of correct. A wider cavity aperture enables a smaller divergence, but it does not guarantee it. The focusing needs to be right, and the laser needs to be clean enough for it to work. Once your divergence is diffraction-limited, though, only a wider aperture can save you. –  Emilio Pisanty Jul 25 at 23:35

Well a laser source appears to be an "artificial star" ; a diffraction limited source of some small but not zero extent. For example something like a He-Ne gas laser appears to be a source about the diameter of the glowing gas discharge that you see in the tube. The larger that apparent source is the smaller is the angular divergence of the beam. The smaller the source minimum size, the larger is the beam divergence. At the narrowest point of the beam -- the beam waist -- the wavefronts are planar. As you go further from the waist, however, the wavefronts become approximately spherical, and at large distances they are quite spherical.

Small lasers like solid state diode lasers have a very small beam waist, so they have very large divergence-cone angles of many degrees. However, a good clean laser beam can be focused to a fractional wavelength sized spot. Generally the spot will be a clean circle with no surrounding rings, because the beam has a Gaussian intensity cross-section profile, so the beam tapers to zero at the outer edge. It is common practice to use laser optics having a clear aperture of about 1.5 times the diameter where the beam drops to $1/e^2$ of its central intensity. If you crop the beam optics tighter than that so the lens rim has significant beam intensity, then you will get diffraction rings, and the beam quality will degrade. There are standard textbooks, that describe the profiles both laterally and longitudinally for various laser resonator cavity configurations.

It is possible, with a lens, to convert the laser beam mode into a converging beam, so that the beam waist is well outside the laser and you get the smallest possible spot at some distance. The further away that is, the longer will be the distance over which the beam remains almost parallel.

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Following the last paragraph of this answer (and returning to the original question): So can you focus the laser at infinity so that the beam has no divergence? And boy would I love some diagrams to illustrate this answer.... –  feetwet Jul 25 at 22:16

Laser light is collimated because the resonant cavity used in (most) lasers has two parallel mirrors at either end. Light makes many trips between the two mirrors so any light that isn't closely parallel to the axis of the cavity hits the side of the cavity and is absorbed. Only light closely parallel to the axis survives to emerge from the laser.

Actually the mirrors are not exactly parallel. They are normally slightly concave for technical reasons I've never fully understood. However the deviation from a plane is so small that the beam divergence is typically less than a milliradian.

As George says in his answer, the beam will diverge due to diffraction. I calculated this in my answer to Lasers and Collimation. The divergence due to diffraction is also typically under a milliradian.

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First the suggestion that a plane wave must be infinite in extent. Well I suppose that is pedantically correct, even if the wave intensity at the periphery is 10^-(Avogadro's number).

But a single mode fundamental laser mode has a Gaussian intensity beam profile, so the wave amplitude diminishes rapidly with radius, and at the propagating beam waist the wave is exactly plane but essentially zero amplitude at the extreme edge.

Now the parallel mirrors idea. If you have two exactly plane parallel mirrors optically flat to say 1/100th wavelength, and a collimated parallel beam can be generated inside the cavity perpendicular to those ideal mirrors, the beam will bounce back and forth indefinitely.

But suppose the mirrors are not exactly parallel, but have a slight wedge angle, say 10^-100 arc seconds. Well because of that wedge angle, the beam will translate sideways, that tiny angle times the round trip length. So the beam eventually walks off the edge of the mirrors.

So two plane parallel mirrors is an unstable resonator. The beam cannot live inside that cavity. There are actually an infinite number of end mirror configurations, and these can be plotted on a graph of C1 versus C2 where C is the mirror curvature. We optical design types do NOT like radii of curvature. It is some little nit picky thing like my laptop keyboard does not have an infinity key to use for the radius of curvature of a plane surface. And in imaging lens situations it is 1/r or c which determines the focusing effect, and curvature (powers) just add algebraically. So mathematically we deal in curvatures.

A very common stable laser resonant cavity, is a single plane mirror, plus a single concave mirror. The mirror radius of curvature is twice the mirror separation. Actually, this is just half of a "confocal" cavity, where two spherical mirrors each have their center of curvature on the opposite mirror.

Now just think about that. Any line from the center of curvature, is a radius of the sphere, and many such radii, will hit the other spherical mirror. Now one of those radii, will hit the spot on the second mirror, where the center of the first mirror is.

So it can be seen that the single line joining the centers of curvature of the two mirrors, is a radius of both of them; it is a perfect axis of symmetry. Now a few sketches will convince that the two mirror radii, don't need to be exactly equal. The cavity is quite stable against practical manufacturing tolerances. Well in the confocal resonator, you have spherical wave fronts on each mirror, and exactly half way between, is the smaller planar beam waist.

So hell, why not put a plane mirror there, and dispense with half of the cavity length.

And yes, there will be a normal to the plane mirror that passes through the center of curvature of the one spherical mirror, and therefore is a radius. So the half confocal cavity is a very popular choice, and most He-Ne lasers are built that way. That plane mirror with its smaller beam waist is pretty much always the output mirror, so it has less than 100% reflectance, to let some beam leak out. The back spherical mirror can be 100% reflecting, or you can leak a small back beam, that you can sense, as an amplitude feedback signal to regulate the laser beam power.

The beam divergence is not dependent on the radius of the curved mirror, it is the diffraction limited spread of the beam depending on the diameter of the ;lasing "gain medium". longer tube lengths will result in narrower laser line widths, which also affects the divergence. I'm writing this all from memory, so I would have to dig out some text books to get you the design math; but I suspect that Wiki or some other site would have more details on the resonator stability diagram.

Anna V understands all this beam stability stuff, because the exact same optical beam confinement problems occur in particle accelerators. A circular path in a uniform transverse magnetic field, is also unstable as Anna knows, so your particles will all crash into the walls, if you don't design a stable resonant cavity for your circulating particle beam. so you have to have magnetic lenses with alternating gradients and all kinds of trickery.

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I'm not sure the "infinite plane wave" thing is just pedantry - it's a hopefully simpler way of talking about what happens when you look at the field in momentum space: there will always be a spread of plane wave directions in a wave superposition, thus a tendency to diverge, unless the wave has infinite sideways extent. The Gaussian mode also has a nonzero spread of angles (even though it saturates the Heisenberg inequality). The fundamental issue is that a function and its Fourier transform cannot both have compact support - this is a real and inescapable fact. –  WetSavannaAnimal aka Rod Vance Jul 17 '13 at 2:49
+1 for the lucid discussion of design issues. –  WetSavannaAnimal aka Rod Vance Jul 17 '13 at 2:50