Collinearity of two infrared (10.6um) beams

I'm interested in aligning two infrared (10.6um) beams so that they are as collinear as possible. What is the best way to do this while maintaining the polarization of the two beams? Is there a theoretical limit to how well they can be aligned (eg maybe you can't align them better than their wavelength)? I need them aligned to within about 10um over a distance of about 5 metres for an experiment. Here is how I was thinking of doing this:

• I can think of no theoretical limit to the alignment - just to your ability to measure it - your detection scheme needs to be better than this. A pinhole that can be translated across the path of the beams, and turning the sources on one at a time, should allow the measurement accuracy you need. Could you show with a sketch how you intend to achieve this? Also - "Aligned to within 10 um" over what length (since we are talking colinear presumably you want them in parallel over a certain run distance). – Floris Oct 10 '14 at 23:01
• Good point, I would need them aligned to within 10um over a distance of about 5m. I've edited my question and added a picture of how I was going to attempt to do this. – cpc333 Oct 10 '14 at 23:17
• Are they mutually coherent, or from different sources? – WetSavannaAnimal Oct 10 '14 at 23:37
• I would split the setup into two parts - one that gets the beams parallel, and another that translates one of the beams. The latter is done easily with a non reflective parallel faced plate that you angle slightly - this allow minute shifts without changing direction. – Floris Oct 10 '14 at 23:37
• They are from the same source, then split, then one is frequency shifted by an acousto-optic modulator and rotated by a 1/2-waveplate. – cpc333 Oct 10 '14 at 23:47

Great setup! I don't envy your task: IR is very tricky and frustrating to align.

Do not expect to solve this problem overnight: this is a many month project, unless you can find someone who has done exactly this before, in which case it would have been a many month project for them.

Part of the translation is easy: for fine ajustment, you use an angled plate as Floris suggests: as you change the angle, the sideways translation varies rougly as $(n-1)\,t\,\theta$, where $n$ is the refractive index of the plane, $t$ its thickness and $\theta$ the tilt. You'll naturally need two of these, with orthogonal tilt axes, to get you $x$ and $y$ translation.

But you'll also need the ability to coarsely translate many millimetres: as you adjust the beam's orientation through angle $\delta \theta$, it will shift sideways $5m \times \delta \theta$ at the end of your beams. Two orthogonal, variable length periscopes kitted with micrometers will do this. You may be able to lose the rotating plate if you kit your periscopes with differential micrometers.

The alignment of directions is the tough bit. 10um over 5m is a very tall ask indeed: $2\times 10^{-6}$ radians or half a second of arc. The very ability to rotate something to this precision is a tall ask: well beyond any goniometer you're going to find easily. A differential micrometer will get you between $0.1$ and a half $\mu m$ translation resolution (depending on practice), so I'd be grounding my rotation method on something like this with a half to one metre long lever arm. At this size, you're going to find temperature variations a problem. Another method would be to have a mirror on a leaf spring; you finely vary the force on the latter. You'll naturally need two orthogonal rotating mirrors whichever way you go to get you the $\phi$ and $\theta$ spherical co-ordinates.

I would suggest your "within $10.6{\rm \mu m}$ over 5m" specification needs to be looked at very carefully. Do you really need this accuracy. You see, for a laser beam to be a "beam" within $2\times 10^{-6}$ radians, this implies a very wide beam. Let's look at a Gaussian beam: one whose transverse variation goes like

$$E(x,\,y,\,z=0) \propto\exp\left(-\frac{x^2+y^2}{2\,\sigma^2}\right)$$

where $\sigma$, the spotsize, measures the beamwidth. A common beamwidth definition is the Petermann II diameter $\mathscr{P} = 2\,\sqrt{2}\,\sigma$. The Fourier transform of this beast goes like:

$$\tilde{E}(k_x,\,k_y) \propto\exp\left(-\frac{1}{2}\sigma^2\left(k_x^2+k_y^2\right)\right)$$

where $k_x,\,k_y$ are the transverse wavenumbers, so that

$$\theta\approx\frac{\sqrt{k_x^2+k_y^2}}{k} = \frac{k_\perp}{k}$$

is the skew angle relative to the nominal propagation direction of the plane wave component in question. Now, if the beam's angles are confined to skew angles of say $10^{-5}$ radian, this means that $k_\perp\,\sigma \approx 2$ when $k_\perp/k\approx 10^{-5}$, that is, $10^{-5} \,k\,\sigma\approx 2$ or $\sigma \approx 10^5\,\lambda/\pi$, which, at $10.6{\rm \mu m}$ wavelength, works out to be $0.337{\rm m}$. That is, your beam is $\mathscr{P}=2\sqrt{2}\sigma \approx 1{\rm m}$. Your beam is a METRE wide! So do you really mean to align your beam to $10.6{\rm \mu m}$ accuracy? (even with such a wide beam, its centroid accuracy can still be microns).

Now, onto the optics. You adjust your beams so that they meet at the 5m target. You'll need to do this with a sense card, pinhole or CCD (I'm not sure what technology will work here at $10.6{\rm \mu m}$). Then, with the beams meeting at the target, you check how far they are apart at the beginning of their 5m run by the same method. This lets you work out their alignment, so you now rotate the beams to meet at the beginning of their run. This rotation will split them apart at the far end, so you translate your beams with the periscope / angled glass so that they are back together at the far end. Then you repeat the process. Hopefully iterating will bring you the result you need.

Another device which will test alignment I believe very effectively in your case is the point diffraction interferometer. It won't get your final alignment, but it will help you find the neighbourhood of where you need.

This is a way cool device. The configuration I show is a little different from the Wiki page and also from the point diffraction interferometers in the literature. It has a subwavelength pinhole P (get a 1um or 2um electron microscopy pinhole) and a collimating, high quality objective L focussing on the pinhole. In the other arm of the Mach Zehnder is a neutral density wheel - a variable attenuator. The principle is this: one copy of the beam is passed through the subwavelength pinhole, which becomes a point, almost isotropic source independent of the aberation or tilt of the beam. This aberration free source is collimated by the objective whose beam diameter will have to be equal to or wider than that of the laser beam. This is OK, most objectives have infinity conjugate beamwidths of between 6mm and 12mm - near enough for your application. See this one http://www.edmundoptics.com/microscopy/infinity-corrected-objectives/zinc-selenide-infrared-ir-focusing-objectives/3423 for example. Anyhow, you now have an aberration free, reference beam which does not tilt with the input beam - all directional and wavefront information is destroyed by the subwavelength pinhole (it becomes part of the evanescent field, which does not propagate). This constant direction reference is interfered with the other arm of the Mach Zehnder. Obviously, the pinhole throws a huge amount of the light away, so you must now adjust the neutral density wheel (I've also used a rotating polaroid) to match the power of the two beams and get high fringe visibility. So now, as you tilt a beam slightly at the input, this tilt will show up as a relative tilt of the input beam and the reference beam. You'll see fringes: one fringe across a 6mm beam corresponds to a tilt of 10.6 / 6000, or about $10^{-3}$ radians, and you can probably see about a 10th of a fringe with practice, so that resolution will be $10^{-4}$ radians. If you make one of the mirrors creep back and forth a wavelength or so with a piezotranslator at a frequency of a few hertz, you'll get a shimmering pattern that will help you null the fringes accurately.

So you set this device up on a two tilt axis mount (or two tilts + X-Y, if you can't put the centre of rotation at the device's input) and you adjust the beam's orientation to null the fringes for one of the beams. Do this near the beginnings of the beams so that tilts of the beams only beget small translations. Then you translate the second beam to enter the device and adjust its tilt to again null the fringes. The beams are now aligned and when you translate them together, they should be very near to the configuration you need.

WARNING: You'll hate me for saying this, but I'm trained as a laser safety officer, so I feel I have to say it even though you and your alma mater are likely well grounded in the principles: one cannot overstate the dangers of IR light. Your body lacks the blink relfex which will probably render even an accidental class 3R viewing of visible light quite harmless in the long run. Not so with IR, severe damage can be quite painless and imperceptible at first. One of the universities I worked for had a case where an experimenter lost the sight in one eye (temporarily, although there was some residual permanent damage) and yet was unaware of it (the brain does amazing fill-in tricks) and was run over by a car (luckily not fatally) when he went for lunch. Make sure you know the IEC 80625 maximum permissible exposures and keep within them. You can't see the beams, so your saftey goggles are not going to be a hindrance.

Good luck, and be sure to write up whatever you come with as an answer to your own question.

• While you're on the subject of safety, I'd suggest having the combined beam travelling horizontally, not upwards! – ptomato Oct 11 '14 at 4:13
• @ptomato Good point. – WetSavannaAnimal Oct 11 '14 at 4:37
• I don't need the divergence of the beams to be low, I just need the centre of the two beams to line up to within 10um and to stay lined up to within 10um for a distance on the order of 5m. – cpc333 Oct 11 '14 at 16:58
• @cpc333 That's OK. I'm just checking your needs. Have you calculated what your beam sizes will be over the five metres? – WetSavannaAnimal Oct 11 '14 at 23:25
• 5mm diameter or so. The 5m is on the long side to be safe. I need to walk the beams over to the plasma. It's true I should make that distance as short as possible. I expand the combined beams to lower their divergence. – cpc333 Oct 11 '14 at 23:56

Although the method illustrated in the pic below can in principle assure an unprecedented level of collinearity of the beams in a rather uncomplicated manner, in practice it is subject to the availability as well as a reasonable level of performance in usage of fiber-optical components in the $10\, \mu$m regime (I had written a long and complete answer, and just before posting that, I realized that the wavelength of interest is $10.6\, \mu$m instead of $1.06\, \mu$m).

And unfortunately, it does seem harder to find fiber components. A quick search on google did show up some components like fibers, fiber couplers, fiber collimators but I could not find polarization controllers. Besides these components may also not only be quite expensive but also rather lossy.

Nonetheless, if you see any potential in this method, you could try to extend the survey, get quotes from companies, and see if the time & money estimates may prove not too bad?

• Yes, the wavelength makes this really hard. You might like to check my divergence calcs in my updated answer: I'm not convinced the OP needs the accuracy stated. – WetSavannaAnimal Oct 11 '14 at 10:28
• Lossy would not matter too much. As long as I'm not losing 99%. The bigger problem is the porlarization.I need one beam left circularly polarized and the other right circularly polarized. Also, because one beam is frequency shifted by 25MHz by an AOM, the end result of the superposition is a linearly polarised beam rotating at 25 MHz (I think).So if there is a way to make that with a single beam I think that would also work. – cpc333 Oct 11 '14 at 16:37
• I would be inclined to say that the polarization may turn out to be okay if you use short fibers that are not subject to high environmental fluctuations, such as vibrations or temperature variations. Polarization controllers use birefringence and at least the basic principle should work. But it may require bending of the fiber which might add to your losses. So power budget might really be critical in the end. – jayann Oct 11 '14 at 17:51