PRINCIPLES OF SPATIAL FILTERS The input Gaussian beam has added to it, spatially varying intensity "noise". When a beam is focused by an aspheric lens, the input beam is transformed into a central Gaussian spot (on the optical axis) and side fringes which represent the unwanted "noise". The radical position of the side fringes is proportional to the spatial frequency of the "noise". By centering a pin hole on the central gaussian spot, the "clean" portion of the beam can pass while the "noise" fringes are blocked. The diffraction limited spot size at the 99% contour is given by: D= f/r where =wave length, f=focal length and r=input beam i/e2 radius. A pin hole that is approximately 30% larger is chosen to allow the focused Gaussian spot to pass while blocking the "noise" fringes that are shifted off axis.
As a side note, spatial filters represent sort of the 'high school' level of holography sophistication - most start out without them, but once you go past the 'Wow I made a hologram' stage and want to make them look really good, a spatial filter is a must have. I just replaced my aluminum foil pinhole with a real one, and even that has made a significant difference.
Of course, when you get to variable beam splitters, that's college level...
Tony, how much did you pay for you pinhole(s)? Do you have a source. Most are very expensive.
Also, there are lot of factors that we did research on at Lenox Laser when I worked there concerning pinholes. The optical roundness varies from one manufacturer to another and Lenox Laser makes the roundest and most tightly toleranced pinholes I have even seen. Another variable with pinholes is the substrate thickness. 1/2 mil, 1 mil, 2 mil all have different properties when used with different objective lenses. Because the actual focal spot is smaller for a higher number objective you want a thinner substrate. Slag and burrs are another variable. And so it substrate material (Gold, Stainless Steel, etc) for higher power lasers. A black coating on the side of the pinhole that the beam enters may save your eyeballs also when tuning.
As you can see from the photo a 40x objective would be more efficient with a thinner substrate as a lot of the "good" beam would get blocked with a thicker substrate. My diagram is not to scale and is just a representation of the idea. If you mathematically figure in the diameter of the beam with the properties of the objective you will find what the maximum tolerable thickness is. We found that 1/2 mil (.5 thousandths of and inch) is good for most applications.
"how big is the beam with 45x objective in 1m distance?
"
You need to define the diameter of the input beam in order to work this out. The focal length of the objective is approximately 160mm/45 = 3.56mm.So if rin is the input beam radius, then the output beam radius 1 metre away will be 1000rin/3.56 = 281rin.
John- Unfortunately I don't really have a 'source'. I got my pinhole along with other stuff from 'hologuy', Robert of Sterling Resale Optics. It was used, dirty, and cheap. I had it ultrasonicly cleaned and now it's used, clean, cheap, and better than aluminum foil!
BTW, hologuy/SRO is great to do business with, professional, fast, honest, good values.
>BTW, hologuy/SRO is great to do business with, professional, fast, honest, good values.
I fully agree on this. I bought much of my optics stuff there, incl
spatial filter and waveplate and mounts, and it was always very
positive. In fact I was once visiting him at his home and got a
good deal for some things he had stored his garage - I highly
recommend !
I don't use a spatial filter yet, but thinking about to buy / make one. With making I mean I buy the objective and the pinhole, and do the mounting myself.
How clean must the beam and the lens be, to get good results with a spatial filter? Do I have to clean the lens as good as before (this is often a task of nearly an hour). if I have two beams, which are very near (e.g. from double reflection on glass) , so that I get small interference at the lens, can i get rid of this by using a spatial filter and get a perfect clean beam?
where did you buy your filter?
If the filter is working well, it drasticly reduces the need for cleanliness of upstream optics, including the lens. My lens is a 40x which has lots of specks and flaws, and the filter cleans up most of them. After that, it's a matter of positioning the beam to use a clean area.
I'd urge you to experiment with a foil pinhole. It's tricky, and not as good as a 'real' setup, but it's waaaaayyyyy better than nothing - you'll be amazed at what it will do for your holograms!
My latest spatial filter is home made from square steel tubing and aluminum angle - has X,Y, and Z adjustment screws. Cost me next to nothing, other than a trip to Home Depot and an evening on the drill press.
" if I have two beams, which are very near (e.g. from double reflection on glass) , so that I get small interference at the lens, can i get rid of this by using a spatial filter"
Probably not. Spatial filters are great at getting rid of rings and spots in the beam, but not superimposed low frequency fringes.
The idea is that when you focus light through a lens, the focal point is a small circle of light which is the Fourier Transform of the light that went into the lens. If the light that went into the lens is clean, the focused circle is clean, but if the light that went into the lens has a superimposed pattern of some kind, the circle of light is the Fourier Transform of that pattern. If the superimposed pattern on the light is black rings and circles (usually caused by scattering off dust particles in air), the Fourier pattern is a series of rings surrounding the central, clean focused spot. It doesn't matter where on the input light the rings are, the focused light will always have the rings surrounding the focused spot. If you put a small hole at the focused spot to allow the clean focused spot to pass but block the space surrounding the focused spot to filter them out (Hence "Spatial Filter"), the expanded beam no longer has rings and spots. However, the Fourier Transform of a series of lines is also a series of lines, so the central focused beam will be a tiny spot with a series of lines. Unless you get a kind of flat spatial filter and fit it ju-u-u-st between the fringes, you live with the fringes.
It should also be emntioned that while the spatial filter cleans up the rings from the beam it is still advisable to clean the optics ahead of the pin hole because the more the filter has to filter the less power reaches the plate.
I was hoping nobody would ask! It can be shown (but don't ask me to do it here!) that the FT of a circularly symmetric function is also circularly symmetric. I think, without calculation, that the asymmetry due to the ellipticity would create additional sinusoidal terms that would be seen as rings whose diameters would depend on the eccentricity b/a. In the limit, as b->a, the 'new' rings would slowly vanish leaving the Airy pattern. In addition, there would also be the rings that came from superimposed pattern on the beam due to dust etc., but these are there in the circular beam as well. So there would be a smaller series of rings within the standard rings in a circular beam which would kill the light even more.
Hi everyone:
I have a couple of questions regarding spatial filters.
I have been trying to use a spatial filter to clean up the output of a laser diode. Newport model with a 25um pin hole and 10x objective. I can't seem to get rid of the airy rings formed by the spatial filter and I seem to be losing a huge amount of power, in other words the filtered light is very faint and contains diffraction rings.
My questions are:
1. Should I DEFINITLY be able to remove the airy rings regardless of the pin hole size and objective? Or are they sometimes unavoidable regardless how much time you spend tweaking the xyz adjustments. And how much power is usually lost in the filtering process?
2. The output from the diode is highly elipitcal, nearly 2:1, does this significantly reduce the performance of the spatial filter?
I am assuming that if I used a larger pin hole or a different objective I should get more power through the filter and potentially remove the Airy rings but I would like some advice before I purchace anything. What "pin hole size- objective" combinations work well for you guys? From what I've read the output from dioes is less colminated and less spherical than HeNe's
and don't filter as well can anyone comment on this?
Using an ordinary diode laser with an circular pinhole, you'll never be able to get rid of the airy rings. It's because one axis of the laser profile will always be hitting the edge of the pinhole if the other is passing cleanly through.
You need to either circularize the beam with additional optics (less expensive to do) or locate a manufacturer that makes elliptical pinholes (I don't know of any but I've read a post some time ago that mentioned that someone had approached a company about making one for about $450).
I don't have any specific recommendations about the optics to use to circularize the beam but I imagine that Colin or TomB will probably reply as well (they're our resident beam circularizers).
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If someone says it can't be done but they haven't tried it, don't believe them. http://www.dragonseye.com/Holography
The pinhole to objective sizing depends on a few things. Number one, as Michael stated you want a round beam that is collimated. Lenox Laser would make you an oval pinhole that would not be as expensive as mentioned above but the elliptical pinhole would have to be oriented to the elliptical beam. Any deviation would create rings and loss of power. Also the diameter of the beam is also a factor in pinhole to objective sizing. The following is from Newport's site and I found it to be very accurate;
Objective Lens Recommended Max. = w
Input Beam Diameter(mm) = x
Calculated Pinhole Diameter* (µm) = y
Recommended Pinhole Diameter (µm) = z
w x y z
---------------------------------
M-5X 5.0 32.3 50
M-10X 5.5 18.7 25
M-20X 5.0 10.5 15
M-40X 4.0 5.4 10
M-60X 3.5 3.7 5
Also, I have seen pinholes act as you described if they are dirty, try to blow it out with clean air. I use the can of air for computers that comes with a straw. If the pinhole is in a mount simple blow out the hole with the straw as close to the pinhole without touching it. Allow an opening out the back so the air can pass through. Most pinholes are either drilled with a laser which puts slag on the edge or reamed with a pin which causes a little burr material just outside the hole edge. Touching the pinhole hole may bend this slag or burr into the hole and it may still be attached thus even air will not blow it out. An ultrasonic cleaner may free the particle(s) but I have found that air will do the same thing if done properly.
Also, if you shine a laser though the pinhole without an objective in a dark room and have a white card about 6 inches from the pinhole you should be able to see rings that are not round or deformed where the dirt is. If you get really clean, round, undeformed rings the pinhole should be clean.
Also, pinholes down to 25 microns can be looked though to see if there is total blockage. Hold it up to a standard light bulb and look through to see light coming through. A 10 micron is barely visible with this technique but you should be able to see a glimmer of light coming though.
"1. Should I DEFINITLY be able to remove the airy rings regardless of the pin hole size and objective?"
Yes. The Airy rings are diffractive patterns around the central beam and are circular. They're caused by the rings and whirls on the input beam, as I mentioned in an earlier post. If you have a circular beam, the Airy rings are circles and you can always find an appropriate aperture that'll block the rings and let through the central bright beam. In this case, you generally lose about 10% if you're aperture size is right, actually thepretically you only lose 4%, but usually commercial apertures are not made to theoretical radii exactly. If you calculate the theoretical aperture size, as per John's formula, you gemnerally don't find a commercial aperture of this size. However, if your beam is elliptical, then the Airy rings are circles around the center of the ellipse and the elliptical beam is Gaussian. This means that you'll lose more power because the power is not centered any more but spread out in the 'lobes' of the ellipse.You'll also need a smaller aperture becuase I think the Airy rings will be more diffused and flattened. This is more a gut feeling then calculation, though.
When using John's formula for an elliptical beam use the major diameter of the diode beam. Also for some reason it seems to help to have the beam not colminated but expanding ever so slightly.
I circularize my beam with cylindrical lenses but I would advise anamorphic prisms. If you decide to go the lens route then you would need to make the ratio of the focal lengths the same as the ration of you major to minor axis. So for you (2 to 1), you could use a 25mm fl and a 50mm fl spaced 75mm apart. Depending on which axis you orientate them you can either widen the small axis or narrow the wide axis. This will change which pin hole you will use.
The expert on spatial filters for elliptical beams is Jonathan H. Perhaps he will read this soon, I know he does not get to participate as much of late. He was the first person I enountered using eliptical beams with spatial filters and has many years of experience over 100s of holograms.
I don't know why it's so easy to go off on a tangent during these threads. Maybe we are reminded about other things when we read replies...
Do you think that changing the diode laser beam to a circle helps to manage the beam easier or does it just help with a spatial filter? It seems that a large part of the beam is wasted when I am using 2.5 X 2.5 plates and to a lesser extent when I am using 4 X 5 plates.
Actually, I have been working a lot with bypass setups lately and I like about 2.5 to 1 ellipse beams except I need the polarization to reverse. So I have been using a 1/2 wave plate and not circularizing the beam. I have just been taking the collimating optics off the diode and passing it directly through a very clean 1/2 wave plate to rotate the polarization and sending it to a 1' x 1' table. I have been able to make every kind of H1 and H2 that a splitbeam set can make with much less money invested and with a set up that can be done on my kitchen floor with three sorbothane hemispheres for isolation.
I think this is a great way to learn about holography without spending lots on a table and equipment. The only drawback is there is not a way to make multiple object beams for display work. This is a small price to pay for the beginning holographer who would be happy to make a rainbow H2 for less than $100!
Colin, This is just a thought out loud. Go to overhead lighting in your bypass setup, get the polarization correct with rotating the diode. I understand that this will leave the long elliptical axis opposite to the object. But then add two additional front surface mirrors to capture this "wasted" light for additional lighting. The drawling is not perfect and not to scale but it represent the idea. In this case the polarization is vertical with respect to the table thus Brewster's to the plate is preserved. The long elliptical axis is side to side thus the two mirrors take this light and redirect it to the object. You would get overhead, and both sides object lighting. Have you tried this yet?
"Do you think that changing the diode laser beam to a circle helps to manage the beam easier or does it just help with a spatial filter?"
I hope to answer this question in the next couple of weeks. I have been spending hours with a new spatial filter and playing with different objectives and pinholes as well as taking my cylindrical lenses in and out of the set up. I hope to have some real answers soon.
From various sources - couldn't find this all summarized in one place.
1) DIN Microscope objective back focal length in mm (for standard 160 mm DIN tube length):
f=160/M
Example: M = 10X is 16mm focal length nominal (must consult objective maker's specs for the exact value)
2) Beam diameter D for a collimated laser diode:
D = 2*f*sin(theta/2), where f is collimator lens focal length and theta is the divergence angle (get this from the laser diode specs)
Example: for f = 5 mm, 8 degree theta, D = 0.698 mm. For 24 degree theta, D = 1.908 mm.
So a laser diode with typical 8 X 24 degree divergence should give a 0.7 X 3 mm elliptical beam with this collimator. There may be some subtleties regarding beam waist position, lens aberrations etc. that I have not accounted for (or understood yet)
3) Focused spot diameter D2 for a collimated input beam diameter D, wavelength W, focusing lens focal length F:
D2 = (4*W)/Pi * F/D
Example: for F = 16mm (10x objective), W = 660 nm and a 0.7 X 3 mm beam, the spot size should be 19.275 x 7.047 um, which would easily fit through a 25 um diameter pinhole, but the filter would be less effective on the 7 um axis, letting more garbage through. Enough to be seriously objectionable? I don't know, but I doubt it ... Note that when the input beam diameter increases, the focused spot size gets smaller, which is why a beam expander is often used in front of spatial filters and lens-type fiber couplers.
How do I calculate what lens or objective I should use if I want the diverged beam to cover a 14" f/4.9 collimation mirror? I need to put the spatial filter in the focal length of the mirror.
I found a pretty good spatial filtering app note here: http://www.oriel.com/down/pdf/09040.pdf
The "Freely diverging beam" section shows how to calculate the output cone angle, from which the beam diameter at a given distance can be calculated by simple trigonometry which is beyond me at the moment. (exercise left to the student). So, you could start with the collimator diameter and focal length which determines the require cone angle, then work backwards to find a feasible combination of objective, pinhole size, and input beam diameter that illuminates the collimator well. It would probably be easiest to put all the math in a spreadsheet or MathCad type program and fiddle with parameters until a solution was found.
Re spot size calculations, I'll have to compare their math with mine to see if the results come out the same and if not, why not.
Without complicated math, here is how I do it. First place your spatial filter at the focal length of the collimator. From your statement this is 14" * 4.9 = 68.6". This is where it has to be no matter what the calculations come out to be if you want the diverging beam to be collimated. So that is a constant. Then, simply put an objective (no pinhole) in the spatial filter and see what you have. Then go up or down. You will probably need a 20x to cover the entire mirror at that distance but check it out. Remember that the focus spot of the objective has to be 68.6 inches away from the collimator not the objective itself. With my spatial filter I put exactly where the pinhole plane is. The objective moves in and out. The pinhole remains fixed and will always be at the focus of the objective when tuned.
Let me ask you a question. Are you making a 12" or 13" hologram? If you are only making a 4" x 5" hologram you do not need to cover then entire collimator, just as much as you need which will probably only use a 10x or 5x. Otherwise you would waste a lot of energy. But no matter what objective you use, put it 68.6" away from your collimator.
My mirror is 14" diameter and focal length of 75". I have never needed a 40x. Just 5x, 10x, and 20x for different size plates.
