Red Green Expansion Difference

[Alan Sailer]

I have been wasting a lot of film (!) messing with red/green exposures in a simple denisyuk set-up.

I am happy that I was able to avoid buying an expensive dielectric mirror to combine the beams. I'm
using a dielectric filter from on old LCD projector (rotating RGB filters to strobe the colors). It seems
to work well.

However I am puzzled as to why the 40x microscope objective is expanding the red and green lasers
differently. For example when the green beam is one inch in diameter the red beam is 1.7 inches. The red laser
is a 24 inch long HeNe and the green is a CrystaLaser DPSS unit. I'd expect the longer HeNe to be better
collimated than the green.

The end result is that I can't get a color balance across the entire film plane. The edges are red when
the center is closer to balanced.

Cheers.

[Din]

However I am puzzled as to why the 40x microscope objective is expanding the red and green lasers
differently. For example when the green beam is one inch in diameter the red beam is 1.7 inches. The red laser
is a 24 inch long HeNe and the green is a CrystaLaser DPSS unit. I'd expect the longer HeNe to be better
collimated than the green.

The end result is that I can't get a color balance across the entire film plane. The edges are red when
the center is closer to balanced.

Cheers.

It's because the beam waist, W(0), for each laser is different.

The beam divergence of a Gaussian (laser) beam is not collimated at the output. It initially converges, reaches a point of minimum convergence, then diverges. The beam radius at minimum convergence is known as the beam waist, W(0). Assuming the beam is propagating along the z axis the divergence of the laser beam downstream, W(z), is dependent on the beam waist, W(0), in the following manner:

W(z) = W(0)√[1 + (z/z(0))²]

where z(0) is the point of minimum convergence. In other words, if the waist is large, then the beam divergence is large for any given point z along the beam. The beam waist is usually quoted in the laser spec. To be completely accurate, passing a beam through a lens creates a new beam waist W'(0), but this parameter is also dependent on the laser's "built-in" beam waist.

If you're concerned, you might look into placing an iris in the beam and widening it enough to block the "overspill" red light. Thor labs ( https://www.thorlabs.com/newgrouppage9. ... oup_id=206 ) has iris's, but it may be expensive; it's the "go-to" company for a lot of technical holographers. There are probably cheaper alternatives. It may be possible to make your own iris if you don't need a variable open diameter.

[Joe Farina]

As Din explained, the difference in expanded beam diameters is in the nature of things, and that's what you want to see. The input beams to the spatial filter should be smallest for blue, larger for green, and still larger for red. Likewise, the expanded beams coming out of the spatial filter (or just the objective) will smallest to largest in that same order. Just use the uniform middle part of the expanded beam for the hologram. Inevitably, there will be wasted light for green and red.

[BobH]

Beam diameter AND divergence need to be the same for all colors to get the best result when going through a spatial filter. It was easier when all the beams came from gas lasers which had similar divergences. DPSS laser beams have generally much smaller diameter and much larger divergence than gas lasers. I'd recommend an up-collimating telescope for the DPSS lasers to match the gas laser's beam. That's the only way you'll get the collinear beams to go through a pinhole at the same time, and the uniformity of color you want at the recording plane.

[Joe Farina]

Beam diameter AND divergence need to be the same for all colors to get the best result when going through a spatial filter.

That hasn't been my experience. To get a clean spatially filtered beam with blue, green, and red, I've had to use a specific size input beam for blue, green, and red, and the diameter needed to be smallest for blue, bigger for green, and still bigger for red. Otherwise, it wouldn't spatial filter properly. I would be extremely happy if this wasn't necessary, and there is a way to make the output (spread) beams closely matched in size. I've read (I'm pretty sure it was in one of Bjelkhagen's papers, or maybe during a tutorial in Lake Forest) that the input beams should not have the same diameter, and there will be some wasted light at the edges for green and red. I think this has also been discussed on this forum, or the old one. I do modify the divergence of the blue and green using 3-lens telescope arrangements. High-quality objectives are supposed to have high levels of correction for both chromatic and spherical aberrations, but even with my PLAN objective in a Newport spatial filter, the spread beam won't be uniform unless the input beams are different diameters.

[BobH]

To get a clean spatially filtered beam with blue, green, and red, I've had to use a specific size input beam for blue, green, and red, and the diameter needed to be smallest for blue, bigger for green, and still bigger for red.

This would be consistent with the blue and green lasers having greater divergence than the red. An oversized pinhole might allow that, and you could get a reasonable match at a specific distance away from the spatial filter, but diameter and divergence must both match for accurate results. All the literature supports that.

[Joe Farina]

diameter and divergence must both match for accurate results. All the literature supports that.

I tried to find some references on the subject, but the only thing I found so far was a 2006 post on the old forum. Unfortunately, the link to Edmund doesn't work anymore. I'm not sure where Colin got his information from.

https://www.holographyforum.org/forum/v ... php?t=5362

Say we had three (RGB) beams that in practice (maybe not so easy to do accurately) had the same diameter and divergence before the spatial filter. Would the spread beam be uniform with no wasted light at the edges for green and red?

[BobH]

Yes.

[Joe Farina]

Ok, thanks, I didn't know that. It would be a big benefit to have no wasted red light, which is much more difficult to find in high powers except Cobolt. But the practical realization of this matched diameter/divergence may not be so easy. Say we call the entrance aperture to the microscope objective point "X" which is some distance away from the lasers. At that point, (point X) I assume the diameters/divergences must match. It seems to me that it might take a (very) large amount of fiddling with the telescope lenses (at unknown positions) to get this ideal situation. My lab layout would complicate this further, since the lasers are in another room.

[Din]

Briefly, when a coherent beam (or any beam, really, but the rings are most prevalent the narrower the bandwidth) passes through a biconvex lens, it does not focus to a point. It forms a series of circles, known as Airy rings because the focus of a thin, biconvex lens is the Fourier transform of the input plane of the beam to the lens, along with all the non-uniformities of the beam such as dust etc. The circles in the Airy ring are caused by the non-uniformities - they're known as the higher frequencies encoded in the beam. If you could block all the Airy rings, the output of the lens would be due to a completely uniform input plane to the lens - this is spatial filtering, you're 'filtering' the light in space by, effectively, putting an iris in the beam at some point in space to get rid of some spatial part of the beam. This is why when you mis-align a pinhole, you get a series of rings, you're allowing Airy rings through.

At any rate, the radius of the first Airy ring is given by (this is where Colin's formula comes from, but he uses 1.27 as the coefficiant which is incorrect - the number comes from the first zero of the sinc function)):

r = 1.22*f*λ/D = 1.22*F#*λ

Thus, in order to perfectly pinhole a focused beam, the pinhole has to have a radius equal to the first Airy ring, as per above equation.
But, a few points:
1. commercially available pinholes will probably not come in exactly the calculated first Airy radius, so, you have to approximate.
2. there's no such thing as a beam consisting of just one wavelength, which is impossible for various reasons. Therefore, even the best laser has a bandwidth, and so no pinhole, calculated for just one wavelength, will completely block off the first Airy ring. However, the narrower the bandwidth, the more prominent the central wavelength. This is not too bad for display hologram - as Kaveh used to say: "The eye forgives a lot". It is important in technical work, where the hologram is "read" by a machine.
3. The Airy radius is based on the fact that the input beam is completely collimated, impossible in a Gaussian beam. As a matter of interest, Airy was an astronomer, and so was looking at light from distant stars, whose light is pretty well collimated.
4. The actual focal length is also based on a laser parameter known as M², usually given in most laser specs. If M² = 1, you have a perfect Gaussian (laser) beam, but in practice it's usually higher. The closer to M² is to 1, the better.

Having said this, it should be obvious that several wavelengths in an input beam cannot all pass through the same pinhole and have all the Airy rings filtered out - a point that Colin makes. The divergence of a beam, having passed through a lens is based on the NA (numerical aperture) of the lens, which is usually printed on the side of the objective, but only if the lens aperture is filled. If the aperture is not filled, the beam divergence is given by the beam diameter at the lens, and it's focal length (see below). I assume this is the basis of Colin's calculation of increasing the beam diameter. In the case the beam fills the diameter, the divergence θ = arcsin(NA), in the case, it does not fill the aperture, θ = arctan(D'/(2f), D' < D. So, by manipulating the input beam diameter, you may be able to have roughly equal beam divergencesrs for all three beams (another point Colin makes), but you cannot filter out all the higher frequencies in all the beams. However, remember again, the eye forgives a lot!
Airy rings: https://www.edmundoptics.com/knowledge- ... e0lgqW_Qh0

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