Patterns

[Din]

Alan
A couple of points, if I may.

If you notice some effect, and you want a reason for it - lines on the image, for example - you are going down a technical rabbithole. There are too many narcissists with giant egos who'll throw cliches at you like "Fresnel reflection" and "p polarised", without themselves actually understanding these terms. Cliches don't create understanding. If you want to understand, you do have to dig into the more technical aspects of holography, and understand them beyond spouting cliches. Unfortunately, holography is a branch of the physics of optics, and therefore understanding requires knowledge of the physics of optics. Also, there are multiple reasons for an effect. In the case of the lines on the image, it could be object motion, or, it could be mode hopping of the laser. In either case, you have to understand what laser modes are; simply saying "instability" is not nearly enough. If it is object motion, you have to understand phase relationships. Again, down the technical rabbithole! You have to eliminate each possible reason, one by one. But, as I've been saying for well nigh twenty years now, you don't need theory to make a hologram. There are many practical strategies where you don't have to understand what you're doing, you simply have to follow the procedure. Some people are comfortable with that, and some are not.

The other issue is that the more sophisticated your hologram, the more equipment you need, and the more you have to understand the equipment. Unfortunately, holography is not well supported in terms of equipment, you have to scrounge. Forty years ago, I worked with a holographer called Jeff Blythe who used to literally dumpster dive! You have to adapt. Joy used to go to Ace Hardware for some equipment because nothing existed for the particular application at hand. The staff at Ace had much joy asking her what the L bracket was for, and she'd say, "to hold a half-wave plate".

So, I'd start off simple. Fred's book "Holography Handbook" ( https://www.amazon.com/Holography-Handb ... 0894960172 ) is a very simple book on basic methods. It's a little dated and you have to adapt it to modern day materials, but the basic ideas still hold. It simply tells you what to do to make a hologram, not why you have to do it. There are no involved explanations about 'fringes', or 'interference' or 'diffraction' (usually explained in terms of cliches to make people sound "smart"). You don't need those terms.

About three years ago, I was contacted by a lady in her late 60's. She wanted to learn how to make a hologram, called Martina Mrongovius, who then referred her to me since she lived in LA. She was in her 60's, as I said, and had been trying to learn how to make a hologram since she was 16, with no luck. She had all these cliches thrown at her - interference, diffraction, fringes etc etc, and had not understood any of it. We simply set her up with a Denisyuk, told her not to worry what any of it means, and she was thrilled to produce her first hologram for 60 years.

[Alan Sailer]

BobH,

My set-up is simple single beam. I have been getting decent results. I'm just seeing stuff now that
I ignored in the initial rush of "OOOh a hologram". For instance, I'm seeing the surface bulleye targets
pattern that I am assuming will be be eliminated by spatial filtering. And the lines I am asking about here.

I'd like to make a few final holograms that eliminate these defects (as much as I can). Then I quit...

Din,

I believe you. As I mentioned earlier, my tolerance for math and calculations has evaporated over the years.
For the last thirty odd years I worked in a sea of PhDs who could do all the calculations. I specialized in making
the results of those calculations work on the bench. I built shit and was good at it. Now that I am retired
I still build shit. Last year it was a demo of the single photon double slit experiment. It was great watching those
phosphor hits build up a full interference pattern over time.

A funny story. I was in Home Despot with some plumbing parts to build a light tight enclosure for that project. The
guy in front of me in line asked what project I was building with such a ragtag assortment of parts. I told him I
was trying a classic physics experiment ie the single photon double slit....imagine my shock when he said "I love
the single photon double slit experiment!".

You never know...

Cheers.

[Din]

Actually, I got my first job as a professional holographer thanks to Guinness. In 1981 I happened to be at a pub with friends, having just made a Denisyuk of some small porcelain figure. While at the bar, looking at my hologram, the gent standing next to me looked over, asked me if that was a hologram, and did I make it. Yes and yes. Then he beckoned to the barkeep to pour me a Guinness. So, we continued to talk, liberally tanked up on Guinness. I told him I had just dipped my toe in holography and were there actual jobs. It turned out that he was a Guinness rep, and had just been at one of his clients in a pub in the the village of Braxted, where Applied Holographics was situated. He'd started up a conversation with the Applied holographers, who told him that they were looking for a holographer. So, I got an introduction, went to Braxted for an interview and got hired immediately by Hamish Shearer.

Sometimes you gotta believe in synchronicity!

[Alan Sailer]

Din,

And they say alcohol is bad for you....

BobH,

Mode hopping is it. The He-Ne is crystal clear.

Cheers.

[BobH]

However, there is no such thing as "Fresnel reflection", as opposed to any other kind of reflection; the phenomena of reflection is defined by the equation:i = r ( angle of incidence = angle of reflection ).

"No such thing as "Fresnel reflection"?!? You didn't think so almost three years ago when you said the following here:

"BobH wrote: ↑Sun Jan 23, 2022 7:50 am
"Beam ratio is independent of the orientation of a recording plate, as it determines the fringe visibility in space."

Not true. The beam ratio is determined by the actual actinic radiation actually entering the plate. Therefore, Fresnel reflection need to be considered, since such reflection diminish the actual radiation entering the plate. So, for example, if you were recording at angles of 0 and 80 degrees, then very little of the beam power measured at 80 degrees is actually entering the plate. The meter reading itself is independent of polarisation and therefore cannot determine how much power is entering the plate."

Pure evidence that you'll say whatever you think is necessary to troll and argue with anyone who challenges your generally poor advice (despite your correct point about the real value of the beam ratio above).

[Din]

Not true. The beam ratio is determined by the actual actinic radiation actually entering the plate.

No true, the fringe structure exists in space whether or not there's a plate. At any given point in space, in one dimension, the fringe structure is given by the visibility function:
I = I(1) + I(2) + 2√(I(1)I(2)) cos (φ), where I(1) and I(2) are beam intensities and φ is the phase difference .
The Fringe Visibity function is given by:
V = {I(1)(max) - I(min)}/I(max) + I(min). (https://phys.libretexts.org/Bookshelves ... Visibility)
These functions have nothing to do with any plate, you can shine these light on a wall, independent of any 'plate'. I'm not sure what you're getting at, since I don'tunderstand "wooly thinking"

since such reflection diminish the actual radiation entering the plate.

Exactly! That's why efficiency is not polarisation dependent. The only parameters are the light going in and the light going out. Any light that does not go into the material, whether because of the Fresnel equations, or just dust in the room, does not take part in the efficiency calculation.

Pure evidence that you'll say whatever you think is necessary to troll and argue with anyone who challenges your generally poor advice (despite your correct point about the real value of the beam ratio above).

Ad hominem attack, which I'm reporting. By the way, all this "evidence" is purely hand-waving. Do you a mathematical argument?

[BobH]

"Not true. The beam ratio is determined by the actual actinic radiation actually entering the plate. Therefore, Fresnel reflection need to be considered, since such reflection diminish the actual radiation entering the plate. So, for example, if you were recording at angles of 0 and 80 degrees, then very little of the beam power measured at 80 degrees is actually entering the plate. The meter reading itself is independent of polarisation and therefore cannot determine how much power is entering the plate."

Those were your words in the thread of Sun Jan 23, 2022 7:50 am. Something went wrong with my quoting it from there.

You're now actually arguing with yourself.

[Din]

"Not true. The beam ratio is determined by the actual actinic radiation actually entering the plate. Therefore, Fresnel reflection need to be considered, since such reflection diminish the actual radiation entering the plate. So, for example, if you were recording at angles of 0 and 80 degrees, then very little of the beam power measured at 80 degrees is actually entering the plate. The meter reading itself is independent of polarisation and therefore cannot determine how much power is entering the plate."

Those were your words in the thread of Sun Jan 23, 2022 7:50 am. Something went wrong with my quoting it from there.

You're now actually arguing with yourself.

Oh my Gawd!
The meter reading has nothing to do with the mechanism of the DE. Whatever method light is prevented from entering the material, only light that has entered the material counts toward the DE of the hologram.If light does not go into the medium, then any light that doesn't go into the medium is irrelevant. In this case, any light that diffracts out of the medium is what the DE is based on, DE, η = |S(d)|²/|R(0)|² (for a transmission hologram). The action of the complex function S(d) - light leaving the medium - as a ratio of light eentering the medium, the complex field R(0) gives the DE. The intensity of the light exiting the medium, as a ratio of the light entering the medium.

Thus, if light is prevented from entering the medium the DE of the medium is superfluous to the light that has not entered and R(0) = 0 for superfluous light.

The meter reading itself is independent of polarisation and therefore cannot determine how much power is entering the plate.

.
It could be dust in the room, it doesn't matter; if light does not enter the medium, it doesn't count as part of the DE. This is what the DE is.

I can see how you might be confused (happens a lot, doesn't it). I'll try and keep it really, really simple. !0 units of light approach the material. Through various mechanisms, 4 units of light do not enter the medium - it could be polarisation, it could be scattering (oops! Sorry! I said I'd keep it simple) So, if you lose 4 units of light through various mechanisms (aliens, scattering, The Great Kogelnik in the sky - whatever). Now, 6 units of light enter the medium. You can measure this with a meter, even a meter with a polartiser on the head. Now, provided you keep the polariser in place you measure the light exiting the medium. will be about 6 units. Now, if the light leaving the diffracted light in the direction is measured at 3 units, the hologram is 3/6 = 0.5, and your hologram is 50% efficient.

[Din]

In re-reading this explanation of diffraction efficiency, I overlooked the obvious, viz the difference between the recording of a hologram and the reconstruction of it I was pretty sick last week, so the DE of my brain function was compromised. Yesterday morning, Joy saw me picking up where I left off reading "The Birth of Classical Europe. A history from Troy to Augustine" and quipped, "Brain's back online I see".

At any rate, when you're recording a hologram, the beam ratio is important, as is polarisation because you're creating a diffractive structure. You're creating a structure - the diffractive structure - based on the modulation, the variation between the actinic to non-actinic reaction. Polarisation is important in these circumstances because the modulation is proportional to the cosine of the polarisation angles of the two beams. If, for example, the polarisation of the (two) beams were perpendicular, then, since cos(90) = 0, there'd be no modulation. Also, if either beam is elliptically polarised there'd be no modulation. But, the important point here is that you're creating a diffractive structure, there is no diffractive structure until the process is complete. Even then, modulation is very low if further processing is required. So, for example, the modulation for unprocessed silver and DCG holograms is very, very low and is known as the 'latent image'. In silver halide, for example, development amplifies the modulation by some 10,000,000 times. All of this is not quite true for ferro-electric crystals like lithium niobate, In which case, the actinic reaction is one of electron mobilisation - the energy from the light causes electron movement, which in turn causes a variation in charge density, and, the refractive index is proportional the charge density, or the modern photopolymers, where the light causes monomers to polymerise, and density difference between the polymers and the monomers cause an index modulation; details on the Bayer photopolymer here: https://pmc.ncbi.nlm.nih.gov/articles/PMC6418958/. But, let me again stress, these parameters determine the modulation of the diffractive (Bragg) structures.

However, this is very different to reconstructing an already recorded hologram. In this case, the modulation is already set in stone, as it were. Now, we have a finished diffractive (Bragg) structure. As explained, once light enters such a structure, there is an interchange of energy between the entering light and the (already finished) diffractive structre. This energy interchange "creates" new light fields - the diffracted light field - with new properties of intensity and direction. The only parameters in this interchange is the energy of the light and the modulation of Bragg structure. The efficiency of the energy interchange - the 'diffractive efficiency' - is based on how much energy goes to the new light field and is defined as the ratio of the energy of the original lightfield and the new lightfield.

So, to recap, while recording a hologram, you're creating a modulation. But, you cannot determine the efficiency of that modulation while you're creating it; you cannot lift yourself by tugging at your boots. Once created, the modulation is built-in. Now, the efficiency of that modulation is determined by an energy interchange between the structure and the incoming (scalar) light field. Details here: https://onlinelibrary.wiley.com/doi/abs ... .tb01198.x

[BobH]

In re-reading this explanation of diffraction efficiency, I overlooked the obvious, .... Details here: https://onlinelibrary.wiley.com/doi/abs ... .tb01198.x

Just more obfuscation to distract from your misunderstanding of the topic of this thread, the nature of Fresnel reflections, and the dependency of diffraction efficiency on polarization of the reconstruction light in thick volume materials like covestro's photopolymer.