I am not up the latest processing techniques, but whatever you can do to reduce the scattering is beneficial, and increases the signal to noise ratio.Martin wrote:I am wondering about the effect of grain size: assuming you have a very fine grain emulsion, say "Ultimate Ultra Fine" (which is said to have 8nm grain size). What happened if you were able to process that emulsion in a way you would get even smaller grains?
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Kaveh
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Colin Kaminski
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Can you expand on why this is true?Kaveh wrote: In reflection you don't have intermodulation noise around the image.
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Dinesh
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[quote:acffec74aa]I am wondering about the effect of grain size: assuming you have a very fine grain emulsion, say "Ultimate Ultra Fine" (which is said to have 8nm grain size). What happened if you were able to process that emulsion in a way you would get even smaller grains? [/quote:acffec74aa]
There are two main scattering mechanism - Rayleigh and Mie. In Rayleigh scattering, the individual atoms are seen as oscillating structures where the incoming wave causes a charge seperation of the positive nucleus from the center of the negatively charged electron surfaces. This creates an electric dipole which is driven into oscillation and radiates outward spherically. In this situation the scattering centeres are very, very small (atomic dimensions) and sparse - the seperation between individual scattering centers is much greater than the incoming wavelength. In biological molecules such as proteins, there is a molecular structure that simulates this condition where electrons are distributed along a "molecular spine" - I believe these are called chromatophores? This is responsible for eye color and peacock feathers. The prevailing opinion is that scattering inside a holographic medium is primarily Rayleigh scattering. If such were the case, the grain size could go down to about 1 Angstrom and still scatter, however, the smaller the particle size the more blue shifted the scatterinf radiation. If there's no blue in your beam, then the smaller the particle size, the less scatter you'd have (This is conjecture!) I personally don't believe that scattering inside an emulsion is Rayleigh and I'm trying to come up with a mechanism that takes into account the helicity of the gelatin molecule and the resulting birefringence (not succeeding for apathy mainly!).
In Mie scattering, the scattering wavelength depends on the size of the scattering center. Mie, if you like, is the upper limit of Rayleigh because as the molecule grows in size, it becomes less and less of a dipole and more of a string of dipoles. The scattering wavelength now depends on the increasing size of the molecular structure and as the molecule grows towards 700nm, the scattering radiation shifts towards white because more and more of the spectral components get scatttered. This is why clouds are not only white but bright enough to be seen against a bright blue sky. In this case, the larger the grain the more Mie scattering will be present and the more red shifted will be the scattering radiation.
Since holographic grain sizes are usually in the sub 50nm region, Mie scattering is not seen. I have sometimes wondered whether "continuous" material like DCG actually is a clump of large molecules with higher densities immersed in a lower density continuous medium. In this case Mie scattering will occur off the the crosslinked centers and the hologram will be noisier in the red than in the blue. I think I've seen this, but maybe because I've wanted to since you always fall in love with your own theories!
[quote:acffec74aa]Can you expand on why this is true?[/quote:acffec74aa]
Because of the low spatial frequencies. Assume an object dimension of x ata distance d from the plate. Then the angle subtended by the object at the plate, phi = (inv)tan(x/2*d). Putting in typical numbers, say x=3 in and d=8in, then
phi = 2*(inv)tan(3/16) = 2*10.6 degrees = 21.2 degrees
If you're shooting with HeNe, the spatial frequency
nu = sin(21.2)/0.000633 = 571 l/mm
If your ref is at Brewsters, the spatial frequency is about 1400 l/mm.
Unless the emulsion were bloody thin, the spatial frequency of 570 compared with one of 1400 will never get translated into volume fringes. At the risk of raising ire, the 'Q' factor for the 570 l/mm fringes is too low for Bragg (volume) diffraction and will be primarily Raman-Nath. Of course, the situation is even better for reflection geometries. In this case, thefrequencies due to intermodulation noise will form a ripple on the surface of the emulsion as opposed to volume fringes caused by the main beams.
There are two main scattering mechanism - Rayleigh and Mie. In Rayleigh scattering, the individual atoms are seen as oscillating structures where the incoming wave causes a charge seperation of the positive nucleus from the center of the negatively charged electron surfaces. This creates an electric dipole which is driven into oscillation and radiates outward spherically. In this situation the scattering centeres are very, very small (atomic dimensions) and sparse - the seperation between individual scattering centers is much greater than the incoming wavelength. In biological molecules such as proteins, there is a molecular structure that simulates this condition where electrons are distributed along a "molecular spine" - I believe these are called chromatophores? This is responsible for eye color and peacock feathers. The prevailing opinion is that scattering inside a holographic medium is primarily Rayleigh scattering. If such were the case, the grain size could go down to about 1 Angstrom and still scatter, however, the smaller the particle size the more blue shifted the scatterinf radiation. If there's no blue in your beam, then the smaller the particle size, the less scatter you'd have (This is conjecture!) I personally don't believe that scattering inside an emulsion is Rayleigh and I'm trying to come up with a mechanism that takes into account the helicity of the gelatin molecule and the resulting birefringence (not succeeding for apathy mainly!).
In Mie scattering, the scattering wavelength depends on the size of the scattering center. Mie, if you like, is the upper limit of Rayleigh because as the molecule grows in size, it becomes less and less of a dipole and more of a string of dipoles. The scattering wavelength now depends on the increasing size of the molecular structure and as the molecule grows towards 700nm, the scattering radiation shifts towards white because more and more of the spectral components get scatttered. This is why clouds are not only white but bright enough to be seen against a bright blue sky. In this case, the larger the grain the more Mie scattering will be present and the more red shifted will be the scattering radiation.
Since holographic grain sizes are usually in the sub 50nm region, Mie scattering is not seen. I have sometimes wondered whether "continuous" material like DCG actually is a clump of large molecules with higher densities immersed in a lower density continuous medium. In this case Mie scattering will occur off the the crosslinked centers and the hologram will be noisier in the red than in the blue. I think I've seen this, but maybe because I've wanted to since you always fall in love with your own theories!
[quote:acffec74aa]Can you expand on why this is true?[/quote:acffec74aa]
Because of the low spatial frequencies. Assume an object dimension of x ata distance d from the plate. Then the angle subtended by the object at the plate, phi = (inv)tan(x/2*d). Putting in typical numbers, say x=3 in and d=8in, then
phi = 2*(inv)tan(3/16) = 2*10.6 degrees = 21.2 degrees
If you're shooting with HeNe, the spatial frequency
nu = sin(21.2)/0.000633 = 571 l/mm
If your ref is at Brewsters, the spatial frequency is about 1400 l/mm.
Unless the emulsion were bloody thin, the spatial frequency of 570 compared with one of 1400 will never get translated into volume fringes. At the risk of raising ire, the 'Q' factor for the 570 l/mm fringes is too low for Bragg (volume) diffraction and will be primarily Raman-Nath. Of course, the situation is even better for reflection geometries. In this case, thefrequencies due to intermodulation noise will form a ripple on the surface of the emulsion as opposed to volume fringes caused by the main beams.
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Kaveh
BEAM RATIOS
The source of intermodulation noise is interference between different parts of the object, leading to low frequency fringes which are not really wanted. Now with an amplitude transmission hologram, the result is low frequency diffraction, i.e. around the reconstruction beam. This won't normally bother you. But with a phase or bleached transmission hologram, due to their inherent 'nonlinearity', it turns out there is also some unwanted diffraction around the image.Colin Kaminski wrote:Can you expand on why this is true?Kaveh wrote: In reflection you don't have intermodulation noise around the image.
Now the same 'self-interference' will occur with reflection holograms, but of course the unwanted hologram elements are all transmission, not reflection. So you will only see them if you are illuminating the hologram as a transmission one. This is why you can use 1:1 beam ratios in reflection.
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Colin Kaminski
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OK, so in simple terms it is not only important that the noise is low spatial frequency but also that the noise is from two points coming from the object, therefore on the same side of the plate, so the angle of the fringes is perpendicular to the plate and not parallel. This then makes the noise a transmission image not seen on the reflection hologram.Kaveh wrote:The source of intermodulation noise is interference between different parts of the object, leading to low frequency fringes which are not really wanted. Now with an amplitude transmission hologram, the result is low frequency diffraction, i.e. around the reconstruction beam. This won't normally bother you. But with a phase or bleached transmission hologram, due to their inherent 'nonlinearity', it turns out there is also some unwanted diffraction around the image.Colin Kaminski wrote:Can you expand on why this is true?Kaveh wrote: In reflection you don't have intermodulation noise around the image.
Now the same 'self-interference' will occur with reflection holograms, but of course the unwanted hologram elements are all transmission, not reflection. So you will only see them if you are illuminating the hologram as a transmission one. This is why you can use 1:1 beam ratios in reflection.
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Martin
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Though I am not sure, I believe Mie scattering may also involve very small particles (<10nm probably).
"If there's no blue in your beam, then the smaller the particle size, the less scatter you'd have (This is conjecture!)"
The smaller the particle size (= silver halide grains), the less scatter you'll get. And this also applies to blue radiation. Actually it's even more pronounced for blue wavelengths.
The larger the grains the more scatter is produced. Say you have a reflection hologram recorded on 8E75, exposed at 633nm. If you reconstruct that hologram with a white light (halogen lamp), most of the scatter will be caused by the blue part of the spectrum. So putting a yellow absorption filter (cutting blue), you will substantially reduce scatter.
"If there's no blue in your beam, then the smaller the particle size, the less scatter you'd have (This is conjecture!)"
The smaller the particle size (= silver halide grains), the less scatter you'll get. And this also applies to blue radiation. Actually it's even more pronounced for blue wavelengths.
The larger the grains the more scatter is produced. Say you have a reflection hologram recorded on 8E75, exposed at 633nm. If you reconstruct that hologram with a white light (halogen lamp), most of the scatter will be caused by the blue part of the spectrum. So putting a yellow absorption filter (cutting blue), you will substantially reduce scatter.
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Kaveh
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You hit the nail on the head.Colin Kaminski wrote:OK, so in simple terms it is not only important that the noise is low spatial frequency but also that the noise is from two points coming from the object, therefore on the same side of the plate, so the angle of the fringes is perpendicular to the plate and not parallel. This then makes the noise a transmission image not seen on the reflection hologram.
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Dinesh
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[quote:4d1a75928d]Though I am not sure, I believe Mie scattering may also involve very small particles (<10nm probably). [/quote:4d1a75928d]
Yes, true. In Mie scattering, the scattering intensity varies as the size of the scattering object. If you have small particles, you'll get predominantly Rayleigh and some Mie. As the particle size increases, Mie scattering predominates and Rayleigh scattering dies down. In Mie scattering, all the particles of size equal to,or less than 700 nm contribute to scattering in the visible so the scattering intensity is that much greater.
[quote:4d1a75928d]The smaller the particle size (= silver halide grains), the less scatter you'll get. And this also applies to blue radiation. Actually it's even more pronounced for blue wavelengths.[/quote:4d1a75928d]
I think that the smaller the particles, the more blue scattering you get. The smaller the particle, the more you get Rayleigh scattering. In Rayleigh scattering, the scattering intensity is proportional to the 4th power of the frequency, therefore scattering in blue overwhelms scattering in red. So if you shoot in red, the smaller the particle, the less the scattering - assuming Rayleigh scattering only. The more important factor may be the ratio of Mie to Rayleigh scattering across all wavelengths and across varying particle sizes.
[quote:4d1a75928d]OK, so in simple terms it is not only important that the noise is low spatial frequency but also that the noise is from two points coming from the object, therefore on the same side of the plate, so the angle of the fringes is perpendicular to the plate and not parallel. This then makes the noise a transmission image not seen on the reflection hologram.[/quote:4d1a75928d]
Like everything else that's "simple", it's really not that simple. Ok, let me not use the R-N words or the B word. I promise: no mathematics!
All holograms are a mixture of the so-called "thick" variety and "thin" variety. By "thick" is meant that the fringe structure exists as planes within the emulsion and by "thin" is meant variations in surface profile. Let me repeat that no hologram is ever one or the other. This difference in types determines which type of dioffraction occurs and in all holograms, both types occur. As the emulsion depth increases, so does the "thick" type of diffraction start to dominate, but the "thin" type does not disappear, it's still there. The shooting geometry - "Transmission" or a "Reflection" - is independant of the type of hologram and the type of diffraction. You can see this if you shot a hologram of an object with a reference of 60 degrees using a 457nm HeCd. In Dichromate (usual emulsion depth is 12- -15 microns), you'll get a lot of the "thick" type of diffraction and almost none of the "thin" type. If you shot the same hologram in Photoresist (usual emulsion depth is about a micron or so), you'll get all of the "thin" type and none of the "thick". In Silver Halide (usual emulsion depth is about 8 microns), you'll get mostly "thick" and some "thin". You can also see this if you slit an H1 as if you were making a rainbow. However, if you now bring the H2 reference from the "other" side - a la "reflection" geometry, you'll get a "reflection rainbow" (I did this once to convince Mitch it was true - wierd effect). You now have a "reflection" hologram replaying as a "transmission" hologram.
Insofar as intermodulation noise is concerned, it depends on the medium, on the size and position of the object, the make-up of the object and on the medium in which you shoot. If you make a DCG emulsion that's only about 5 microns or less thick, and shoot a highly reflective and very large object very close to the plate, you'll get intermodulation even in reflection. This is because the emulsion is thin enough that the "thin" type of hologram will exist in such an emulsion thickness, despite the fact it's DCG. As I said the intermodulation frequency will appear as a "ripple" on the surface of the DCG emulsion and will contribute to the diffraction - the more so the thinner the emulsion depth. If the object is large enough and close enough to the plate, you may even get "thick" fringes" due to self interference of the object. The angle that the extremes of the object makes with the plate may be greater than the angle that the reference makes. In such a case, the 5 micron emulsion will cause the intermodulation to predominate and you'll lose all image contrast. A very bright object point against a dim object (a high dynamic range) shot on a soft emulsion will also cause intermodulation because the surface profile will be exaggerated by the higher intensity of the bright part. Since this surface profile diffraction is angle independant on replay, the noise will appear extended.
I'm not even going to get into back reflection of the reconstruction beam diffracting off surface profile structures of reflection holograms! But ask yourself, is your Visa hologram a reflection or transmssion? Even if you take off the reflective backing, you won't see much by reconstructing it from a back light a la transmission replay.
Yes, true. In Mie scattering, the scattering intensity varies as the size of the scattering object. If you have small particles, you'll get predominantly Rayleigh and some Mie. As the particle size increases, Mie scattering predominates and Rayleigh scattering dies down. In Mie scattering, all the particles of size equal to,or less than 700 nm contribute to scattering in the visible so the scattering intensity is that much greater.
[quote:4d1a75928d]The smaller the particle size (= silver halide grains), the less scatter you'll get. And this also applies to blue radiation. Actually it's even more pronounced for blue wavelengths.[/quote:4d1a75928d]
I think that the smaller the particles, the more blue scattering you get. The smaller the particle, the more you get Rayleigh scattering. In Rayleigh scattering, the scattering intensity is proportional to the 4th power of the frequency, therefore scattering in blue overwhelms scattering in red. So if you shoot in red, the smaller the particle, the less the scattering - assuming Rayleigh scattering only. The more important factor may be the ratio of Mie to Rayleigh scattering across all wavelengths and across varying particle sizes.
[quote:4d1a75928d]OK, so in simple terms it is not only important that the noise is low spatial frequency but also that the noise is from two points coming from the object, therefore on the same side of the plate, so the angle of the fringes is perpendicular to the plate and not parallel. This then makes the noise a transmission image not seen on the reflection hologram.[/quote:4d1a75928d]
Like everything else that's "simple", it's really not that simple. Ok, let me not use the R-N words or the B word. I promise: no mathematics!
All holograms are a mixture of the so-called "thick" variety and "thin" variety. By "thick" is meant that the fringe structure exists as planes within the emulsion and by "thin" is meant variations in surface profile. Let me repeat that no hologram is ever one or the other. This difference in types determines which type of dioffraction occurs and in all holograms, both types occur. As the emulsion depth increases, so does the "thick" type of diffraction start to dominate, but the "thin" type does not disappear, it's still there. The shooting geometry - "Transmission" or a "Reflection" - is independant of the type of hologram and the type of diffraction. You can see this if you shot a hologram of an object with a reference of 60 degrees using a 457nm HeCd. In Dichromate (usual emulsion depth is 12- -15 microns), you'll get a lot of the "thick" type of diffraction and almost none of the "thin" type. If you shot the same hologram in Photoresist (usual emulsion depth is about a micron or so), you'll get all of the "thin" type and none of the "thick". In Silver Halide (usual emulsion depth is about 8 microns), you'll get mostly "thick" and some "thin". You can also see this if you slit an H1 as if you were making a rainbow. However, if you now bring the H2 reference from the "other" side - a la "reflection" geometry, you'll get a "reflection rainbow" (I did this once to convince Mitch it was true - wierd effect). You now have a "reflection" hologram replaying as a "transmission" hologram.
Insofar as intermodulation noise is concerned, it depends on the medium, on the size and position of the object, the make-up of the object and on the medium in which you shoot. If you make a DCG emulsion that's only about 5 microns or less thick, and shoot a highly reflective and very large object very close to the plate, you'll get intermodulation even in reflection. This is because the emulsion is thin enough that the "thin" type of hologram will exist in such an emulsion thickness, despite the fact it's DCG. As I said the intermodulation frequency will appear as a "ripple" on the surface of the DCG emulsion and will contribute to the diffraction - the more so the thinner the emulsion depth. If the object is large enough and close enough to the plate, you may even get "thick" fringes" due to self interference of the object. The angle that the extremes of the object makes with the plate may be greater than the angle that the reference makes. In such a case, the 5 micron emulsion will cause the intermodulation to predominate and you'll lose all image contrast. A very bright object point against a dim object (a high dynamic range) shot on a soft emulsion will also cause intermodulation because the surface profile will be exaggerated by the higher intensity of the bright part. Since this surface profile diffraction is angle independant on replay, the noise will appear extended.
I'm not even going to get into back reflection of the reconstruction beam diffracting off surface profile structures of reflection holograms! But ask yourself, is your Visa hologram a reflection or transmssion? Even if you take off the reflective backing, you won't see much by reconstructing it from a back light a la transmission replay.
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Kaveh
BEAM RATIOS
Well, I think it is, Dinesh.Dinesh wrote:Like everything else that's "simple", it's really not that simple.
I think I follow what you say, but isn't the fundamental point that any intermodulation noise in a reflection hologram is actually a transmission component, not a reflection, so when you reconstruct the image in a reflection mode, you won't see the noise?
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Dinesh
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[quote:35a48e9822] isn't the fundamental point that any intermodulation noise in a reflection hologram is actually a transmission component, not a reflection, so when you reconstruct the image in a reflection mode, you won't see the noise?[/quote:35a48e9822]
In most cases, yes it's true. But in DCG this can trip you up. The emulsion in DCG is much, much softer than commercial Silver Halide. Because of this very soft emulsion, when shooting with the emulsion towards the object, transmission gratings due to intermodulation, bad baffling etc take on a surface profile on the emulsion surface. If you now reconstruct with the emulsion facing you, the reconstruction beam hits the emulsion surface and there's diffraction off this surface profile even before the reconstruction beam enters the emulsion and hits the Bragg fringes within. If the emulsion is deep, the "surface component" of the diffraction is pretty small compared the Bragg diffraction, but there's also a balance between the amount of noise, the softness of the emulsion and the emulsion depth. There have been times when I've been convinced I had a 15 micron emulsion which turned out to be 5 micron emulsion. When shooting large metallic objects in single beam Denisyuk, you do see noise from such a shallow, soft emulsions if you flip it and see the pseudoscopic. Usually, either the object is a long way away from the emulsion or you shoot single beam Denisyuks with the emulsion towards the object and never flip the plate to reconstruct with the emulsion facing you. At least, you never display it that way.
Photoresist also can trip you up. The only way to reconstructt on a photoresist is to have the emulsion facing you and reflect the reconstruction beam. The diffraction occurs by reflection off the surface profile. If you tried to back light a resist plate, you'll see almost nothing. This also trips up people who think that the difference between reflection and transmission is how you reconstruct. I've heard of a number of cases where an interviewee was asked this question, replied that the difference was how you viewed the hologram and was shown a resist plate. Sneaky! Anyway, in most cases, almost all cases, resist holographers make an H1 in Silver and an H2 in resist. In this case, there is no intermodulation noise. However, I've heard of at least one case where, to save themselves money, the company decided to shoot the H1 on resist as well. They coated their own resist. Since the person in question was an old friend of mine, he asked if I could figure out the "wierd glow" around the hologram!
You're right it's almost impossible to get this in Silver Halide because of it's much greater hardness. I've not seen this in Slavich yet, so even a "soft" emulsion such as Slavich is too hard for surface profile noise.
In most cases, yes it's true. But in DCG this can trip you up. The emulsion in DCG is much, much softer than commercial Silver Halide. Because of this very soft emulsion, when shooting with the emulsion towards the object, transmission gratings due to intermodulation, bad baffling etc take on a surface profile on the emulsion surface. If you now reconstruct with the emulsion facing you, the reconstruction beam hits the emulsion surface and there's diffraction off this surface profile even before the reconstruction beam enters the emulsion and hits the Bragg fringes within. If the emulsion is deep, the "surface component" of the diffraction is pretty small compared the Bragg diffraction, but there's also a balance between the amount of noise, the softness of the emulsion and the emulsion depth. There have been times when I've been convinced I had a 15 micron emulsion which turned out to be 5 micron emulsion. When shooting large metallic objects in single beam Denisyuk, you do see noise from such a shallow, soft emulsions if you flip it and see the pseudoscopic. Usually, either the object is a long way away from the emulsion or you shoot single beam Denisyuks with the emulsion towards the object and never flip the plate to reconstruct with the emulsion facing you. At least, you never display it that way.
Photoresist also can trip you up. The only way to reconstructt on a photoresist is to have the emulsion facing you and reflect the reconstruction beam. The diffraction occurs by reflection off the surface profile. If you tried to back light a resist plate, you'll see almost nothing. This also trips up people who think that the difference between reflection and transmission is how you reconstruct. I've heard of a number of cases where an interviewee was asked this question, replied that the difference was how you viewed the hologram and was shown a resist plate. Sneaky! Anyway, in most cases, almost all cases, resist holographers make an H1 in Silver and an H2 in resist. In this case, there is no intermodulation noise. However, I've heard of at least one case where, to save themselves money, the company decided to shoot the H1 on resist as well. They coated their own resist. Since the person in question was an old friend of mine, he asked if I could figure out the "wierd glow" around the hologram!
You're right it's almost impossible to get this in Silver Halide because of it's much greater hardness. I've not seen this in Slavich yet, so even a "soft" emulsion such as Slavich is too hard for surface profile noise.