by Dinesh » Sun Aug 05, 2012 3:44 pm
Not true, at least, not completely true. It may be true, sort of, in a display hologram.
Joe Farina wrote:"Depending on pH, temperature, and hardness, some gelatin will be dissolved at all stages of processing, primarily at the surface and internally in a pattern corresponding to exposure." (By "a pattern corresponding to exposure," I assume he means the unexposed areas will dissolve out more.)
The gelatin in the "unexposed areas" does not dissolve (how hot were their development liquids anyway. Gelatin usually dissolves around 30 deg C in the free state. Trapped inside a multiple density matrix, it'd be quite a bit higher!)). If it did, then the Bragg structure would collapse. Besides which, there are no "unexposed areas". The popularised books always show the Bragg planes as distinct planes, whereas they're sinusoidal variations in density. Thus, there is a continuity of "unexposed areas" based on the value of the sin function. If you take an arbitrary point on the sin function, say phi, for example phi = (arcsin)(pi/6) (where the value of the sin function is 0.5), then the ratio of the "unexposed areas" to the "exposed areas", R, is
R = (2* <int>{sin(theta)} [lim 0 -> phi])/(<int>{sin(theta)} [lim 0 - > pi)
<int> stands for integral.
In other words, you integrate the sin function from 0 to phi and double (to catch both lobes of the sin) and divide by the same integral of the sin function from 0 to pi. This would then be the ratio of areas of "unexposed areas" to "exposed areas". But, even now, you've arbitrarily determined areas of lower exposure (still not zero) as opposed to areas of higher areas.Actually, you'd have to carry out this analysis for a sin^2 function, since you cannot record the negative areas of the sin function. In a real hologram, the sinusoidal functions of many thousand of rays are recorded in every possible direction from 0 to 2pi. So, you'd have to replace the (rather simple!) sin function in the above integral by the two-dimensional Fourier Transform of the object beam, appropriately truncated to a solid angle subtended by the plate and the object. This means that there will probably be no real "unexposed areas" since what is "unexposed" for one set of sinusoids will be "exposed" for another set. Remember also that the gamma of the material will "flatten out" the sin^2 function, so additional unwanted terms will appear.
Joe Farina wrote:but as it is hardened it also narrows and at some point it crosses into the no scatter zone quite suddenly
It's not "sudden". The reason it may seem "sudden" is that at some point the bandwidth crosses the photopic threshhold and so there is an apparently abrupt change. For example, when you look at the sunset, it appears there is a clear demarcation in the colours of the sky. If, however, you pointed a spectrophotometer at the sunset, you'll not notice any sudden change in the spectrum, the spectrum of the sky will be uniform. If the colours did change suddenly, it'd mean a sudden change of index at the demarcation points, which would give multiple reflections at distances of shorter than lambda/4; you'd effectively see a number of images of the sun next to each other.
What both Rallison and McGrew are noticing is that softer gelatin produces wideband holograms. These wide band holograms give a "soft" muted look to the colours, as opposed to the more narrow band silver halide imagery. In addition, the "flattening" effect of the sinusoid causes additional Fourier terms, which, in turn, cause spurious gratings and so appear to cause noise. However, this flattening of the sinusoid is very much a combination of exposure and development, not the gelatin itself. it also happens, to a lesser extent in silver If you put one of these apparently "noisier" holograms into a spectrophotometer, you'll notice no sudden changes and no spectral peaks off the main image peak. However, I very much doubt that either Rallison or McGrew ever put one in a spectrophotomer.
Joe Farina wrote:at some point it crosses into the no scatter zone quite suddenly
There is no "no scatter zone". Scatter is the spurious reflection of light from random elements in the gelatin matrix. If the elements exist and are larger than lambda/4, they will scatter as per Rayleigh's law (although I think it's possible to make an argument that scatter in silver is a Rayleigh effect and scatter in dcg is a Mie effect).
By the way, apropos of nothing in particular (except perhaps the "expert syndrome effect"!) I was with a very prominent holographer who told me of light bulbs that apparently gave off the same spectrum as the sun. These bulbs were quite expensive because they were "full spectrum" (whatever that means!). At any rate, I told him that this was impossible. He however insisted and backed up the contention with an impressive list of other experts who apparently also believed in the "full spectrum" light bulbs. "Easily settled", I said, "you have a spectrphotometer in you lab. The spectrum of the sun follows the photopic curve pretty closely. Let's see if the spectrum of this bulb also follows the photopic" Well, it didn't, much to the "expert holographer's" amazement!
Not true, at least, not completely true. It may be true, sort of, in a display hologram.
[quote="Joe Farina"]"Depending on pH, temperature, and hardness, some gelatin will be dissolved at all stages of processing, primarily at the surface and internally in a pattern corresponding to exposure." (By "a pattern corresponding to exposure," I assume he means the unexposed areas will dissolve out more.) [/quote]
The gelatin in the "unexposed areas" does not dissolve (how hot were their development liquids anyway. Gelatin usually dissolves around 30 deg C in the free state. Trapped inside a multiple density matrix, it'd be quite a bit higher!)). If it did, then the Bragg structure would collapse. Besides which, there are no "unexposed areas". The popularised books always show the Bragg planes as distinct planes, whereas they're sinusoidal variations in density. Thus, there is a continuity of "unexposed areas" based on the value of the sin function. If you take an arbitrary point on the sin function, say phi, for example phi = (arcsin)(pi/6) (where the value of the sin function is 0.5), then the ratio of the "unexposed areas" to the "exposed areas", R, is
R = (2* <int>{sin(theta)} [lim 0 -> phi])/(<int>{sin(theta)} [lim 0 - > pi)
<int> stands for integral.
In other words, you integrate the sin function from 0 to phi and double (to catch both lobes of the sin) and divide by the same integral of the sin function from 0 to pi. This would then be the ratio of areas of "unexposed areas" to "exposed areas". But, even now, you've arbitrarily determined areas of lower exposure (still not zero) as opposed to areas of higher areas.Actually, you'd have to carry out this analysis for a sin^2 function, since you cannot record the negative areas of the sin function. In a real hologram, the sinusoidal functions of many thousand of rays are recorded in every possible direction from 0 to 2pi. So, you'd have to replace the (rather simple!) sin function in the above integral by the two-dimensional Fourier Transform of the object beam, appropriately truncated to a solid angle subtended by the plate and the object. This means that there will probably be no real "unexposed areas" since what is "unexposed" for one set of sinusoids will be "exposed" for another set. Remember also that the gamma of the material will "flatten out" the sin^2 function, so additional unwanted terms will appear.
[quote="Joe Farina"]but as it is hardened it also narrows and at some point it crosses into the no scatter zone quite suddenly[/quote]
It's not "sudden". The reason it may seem "sudden" is that at some point the bandwidth crosses the photopic threshhold and so there is an apparently abrupt change. For example, when you look at the sunset, it appears there is a clear demarcation in the colours of the sky. If, however, you pointed a spectrophotometer at the sunset, you'll not notice any sudden change in the spectrum, the spectrum of the sky will be uniform. If the colours did change suddenly, it'd mean a sudden change of index at the demarcation points, which would give multiple reflections at distances of shorter than lambda/4; you'd effectively see a number of images of the sun next to each other.
What both Rallison and McGrew are noticing is that softer gelatin produces wideband holograms. These wide band holograms give a "soft" muted look to the colours, as opposed to the more narrow band silver halide imagery. In addition, the "flattening" effect of the sinusoid causes additional Fourier terms, which, in turn, cause spurious gratings and so appear to cause noise. However, this flattening of the sinusoid is very much a combination of exposure and development, not the gelatin itself. it also happens, to a lesser extent in silver If you put one of these apparently "noisier" holograms into a spectrophotometer, you'll notice no sudden changes and no spectral peaks off the main image peak. However, I very much doubt that either Rallison or McGrew ever put one in a spectrophotomer.
[quote="Joe Farina"]at some point it crosses into the no scatter zone quite suddenly[/quote]
There is no "no scatter zone". Scatter is the spurious reflection of light from random elements in the gelatin matrix. If the elements exist and are larger than lambda/4, they will scatter as per Rayleigh's law (although I think it's possible to make an argument that scatter in silver is a Rayleigh effect and scatter in dcg is a Mie effect).
By the way, apropos of nothing in particular (except perhaps the "expert syndrome effect"!) I was with a very prominent holographer who told me of light bulbs that apparently gave off the same spectrum as the sun. These bulbs were quite expensive because they were "full spectrum" (whatever that means!). At any rate, I told him that this was impossible. He however insisted and backed up the contention with an impressive list of other experts who apparently also believed in the "full spectrum" light bulbs. "Easily settled", I said, "you have a spectrphotometer in you lab. The spectrum of the sun follows the photopic curve pretty closely. Let's see if the spectrum of this bulb also follows the photopic" Well, it didn't, much to the "expert holographer's" amazement!