Color Holography Optics setup

Light and its behaviour and properties
Dinesh

Color Holography Optics setup

Post by Dinesh »

I think the idea is to form a mask over the emulsion so that any given part of the emulsion receives only one colour (wavelength). One method of doing this is to image a mosaic of colour filters onto the plate while recording. This mosaic can consist of a filter with a series of monochromatic cells so that there is only one colour allowed through over each cell of the mosaic. When reconstructed, there is no cross talk because each cell will only reconstruct the colour recorded by the transmission colour of that specific cell. So, the mosaic can consist of very small cells of RGB filters, or an lcd monitor, as Danny suggested, though I'm not sure how the lcd would allow filtration of any specific colour, since I understand the colour transmission properties of an lcd monitor depend on electronic excitation of the sub-pixel so you'd have to electrically excite the subpixels while recording to get the right ratio. At any rate, when reconstructing the entire mosaic pattern will integrate to a colour image, provided that the cell size is reasonably small. The phrase "spatial multiplexing" comes from the fact that the various colours are multiplexed by spatial separation of individual cells.
jsfisher wrote:The second technique is more curious, at least from my point of view. It "involves coding the reference beam so that each colored reference wave varies in a unique manner over the hologram plate." The author used a ground glass screen as his reference beam coding plate.
The idea is that when a multicoloured beam of light hits the ground glass screen, the light scatters. However, each colour component is scattered through different angles and so hits the plate at different angles. The angular separation creates different spatial frequencies when recording and so "spreads out" the false images. The technique only works for very small angles (I believe the authors used 15 degrees), because if the light scattered over a wide range of angles most of the light would not hit the plate (remember the scattering is random). Also, the more diffuse is the screen, the greater is the noise. This technique is very prone to noise. We've made transmission holograms of actual models using an interbeam separation of about 25 degrees with no coding, which does eliminate cross talk over quite a large area with very little of the noise associated by coded reference beams, but involves quite a complex recording geometry.

Illustration of a colour hologram made by the "coded reference" method. The first is without any coding and the second is by coding.
coded.jpg
And here's one without coding, but with an interbeam angle of 25 degrees (the famous "Bus Stop Babes" that I do all my colour testing on!)
bus stop h1.jpg
bus stop h1.jpg (44.07 KiB) Viewed 18248 times
MilanKarakas

Color Holography Optics setup

Post by MilanKarakas »

Dinesh wrote:
kaveh1000 wrote:Of course color holography is additive by its nature anyway.
Is it?
This is so good question!

I am not an expert, but I have my own 'point of view', which may differ from others. And, this post is not by any means intended to offend anyone, All I need is just clarification of some terminology, for which I consider that is very important, especially in such 'tough' stuff as is hologram and it's properties (at least for beginners, like I am).

I will be rather careful calling it either 'additive' or 'subtractive'. I will rather prefer word 'selective' (I will explain why below).
This is something I have been thinking about for a while now. Consider that an additive system, such as a colour TV, emits R,G and B in appropriate quantities to achive a particular colour balance. However, the TV itself emits the appropriate "wavelengths". If you illuminate a printed sheet with a white light source, then any colour is achieved - "emitted" - by selective absorption of the "unwanted" colours.
I will be careful here calling it 'emitted' too. Maybe re-emitted, or 'selectively reflected' to be more precise. While CRT TV (or monitor) really emit light by its phosphors, which are excited by electron impact, light which goes through color pigment is 'filtered' out only. All reflection which come from surface of pigment is NOT related with pigment's 'filtering' ability. Consider printing on black paper - no image, or image looks faint and weird.

Color chalk on the other hand working on black and white paper, that is because it is white chalk with added pigment(s), so each pigment has its own reflecting surface.

So, it is more complex; transmission with filtering, reflecting, and again transmission by another filtering.

Also, I have problem with understanding 'subtractive' pigments.

While printing is done by pigments of various colors (each one has different ability to selectively filtering, including black one - filtering out all light), pigments which come on top of each other looks dark (what remains is 'subtracted', so yes - subtractive).

But how to call pigments which are next to each other? Each filtering its own color, but what remains gives us impression of different color than each pigment. For example, blue and yellow pigment gives me impression of green(ish) color, but there are no mutual 'subtraction' between each pigment. Then, how to define this effect? Is it 'addition by filtered light' ? Or, such 'printing scheme' should to be avoided?

Our brain is 'fooled' all the time, no matter is it additive or subtractive, or whatsoever it is.
This is a subtractive colour system, since it "subtracts" by absorption all the undesired "colours" (wavelengths). Now consider the colour of a coloured object. When an object is illuminated by white light , then again, certain "colours" are absorbed by scattering processes and whatever's left is emitted in order to give a colour to the real-life object. Is this additive or subtratctive? The coloured object does not emit it's own radiation, but radiates the illuminant after selective absorption. Consider now a colour hologram. In this case, there are two processes - the recording and the reconstruction.
So far, so good.
In recording the hologram, the illuminant is some mixture of RGB (and I'm ignoring the fact that the object in a colour hologram is illuminated by coherent radiation, since the coherence of the illumination is required only for fringe stability and not for the colour characteristics per se)
Wait!

We can't ignore coherent radiation. We can't ignore relationship between 'the illuminant' and the 'reference beam' whatsoever this means. All what we can ignore (maybe) is object if we have intention to make HOE (or, the 'object' is a mirror, or another HOE? ). Then, there is no object - just two beams in exactly defined (very important!) relationship, by means of angle(s) and by means of intensity.

By plural of angle, I mean situation where two beams are not with identical angles in respect to normal incidence. The result is diffraction grating with different properties (orders of diffraction are not 'symmetric' to the hologram's plane). I think the result will be gratings which may refract light at different angle differently (intentionally avoiding words 'transmit' and/or 'reflect'). Isn't it the case of shifting 'position' of the color(s) of the monochrome hologram when we changing angle of the light or when we changing the plate in respect to light or to us?

By means of intensity, if two beams has not the same intensity - the result may give us also different properties of such HOE, for example lower diffraction efficiency. Isn't it the case of bright and dark holograms or part of the holograms?

I am still in 'monochrome mode', but still getting full color spectra of refracted white light - only lacking of proper 'alignment' of such color(s) with the ' object's ' color(s). By alignment, I mean not getting exact color of the object where it was, but rather a 'false colors'. And, I am getting blue hologram, but laser was green. Or variation of colors, depending where hologram is overexposed or underexposed, and also where emulsion is thicker or thinner.
whose ratio is given by some appropriate colour map - usually the CIE 1931. If the object re-emits this with no absorption, then the object is, in effect, a mirror, which brings up the point: Is a mirror illuminated by coloured light an additive or subtractive system?
So tough question! And good one!

I think a mirror is a 'selective' by nature. And, mirror can be either additive and/or subtractive (depend of type).

Gold mirror, for example, reflect part of the visible spectra(green, yellow, red, IR...), while another part transmit (blue, violet, etc.). Also, depending of the thickness of the gold. Very thin layer of the gold on the substrate absorb light, while transmit whole visible spectra.

While trying to deposit silver onto glass, I observed at fist black deposit - until certain thickness is not achieved, then it reflect most of the visible spectra. By looking through such mirror, blue and violet light is transmitted.

Dielectric mirrors are very different from metallic mirrors - it can selectively transmit or reflect light. How much, depend of thickness of 'colorless' layers, number of such layers, it's difference in refractive indexes, and so on.

All in all, I will avoid using words 'additive' or 'subtractive' in holographic world. That is because light itself is additive and/subtractive itself. This phenomena is actually responsible to our dark and bright fringes, and essential for holograms as well.

And, it may be confusing too - addition we call constructive, while subtraction we call destructive interference, while the phenomena is not strictly related by any perceptive 'color'.

We actually dealing with two different phenomena: perception of the colors, and properties of the light. And, yet it is very closely related when our perception is in question.

Also, what gives some substance its properties at atomic/molecular level is responsible for its appearance at macro level. If anyone look at the pigment under the microscope with really good magnification, may be surprised that there is no color at some point of the magnification - just structure. Yet, colorless substances may act as a good filter/reflector if properly 'stacked'.
Now consider the reconstruction of a colour hologram. In this case, the hologram is illuminated by white light and emits colour by Bragg selection. This means that unwanted colours - wavelengths - are absorbed, and what's left is emitted to give a specific colour. In other words, the colour of a reconstructed colour hologram depends on the colours in the illuminant that are absorbed. This sounds like an subtractive system.
I think I am covered this subject already above, but I will try to add some of my 'weird' point of view. I think that holograms doesn't require absorption, nor reflection - but rather 're-direction' of the light.

Yes, AgX hologram may absorb some light if silver grains remained (as is it case of not completely 'silvered' mirror), and produce image by diffraction of the light. But, if bleached, such hologram may work on different principle - instead blocking part of the light and causing diffraction, it rather refract light (but not exclusively).

And, reflection hologram doesn't require any 'mirror' at all. Diffracted/refracted light may experience multiple diffraction and refraction until reach 'critical point', where part (if not whole) of the spectra is reflected by total internal reflection.

//Edit: Now I found myself in great doubt about statement above. By observing one of the poor looking reflection hologram, one 'patch' of the hologram showing some reflection looks completely transparent at first glance. But, careful re-inspection showing faint 'suppression' of the green color, exactly where hologram showing greenish reflection. So, what remains: various thing happening 'inside' hologram, including diffraction, refraction and reflexion (and maybe something else, for which I am not aware at this moment).//

For this reason, I am thinking of hologram NOT by means of small mirrors and/or lenses, but rather of small diffractive gratings (with 'repeating', refractive elements).
Therefore, I would argue that the recording of a colour hologram may be an additive system (what colour is a chameleon on a mirror?),
Depend of its mood, may be green or orange/red (if threatened):

http://www.youtube.com/watch?v=JaKk5dDW ... =endscreen
but the reconstruction of a colour hologram by selective absorption of not-colours is a negative system.
I think I am answered above to that issue too. Maybe I am not so clear, maybe my English is not the best, and my understanding of light and optic is not the clearest as well. But, I feel that some of the terms in holography are not properly used most of the time. For that reason there are confusions from time to time.

Best wishes--
milan--
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MilanKarakas

Color Holography Optics setup

Post by MilanKarakas »

Dinesh wrote:I think the idea is to form a mask over the emulsion so that any given part of the emulsion receives only one colour (wavelength). One method of doing this is to image a mosaic of colour filters onto the plate while recording. This mosaic can consist of a filter with a series of monochromatic cells so that there is only one colour allowed through over each cell of the mosaic. When reconstructed, there is no cross talk because each cell will only reconstruct the colour recorded by the transmission colour of that specific cell. So, the mosaic can consist of very small cells of RGB filters, or an lcd monitor, as Danny suggested,
This is very interesting topic. One day I will (maybe) able to assembly panchromatic holo-recording system. Currently, I am not even able to do monochromatic holograms (my laser is in re-building phase, making thermostat, current regulator, etc.).
though I'm not sure how the lcd would allow filtration of any specific colour, since I understand the colour transmission properties of an lcd monitor depend on electronic excitation of the sub-pixel so you'd have to electrically excite the subpixels while recording to get the right ratio.
Liquid crystals are in a 'random' state, not rotating light when not under certain voltage. I believe that each sub-cell consist of RGB filters close to such crystals. Once 'electrified', this crystals begin to rotate light, depending of voltage - more or less, thus permitting light passing through such filters.

I think that a 'plain' LCD screen with removed polarizers will be okay for its purpose, but not sure of 'side effects': since there are fine repeating structure, does it will act like diffraction grating?

I am asking that, because noticed that projected beam on the screen sometimes showing interferences from obstacle if such obstacle is not very close to the screen. If that is the case, does such RGB filters should be as close as possible to the emulsion?
At any rate, when reconstructing the entire mosaic pattern will integrate to a colour image, provided that the cell size is reasonably small. The phrase "spatial multiplexing" comes from the fact that the various colours are multiplexed by spatial separation of individual cells.
jsfisher wrote:The second technique is more curious, at least from my point of view. It "involves coding the reference beam so that each colored reference wave varies in a unique manner over the hologram plate." The author used a ground glass screen as his reference beam coding plate.
The idea is that when a multicoloured beam of light hits the ground glass screen, the light scatters. However, each colour component is scattered through different angles and so hits the plate at different angles. The angular separation creates different spatial frequencies when recording and so "spreads out" the false images. The technique only works for very small angles (I believe the authors used 15 degrees), because if the light scattered over a wide range of angles most of the light would not hit the plate (remember the scattering is random). Also, the more diffuse is the screen, the greater is the noise. This technique is very prone to noise. We've made transmission holograms of actual models using an interbeam separation of about 25 degrees with no coding, which does eliminate cross talk over quite a large area with very little of the noise associated by coded reference beams, but involves quite a complex recording geometry.

Illustration of a colour hologram made by the "coded reference" method. The first is without any coding and the second is by coding.


And here's one without coding, but with an interbeam angle of 25 degrees (the famous "Bus Stop Babes" that I do all my colour testing on!)
What is exactly 'interbeam angle'? Two different wavelength reference beams not emerging from the same spot?

Thank you in advance for clarification.

milan--

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Dinesh

Color Holography Optics setup

Post by Dinesh »

MilanKarakas wrote: light which goes through color pigment is 'filtered' out only. All reflection which come from surface of pigment is NOT related with pigment's 'filtering' ability.
Actually, it is related to the pigment's filtering ability. The colour gamut for a printed page is the CMYK (Cyan, Magenta, Yellow and Black) gamut - a subtractive system. The reason it's called "subtractive" is the the pigments selectively absorbs certain colours . The light then passes through to the paper (usually white) and and then reflects back again through the pigment. Now, the colour left over after the absorption of the pigment is what is seen. The pigment has, in effect, "released" the leftover colours. The body of the pigment - the "inside" of the pigment" - absorbs the negative colour (called "complementary" colour by printers). For this reason, you print in the "opposite", complementary, colours from the colour you want the reader to see. CMYK is the "opposite" of RGB. This is the same system in colour photogrtaphy, where there is a CMYK filtering system in front of the film. One problem is that the gamut of CMYK is smaller than the RGB gamut, so that the intense colours seen on a TV cannot be accurately reconstructed by printing.
MilanKarakas wrote:But how to call pigments which are next to each other? Each filtering its own color, but what remains gives us impression of different color than each pigment. For example, blue and yellow pigment gives me impression of green(ish) color, but there are no mutual 'subtraction' between each pigment. Then, how to define this effect? Is it 'addition by filtered light' ? Or, such 'printing scheme' should to be avoided?
If you have two (or more) pigments on top of each other, then each pigment selectively absorbs the complementary colour and you see what's left over. If you're interested, there are probably a lot of places on the internet where you can look up CMYK.
MilanKarakas wrote:Wait!

We can't ignore coherent radiation. We can't ignore relationship between 'the illuminant' and the 'reference beam' whatsoever this means.
When I say, "ignore coherency", I mean I'm looking at the colour aspect of the object beam. Sorry if I didn't make it clear, but "illuminant" is used by colour scientists to define the colour profile of a source of light in terms of some colour map. Thus, for example, the light from daylight under a clear cloudless sky is called D65, based on the 1931 CIE and on cone respones in the retina. You can look up D65 in the CIE website. If you want to make an object indoors appear (such as in a store) the same as if you're looking at it outdoors in standard daylight conditions, you light the object with a source that simulates D65. The combinations of laser light illuminating the object also has an "illuminant profile", so for example, if you want to make your colour hologram appear as if you're looking at the object under daylight, you use D65 to illuminate the object. If the reproduction is faithful, the final hologram colour will be the same colour as if you looked at the original object in daylight. This may not be easy, in fact, it may not be possible. This is because D65 is reproduced by a thermal source with a certain colour temperature and then filtered by a specific set of chemicals. To reconstruct D65 under laser illumination will require more than three lasers.
MilanKarakas wrote:All what we can ignore (maybe) is object if we have intention to make HOE (or, the 'object' is a mirror, or another HOE? ). Then, there is no object - just two beams in exactly defined (very important!) relationship, by means of angle(s) and by means of intensity.
Yes, HOEs are different. HOEs don't really have a "colour". This is a difference between radiometry and photometry.
MilanKarakas wrote:Gold mirror, for example, reflect part of the visible spectra(green, yellow, red, IR...), while another part transmit (blue, violet, etc.). Also, depending of the thickness of the gold. Very thin layer of the gold on the substrate absorb light, while transmit whole visible spectra.
The optics of metals is quite different from the optics of insulators. In a metal, there are a large number of free electrons (which gives the metal it's properties). Theses free electrons effectively absorb light pretty uniformly. So as light enters the metal, in the first few microns of the metal, the bound electrons scatter the light back. As the light further penetrates the metal, the free electrons then absorb all the light. This distance at which light is scattered back is called the "skin depth" of the metal. The skin depth for gold is about 8 microns, for example. So a piece of gold less than 8 microns will transmit, but any deeper and there will be no light passing through. The bound electrons in the gold atom give it that gold colour. If a piece of any metal is very, very small it will appear black because the size is of the order of a wavelength so there's not enough penetration for the bound electrons to scatter.
MilanKarakas wrote:We actually dealing with two different phenomena: perception of the colors, and properties of the light. And, yet it is very closely related when our perception is in question.
Very, very true. I think a lot of people don't really understand this. Colour is a psychological phenomena based on perception. It is not absolute. Nothing has an absolute colour, but everything has an absolute spectral reflectivity and an absolute spectral absorptivity. The art and science of colour reproduction is to try and fool your brain to reconstruct the psychological cues that it goes through when you observe something in nature.This is as true in colour printing as it is in colour holography. In both cases, the actual wavelengths trigger responses in the retina, but the perceived colour is due to processing in the brain (there is some "pre-processing" in the retina also - it's a little like an ALU in a microprocessor). The absorption spectrum of the cones in the retina already alter the perception of the wavelengths entering the eye. If you're interested, I wrote an essay on colour perception ( http://www.triple-take.com/publications/colour.pdf )
MilanKarakas wrote:But, I feel that some of the terms in holography are not properly used most of the time. For that reason there are confusions from time to time.

Yes, also very true. There are a lot of people who feel that the terms used are absolute. You should hear the controversy about "achromatic" holograms and "digital" holography!
MilanKarakas wrote:And, this post is not by any means intended to offend anyone, All I need is just clarification of some terminology
I'm afraid I do not understand why people are offended when standard "known" things are questioned! Surely all human knowledge from art to science to religion to politics progresses by questioning the standard assumptions. Everyone "knew" that heavy objects fell faster than light objects, until Galileo asked "Why?". By the way, a lot of people were offended when Galileo asked "Why?" Everyone knew that to cure a disease, you prayed to the gods, until Galen asked "Why?" (and a lot of people got mad at him too!). If we did not question standard wisdom, we would still be gnawing on bones in front of a fire, dressed in animal skins and living in caves! But, a lot of people react with anger, go silent, refuse to accept or roll their eyes when a standard assumption is questioned; and these are the so-called "smart" people! I think that there is an intellectual conservatism that pervades certain fields of human activity, and I think that if holography is not the worst, it's certainly pretty bad. Everything discovered over a half century ago is set in stone and repeated by the "holographic cognoscenti" as the high priests of a religion. No one dare to question these high priests!

You should question! Question everything! Question everyone! If someone says 2 + 2 = 4, ask "why?!". But, I accept it takes a certain courage, because when you question, you open yourself to the accusation of anger. By the way, Einstein gave his first lecture on relativity to a group of engineers. These engineers boo-ed him and shouted, "what rubbish!". Eventually, as Einstein slid the blackboard across the room, one engineer shouted, "You see, gentlemen, the board moves against the room. The room does not move against the board".
Dinesh

Color Holography Optics setup

Post by Dinesh »

MilanKarakas wrote:What is exactly 'interbeam angle'? Two different wavelength reference beams not emerging from the same spot?

Thank you in advance for clarification.
The interbeam angle is the angle between the object beam and the reference beam. In a HOE, there are usually only two beams, so there is a definite interbeam angle. In a display hologram, the interbeam angle is not so well defined because the object throws off spherical waves. However, if the object is small compared to the plate size, then the angle between the centre of the object to the plate and the reference angle to the plate is the interbeam angle. If the object is large, then you can still take the centre of the object to the plate as the "object angle", but you have intermodulation noise, where the two ends of the object create a grating also and this shows up as noise.

In transmission colour holography, the multi-wavelength gratings are independent of each other, assuming RGB recording, the red laser creates 'red' fringes independent of any other laser hitting the plate (this is not completely true because it depends on the spatial frequency and the resolution of the medium, but it's close enough). However, in a transmission hologram, the diffraction is Raman Nath. This means that all light will diffract from all fringes, there's no Bragg selectivity. So, the red light will diffract from the "green" and "blue" fringes as well. This is known as "cross talk". However, The green image of the green light hitting the red fringes (for example) will shift the green image off to one side of the "true" position.Thus, there is a "true" colour image, positioned where the original object was, and there are "ghost" or "false" images on either side of the "true" image. For M recording wavelengths, there are [M^2 - M] "false" images.

One way of getting rid of these "false" images is to place them a long way away from the "true" image; the position of these "false" images depends on the interbeam angle of each recording wavelength and the difference between them. So, if the distance between the "true" image and the "false" image is d(w1, w2) for a specific "false" image for a specific correct wavelength and a specific incorrect wavelength (so that, in the example, if the green "false" image from the "red" fringes is 5cm, then d(red,green) = 5cm) and if the angle difference between the correct wavelength and the incorrect wavelength is [theta(w1,w2)] ( if the red wavelength is recording at 20 degrees and the green wavelength is recording at 30 degrees, then theta(red ,green) = 10 degrees ), then there is a relationship between d(w1,w2) and theta(w1,w2). This means that the greater the angles between the recording wavelengths, the further away the "false" images. If you're shooting on a 10cm x 12.5cm plate and you can arrange for the "false" images to be greater than 12.5 cm away from the "true" image, the cross talk will not be seen.

In the coded reference system, the interbeam angle is the same for all wavelengths (in the hologram shown, this was 15 degrees). However, when this combined beam hits the ground glass, the light scatters. This means that on the other side of the ground glass, the angles of the RGB components will all be different. That is, theta(w1,w2) will be different, but random for all w's. So, the plate "sees" all the components at different angles and shifts the "false" images far away (or rather further away) from the "true" image and so, if the plate is smaller than the "false" image distance, plate size < d(w1,w2), the "false" images will disappear.
Ed Wesly
Posts: 513
Joined: Wed Jan 07, 2015 2:16 pm

Color Holography Optics setup

Post by Ed Wesly »

Bragg diffraction reflects a certain color strongly thanks to constructive interference but it doesn’t necessarily absorb the other colors. If you’ve ever made a bright green reflection hologram, you would observe the undiffracted zero order passing through the hologram will cast a magenta shadow from the unused blue and red being transmitted and not reflected. Kind of like a dichroic filter.

"but the reconstruction of a colour hologram by selective absorption of not-colours is a negative system."

The correct term is subtractive, not negative. The rest sounds like pure nonsense.

"Is a mirror illuminated by coloured light an additive or subtractive system?" The mirror is passive, but the image of the light sources it reflects retain their additive properties.

If one considers a reflection hologram a type of modulated mirror, wherein each object point is reconstructed as a reflection of the replay light at various depths, then it could be argued that a hologram is an additive system, especially coupling in the fact of the color being generated by Bragg diffraction.

"This is the same system in colour photogrtaphy, where there is a CMYK filtering system in front of the film." Absolutely not true in the chemical or digital world. Analysis of the scene in photography begins with the RGB primaries, just like our eyes do. Photographic films have a top layer that only sees blue, middle layer sees blue and green thanks to its sensitizing dye; bottom layer sees blue and red thanks to its sensitizing dye. There is a yellow layer under the blue one to eliminate blue light from reaching the middle and bottom layers, as the silver halides cannot be made to not see blue. So there is only one subtractive primary involved, but it's role is not as a primary.
"We're the flowers in the dustbin" Sex Pistols
Dinesh

Color Holography Optics setup

Post by Dinesh »

Ed Wesly wrote:Bragg diffraction reflects a certain color strongly thanks to constructive interference but it doesn’t necessarily absorb the other colors.
There is absorption. Equations 21 and 22 in the Kogelnik paper show the amount of absorption.
Ed Wesly wrote:The correct term is subtractive, not negative
Or complementary, it depends on what you mean by "correct". A physicist would have a different "correct" to a chemist and a mathematician would probably have a still different "correct". Sometimes, it's more useful to use a colloquialism than be overly pedantic
Ed Wesly wrote:The mirror is passive, but the image of the light sources it reflects retain their additive properties.
The mirror is not passive. The bound electrons in the mirror oscillate by pi out of phase to the incoming E vector. These charged oscillators then re-emit, which gives rise to emitted spherical wavefunctions. Since the wavefront from each atomic oscillator is spherical, there needs to be some form of destructive interference system otherwise it would be difficult to derive the law of reflection theta(i) = theta(r). These are active atomic processes governed by Maxwell's equations. As Milan pointed out, a gold mirror does not act passively, the skin depth determines the level at which atomic procceses overcome resonance effects.
Ed Wesly wrote:If one considers a reflection hologram a type of modulated mirror, wherein each object point is reconstructed as a reflection of the replay light at various depths
But, if a mirror is passive, how do you get Bragg selectivity?

The reconstruction of a Bragg hologram is an integral function of the plane distribution. In other words, it's a statistical averaging over the Fourier transform of the Bragg plane vectors. There are no Dirac reflectors otherwise you'd have the same problem as the mirror; it would not be possible to derive the Bragg condition from a series of spherical wavefronts. The final colour is a funtion of colour mixing from the specific wavelengths resulting from the Bragg condition. The Bragg condition "removes" certain components of the reconstructing light. Thus, if the hologram were reconstructed with a set of wavelengths for which there were no corresponding Bragg planes, then those colours could not be removed since they were never there in the first place. This occurs when the colour temperature of the reconstructing source does not match the Bragg plane distribution.
Ed Wesly wrote: wherein each object point is reconstructed as a reflection of the replay light at various depths
What propagates the wave vector? If each object point is reconstructed, then these "various depths" would emanate spherical waves since they would be point sources.
Ed Wesly wrote:"This is the same system in colour photogrtaphy, where there is a CMYK filtering system in front of the film." Absolutely not true in the chemical or digital world. Analysis of the scene in photography begins with the RGB primaries, just like our eyes do. Photographic films have a top layer that only sees blue, middle layer sees blue and green thanks to its sensitizing dye; bottom layer sees blue and red thanks to its sensitizing dye.
Perhaps you might argue with Wikipaedia ( http://en.wikipedia.org/wiki/Color_photography )
In photography, the dye colors are normally cyan, a greenish-blue which absorbs red; magenta, a purplish-pink which absorbs green; and yellow, which absorbs blue. The red-filtered image is used to create a cyan dye image, the green-filtered image to create a magenta dye image, and the blue-filtered image to create a yellow dye image. When the three dye images are superimposed they form a complete color image.
Or, you might argue with Kodak ( http://motion.kodak.com/motion/uploaded ... -image.pdf ) who state they use a cyan dye (C), a magenta dye (M) and a yellow dye (Y).

In the digital world, no dyes are used, since the colour is separated by optical or electronic methods and captured by ccd's. Thus the question of dyes in digital photography does not arise.
Ed Wesly
Posts: 513
Joined: Wed Jan 07, 2015 2:16 pm

Color Holography Optics setup

Post by Ed Wesly »

I won’t argue with wikipedia or Kodak, just you. You said that the CMYK is in front of the film. The CMY is in the film, and there is no K in color photographic film as dyes, it comes about as the superimposition of the CMY. This is in the synthesis step of the color photograph, not the analysis, which is what I believe was being discussed. Plus the dyes do not exist as C, M and Y in the emulsion until after exposure and processing, only as colorless precursors in the case of transparency films, or orangey looking couplers in the case of color negative stock.

If you have your copy of Kogelnik’s equations 21 and 22 handy, why not post them and explain to us how there is absorption? And how come you didn’t stick around at the last symposium and debate David Radcliff’s deconstruction of the Koglenik equations?

One gets Bragg selectivity by having layers of mirrors, correct?

“Sometimes, it's more useful to use a colloquialism than be overly pedantic”. Or most likely is that you do not use the correct terminology as you are not practiced in the art and technique of photography.

In the world of optics I come from, a mirror is considered a passive device, as opposed to something like an acousto-optic modulator, which is an active device.
"We're the flowers in the dustbin" Sex Pistols
Dinesh

Color Holography Optics setup

Post by Dinesh »

Ed Wesly wrote:One gets Bragg selectivity by having layers of mirrors, correct?
No, the mirror concept is a model. Bragg selectivity occurs because of the Bragg equation. The Kogelnik theory uses selective interchange of energy through the planar structure, hence it's a couple wave theory. It is possible to model this as a set of mirrors and this has been done many times in the past, but the paradigm breaks down when you try and derive the zeta function - the dephasing measure. The paradigm also breaks down when you have a non-linear material such as LiNiobate or the new Nitto Denko material. The mirror metaphor would also limit the spatial frequency, and the fringe structure itself, especially when applied to a polymeric substance. In a polymeric substance the molecular energy levels may have resonance absorption in some parts of the visible.
Ed Wesly wrote:explain to us how there is absorption?
Well, the alpha parameter carries the absorption. It's one of the central assumptions of the coupled wave theory. It's what enables the equations to be first order, and so to be readily solvable as a pair of coupled first order differential equations. If alpha were zero, there'd be no coupling
Ed Wesly wrote:In the world of optics I come from, a mirror is considered a passive device, as opposed to something like an acousto-optic modulator, which is an active device.
You need to define "passive" and "active". Just because you "consider" it passive, does not actually mean that it is "passive". However, to follow the analogy further, the Nitto Denko material is electro-optic; it needs some 5 - 10 kV across it in order to achieve a delta n. However, if a mirror is "passive" (presumably in the electrical sense), then either the Nitto Denko material does not contain Bragg planes, or they consists of Bragg planes that are not mirrors, or mirrors are "active". Since I worked on the development of the material, I do know that the Bragg planes do exist within the material.
Ed Wesly wrote:And how come you didn’t stick around at the last symposium and debate David Radcliff’s deconstruction of the Koglenik equations?
Well, for one thing, I don't "believe" in "experts". I derive the equations myself, and I'd already derived the Bragg-planes-as-mirrors equations myself several years ago. For what it's worth, the way I see it, epistemology is independent of personality. For another thing, the mirror analogy is old, the Bragg-planes-as-mirrors metaphor goes back to the 60's.
Dinesh

Color Holography Optics setup

Post by Dinesh »

So, getting back to the central concept, Milan, I hope I answered some of your questions.

Danny, I thought your scheme would have some problems because you'd need to energise the lcd's, but Milan explains that the electrical energy is a control feature, not an enabling feature. There are apparently filters behind each liquid crystal that carry out the enabling feature. In this case, it would depend on the transmission bandpass of the filters. If the filter for "red" were, eg, 650 - 700nm, then this would not pass the 633nm of HeNe.
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