RGB and persistance of vision

Holography related topics.
hjalmar1234
Posts: 4
Joined: Tue Jul 19, 2016 12:10 pm

RGB and persistance of vision

Post by hjalmar1234 »

Would it be possible to record three seperate holograms, one r, one g, one b stitch the developed film together and use something similar to a movie projector (just the mechanism, not the lenses etc.) to play the holograms back in a loop at 60fps and open and close the appropriate laser shutter at the appropriate times. Has this been done before? If not, why not? Please excuse my lack of knowledge and subsequent mistakes, I haven't managed to get hold of holography books yet and have only recently developed a interest in the field. I have however been burning to ask this question. Also, seeing as wikipedia is useless, can someone please explain the basics of normal colour (not rainbow) holography to me.
lobaz
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Joined: Mon Jan 12, 2015 6:08 am
Location: Pilsen, Czech Republic

Re: RGB and persistance of vision

Post by lobaz »

Hologram is a very fine pattern that causes diffraction of ANY light. That is, if you prepare a "red hologram" and illuminate it with green light, it will still produce the image. This image will be green, of course (hologram does not change wavelength of reconstruction light). Moreover, its size and position will be also different compared to proper red illumination.
Your idea is to have three stitched holograms (R, G, B) and illuminate them sequentially with R, G, B lights. Thus, each illumination will produce not just the right image, but two unwanted wrong images as well, because all holograms are still in place.
It is thus necessary to create e.g. a red hologram that produces no image when illuminated with green light. There are ways how to do that, for example using thick reflection holograms. But now, as the red hologram does not produce unwanted green image, there is no need for time sequential illumination - red, green and blue lights can be on simultaneously.

A different situation is in digital holography, where the hologram is displayed using a very fine microdisplay (usually called spatial light modulator, SLM for short). In this case, you can display red/green/blue holograms sequentially and illuminate them with proper light, just like in single-chip DLP beamers. This is quite common practice. However, current SLM's are still too coarse so the results are far from impressive.

Petr
Din
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Joined: Thu Mar 12, 2015 4:47 pm

Re: RGB and persistance of vision

Post by Din »

Is your intention to make a colour hologram, or a holographic movie?

In a colour hologram, there is no need to multiplex it like this. You record a red hologram, a green hologram and a blue hologram, all on the same emulsion, in reflection mode. In transmission mode, as Lobaz says, you get what's called "cross talk" and all three wavelengths will create all three images. You get nine images and only one will be the correct image. But, in reflection mode, there is no cross talk. For technical reasons, it's best to shoot this as a Denisyuk. So, you do not split the beam. You simply place your object behind the film, and hit the film from the front with all three wavelengths at some angle.
hjalmar1234 wrote:can someone please explain the basics of normal colour (not rainbow) holography to me.
In a thick, reflection, hologram, the recording of the hologram is a set of planes within the hologram. This is like sheets of thicker material inside the emulsion. The separation of these sheets is a function of the wavelength. So that, if you record with red at 633nm, the sheets will be separated by 314nm (half the wavelength). Similarly, recording with green at 532 will give sheets separated by 266nm and blue at 457 will give a separation of 218nm. All three sets of sheets will exist simultaneously.

If you record with a single wavelength, only one set of sheets will be recorded. If you now reconstruct with white light, the sheets within the hologram will react only to the sheets recorded, all other wavelengths will be lost. So, if you record with 633nm, the sheets are separated by 314nm. If you hit this with white light, the sheet separation of 314nm will react to red light (2xlambda) only, all other components of the white light will be lost.

If you record with three wavelengths of red, green and blue at 633, 532 and 457,then you get three sets of sheets at 314 (red component), 266nm (green component) and 218 (blue component). When you hit this with white light, the red component "picks off" the red component of the white light, and similarly, the green and blue component pick off the green and blue component. This is done simultaneously, so there's no need to multiplex the reconstruction white light beam
hjalmar1234
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Joined: Tue Jul 19, 2016 12:10 pm

Re: RGB and persistance of vision

Post by hjalmar1234 »

Thanks for answering so soon, Din and Lobaz.

Lobaz, the idea was not to have all the plates in place at the same time, but to have them on a motorized 'conveyer belt' thus eliminating crosstalk.

Din, I realise that I did not properly state my question, but you answered exactly what I wanted to know. My plan was to eliminate the problems with continous white light, but I never considered combined rgb light. Mind you, I did not know about interference planes so I did not fully understand the problem.

Would my idea resolve the problems with transmission holograms? Why does crosstalk only occur with transmission holograms? Sorry to waste you time with more questions, but I'm really excited about holography and the local library opens on Monday which means another week before books arive from Cape Town.
Din
Posts: 402
Joined: Thu Mar 12, 2015 4:47 pm

Re: RGB and persistance of vision

Post by Din »

hjalmar1234 wrote:Why does crosstalk only occur with transmission holograms?
Because in transmission holograms the diffraction mechanism is different from reflection holograms. There are two diffraction mechanisms. Technically, one is called Raman Nath, and one is called Bragg. In the Raman Nath type of diffraction, everything diffracts, but at different angles. Transmission holograms diffract by Raman nath diffraction and reflection holograms diffract with Bragg diffraction.

A transmission hologram has what's called a "spatial frequency", which is the number of lines/unit distance - generally, in holography, this is given by lines/millimetre - and is determined by the angles between the reference and the object. When you reconstruct the hologram, this frequency determines the angle through which any wavelength diffracts. I don't know how much maths you have, but the angle through which any wavelength diffracts is given by the "grating equation" (lambda)*f = sin(theta), where lambda is the wavelength you're hitting the transmission hologram with, f is the spatial frequency already recorded in the hologram, and theta is the diffraction angle. So, for example, if the spatial frequency of your hologram is 1000 lines/millimetre, and you hit the hologram with green light (lambda = 0.000532, the extra number of zeroes to keep the units consistent), you have sin(theta) = (1000)*(0.000532) = 0.532. So, the angle whose sin is 0.532 is 32.14 degrees. So, hitting this hologram with a wavelength of 532nm will cause a diffraction through 32 degrees. It doesn't matter what wavelength you hit the transmission hologram with, everything diffracts. So, if you record a transmission hologram with any single wavelength,ie recording only one spatial frequency, then all wavelengths will diffract. If you then reconstruct the hologram with three specific wavelengths, say 633nm, 532nm and 457nm, they'll all diffract and give you three images. If you make the transmission hologram with three wavelengths, then you'll record three spatial frequencies. If you now hit the transmission hologram with three wavelengths, any one frequency will give you three images, all three frequencies will give you nine images.

In Bragg diffraction, the planes, or sheets, within the hologram will function as filters, filtering out all wavelengths except one, at a specific angle.These planes are tilted by some angle and separated by some distance. This tilt angle and this separation determines which angle and which wavelength will diffract. This is called "Bragg selection". Bragg selection allows diffraction for only one wavelength at only one angle. No other wavelength will diffract. So, if you have the right wavelength at the wrong angle, there will be no diffraction. So, if you record a reflection hologram with 633nm, 532nm and 457nm, at some angle, then Bragg selection ensures that the hologram will only diffract if it's hit with those wavelengths and at the angle with which you shot it.
lobaz
Posts: 280
Joined: Mon Jan 12, 2015 6:08 am
Location: Pilsen, Czech Republic

Re: RGB and persistance of vision

Post by lobaz »

hjalmar1234 wrote:Would my idea resolve the problems with transmission holograms?
To make it clear: you have a film strip with R/G/B/R/G/B/... transmission holograms. You feed it to a "projector" that illuminates it with matching laser light.
Yes, it would definitely solve the crosstalk problem. On the other hand, you have plenty of other, mostly mechanical problems. For dark environments and small viewing angles, the flicker-free frequency is about 50 Hz, which means you need to perfectly register 150 holograms per second. But yes, it can be done.
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