Talk:Holography

From Wikipedia, the free encyclopedia

This article is part of WikiProject Media, an attempt to better organize information in articles related to media. If you would like to participate, you can edit the article attached to this page, or visit the project page, where you can join the project and/or contribute to the discussion.

"The first holograms which recorded 3D objects were made by Emmett Leith and Juris Upatnieks in Michigan" - No. Gabor made the first holograms.Restname 23:19, 18 July 2006 (UTC)restname

Gabor's holograms were not 3D. Among other things, he lacked a light source with sufficient coherence (e.g. a laser) to record 3D holograms. -eyliu 24 July 2006

I am removing the reference to the CLARO TV, since it has nothing to do with holography 84.154.234.43 16:48, 11 July 2006 (UTC)

"An alternate method to record holograms is to use a digital device like a CCD camera instead of a conventional photographic film. This approach is often called digital holography. In this case, the reconstruction process can be carried out by digital processing of the recorded hologram by a standard computer. A 3D image of the object can later be visualized on the computer screen."

The above statment is untrue. A CCD does not have the spatial frequency to record a hologram. The requirements for a transmission hologram are at least 700 lines per mm.

Actually, CCDs can and routinely are used to record holograms. It is true that high-resolution single-exposure off-axis holography can require a lot of spatial bandwidth to avoid aliasing. The spatial bandwidth requirements can be reduced by bandlimiting the object (by spatial filtering the scene, for example), at the price of reduced spatial resolution. Alternatively, one can record an on-axis hologram and use phase-shifting (instead of the off-axis technique) to suppress the conjugate image, essentially trading spatial bandwidth for temporal bandwidth. -eyliu 14 Mar 2006

Some recommended corrections:

"These versions of the rainbow transmission holograms are formed as surface relief patterns in a plastic film, and they incorporate a reflective aluminum coating which provides the light from "behind" to reconstruct their imagery. Another kind of common hologram (a Denisyuk hologram)is the true "white-light reflection hologram" which is made in such a way that the image is reconstructed naturally using light on the same side of the hologram as the viewer.SPM

I disagree with the premise that Denisyuk holograms are "true" white-light holograms and rainbow holograms are not. It is true that Denisyuk holograms have the potential to reproduce color more accurately than film holograms, but this is highly dependent on the experimental methods. As the thickness of the recording medium tends towards zero, Denisyuk holograms become equivalent to embossed rainbow holograms. -eyliu 14 Mar 2006

The following statement is not true:

"The coherence length of the beam determines the maximum depth the image can have."

It is actually fairly easy to make a hologram containing an image that is far deeper than the coherence length of the laser. The coherence length of the laser determines the maximum path length difference between the reference beam and the object beam in making a hologram, but different parts of a deep scene can be illuminated with different portions of a beam, with each portion delayed by an appropriate amount to ensure that the object beam from that part of the scene has traveled the same distance as the reference beam.SPM

While it is true that you can make deeper holograms than the coherence length of the laser the idea of "Multiple Coherence Volumes" is a bit too technical for this discussion. IMHO

Probably the simplest correction is to change "coherence length" to "coherence". The counterexample trades temporal coherence (coherence length) for spatial coherence (the counterexample requires that different parts of the beam be correlated, which is the definition of spatial coherence). Generalizing the coherence requirement closes that loophole. -eyliu 14 Mar 2006

Has an analog form of electronic holography ever been tried?203.26.37.35

This article is yet to mention Zebra Holograms, which can display "full colour" (RGB and all combination of them) and full parrelex. They are made up of what are called hogels (holographic elements, equivelet to pixels in three dimensions). Each hogel conatins a 1280 line RGB image with 1024 pixels per line. It is my understanding that each hogel is a two dimensinal holographic reprisentation of the entire image. Is my information correct?203.26.37.35

Contents 1 Are rainbows holographic? 2 Example 3 Real-time holography 4 pictures!!! 5 question 6 depth perception from phase or from stereoscopy ? 7 Specular reflection. 8 Do holograms record the phase? 9 Inconsistent due to bickering... 9.1 Farewell 10 some thoughts 11 Hologram store data in 3d space? 12 perspective, moving left/right and cutting film 13 Added to Hologram Recording section 13.1 Window analogy

[edit] Are rainbows holographic?

This is perhaps a naive question, or perhaps it has no answer. But I wonder, do rainbows qualify as free-standing holographs? (I suppose the 'hologram' or analog to it would be the rain drop or whatever is bending the light.) oneismany 16:59, 26 February 2006 (UTC)

No, the drops in rainbows act like little prismas, please see rainbow. --danh 13:37, 27 February 2006 (UTC)

Yes, rainbows are "holographic" to the extent that the entire rainbow image is contained within each spot on the film plane (i.e. within each raindrop.) On the other hand, one could argue that rainbows are not reconstructed via diffraction effects, therefore they are not holograms. But rainbow holograms do not employ diffraction either. In rainbow holograms the spacing of the interference fringes can only control the position of the colored bands, and has no effect on the 3D structure of the reconstructed image. --Wjbeaty 02:50, 29 March 2006 (UTC)

So is there any isomorphism between prismatic effects and holographic effects? E.g., a description of orthoscopic imagery composed of light that is equally true of holographs and prism effects? oneismany 21:13, 28 February 2006 (UTC)

On a related note, can a holographic prism refract light? Could holographic raindrops produce a rainbow? If they could, would the rainbow be a real rainbow or a holographic rainbow? Would a holographic rainbow be stationary like a holograph, or would it vary according to the position of the viewer like a real rainbow? oneismany 18:42, 2 March 2006 (UTC)

Oneismany, I'm not sure what you mean by a "holographic prism", but if you mean the image of a prism in a hologram (a holograph is a handwritten document), then, no. The image of a prism in a hologram is no more physical than the image of a prism in a mirror, and it doesn't affect the light or matter around it.

But if you mean, "can a hologram be made to act like a prism?", then, yes. A hologram is basically a complex diffraction grating, and a diffraction grating will bend different colors of light at different angles, like a prism.

-- The Photon 04:43, 3 March 2006 (UTC)

I would add one caveat to the comparison of refractive optics (such as a prism) and diffractive optics (such as a grating). The monochromatic behavior may appear similar, but the polychromatic behavior is generally different. For example, refraction angle generally decreases with increasing wavelength, but diffraction angle increases with increasing wavelength. -eyliu 14 Mar 2006

[edit] Example

This article would benefit from at least one example image. --SparqMan 05:11, 4 March 2006 (UTC)

[edit] Real-time holography

I recommend that the section on real-time holography be deleted. The principles of real-time holography are no different than conventional two-step (recording and reconstruction) holography and the concept emerges naturally from the use of a recording mechanism (the photorefractive effect, for example) which does not require photographic developing. -eyliu 14 Mar 2006

May I suggest that this section be retained for the following reasons: 1. Real-time holography (RTH) is a fundamentally different physical process than "conventional" holography, in that film is replaced by a nonlinear optical material, a spatial light modulator, etc. 2. The notion of RTH provides one with a very elegant way to compare/connect nonlinear optical interactions with that of conventional holography, except, that, in the the case of RTH, ALL the beams can interact essentially simultaneously. 3. The notion of ALL beams interacting at the same time has no obvious connection with conventional holography; i.e., why would one (and, for that matter, how could one) have all beams interacting in a given material at the same time. 4. Granted, some of the "formality" of conventional holography has parallels to RTH, but, in fact, it is precisely that parallel connection that provides a heuristic way to relate holography with nonlinear optics. 5. When the connection was made in the early days of phase conjugation, it provided a means by which to enlighten the research community to consider how other classes of NLO interations can be viewed as a RTH picutre: from "conventional" NLO processes (i.e., a third-order nonlinear polarization) to optical coherence processes (e.g., photon echoes, etc.) to simulated scattering processes (SBS, SRS, etc.), to novel classes of materials such as photorefractives, which involve a combination (multi-step) optical processes (involvong space-charge fields that, via the E-O effect, modulate the refractive index spatially, etc.). So, for purely historical reasons alone, this connection is very important in the evolution of the field. I speak from personal experience, being one of the initial group of researchers that explored this field... LASERMAN; APRIL 20, 2006

[edit] pictures!!!

bla bla ,I scrolled all the way down and..NOT A SINGLE IMAGE??

one image is worth a 1000 words..

WHATS DOES IT LOOK LIKE ? --Procrastinating@^talk2me 14:34, 26 March 2006 (UTC)

Two-dimensional photographs do a poor job of illustrating the three-dimensionality of holograms. Also, many photographs are copyrighted. -eyliu 31 Mar 2006

[edit] question

okay, I read almost all of that page, but I didn't understand any of it, so I don't know if the information I was looking for was there or not. However, I really want an answer to my question, so I'm going to ask it despite the risk of being repetitive. I am reading about Holomovment and the theory is associated with holography. I read that if a piece of holographic film is cut in half, both pieces will still contain all of the original data. How is this possible?

it may work kind of like a mirror, using reflection off the bumpmap to view a 3D image, or like this thing http://www.wisinfo.com.tw/wistek/mirage/mirage_model_22__gigantic_3d_hol.htm , so like a mirror, you would still see the image but it would be smaller, thats my theory i bet its wrong though. but i also have a Q, could you make the hologram larger than the orginal object? 24.3.56.115 15:53, 16 May 2006 (UTC)

HERE'S ONE WAY TO LOOK AT THIS RATHER CONFUSING NOTION (NAMELY, WHY DOES ONE SEE A RECONSTRUCTED HOLOGRAM FROM PIECES OF THE ORIGINAL HOLOGRAM?) If one assumes an object with a diffusely scattering surface (that is, an object whose surface features are not all specular [mirror-like], but, instead, scatters the "reflected light into a large cone of angles), then, the laser light that illuminates the object will scatter back toward the film plane (or, other static or dynamic holographic recording medium) and will, in essence completely illuminate the film. Since each fearute on the object has microscopic surface imperfections (recall, that it is assume to not be "mirror-like"), then each feature will diffusely scatter the incident laser light into a large range of angles, "filling" the film plane. By this line of reasoning, all the features on the object's surface will scatter the incident light across the entire film plane. Thus, each piece of the film will have information from the entire surface of the object. Now, the reference laser beam illuminates the film plane also. This combination of beams forms a spatially complex set of "gratings," or, interference patterns, across the film plane. Hence, at each "patch" on the film plane, information from all features of the surface will coherently combine with the reference beam. The "price" one pays for looking at the hologram using only a piece of the original hologram is that the spatial resolution is degraded; that is, the sharpness of the object's detailed features, edges, etc will become blurred, as if it slightly out of focus. The degradation results from diffractive spreading of the reconstruction beam that diffracts from the complex grating formed in the piece of the larger hologram. The diffraction spreads the readout beam (given by the ratio of the optial wavelength divided by the scale size of the piece); the entire hologram has the greatest lateral dimension, hence the least diffractive spread, whereas pieces of the hologram result in greater spreading of the readout beam, thus, degradting the sharpness of the reconstructed image.

By the way, if one were to perfectly image the light scattered from the object onto the film plane, then, there will be a one-to-one "mapping" of each pixel of the object to a single location on the film plane. In this case, every element on the film plane contains only information which has been imaged there. So, if one breaks the developed hologram in this case, the pieces will each contain different aspects of the surface. This is why one typically lets the scattered light "spray" all over the film, and, may employ a simple lens system to help collect the light, but, not to image the light, onto the film plane.

Hope all this helps (and, hopefully, my take on this is correct!!)... GOOD QUESTION!! --- Laserman; June 13, 2006

The assertion that "all the information of the scene is contained in each bit of a hologram" should be completed with "if the scene permits". It is likely that the right side of the plate does not "see" the left side of an object (if this side is occulted by the object itself). A bit of hologram from this region will not allow to see the left side of the object. LPFR 06:58, 18 July 2006 (UTC)

Who really cares about which space television programs use a holography-like tool (end of 1st paragraph)? Maybe that should go.

[edit] depth perception from phase or from stereoscopy ?

From the article it seems that the 3D impression in holograms comes from the fact that they capture not only the wavelength (color) but also the phase of light (section 'Technical description'). This is contrasted with so-called "holograms" on identity documents, which achieve a 3D impression by stereoscopy, i.e. different apparent viewing angles of the two eyes (main section 'Holography'). I think this is probably incorrect, for two reasons. First, our eyes are to my knowledge not sensitive to phase, and certainly do not use it to reconstruct depth information. Second, when one changes the viewing angle of a true hologram (contrary to e.g. a photograph), one sees the depicted object at a different angle. This implies that when watching from a constant position, both eye are seeing the object from different angles because they have different vantage points. This would classify as stereoscopy, I'd think. I do not know the details of producing holograms and I'm sure the light's phase information plays a role, but I'd doubt if the perception of depth in the holographic image itself came from anything but stereopsis. 29 Aug 2006

• Right, the eye cannot not see the phase of the light. However, the fact that the phase of the light is recorded in the hologram means that when re-illuminatated by the reference beam, the original wavefronts from the object are (ideally) completely reconstructed. This means that as the viewpoint is changed, the object's image rotates just as the real object would (up to the limit of the field of view of the hologram, of course). Thus, with two viewpoints (stereoscopic eyes), the object appears in 3D. The light phase isn't responsible for the 3D appearance, but the recording of the phase allows the reconstruction of a 3D appearance. --Bob Mellish 19:47, 29 August 2006 (UTC)

Thanks for your response. So we agree on the role of phase in constructing holograms. I think the article on the other hand gives the impression that, rather than being a technical tool that allows one to reconstruct the light coming from the source, phase plays a role in the appearence of the holographic image itself, particularly its depth. Consider for instance: 'both the amplitude and the phase of the light (usually at one particular wavelength) are recorded. When reconstructed, the resulting light field is identical to that which emanated from the original scene, giving a perfect three-dimensional image.' and, regarding false holograms: 'All depth disappears if you turn the hologram 90° or if you look at it with just an eye. This is not the case with true holograms, which are not based on binocular vision'. The depth in true holograms does disappear when looking with just one eye (unless you compensate by moving it) because their depth illusion is based on binocular vision. Right? Thanks, Jan.

A hologram does not record the phase. There is no way to record the phase of light. The retina, photographic emulsion, photodiode, photomultiplier, etc. are just sensible to intensity (power per unit surface). The only way to measure phase is to add the wave to be measured and the reference wave and measure the intensity. But this measure does not even give directly the phase. As the phase cannot be recorded, what a hologram does is to record the place where the phase is "good" as I tried to show in "working principle...".

What you get with a stereoscopic pair is very different of what you get wiht a true hologram. Assume that in a scene there is an object F in the foreground that partially hides an object B in the background. In a stereoscopic pair, when you shift your head "to see what is under the object F", the object F will follow the movement of your head and what was hidden will remain hidden and what was visible will remain visible. Simultaneously the 3D deep seems to flatten. Buy a 3D comic: the experience is worth the bucks. This is not the case with true holograms. When you shift your head, hidden zones appear and others are hidden. You can see a series of pictures of a true hologram in this site: http://hyperphysics.phy-astr.gsu.edu/hbase/hframe.html LPFR 08:45, 1 September 2006 (UTC)

[edit] Specular reflection.

I have removed the following from the specular reflection article. Perhaps it would be a better fit here. Specular reflection may be important for holography, but holography is not really that relevant to specular reflection.--Srleffler 04:13, 6 September 2006 (UTC)

Specular reflection is very important for making good scratch holograms, which are optically similar to Benton Rainbow Holograms (AKA: "White Light Holograms"); see also SPIE article and the FAQ, and the main Wikipedia Holography entry.

[edit] Do holograms record the phase?

The assertion by Danh that both the intensity and the phase relative to a reference beam are recorded is an old, hard-to-die, idée reçue. I imagine that this is because it is an easy answer, given to annoying questions, by teachers who had not understood how a hologram works.
Holograms do not record the phase as there is no means to record a phase. All light sensitive surfaces or devices (retina, photographic emulsion, photodiode, photomultiplier, etc) are sensible to intensity of light. Phase is just a difference in timing between two waves. You cannot record these differences in time with a photosensitive surface or device. The only thing you can do is transform these differences of phase in amplitude, adding the two waves. But even then, the only thing you get is something like . You get something related to de modulus of the phase difference (the sign of the difference is lost). However, as the squared cosine is multiplied by the amplitude, you cannot know if a zero is due to phase difference of 180° or to a zero of amplitude.
An hologram is just a diffraction image recorded on a photosensitive support. Would the same teacher as above say that the diffraction image of Young slits or Fresnel biprism has recorded the amplitude and the phase? No, because now the teacher has really understood the problem and does not need evasive (and incorrect) answers. If holograms recorded the phase, so would all diffraction images, including current photographic images that are Fraunhofer diffraction patterns of incoherently illuminated scenes.
User Danh also wrote When reconstructed by illuminating with the appropriate light, the resulting light field is identical (up to a constant phase shift invisible to our eyes) to that which emanated from the original scene. This is wrong. Neither the amplitude nor the phase, are the same. The phase of light exiting the hologram is identical to the phase of the illuminating beam (only the amplitude is modulated by the hologram). This is not the case when recording the hologram, where the phase is the result of the sum on the reference beam plus the light coming from the scene. Another proof that the reconstructed light is not identical to the original, is the formation of real (in the optical meaning) images in the wrong side of the hologram. This is the reason why the normal illumination is seldom used. The more common "of-axis holograms" separate real from virtual images. Real images are always there but not in the same line of sight. As I wrote (and Danh "corrected"), the reconstructed wave front are similar to the originals, but they are far from identical.
I tried to explain the working principle of holograms in the paragraph so named (the two images are computed, not draw). I am afraid that I did not succeed.
I will not restore what Danh wrongly "corrected". This is the inconvenient of wikipedia. Anyone who thinks to know the truth can "correct" you without any verification or reading of a good book (as Jenkins & White "Fundamentals of optics" 4th ed. McGraw-Hill 1981 for example). I will not enter a guerrilla of correcting corrections. LPFR 12:39, 17 September 2006 (UTC)

Hi LPFR. As I stated in the edit summary, your correction were technically absolutely correct. Just, this is the first section that explains the working of a hologram. It should do it as simply as possible, so that also an interested layperson gets out some information. Sorting out the details here makes it only more difficult to understand, for those that do not know it yet, what a hologram is. A better location would be a bit down in the section "Hologram engraving" or in "Holographic recording process", which BTW duplicate each other and should IMO be merged.

For the things you call "wrong", they are mostly compromises between rigorosity and usability for explaining. Of course Newton's second law is wrong, and perhaps even general relativity. But they are nevertheless very useful for understanding.

So let's resolve the things, if I make an edit, it doesn't mean you are not welcome any more to contribute. a) The phase: I think we agree that the important thing is only the relative phase, this lets you see in 3D. And the relative phase of two points you get looking at a hologram is similar to that of the real image. So what do you think about changing both the intensity and the phase relative to a reference beam are recorded to a more general both the intensity and some phase information are recorded? Keep in mind that "recorded" is intended in the sense "can be obtained back", not that it is physically present unencoded in the hologram. b) Identical: this was the wording of before and I don't like it either. Proportional (not really correct)? Simililar I wouldn't recommend, because it would include also distorted or morphed images. Anyway, the meaning should be something like "retaining the intensity proportions and the relative phase". Suggestion from anybody? --danh 22:49, 18 September 2006 (UTC)

I think I'll venture a comment, although this is a fiendishly difficult question where it is easy to get things mixed up. I tend to agree with Danh and disagree with LPFR, but here are a couple of important things I think we all agree on: a) The phase is not, and cannot ever be, recorded. Only the intensity of a wavefield is recorded. If I talk about the phase being recorded, I really mean that it can be recovered. b) The "absolute value" of a phase is not important, only relative phases need be considered. One can always change the origin and get whatever absolute value one wishes. When I talk about the phase, I am referring to the relative phase.

With this in mind, I believe that the wavefield exiting the hologram after reconstruction does contain the phase of the original object wave. This following is adapted from A. Tonomura's book on electron holography (Springer-Verlag 1993): We record the interference pattern from a refrence wave φ_r and the wave scattered from an object (the object wave) φ_O. These waves have, of course, both amplitude and phase, but the recorded interference pattern has only the intensity I = | φ_O + φ_r | ², recorded e.g. in a film. If this film is subjected to the (known) refrence wave, the transmitted wave will be

As we see, the two last terms contain the original object wave and its conjugate. These can be separated in off axis holography. Since the refrence wave is known, we now have reconstructed the original object wave, including its phase. If you read Gabor's original paper (Nature, vol 161 pp. 777-778, 1948) this is as he intended: "the new principle provides a complete record of amplitudes and phases in one diagram". So, in conclusion, I believe that the wavefront of the original object, including its phase, is recovered when the hologram is illuminated by the refrence wave. O. Prytz 14:32, 19 September 2006 (UTC)

Hi DANH. There is a frontier between deep simplifications and false statements. Second Newton's Law is not rigorously exact but it is not wrong. "Holograms record the phase" is wrong, and not because of lack of precision but just it is untrue. I do not think that propagating false "idées reçues" as explanations is a good thing. If you cannot respect the verity when giving an explanation, it is better not to give an explanation at all. I wrote "but some information related to phase is recorded", this is as far as you can go respecting the truth. When I added the paragraph "working principle ...", I did not erased or corrected the paragraph "technical description" - I do not like to modify other's work. I hoped that the author would correct it himself. As this did not happen, I corrected the text, trying to conserve as much as I could of the old one.

At the right, you can see the reconstructed wave fronts from the hologram. As you can see, they have nothing to do with the original wave fronts. When I wrote "similar" this was as far as I could go. When you write "identical", you do not know what you are talking about. You can see the formation of the real image and, at the far right, the wave fronts of the lecture beam, but the wave fronts of the virtual image are not even visible.

You seem to think that the most important thing is phase. NO, it is not. If it where so, just a tilt or a bend in a hologram would ruin the reconstructed image. The angle between the hologram and the reconstruction beam do not need to be exactly the same as the original. If the holograms deforms, you get still an image. The phrase "both the intensity and some phase information are recorded?" is still too far from reality. What an hologram record is the places where the phases of the light from the scene and the light from the reference beam where near. And this record of places is modulated by the amplitude. You do not record both of them. You record just the product of them, as I wrote. The "catch" in the holograms is not the phase but the place where the phase "was OK". This record allows - in the reconstruction process - to create wave fronts that are very different (not identical, not proportional) to the original wave fronts. However, they create a virtual image of the scene plus a bothering real one. LPFR 14:49, 19 September 2006 (UTC)

Hi O. Prytz. You do not need a formula to know the phase of light exiting the hologram. If the hologram is a classical photographic one, that is, and amplitude-modulating plate, the only thing the hologram does to the light is modulate its amplitude. Then the phase is the same as if the hologram where not there. The phase is that of the illuminating beam. No formula can change this fact. This demonstrates that you do not get the original phase. LPFR 15:28, 19 September 2006 (UTC)

Ok, perhaps it's bit dangerous for me to venture into this, as it's been about 10 years since I've studied holography, but I think you're getting a bit semantic here, LPFR. I could say the same thing about an amplitude grating: it only modulates amplitude leaving the phase unchanged. But then how does the effect of a grating manage to look so much like a linear phase? Well, I think the answer is that a grating is like two linear phases in cojugate pairs, which is why an amplitude grating has at least two diffraction orders, one positive and one negative. (Which I think is analogous to the issue you guys have discussed above regarding the reasons behind off-axis holography.) But in the case of a grating (which can be considered a very simple hologram) if you restrict yourself to looking at just one order, the effect of the grating in the far field is indistinguishable from a linear phase applied to the illuminating field. So what I'm trying to say is this: (a) when you modulate the amplitude of a field, you ARE changing the far field phase, and (b) if you restrict yourself to the right diffraction order it is fair to say that the phase information was preserved. I think you're getting caught up in semantics by insisting people not say that phase information is recorded or that the field isn't perfectly reconstructed. Sure, there are complicating issues like the fact that the reconstruction also includes spurious fields, but if the phase wasn't recreated in the virtual image, holography wouldn't work. When you bend a holograph, it DOES mess with the image. It's just that bending shifts all the phases in a given region together, resulting in a distortion but not complete destruction of the image. At least when it comes to monochromatic holography, i think it is very fair to say that the phase information is recorded. Recording just means the information is there, and says nothing about reconstruction or accessibility. I'm interested in your response to this. Birge 19:19, 19 September 2006 (UTC)

LPFG, here you talk about the absolute phase of the reconstructed wave. O. Prytz was talking about the relative phase between two points in the hologram. This is what we want (and for this a hologram with one point cannot tell anything). --danh 20:14, 19 September 2006 (UTC)

LPFG, most of your objections come your interpretation of the term "recording", which is different to that of the original author, O. Prytz's and mine. So, let's change its meaning to your stricter one, like in "In a hologram, information from both the intensity and the phase is recorded." or use a totally different word.

Anticipating further comments, with "reconstructed hologram" we refer to just the virtual image, not the real image term, the transmitted or the halo terms, which are always present too. I think it's acceptable. For a homogeneous intensity of the recording reference beam, the same reading reference beam as the recording one, a linear recording material, this term is (up to a constant factor) identical to the original image. But tell this to someone who wants to know what a hologram is :-). --danh 20:14, 19 September 2006 (UTC)

Hi,
I wish to remind the subject of this talk. It is about "How a hologram works" and about two assertions: "A hologram records the phase" and "The hologram reconstructs the wave fronts as they where at the recoding time"

First a wish to tell DANH that I am not as dumb as to confound relative and absolute phases. Incident closed.

To BIRGE I said that, for the simplification of the talk, I do not think a good idea to replace a diffraction grating by "two linear phases in conjugate pairs". The phrase "Recording just means the information is there, and says nothing about reconstruction or accessibility" reminds me an 30 years old hoax of a technical datasheet of a "write-only" memory. I do agree that a hologram is the same thing as a diffraction grating. The difference is that a hologram is not a periodic pattern, but the reconstruction process is the same. The sentences about folding a hologram where just to remind DANH that phase is not the most important thing in life.

I am no sure that you read what I wrote. What I said is that a hologram does not record the phase but the places where the phase of the light coming from the scene is near the same as the phase of the reference beam. In this way, in the reconstruction phase, only the light that has the "correct" phase is allowed to pass. This, I tried to demonstrate in the paragraph "Working principle ..." with the images I computed. Secondly, I maintain that the hologram does not reconstruct the wave front as they where at the recording time. I demonstrate this in three English sentences (see above, my answer to O Prytz). I computed the wave fronts created by the hologram (see image above) to convince you that the wave fronts are very different from the original ones. Is this not suffice as a demonstration, I am afraid that I cannot do any more to convince you. In science, when someone gives a demonstration, either you accept it or you demonstrate that the demonstration is erroneous. LPFR 12:41, 20 September 2006 (UTC)

I think you talk about different things. Birge refers to the virtual image (which has the same phase as the original image, up to a constant shift), while your demonstration contains all other terms too. If you subtract the real image (nicely visible in front of the holographic plate), the transmitted plane wave and the halo term, you recieve the original wave (up to a constant phase shift and a reduced intensity). --danh 14:32, 20 September 2006 (UTC)

Hi, LPFR. Thanks for the response. I think I see one point of contention. When you talk of the hologram "recording where the phase is correct" I understand what you're saying, and agree. But that's only a *binary* hologram, where the material response is either on or off. And in that case, I agree that there will be significant differences, for the same reason a square wave introduces harmonics not present in the sinewave it approximates. But in a continuous hologram, there is a lot more being recorded that just "where the phase is correct". You're getting an analog recording of *how much* the phase is "correct" or not. And the bottom line is that if you're able to isolate the right diffraction order you get exactly what you put in. So, is what you object to that we are ignoring the other components in the reconstructed field, or is it that you disagree with the statement that within the correct diffraction order we exactly reproduce the phase and amplitude of the writing field? I agree that science must rely on demonstration, but I think we're still sorting out where we differ. I'm not sure we disagree as much as we misunderstand each other. Birge 18:21, 20 September 2006 (UTC)

I thought you where old enough to understand the phrase "where the phase is correct" as a shortcut for , as I wrote above a few times. Sorry. LPFR 18:45, 20 September 2006 (UTC)

Mon dieu! Il n'y a pas raison pour que vous soyez si français. It just seemed like you were suggesting information was ALWAYS irretrievably lost because of the translation into a pure amplitude modulation, and it's not always the case. That was my reason for the simple example of a sinusoidal amplitude grating (and whether you like it or not, such a grating does produce two new plane waves with conjugate linear phases relative to the illuminating field). Personally, I think that's a pretty good example of how an amplitude modulation can look just like a phase modulation under certain restrictions. It's entirely anologous to how a signal can be entirely reconstructed from the amplitude of its fourier transform if you are able to make certain assumptions about its support in the time domain. You seem caught up in the fact that it's POSSIBLE for the reconstructed hologram to include spurious fields (as in your illustration) but your illustration is not a correctly done hologram that anybody would ever use. Again, though, I think we all pretty much understand what's going on, and I think we need to agree on how to best explain holography while making it clear that the real situation is complex and care must be taken for the hologram to truly reproduce the original field, and only at certain angles, etc. Would you be happier if we said something along these lines? I agree with you that we shouldn't state things simpler than they really are, but on the other hand I don't think we can go into academic detail about exactly which conditions must be satisfied for the holographic reproduction to equal the original field. Birge 21:08, 20 September 2006 (UTC)

Hi Birge. Usually French physicists favor formulas instead explanations and understanding. I hope it is no my case. First, there is a big difference between the interference pattern of a periodic structure and a non-periodic one. In a periodic structure, as a grating, the number and the amplitude of primary, secondary, etc. depends on the "shape" (with, binary, sinusoidal, etc.) of the slits. I am not sure (that is, I cannot assert) if the diffraction pattern of a non-periodic structure depends also on the "shape". I guess that it is not the case. The case of a sinusoidal grating is a particularly very well behaved one.

Long time ago, I read that you can make diffraction gratings and holograms that just modify the phase and are 100% transparent. The result is about the same as with amplitude gratings or holograms. The way to do this is to use light sensitive gelatin that swells with light. The drawback is that they are not dimensionally stable. Let us come back to the main talk.

I would really like that you read the paragraph "Working principle ... ". I think that I explained the working principle simplifying as much as decency allows but with all the caveats about the explanation and stating that this type of thin hologram with on-axis illumination is not of practical use. If I missed other important caveats, just add them.

Your sentence "for the hologram to truly reproduce the original field" shows that you still believe that a hologram MUST reproduce the original field. A Hologram doest not reproduce the original field. At least, not truly. The demonstration is:

If the hologram is a classical photographic one, that is, and amplitude-modulating plate, the only thing the hologram does to the light is modulate its amplitude. Then the phase is the same as if the hologram where not there. The phase is that of the illuminating beam. In the same place, the phase of the original field was the result of the addition of the illuminating beam and the light coming from the scene and diferent from the phases of the reference beam. Then, the phases are not reproduced as they where. This is valid for any amplitude modulating hologram and not only for thin on-axis holograms.

If the phase at the hologram level are not the same, the far field or the interference pattern can not be the same. This comes from Huygens principle (wavelets). Then, amplitude holograms DO NOT truly reproduce the original field.

But it is not necessary to "truly reproduce the original field" to obtain something related to the original image. When you look (just with one eye!) a picture, you see the same thing as the original scene, and the light fields are different (yes, I know about color. Just illuminate the scene and the picture with monochromatic light).

As I said, in science either you accept a demonstration or you demonstrate that it is erroneous. If you think that my demonstration is false, just prove it.

As for a reasonable simplified explanation of how works an hologram I think that you could say: As phase cannot be recorded, the hologram record the places where the phase of the light coming from of the scene is near (squared cosine if you prefer) the phase of the reference beam. These places are favored by the amplitude of light coming from the scene and by the similitude of phases. When observing the hologram, the places where the phase was the near that of the illuminating beam transmit more light than the others do. This does not reproduce exactly the originals wave fronts (or light field), but is enough to create wave fronts similar to the original. I do not mind exact wording, as long as there are not assertions like the phase is recorded or the light fields are exactly reproduced. LPFR 09:26, 21 September 2006 (UTC)

Hi, LPFR. I would agree with those changes, and I think you should feel free to make them (I don't want to take credit for your work). For the record, I don't think you ever got my point that the hologram CAN exactly reproduce the writing field under certain restrictions, and therefore that ALL of the phase information was preserved modulo a constant. But I also don't disagree with your wording because, in general, you are right that the reconstructed field is not the same. Birge 12:25, 25 September 2006 (UTC)

Birge, it is also my understanding that the virtual image (the third term in the rhs of the equation of O. Prytz) contains all the phase information (modulo a constant).
LPFG, I like your changes, but I have a few comments:
in The hologram reproduces the original light field accurately: FALSE you treat holograms as synonym of amplitude holograms. I think phase holograms are much more important than amplitude holograms (at least in science), so I would welcome if you included them too. The second thing is that I don't really agree with the statement :-). I mean, O. Prytz's equation demonstrates that the virtual term is proportional to the original light field φ_O if the intensity of the reference wave is homogeneous. Do you agree with that equation (up to a complex constant)? Do you call all 4 terms as the hologram or does just the virtual image qualify too? --danh 04:37, 26 September 2006 (UTC)

Hi Birge. I posted my last paragraph (True or false?) at the same time as you wrote your comment in this talk. Anyway, as all this talk does not suffices to convince physicists like Danh, I think that we can keep it as it is, at least for a time.

To Danh I say that I read the "O. Prytz's equation" as "The light exiting the hologram has the same phase as the illuminating light , and its amplitude is modulated by the transparency of the hologram ".

Yes, I did not read the rightmost expression, but as there is an equal sign, the meaning of the last expression must be the same as the meaning of the second one. The interpretation made by O. Prytz is erroneous. Writing an equation in a different form does not allow you to extract terms that are not in the first one. I also think that "O. Prytz's equation" should be written with different names for the engraving and the reading reference beams. In this writing, the third expression is very different and O. Prytz's would not have been misled.

Yes, I do not consider other holograms than amplitude ones. The length of this talk proves that even amplitude holograms are not that simple to understand. When the explanations for amplitude holograms will be so clear that everybody understands them, we can consider adding explanations for other types.

I am almost sure that "phase holograms" are not the same thing for you and me. Would you please tell me what do you mean by "phase holograms" (and if possible, how you record them)? LPFR 08:22, 26 September 2006 (UTC)

Ok, it seems we'll have to do this in some detail. The following is a more detailed examination of the recorded intensity which follows from simple complex calculus:

If we assume and , we can write the recorded intensity as:

The two first terms are just the intensities of the object and refrence waves, while the third term is the interference of these two and contains phase information. In fact, the recorded intensity must contain at least some phase information, think of diffraction: how else should we account for systematic extinction of reflections if not by knowing that there is a phase difference of π. But the genius of holography is that we treat one of the waves as known, and therefore we can fully retrieve the object wave:

which is my original statement from sept 19. I hope you are satisfied with the mathematical validity... I should certainly note that reconstructing the original wavefront exactly probably is impossible due to experimental difficulties (lens aberations if nothing else). But in principle it is certainly possible.O. Prytz 09:35, 26 September 2006 (UTC)

The term contain as much information about the phase as a Young slits diffraction picture. That is, you know when they are in phase or out of phase. But, as the term is multiplied by the amplitude, if you do not know the amplitude you do not know much about each of them.

If you see a Young interference pattern you can say where the waves where in phase or in opposition. However, when the diffraction pattern is not periodic, you cannot tell, given the light level in a point, what the phase or the amplitude was.

In your formula you are using the same beam to record and to read the hologram . This is the case when you use holograms built by non-linear effects for phase conjugate mirrors and the like. I suspect that your formula comes from a text about phase conjugate optics. In the holograms recorded on a sensitive surface, the two are not the same. Try to name one of them and the other and start again with your math. All the information that you can extract from your formula is already in the left term. You do not need lenses to record or observe a hologram? LPFR 15:12, 26 September 2006 (UTC)

So the difference is that LPFG talks about the whole wave , while I talk only about the virtual image .   has the phase of , while has the phase of . But in itself has no useful properties, the 3D image that looks exactly like the original image is the virtual image.

I find the first two "true or false" at least misleading, since, whatever you call a hologram, it yields a virtual image that can accurately reproduce the original light field (even with the relevant phase information).

About phase holograms: In the equation above, the transmission function of the hologram is set to be the intensity . This is a very limited case. In general

T = H(x,y)φ_r,   with a complex transmission function. If is constant you have an amplitude hologram, while if is constant you have a phase hologram. You get phase holograms whenever the holographic material does not change the absorption in response to light intensity, but the refractive index. The table in the holography article displays the most common recording materials. --danh 17:04, 26 September 2006 (UTC)

Hi, Danh. I made a question about what phase holograms are for you, because I am not sure that you are aware that the information recorded in phase holograms is the (squared) amplitude of the light field and NOT the phase of the light field. You can spare the formulas. In physics, you do not need formulas to explain or understand. You need formulas to calculate.

I wrote the demonstration that the light field exiting the hologram is different from the original field. In science, either you accept a demonstration or you prove that the demonstration is erroneous.

You say that a part of creates the virtual image. Fine! We agree. This is what I wrote as "... similar enough to create front waves similar to the original..."

You wrote " has the phase of ". Yes, but if you are taking the formulas of O.Prytz you will see that this term appears added with others terms and there is no way to isolate . The two equations in each side of an equal sign contain the same information. You cannot have more information in one side and less in the other one. Math does not add physics.

I wrote to O.Prytz, that "his" formula was not appropriate for the holograms we are talking about, but for phase conjugate holograms. In "his" formula, he uses the same beam to write and to read the hologram. This is not the case when reading a previously recorded hologram. Instead of using , he should have written , as the write and the read reference beams are not the same (they are the same in conjugate phase devices). LPFR 11:36, 27 September 2006 (UTC)

Hi LPFR. The demonstration and the "=" sign apply to T and not to the virtual image. In normal 3D holograms (off-axis holograms) the virtual image is well separated from the other three terms (and therefore you can see it).

What we call "O. Prytz's equation" is the equation for normal holographic readout, not for phase conjugated readout. And adding a different readout wave than the recording wave just complicates things, without changing them much. The image is somehow distorted, but can be reproduced without distortions by the right wave. It's a bit like looking a photo with say blue light: the colours you see will just be different than what they should be. --danh 14:12, 27 September 2006 (UTC)

Let's stop calling it "O.Prytz's" equation, ok? :) It's not my equation, and since it in reality is the only equation used in this discussion, simply "the equation" should suffice.

You are quite right, LPFR, that math does not add more/new physics. However, it can help us understand the physics that is already there, without having to resorting to the vagaries of ordinary language. In this case, if you accept the left side of the equation you must also accept the right side, given that my derivation is correct (I make no guarantees, please check the math).

I do not understand why you maintain that the refrence and reconstruction waves cannot be the same. There might be something I'm missing, but could we not for example use the same set ut that was used to record the hologram, and just block the object wave before it reaches the previously exposed film? Anyway, I believe that using a different (known) wave would just result in a predictable distortion of the image (e.g. magification).

What I believe we have demonstrated (mathematically) is that the reconstructed wave consists of three parts: the object wave (multiplied by a constant) , the conjugate of the object wave (multiplied with a constant and with its phase shifted I guess, "the conjugate image") , and the refrence beam multiplied by a constant .

Now, as danh states, in off-axis holography the virtual and conjugate images can be separated, so this should not cause a problem. However, I must confess that I don't know how one usually handles the fact that you have a contribution to the wavefront from the refrence beam (multiplied by a constant). (I'm an electron microscopist, and I believe that in nowadays in electron holography one rarely reconstructs the image with a physical refrence wave. It's usually done digitally, and since the refrence wave is know it can be subtracted?) In this light LPFR may be correct that the reconstructed wavefront is not equal to the original object wave, since it contains at least two contribution, only one of which is the original object wave. I will, however, maintain that the original phase is retained in the virtual image. O. Prytz 16:38, 27 September 2006 (UTC)

Hi O.Prytz and Danh. The reference and reconstruction beams are not the same because they are not coherent. Even if you use the same experimental setup, for the beams to be the same would ask for a coherence time of a few hours (needed to develop the film) or a few years if you examine the hologram later. From a mathematical point of view, the term that would be different in "the equation" is . If the two are coherent, the mean value will be non-zero as . If they are incoherent, the mean value is zero as . As you know, holograms can be read with beams that do not have the same wavelength as the recording beam. Then, "the equation" should work also for different beams. Rework "the equation" with and and you will see that you cannot extract more information from the right side of the "=" sign than from the left side.

I do not understand what O.Prytz mean by "the original phase is retained in the virtual image". Neither have I seen the link with my assertion about light field issuing the hologram.

I wrote the demonstration that the phase of light issuing the hologram is different from the phase at the recording time (I do not care if you fraction this light in a number of components). Either you accept my demonstration as true, or you prove that my demonstration is false. Until you prove that my reasoning has a glitch, my demonstration holds. LPFR 08:52, 28 September 2006 (UTC)

The reference and reconstruction beams don't have to be coherent relative to each other since they never appear together. They just have to be close enough in frequency and fairly monochromatic. The absolute phase of light never matters in linear optics. Birge 16:21, 28 September 2006 (UTC)

[edit] Inconsistent due to bickering...

Due to the arguments taking place, this article presently contradicts itself about phase. And someone might want to correct errors like "Holograms record de phase" while they're at it. I don't really want to venture into editing because I don't know enough about the rest of the subject. Queerwiki 02:23, 28 September 2006 (UTC)

Well, I did, and this gave raise to this (very long) argument. LPFR 08:52, 28 September 2006 (UTC)

Okay, well then let's a) get in some hard citations here, and b) put a dispute notice up until we've resolved everything. Oh, and I will correct c) "de" is not an English word!!! Queerwiki 15:59, 28 September 2006 (UTC)

Most of the bickering above comes from one of two things: how to interpret the verb record, and not understanding if the other is talking about the "virtual holographic image" or the whole wave transmitted by the hologram.

I think the record thing is now solved. Using it in the sense of "stores in a coded form" or "allows retrieving" is ok.

And I hope also the other will not be confused anymore. The "virtual holographic image", , is a part of the whole wave transmitted by the hologram, . The "virtual holographic image" has the phase of , and has the phase of φ_r.

The inconsistency of the paragraphs Holograms record the phase: FALSE and The hologram reproduces the original light field accurately: FALSE. comes from the fact, that they speak only about the whole wave transmitted by the hologram . This reflects the viewpoint of (only) LPFR, that is the important wave. But the virtual image that possess these holographic properties, retaining the original light field and its phase (and thus its 3D appearance). And if one speaks of holographic image one certainly thinks of the virtual image and not of the whole wave . So the two statements above do not hold and both paragraphs should go away. Does anybody agree (or disagree)?

BTW, citations were asked:

"The unique characteristic of holography is the idea of recording the complete wave field, that is to say, both the phase and the amplitude of the light waves", Optical Holography: Principles, Techniques and Applications, p. 1, by P. Hariharan (readable on Google Books).

After describing photographs, "Holography also records the intensity distribution of a wavefront; in addition, the local propagation direction (or phase) is recorded through the process of optical interference.", Handbook of Optics, Vol.II, p.23.3, by OSA.

--danh 20:13, 28 September 2006 (UTC)

I agree with your statements above. I would also like to say that regardless of who is correct about the phase being reconstructed, I find the "true or false" section of this article inappropriate for an encyclopedia.

With regard to citations, I'll just repeat the one I gave earlier where Gabor talks of holography ("the new principle"): "the new principle provides a complete record of amplitudes and phases in one diagram". Nature, vol 161 pp. 777-778, 1948. There are tons of other citations where he and other pioneers talk of the object wave being "reconstructed", but I guess those are too vague to use in this discussion. O. Prytz 08:11, 29 September 2006 (UTC)

[edit] Farewell

As no one seems to share my ideas, I infer that it is people, as Danh, who wrote:

"In an hologram, both the intensity and the phase relative to a reference beam are recorded." (sic) and

"the resulting light field is identical (up to a constant phase shift invisible to our eyes) to that which emanated from the original scene" (sic),

who have really understood how a hologram works, and not me. In consequence, I am withdrawing all the explanations and demonstrations that wrote in the article, because they are in contradiction with Danh conceptions. Please be kind enough not to restore partially any of my writings. Farewell. LPFR 11:33, 30 September 2006 (UTC)

Thanks for removing the two contended paragraphs. The rest was an unprovoced decision. --danh 19:16, 30 September 2006 (UTC)

[edit] some thoughts

the way i was taught to think about holography was this: if you throw a stone into a pond waves ripple out from the center. if you throw two in then two circular wave patterns emerge. now if you take a "slice" across the surface where the waves meet you get a standing pattern. if you record that pattern you get a water-wave-hologram. if you place that in the water and then trow one stone in, the "slice" will bend, shape and redirect the waves from the first stone and recreate the pattern that the second stone would have created. in terms of holography the first stone is your normal light source when viewing the hologram and the second wave is the laser used to illuminate the scene.

interestingly what really blew my mind (and is used in airplane HUD setups) is that if you take a hologram of a lens the hologram will allow you to send and focus light through the virtual holographic lens. —The preceding unsigned comment was added by 82.108.42.194 (talk • contribs) 09:43, 22 November 2006 (UTC)

[edit] Hologram store data in 3d space?

well... AFAIK, hologram store image information in 2d. I mean, the interference information is still recorded on a surface and not in a volume, right? So why in the first paragraph in the holographic storage section it said "The advantage of this type of data storage is that the volume of the recording media is used instead of just the surface."?

There exist both types, but only with volume holograms you can record more data than on a magnetic hard disk, see Holographic Versatile Disc or InPhase Technologies. --danh 00:59, 25 November 2006 (UTC)

[edit] perspective, moving left/right and cutting film

When viewing a hologram, you can move and see the image as it looked from your new position.

However, my question is this: how far (left and right) can you move and still see the proper perspective? Is it the same distance as the size of the film used to record the hologram?

If that is the case, then, if I cut a piece of the film, I can still see the entire image, but what I loose is that I can no longer move left and right as far as I used to? Is that correct? Do you also loose resolution when the cut the film? Ariel. 01:04, 8 December 2006 (UTC)

Good question. For a recording process as shown in the article it is like you say, you cannot move as far to the right or to the left as you could before. You don't loose resolution other than that given by Fresnel diffraction (it's as if you see the object trough a window of the same dimension as the film). --danh 15:32, 10 December 2006 (UTC)

[edit] Added to Hologram Recording section

I added some explanation to the "Recording" section (after the diagram), as the existing explanation seemed a bit thin, especially for the lay person. Holograms are the kind of thing average people are curious about, as it seems amazing that a three-dimensional scene can be recorded on a (for all intents and purposes) two-dimensional medium; It's the kind of thing about which people exclaim, "How do they DO that?!?" Yet the existing explanation of the recording process is really not geared towards understanding by the average individual. I remembered reading that section a long time ago and coming away just as confused as when I started. Now that I have a better understanding of holograms and saw the section was still unchanged, I thought I would add something that I think would have originally helped me understand the concept better. Here's what I added:

"It is also important to note that these recorded fringes do not only directly represent their respective corresponding points in the space of a scene (the way each point on a photograph will only represent a single point in the scene being photographed). Rather, an individual section of even very small size on a hologram's surface contains enough information to reconstruct the entire original scene (within limits) as viewed through that point's perspective. This is possible because during holographic recording, each point on the hologram's surface is affected by light waves reflected from all points in the scene, rather than from just one point. It's as if, during recording, each point on the hologram's surface were an eye that could record everything it sees in any direction. After the hologram has been recorded, looking at a point in that hologram is like looking through one of those eyes.

To demonstrate this concept, you could cut out a small section of a recorded hologram; If you then view that cut-out section, you could still see most of the entire scene simply by shifting your viewpoint, the same way you would look outside from a small window in your house, for example."

This may still not be the best way to describe the concept to the lay person, not to mention I sacrificed some technical accuracy in exchange for easy reading, but I feel at least some kind of similar description is needed. I'm fairly new at this, this addition being my first major edit of an article, and I didn't discuss the addition before I posted it in the article; so I apologize if that's contrary to general practice here (I hope someone will tell me whether or not it is). If anyone has any comments or suggestions regarding what I added please let me know.

Thanks.

Equazcion 23:09, 18 December 2006 (UTC)

Hi Equazcion. Thanks for your additions, they are very welcome. It is fine to edit the article directly. Only for very controversial articles (like Israel or Bush...) I would first post to the discussion page. I hope we will see many more edits from you. --danh 23:13, 19 December 2006 (UTC)

Danh, I just noticed your use of the "window" analogy above in your response to Ariel. I honestly didn't see that before I wrote my addition; but nevertheless you did come up with it first, so I thought I should acknowledge that. I don't want you to think I'm trying to take credit for your words.  :-)
Equazcion 06:58, 21 December 2006 (UTC)

No problem. --danh 17:02, 21 December 2006 (UTC)

[edit] Window analogy

I've changed the window analogy wording back to the way it was, more or less, before Netjeff changed it. Here are the two versions:

• Original: To demonstrate this concept, you could cut out a small section of a recorded hologram; If you then view that cut-out section, you could still see most of the entire scene simply by shifting your viewpoint, the same way you could look outside in any direction from a small window in your house, for example.

• Netjeff: To demonstrate this concept, you could cut out a small section of a recorded hologram; you could still see most of the entire scene simply by shifting your viewpoint around the cut-out section. The effect is similar to the way you can see most of the outside world through a small window from inside your house.

While I agree that my wording could use some work, I think Netjeff's is worse (no offense), especially because it makes it sound like we're telling people to look at the hole in the cut-out hologram, rather than look at the cutout itself. If anyone has any thoughts on this or an idea for how to better word the whole thing please let me know. Equazcion 02:12, 17 February 2007 (UTC)

Now that Equazcion points out the possible confusion about vieiwing the hole, I agree that my edit might be no better. No offense taken. I think the difficulty is somewhat caused by the future/passive voice like "you could". How about this version?

• If you have a recorded hologram you don't mind damaging, you can demonstrate this concept by cutting out and viewing just a small piece; you can still see most of the entire scene simply by shifting your viewpoint around the small piece. The effect is similar to the way you can see most of the outside world through a small window from inside your house.

Does the more active voice help with the example? Is the style still appropriately encyclopedic? netjeff 01:48, 18 February 2007 (UTC)

If we already are at it, let's add a few details. What about the following? --Danh 01:29, 19 February 2007 (UTC)

To demonstrate this concept, you could cut out and look at a small section of a recorded hologram; from the same distance you see less than before, but you can still see the entire scene by shifting your viewpoint laterally or by going very near to the hologram, the same way you could look outside in any direction from a small window in your house. What you loose is the ability to see the objects from many directions, as you are forced to stay behind the small window.

Needs to be edited for encyclopedic tone though, Wikipedia:Manual of Style#Avoid second-person pronouns. Femto 12:53, 19 February 2007 (UTC)

I think Danh's version is perfect. It's written in the second person because this is an example to illustrate a point and make it easier to understand, rather than explaining the facts, which we've already done, correctly and in the third person. Equazcion 13:17, 21 February 2007 (UTC)

Then my job here is done. :-) netjeff 23:39, 21 February 2007 (UTC)