After appending my 2019-04-30 update to my much older answer, the above excellent, to-the-point and from-the-trenches expert answer came in. I immediately changed the designated answer from mine to the new one. There's probably still some fun reading below in my old reply and update, though... :)

Alas, I must answer my own question: I found a very explicit example online description of someone who created a thick-film transmission hologram of a convex mirror. She (or he) describes seeing her own face clearly, even if only in monochrome. So, if I accept this description at face value, it clearly *is* possible to create a realistic mirror using only wave-exclusion diffraction effects. Cool!

Also, I am amused (or is it chagrined?) that this reminded me of the importance of reading long articles *all* the way to the end, even if you feel you already got the point. This description of an actual holographic mirror was hidden at the very end of the long posting on I mentioned in my question about how transmission holograms cannot form mirrors.

## 2019-04-30 Update

As noted in the comments below, the above link to an explicit description of a holographic mirror unfortunately is no longer available, not even in Internet archives.

However, this draft book chapter PDF on reflection using Denisyuk transmission holograms seems to provide pretty good coverage of the issues.

Still, as I get older I find I like finding the simplest possible explanations of things. The simplest proof that true holograhic mirrors can exist is this: You can see your own face in a pool of calm water.

Why? Well, the reason why thick film holograms can reflect light at all is because *any* change in refractive index in a transparent medium creates an amplitude -- a *probability* -- for light to be reflected back in the direction in which it came. Metal mirrors are just extreme examples of this effect, since the Fermi surface electrons in metals create a nearly 100% probability that photons will be reflected.

The quantum mechanical details of reflections works in transparent materials are covered delightfully in my favorite Richard Feynman book, QED: The Strange Theory of Light and Matter. In addition to its relevance here for understanding what is possible with holograms, I recommend QED strongly to anyone interested in understanding just how utterly and completely weird quantum mechanics really is.

Feynman discusses how properly space layers of changes in refractive index can create a surface that, at least for certain frequencies, has a nearly 100% probability of reflecting light. A holographic mirror!

Finally, take a contemplative look at this image (or a real example from your kitchen) of a roll of very layers of Mylar film:

Nearly everyone has at sometime noticed at some level of consciousness how remarkably metallic such rolls look, almost like aluminum foil. That is because even though the distances between the film layers are not wave-coherent as they would in a photographic hologram, they do collectively reflect more and more light, until the surface looks remarkably metallic... which is to say, remarkably like a mirror.

Such a roll of Mylar film thus can plausibly be construed as a crude mechanically constructed hologram, and thus a proof that at least at some level of quality, transparent materials can indeed be configured to create plausibly effective, metallic-looking reflective mirrors.

The distance between the typical adjacent lines in a hologram is comparable to or longer than the wavelength of the light we use. After all, the lines arise from interference and the interference depends on the relative phase.

If you consider the distance of points H1, H2 from two generic points A, B and calculate the distances, the difference between H1-A and H1-B distances will differ from the difference between H2-A and H2-B by a distance comparable to the distance between H1 and H2 themselves. So the wave is imprinted in the hologram.

However, when the object we are visualizing is sufficiently far from the screen in the normal direction, the change of the phase will actually be much smaller which means that the lines on the photographic plates will be much further from each other than the wavelength. This should be known from double-slit experiments and diffraction gratings.

At most, you need the resolution of the hologram to exceed one pixel per the wavelength of the light. That's comparable to 0.5 microns. Invert it and you get 5,000 wave maxima per inch. That's close to the dots-per-inch resolution of some best printers.

However, the condition above is one for a really fine hologram. In reality, you can make a hologram even when its resolution is worse than that. Note that when we look at the hologram, in each direction we see the result of the interference of pretty much all the points on the screen - it's some kind of a Fourier transform. Because there are so many points that interfere, they can effectively reconstruct the subpixel structure of the image.

It's also a well-known fact that you may break a hologram into pieces and you may still see the whole object in each piece.

## Best Answer

I'll answer the first question: In-Axis holography refers to objects placed within or nearby an axis normal to the holographic plate. This is usually refered as Gabor's architecture. Off-Axis holography removes the problem of

ghost imagesplacing the object outside the normal. It was firstly introduced by Leith-Upatnieks. Let's first consider the holographic record of apoint source:Record of the hologram:Let $\psi_0$ represent the field distribution of the object wave in the same plane as the holographic plate. Similarly, let $\psi_r$ represents the reference wave. The holographic plate is sensible to incoming intensity, therefore its

amplitude transmittanceis:$$ t(x,y) \propto |\psi_0 +\psi_r|^2 $$

Let's consider an off-axis point source at distance $z_0$ from the holographic plate. According to Fresnel's diffraction, the object wave emerges from the point source and reaches the plate as a diverging spherical wave:

$$ \psi_0=\delta(x-x_0,y-y_0) \ast h(x,y;z_0)= \exp(-i k_0 z_0) \frac{ik_0}{2\pi z_0} \exp(-i k_0 [(x-x_0)^2+(y-y_0^2)] / 2 z_0) $$

where h(x,y;z) is a simplified

impulse response functionsolution to the Green problem for Fresnel diffraction. The simplification consist on taking the paraxial approximation and observing the field at a distance $z>>\lambda_0$.Let's $\psi_r$ be a plane wave whose phase is the same as the object wave at $z_0$. Therefore, the reference wave field distribution is $\psi_r= a \exp(-i k_0 z_0)$. Thus, intensity distribution recorded at the plate (

hologram's transmittance) is:\begin{equation} t(x,y) \propto |\psi_0 +\psi_r|^2 \end{equation}

$$ =\left|a + \frac{i k_0}{2 \pi z_0} \exp(-i k_0 \left[(x-x_0)^2+(y-y_0^2)\right] / 2 z_0)\right|^2\\ = a^2+\left(\frac{k_0}{2\pi z_0}\right)^2 + \frac{k_0}{2\pi z_0} \sin\left(\frac{k_0}{2 z_0} \left[(x-x_0)^2+(y-y_0^2)\right]\right)\\=FZP(x-x_0, y-y_0;z_0) $$

This functions are called sinusoidal Fresnel Zone Plates and they look like this:

The center of the FZP specifies

localizationof the point source: $x_0$ e $y_0$. Spatial variation of the FZP is governed by a sinusoidal function with a quadratic spatial dependence. Now let's place the point sourceon-axis: $x_0=y_0=0$, at some distance $z_0$ from the plate: $$ t(x,y) \propto |\psi_0^2 +\psi_r^2|^2 $$ $$ =\left|a + \frac{i k_0}{2 \pi z_0} \exp(-i k_0 (x^2+y^2) / 2 z_0)\right|^2 \\= a^2+\left(\frac{k_0}{2\pi z_0}\right)^2 + \frac{k_0}{2\pi z_0} \sin\left(\frac{k_0}{2 z_0} [(x^2+y^2)] \right)\\ =FZP(x, y;z_0) $$We can see how the holographic record presents a lens structure. Focal distance is parametrized by spatial frequency of incoming FZP. This systems records on a bi-dimensional surface differences between the FZP that conform our object (since any object can be decomposed as a collection of point sources).

Please note here is the answer to your question:

*Different FZP's have different spatial frequencies according to their distance to the plate. This information is encoded in the hologram and revealed upon illumination. The coding consist of a parametrization of different lenses (focal distances) according to the distance of the individual point-sources which make up the object.

Visualization of the hologramIn order to recover the hologram, we simply need to illuminate the plate with the same reference wave. Let's call it reconstruction wave, whose value at the plate is: $\psi_{rc}=a$ ($z_0=0$) Transmitted field distribution

afterthe hologram will be:$$\psi_{rc} t(x,y)=a t(x,y)$$

Finally, the field at distance $z$, according to Fresnel's diffraction:

$$ a t(x,y) \ast h(x,y;z) $$

$$ = a\left[ a^2+\left(\frac{k_0}{2\pi z_0}\right)^2 + \frac{k_0}{4 i\pi z_0} \left( e^{\frac{k_0}{2 z_0} [(x^2+y^2)]} - e^{-\frac{k_0}{2 z_0} [(x^2+y^2)]} \right) \right] \ast h(x,y;z) $$

Now we can take advantage of the fact that convolution is a linear operator and the field finally decomposes in the following terms:

zero order(This gives some noise)\begin{equation} a \left( a^2 + \left( \frac{k_0}{2\pi z_0}\right)^2\right) \ast h(x,y;z) \end{equation}

real image (pseudoscope)(reversed-phase object information)\begin{equation} \sim e^{-\frac{i k_0}{2 z_0} [(x^2+y^2)]} \ast h(x,y;z=z0)= \end{equation} $$ e^{\frac{i k_0}{2 z_0} [(x^2+y^2)]} \ast e^{- i k_0 z} \frac{i k_0}{2 \pi z_0} e^{\frac{-ik_0(x^2+y^2)}{2z_0}} \sim \delta(x,y) $$

virtual image (orthoscope)("in-phase" object information)\begin{equation} \sim e^{-\frac{i k_0}{2 z_0} [(x^2+y^2)]} \ast h(x,y;z=-z0)= \end{equation} $$ e^{\frac{-i k_0}{2 z_0} [(x^2+y^2)]} \ast e^{- i k_0 z} \frac{i k_0}{2 \pi z_0} e^{\frac{-ik_0(x^2+y^2)}{2z_0}} \sim \delta(x,y) $$

In the last term, we have propagated in inverse direction the field immediately anterior to the plate to demonstrate that a virtual image will emerge

behindthe hologram. This is the wave that preserves the orientation of the object's wave phase. The real part provides this information in reverse order, that is why it's called pseudoscope term. Off-axis architecture eliminates this placing object and plate at different directions.Regarding your

second question:The hologram is recorded by photosensible material and I think the question of resolutions is actually limited by minimum distance between

optical centers, those who contribute to the optical density of the material. Working on the linear regimen for this optical density is crucial to avoid over-exposures which can bring upghost imagesto the hologram. These are caused by distortions on the FZP, which no longer behave like sinusoids and create higher harmonics (rather that +1,0,-1 exclusively).