Fundamentals of Photonics Volume 1

[lobaz]

Not yet! There's no one who can make a million copies of a reflection hologram on photopolymer.

By the way, Dinesh, I think this is no longer true. Volume reflection security holograms are quite common today. Indeed, not as ubiquitous as embossed ones, but I have seen them many times. Although I am not sure, I bet they are in photopoymer (I have seen silver halide as well). I am also not sure if it was produced in a one million run, but difinitely in a high volume run. The only serious problem in producing volume security holograms is their cost - many times higher that embossed. This is why just a few customers order them.

[jrburns47]

It’s sometimes hard to keep in mind today the mindblowing novelty of laser viewable transmission holograms today as seen almost 60 years ago. The fact that a 2D image could be projected from any piece of a laser viewable hologram was truly unique and amazing. People who grew up in physics with holography as a standard part of their curriculum don’t get this uniqueness today at all.

[zipsister]

I'm in the process of working on ideas and practicalities for a large scale project where the projection of 2D laser transmission hologram images (as well as 3D laser transmission holograms) will be a focal point. This will happen after I arrive to live in France permanently in September. Still early days but I have begun buying multi coloured high powered laser for the projection part.

[Din]

Moreover, a hologram can be seen as a perfect, diffraction-limited record of an object.

Is it diffraction limited? As I understand, diffraction limitation is the property of an optical system, not a recording. But, if a hologram is a diffraction limited record, then, when I make a holographic lens, and reconstruct it with exactly the same recon wavefront as the recording wavefront, I should get a focused spot size of w0, the waist diameter, assuming that the laser has M² = 1 (which it never does for a real commercial laser). But, the limitation of the medium must surely limit the diffraction properties of the reconstruction? Usually, the reconstructed spot size is never w0. However, it doesn't matter much for display holography due to psychophysical reasons.

By the way, Dinesh, I think this is no longer true. Volume reflection security holograms are quite common today.

Do you have an example? I've never seen one. This is one problem with AR/VR HOEs. Holography is an elegant solution to the complex optics of AR/VR, and produces a very compact, light, head mounted device. But, the problem for commercialisation is mass production, ie taking it out of the lab.

[lobaz]

Is it diffraction limited? As I understand, diffraction limitation is the property of an optical system, not a recording. But, if a hologram is a diffraction limited record, then, when I make a holographic lens, and reconstruct it with exactly the same recon wavefront as the recording wavefront, I should get a focused spot size of w0, the waist diameter, assuming that the laser has M² = 1 (which it never does for a real commercial laser). But, the limitation of the medium must surely limit the diffraction properties of the reconstruction? Usually, the reconstructed spot size is never w0. However, it doesn't matter much for display holography due to psychophysical reasons.

You are right. "Diffraction limitation" is a property of an optical system. As an illuminated hologram reconstructs a diffraction-limited replica of the object wave, I used a shortcut and said that a hologram is a diffraction-limited recording. And indeed, this holds in theory only. Real recording materials and real light sources always introduce some aberration.
I think that it is OK to say a hologram is a diffraction limited case despite troubles with real world components. The same applies, e.g., to lens aberration theory as well. The classical lens aberration theory also expects a perfectly homogenous glass without any striae, perfectly smooth lens surface and so on. By the way: I think that "diffraction limit" comes from scalar wave theory. Is there any analogy to the diffraction limit in full EM theory and in quantum optics?

Do you have an example? I've never seen one. This is one problem with AR/VR HOEs. Holography is an elegant solution to the complex optics of AR/VR, and produces a very compact, light, head mounted device. But, the problem for commercialisation is mass production, ie taking it out of the lab.

Yes, I have some examples taken from commercial product packaging, but I have no idea who produced those. Quick search showed me that, e.g., Bowater Holographics or Krypten can do it. If you need more info on providers of volume holograms in large quantities, I can ask my friends.

[Din]

Quick search showed me that, e.g., Bowater Holographics or Krypten can do it

Bowater looks like reflection holograms. Krypten looks like embossed rainbow transmission. Luminit has a roll-to-roll mass production system for reflection HOEs for the AR/VR/HUD type holograms, not display. But, my point was that in the 1960's when the article was written, there were no mass produced, reflection security holograms

By the way: I think that "diffraction limit" comes from scalar wave theory.

Yes. It comes from the Fresnle-Kirchoff integral. The diffraction pattern is given by a Jinc function, and the Airy radii are the zeros of the Jinc function. The first zero is the classic 1.22fλ/D.

Is there any analogy to the diffraction limit in full EM theory

Vector diffraction theory pretty much follows scalar diffraction theory for diffractive structures > λ. When the diffraction aperture becomes ~ λ or less, the diffraction limitation takes on a different form. But, vector theory takes polarisation into effect, so the diffraction limit changes, based on polarisation. Also, if there are charges present, scalar theory does not deal with the currents produced by the moving charges. In the vector formulation, the 6 vectors of H and E are replaced with a single Hertz vector.

Nowadays, since about 2000, we can now have "super resolution" with metamaterials and negative index lenses.

and in quantum optics?

I don't think the same concept exists in quantum optics. In quantum optics, the energy is carried by photons, which are assumed to be particle-like. In this sense, they don't really have a wavelength, although you can associate a momentum with photons, through the relativistic equations. If you have a momentum, you also have a wavelength via the De Broglie relationship. Photons are bosons, hence they do not follow Fermi-Dirac statistics; you can have an infinite number of photons in the same state, in principle. The limiting factor would probably be the Uncertainty Principle.

[Brian]

"In the areas where the light waves from S are in phase with those from S1¢, the total amplitude is doubled."
But, the intensity, which is what's being recorded, is quadrupled, assuming equal intensities:

I(t) = I(1) + I(2) + 2√[(I(1)I(2)]cos(φ)

If I(1) = I(2) = I, and φ = 2nπ, then
I(t) = 2I + 2I = 4I
(Does this violate conservation of energy? After all, recording with two units of light produces a disturbance of 4 units of light.)

I'll bite...

First, what does the "t" in I(t) represent? Second, intensity I is proportional to, or a measure of, energy. Then...

the interference does not conserve energy at each spatial point in the interference fringe. However, the energy (summed or averaged) over a complete spatial period of the fringe is conserved...

Iavg = ∫ I(t) dφ evaluated from φ = 0 to 2π = 2I

Effectively the interference decreases energy from some spatial positions and adds it back at other positions (so, for example: all energy is shifted away from φ = π but an equal amount of energy is added back to the position φ = 0). This shifting of energy to create peaks and valleys of intensity is the interference pattern.

[Din]

First, what does the "t" in I(t) represent?

t is total. i represented the total intensity as I(t).

Effectively the interference decreases energy from some spatial positions and adds it back at other positions (so, for example: all energy is shifted away from φ = π but an equal amount of energy is added back to the position φ = 0). This shifting of energy to create peaks and valleys of intensity is the interference pattern.

True, You're already averaging. The intensity (I) is stable because it's the average of the cos squared function. This is why you can record a hologram.

The power in an EM wave is given by the magnitude of the Poynting vector, which is, instantaneously

S = (1/µ)(E x H) = (c²ε)(E x H)
= (c²ε)(E₀ x H₀)cos²(k.r)

The cos²(k.r) term is averaged over several cycles to give

<cos²(k.r) > = 1/2

and so, finally, the power in a beam of light is:

I = <S> = (1/2)cεE₀²

and I is proportional to the square of the amplitude

If you have two beams, call them O (object) = O₀exp(iα₁) and R(reference) = R₀exp(iα₂), then the total amplitude A(t) at any point is:

I(t) = |O + R|² = (O + R)(O+R)* = |O|² + |R|² + OR* + RO* = |O|² + |R|² + ORcos(α₂ - α₂) (using Euler)

This is the total amplitude at any point on the plate. Using the Poynting formulation of the intensity as proportional to the square of the amplitude, we can get the total intensity at any point :

I(t) = I₁ + I₁ +2√(I₁I₁)cos(α₂ - α₂)

Thus, thus overall intensity pattern is sinusoidal, with a noise term I₁ + I₁ (which I call the 'dc term'), which encapsulates both beams. If you now add the reference beam into this, you recover the original object beam, with a noise term based on the reference beam. This is all pretty standard. You can find it in any book on holography or on the internet. There are probably those who quote this with no clue on how to derive the Euler formulation of complex numbers, on which this is based. I recommend Hariharan (https://books.google.com/books/about/Ba ... 58764HVUIC). For a more in-depth, but far more mathematical version, take a look at Collier, Burchardt and Lin "Optical Holography"

[Din]

Oops! slight mistake:

I(t) = I₁ + I₁ +2√(I₁I₁)cos(α₂ - α₂)

should be

I(t) = I₁ + I₂ + 2√(I₁I₂)cos(α₂ - α₂).

where I₁ is the object beam intensity and I₂ is the ref beam intensity, or the other way around, it doesn't matter which is which, all that matters is the cos term.

This is not true anymore. It was true about 40 odd years ago, when all transmission holograms were considered "thin". This particular formulation leads to Raman Nath diffraction, which is not selective - all light diffracts. Today, with modern emulsions, if you record with a ref angle > ~7deg, you'll record a "thick" hologram. A "thick" hologram diffracts in the Bragg regime, and the theory behind a "thick" hologram is Kogelnik, which strongly shows selectivity for both wavelength and angle of reconstruction. But, to within some reasonable approximation, you can model a transmission hologram using this, so long as you're aware that it is an approximation.

However, notice that, in the above, if I₁ = I₂ = I', then:

I(t) = I' + I' + 2I'cos(α₂ - α₂)

the maximum of this quantity is

I(t) = 4I'

In other words, if you throw two beams onto the hologram of 1 unit each, the brightest part of the fringe is 4 units!

[Din]

By the way, Brian, is this one of your "One question every week"?