Holographyforum.org / holowiki.org

Oh, I see, you assume phase holograms. I usually assume amplitude modulation

lobaz wrote:Why Bessel? I thought that Bessel functions emerge just in radially symmetric structures due to properties of the Hankel transform.

For simplicity, assume one-dimensional model. Thus, there is a phase variation, as a function of x, φ(x). Therefore, the transmissivity of the hologram, after recording, is:

t = exp[i*φ(x)]

The developed hologram will have a phase variation:

φ(x) = k{a² + r² + 2a*r*cos(2πξx) = φ(0) + φ(1)*cos(2πξx)

So, the transmissivity of the hologram becomes

t = exp[i*φ(x)]*exp[i*φ(1)*cos(2πξx)] = k'*exp[i*φ(1)*cos(2πξx)] = k'*Σiⁿ Jn[φ(1)]*exp[i*n*(2πξx)]

(ignoring the constant phase factor, summation over n, Jn is the Bessel function of the first kind, nth order)

Thus, the amplitude of a thin phase grating is proportional to the Bessel function.

Why Bessel? I thought that Bessel functions emerge just in radially symmetric structures due to properties of the Hankel transform.

Without actually doing the analysis, I would think that you could decompose the 'sharp' grating into a set of sinusoidal gratings. Then the efficiency of each sinusoidal component will be proportional to the zero order Bessel function of the amplitude of the sinusoidal grating. In other words, each sinusoidal component would be proportional to a zero order Bessel function whose argument would be proportional to the coefficient of the decomposition. So, the first order would be J₀(α*4/π), the second order would be J₀(α*4/(3π)) etc.

Partly true. Sharp edges mean that more light is diffracted away from 0 order. It does not tell how much light gets to +1 order. Usually, diffraction efficiency is defined as the ratio of the intensity of a selected diffraction order (say, +1) and the incident intensity.

Imagine, for example, a sinusoidal amplitude grating, i.e. amplitude transmittance is given as

c(x) = cos(2*pi*x)/2 + 0.5

If the grating is illuminated by a plane wave at normal incidence and unit amplitude, the average intensity just behind the grating is

integrate(c(x)^2, x, 0, 1) = 3/8.

It can be easily found that intensity of the 0 order is 0.25x the incident intensity, and intensity of the 1st order is 0.0625x the incident intensity. Thus, the diffraction efficiency is 6.25%.

For comparison, take a flat amplitude grating, i.e. amplitude transmittance is given as

f(x) = squareWave(x) * sqrt(3/4).

Here, squareWave is periodic with period 1, and squareWave = 0 for 0<x<1/2, otherwise 1. The average intensity just behind the grating is again

integrate(f(x)^2, x, 0, 1) = 3/8.

(If we did not introduce the factor sqrt(3/4) to f(x), the average intensity would be 1/2 instead of 3/8, and it would be unfair to compare diffraction efficiencies.)

Now, intensity of the 0 order is approximately 0.19x the incident intensity, which confirms the idea - a grating with sharp edges diffracts more light away from the 0 order.

Intensity of the 1st order is approximately 0.079x the incident intensity, which means that the diffraction efficiency of the "sharp" grating is higher. However, this "sharp" grating produces more diffraction orders (all odd orders), for example 3rd diffraction order is approximately 0.013x the incident intensity.

To conclude:
1. A "sharp" grating diffracts more light, which could indicate higher diffraction efficiency.
2. On the other hand, a "sharp" grating produces more diffraction orders, which could indicate smaller diffraction efficiency.

My point is that I cannot guess which conclusion is valid without a rigorous analysis.

lobaz wrote:surprisingly, power in the +1 order is about 1.6x higher than for the sinusoidal grating. Although I can calculate it, I still cannot imagine why it happens (without any equation).

Because the edge is sharp. Diffraction occurs in any slit because of oscillations of electrons at the slit edge. If the edge is sharp, the E field is greater, and so electrons have more energy to oscillate.

Thank you both for answering my question; and sorry for my late response, I completely forgot this thread.

I was really interested in any response for "estimating diffraction efficiency without actually calculating it". You probably know that a sinusoidal amplitude grating (transmission) makes just 0, +1 and -1 diffraction orders, with diffraction efficiency 6.25 % (power in the +1 order / incident power). Square-wave grating with 50 % duty cycle makes 0, +1, -1, +3, -3, +5, -5, ... diffraction orders (zero and all odd numbers); surprisingly, power in the +1 order is about 1.6x higher than for the sinusoidal grating. Although I can calculate it, I still cannot imagine why it happens (without any equation).

Din wrote:Petr, yes, some idea.

In a Bragg reflection hologram, the efficiency depends on the modulation, which depends on the ratio, and the hardness of the emulsion, assuming proper exposure and development. But, the planes are sinusoidal. This means that if it's overexposed, or overdeveloped, then the sinusoidal nature becomes more rectangular-wave. The exposure/development is non-linear. This gives rise to additional Fourier components. This means the output bandwidth broadens and flattens, since the output waveform is a Fourier transform of the plane distribution. So, I think of the area under the output curve as a measure of efficiency. The broader the bandwidth, and the flatter the output curve, the less efficient overall. This is a good thing for display holograms, since the broader the bandwidth, the brighter is the hologram (The brightness is the integral of the bandwidth multiplied by the photopic function, or, broadly, the area under the output curve). But, for HOEs, it wastes energy into less desirable wavelengths. So, for display, I try and make it broadband, and for HOEs, I try to make it narrowband.

In the photo-polymer this ratio depends on the ratio of the correct local developed polymerized photosensitive monomers to the diffused oligomers after exposure, I think the planes are still sinusoidal. As we may have the same broadband regime in some "developed" holograms I think in a flattening of fringes, not a rectangular-wave.. Direct exposure without post heating gives a much more pure distribution, even overexposed.

Petr, yes, some idea.

In a Bragg reflection hologram, the efficiency depends on the modulation, which depends on the ratio, and the hardness of the emulsion, assuming proper exposure and development. But, the planes are sinusoidal. This means that if it's overexposed, or overdeveloped, then the sinusoidal nature becomes more rectangular-wave. The exposure/development is non-linear. This gives rise to additional Fourier components. This means the output bandwidth broadens and flattens, since the output waveform is a Fourier transform of the plane distribution. So, I think of the area under the output curve as a measure of efficiency. The broader the bandwidth, and the flatter the output curve, the less efficient overall. This is a good thing for display holograms, since the broader the bandwidth, the brighter is the hologram (The brightness is the integral of the bandwidth multiplied by the photopic function, or, broadly, the area under the output curve). But, for HOEs, it wastes energy into less desirable wavelengths. So, for display, I try and make it broadband, and for HOEs, I try to make it narrowband.

The same holds for transmission holograms, but transmission holograms have one advantage over reflection holograms. They are dispersive. That means if, for some exposure and development, you get orders that diffract in different directions, then the fringes are non-linear. Energy is going into directions that you don't want. So, for a particular exposure/development scheme, I look for orders. Sometimes, in a resist hologram, I develop too aggressively, and I see third orders, and even dim fourth orders. I know I've developed too aggressively, and I dilute the etch.

In both cases, I try and think of the fringe/plane structure. I've found that almost everyone thinks of the hologram in terms of the output, be it display or HOE. I try and think of the Bragg planes (for a volume hologram) or the surface profiles (for a surface hologram) inside the emulsion and the photons interacting with the structure. I then try and think of boundaries. When the light gets into the emulsion, it starts electrons oscillating. If there is a sharp distinction between an interference bright area and an interference dark area, then you get a sharp change in the density of the emulsion and so a sharp change in the number of electrons at the barrier. I think this means that polarisation effects are beginning to happen, especially if the change is rapid and large. So, if these sharp boundaries border small changes in density, ie the amplitude of the rectangular wave is small, then I think there are no polarisation effects. But, if the sharp changes are high, ie the amplitude of the rectangular is high, then I think polarisation effects begin to happen. I think this is the basis of the Moharam and Gaylord paper.

It's difficult to differentiate between "holographic" and "other" grating

A side note: there are sub-wavelength features in e-beam mastered embossed holograms, such as sharp boundaries. I think I have heard (although I am not sure) that a proper estimate of their diffraction efficiency requires full electromagnetic field modeling, including polarization. However, I cannot tell how much full calculation differs from scalar theory estimate. And I have no intuitive mental model how to think about polarization effects in such holograms.

By the way, Dinesh, do you have any mental models that allow you to guess diffraction efficiency without actually calculating it?