Digital Holography

[lobaz]

I understand the total disturbance on the hologram is the sum of individual disturbances caused by individual points. Basically, it's a sum of sinusoidal components for each point. Isn't this just a Fourier sum?

Not exactly. Let's switch to continuous domain for simplicity.

The Fourier transform G(fx, fy) of a function g(x, y) is given by

G(fx, fy) = \int \int_{-∞}^{∞} g(x, y) exp(-i 2π [x fx + y fy]) dx dy

and the inverse Fourier transform is given by

g(x, y) = \int \int_{-∞}^{∞} G(fx, fy) exp(i 2π [x fx + y fy]) dfx dfy

This sum can be interpreted as a sum of plane waves in the plane xy. Each plane wave has complex amplitude given by G(fx, fy) and direction given by a unit vector

[λ fx, λ fy, sqrt(1 - {λ fx}^2 - {λ fy}^2)]

Let us also assume that G(fx, fy) = 0 for fx^2 + fy^2 > 1 / λ^2. That is, if the square root in the z component of the direction vector gets imaginary, G(fx, fy) is zero anyway (i.e. no evanescent waves).

On the other hand, if we have point light sources located at z = 0, light in z = z0 can be expressed by

U(x, y, z0) = \int \int_{-∞}^{∞} U(ξ, η, 0) exp(i k r) / r dξ dη

where

r = sqrt[ (x - ξ)^2 + (y - η)^2 + z0^2]

The difference is clear:

• The inverse Fourier transform represents a sum of "sinusoidal" patterns. Each pattern is a "sinusoid" looks like corrugated metal sheet, each component has certain frequency, amplitude, phase and angle of rotation.

  

  "Linear" sinusoidal pattern present in the Fourier transform.

  linear.jpg (8.68 KiB) Viewed 3978 times

• Summation of point light sources is a sum of radially symmetric "sinusoidal" profiles, each profile looks like a wave on a water surface whan you throw a pebble inside. Each profile is shifted in xy, has certain amplitude and phase.

  

  "Radial" sinusoidal pattern present in a sum of spherical waves.

  radial.jpg (18.7 KiB) Viewed 3978 times

Indeed, if z0 is large, then circles of the radial pattern become flatter locally (in some small area covered by holographic plate) and they start to behave like "corrugated sheets". If z0 is infinitely large, then spherical waves become plane waves and the Fourier transform can be used.

(To be continued...)

[lobaz]

As you mentioned, if you have a purely sinusoidal grating with frequency f(x), you'll get three points of light reconstructed.

Just in the far field. If we have a sinusoidal grating of size 10x10 mm, frequency f = 5 cycles/mm, and illuminate it by a collimated wave λ = 500 nm at normal incidence, we get three diffraction orders. If we put a screen to z = 5 meters away from the grating, we should observe three squares damaged by diffraction, not three points:

Intensity on the screen

grating_propagation.jpg (18.13 KiB) Viewed 3978 times

Indeed, the distance between centers of the middle and a side square should be

dist = tan(asin(lambda f)) * z ≈ 500e-9 * 5e3 * 5 = 0.0125 m = 12.5 mm

[lobaz]

So, it seems that all you need is the phase difference, and you can use the classic:
I = (U+R)(U+R)* = (1/r²)U² + R² + 2 cos[φ(r) + φ(o)] = (1/r²)I(O) + I(R) + 2 [cos((φ(o)) - (φ(r))]

So, why plot the complex part of the object wave?

Also, the above indicates that there is an overall increase in amplitude of U² + R² (what I call the 'dc level'). But, if you simply poke holes in the resist at positions given by the phase difference, what simulates the dc level? I assume that the holes from the objective lens should not be developed/etched all the way to the glass, thus giving you an overall phase delay, which corresponds to the dc level.

You are right. If you assume amplitudes of the object and the reference wave equal, you can work with just phase difference. Moreover, you do not need to bother with complex numbers at all, you just calculate the phase difference, take its cosine and you are done. If you have several point light sources, you can calculate cosine of the phase difference between the first point and the reference wave, add it to the cosine the phase difference between the second point and the reference wave, add it to...

In the end, you get a function that has both negative and positive parts. To convert it to amplitude pattern, you must add the dc term, and possibly binarize it. You have correctly mentioned that in classical holography, the dc term is given by |U|² + |R|². Unfortunately, this classical dc term has some internal structure if U is not simple, i.e. the classical dc term causes some noise. In digital domain, you can just transform your function to range [0, 1], i.e. to multiply it with a constant and add another constant. In this case, dc term is perfectly flat, which reduces noise. This approach - calculating cosine of the phase difference - is usually called 'bipolar intensity'.

On the other hand, if you need more freedom, such as variable object beam/reference beam ratio, or you need to manipulate with optical field, you should return to complex numbers.

Note: Let us denote the object wave that includes everything including 1/r term by O. Thus, your equation becomes

I = (O+R)(O+R)* = O² + R² + 2 cos[φ(r) + φ(o)] = I(O) + I(R) + 2 [cos((φ(o)) - (φ(r))]

[Din]

OK, got it. Thanks, Petr.

[Din]

Summation of point light sources is a sum of radially symmetric "sinusoidal" profiles, each profile looks like a wave on a water surface whan you throw a pebble inside. Each profile is shifted in xy, has certain amplitude and phase.

Can you use the Bessel /Hankel functions in a computer for the radially symmetric profiles?

[lobaz]

I am not aware of its use in computer generated (display) holography.
I know just one-dimensional Hankel (Fourier-Bessel) transform that is nice for radially symmetric functions. However, a general optical field is not radially symmetrical, only the impulse response of a point light source is.
I think the Hankel transform could be used for some special purpose CGH, such as for abberation correction, but I am not an expert in this area.

What is actually often used:

• The sum of radially symmetric impulse responses that are shifted in xy plane (i.e. optical field of a 2-D image) can be written in a form of convolution. Convolution can be expressed and calculated using Fourier transforms.
• The other way is to employ the Fresnel approximation of the impulse response. The convolution then reduces to the Fresnel transform, i.e. the Fourier transform of a function multiplied by a complex chirp function.
• Sometimes, the fractional Fourier transform or the linear canonical transform is employed in computer generated holography.
• Recently, Shimobaba ("Fast generation of computer-generated holograms using wavelet shrinkage", Opt. Express 25/1) showed a nice trick how to employ the wavelet transform. I am also aware that the wavelet transform is sometimes used in theoretical reasoning (Liebling, Onural and others).