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Suppose one knows in full detail the phase and intensity of monochromatic light in a plane. This is basically what a hologram records, at least for some section of a plane. By using this as the boundary condition for the wave equation, you should in principle be able to solve for the phase and intensity of light at every point in space (up to some global phase factor that changes in time). Now, how do you translate this into an image? Say, if the hologram is given by a complex field $\phi(x,y)$ on the $XY$ plane, the light has wavelength $k$ and I have a pinhole camera located at position $\vec{p}$ pointing in direction $\vec{r}$, is there a simple formula to translate this field into an image intensity?

In more detail: think of the light at each point in space as being given by a complex number, where the intensity is the magnitude squared and the phase is, well, the phase. I'm leaving out the $z$ coordinate of the field because it shouldn't matter for this problem. The Green's function for the wave equation (basically, the field emitted by a point source located at $\vec{s}$) is given by

$G(\vec{x},\vec{s},t) \sim e^{i(k||\vec{x}-\vec{s}|| - \omega t)}$

where $\omega$ is the frequency of the light, and I've dropped a constant term that doesn't really affect the answer. Also, the $\omega t$ term in the solution shouldn't matter either, since it will be the same globally and doesn't change the magnitude of the field. We can solve for the field $\psi(\vec{s},t)$ at any point in space by taking the integral

$\int \int dx dy \phi(x,y)G(\vec{s},(x,y,0)^T,t)$

but how do we go from this field to an image? Basically, how is the light transformed on passing through the pinhole?

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I cannot follow your description of the experiment with the pinhole camera. Maybe you can be more descriptive. But in short and in general,the image intensity is the modulus square of the complex field $\phi$. – elcojon Jan 13 '13 at 19:43
Actually the proper Green's function for the electromagnetic wave equation is $\frac{\partial}{\partial z}\frac{1}{2\pi}\frac{e^{ik\|{\bf{R}}\|}}{\|{\bf{R}}\|}$ where ${\bf{R}} = (x,y,z)$ and you are evaluating the field on some plane parallel to the $xy$-plane at some distance $z$ from the source. – daaxix Jan 14 '13 at 0:11
As stated by @elcojon, the image irradiance will be the proportional to the field multiplied by its complex conjugate, at each point on the plane. In a real detector (at optical wavelengths) it is actually $\langle u({\bf{R}},t)u^*({\bf{R}},t)\rangle_t$ where the angle brackets are a time average integral over the integration time of the detector. – daaxix Jan 14 '13 at 0:17
do you know how to form an image with a pinhole? it's just a convolution. if you consider the coherence of light things get more complex. but I don't see where you are heading with this. – elcojon Jan 14 '13 at 1:42
I think the pinhole is a wrong analogue. The image will appear from the coherent superposition of light passing through the whole plane of the hologram with the phases necessary to reconstruct the image. A pin hole cannot retain the phase information which even a small area of the hologram has. – anna v Jan 14 '13 at 7:04

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