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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?

For those less familiar with the physics, we can 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 don't know much about physics, but I think if you provided some mathematical relations between image intensity and the other variables in your problem, it would be easier for others to help you. –  Ethan Jan 13 '13 at 4:08
    
This is more suited for Physics SE. I think it should be migrated? –  ramanujan_dirac Jan 13 '13 at 17:16
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