Fresnel diffraction

Hi Everyone,

I am having trouble with my Fourier optics. I am trying to simulate Fresnel diffraction using the following form of the equations:

Li(xi, yi) = K * magnitude ( FT{ P(xp, yp)E(xp, yp) } p= xi/ld ;q= yi/ld ) ^2

K = 1/(ld)^2
E(xp, yp) = e^(i*(PI/ld)*(xp^2+yp^2))

P is the aperture function (0 for transparent, 1 for opaque)
l is the wavelength of light, in my case 575e-9 (middle of the visible spectrum)
d is the distance between the aperture and image planes (17mm in my case)
FT is the Fourier Transform

This equation is slightly simplified because I am only interested in the radiance.

My aperture function P and image function Li are on the scale of 1um per pixel. This means that for every pixel increment in xi the increment in xp is 1e-6/(17e-3 * 575e-9) = 102m. This must be wrong. The image should be expanded onto the image plane not shrunk. Can anyone see my mistake?
Wow! "Way beyond the talents of a fourth-year," but I'll tackle it.

Looks like your PI/ld function goes to zero, making everything else go to zero (or e0, whenever your working with transparent materials.
oops sorry, my scruffy equations PI in this case is pi (3.14....)

Equation 3 from this paper is probably easier to read$FILE/TemporalGlare.pdf

Andy Resnick

Science Advisor
Education Advisor
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I don't exactly understand what you mean by 'the pupil function is on the order of 1 um per pixel'.
In that case let me try a little harder.

My input to this equation is P(x,y), which in my case is a diagram of the human eye viewed from front to back. This is why I call it the 'pupil function'. For practical implementation reasons I have had to limit the resolution of this diagram to 1 pixel = 1 micron. This means with a maximum pupil diameter of 9mm I have a diagram 9000 pixels high and wide. I dont need to include anything outside of the pupil in my diagram because that part is already occluded for being outside the pupil.

Hope this makes a little more sense?

Andy Resnick

Science Advisor
Education Advisor
Insights Author
I'm a little confused- the 'pupil function' is not the optical system; it's the transmission of the exit pupil only. The transmission of the exit pupil can have amplitude (i.e. a circular hole) and phase (that is how aberrations are assigned).

I forget where the ext pupil of the eye is located (it does not have to be physically located within the eye- the entrance pupil is slightly away from the cornea).

In any case, in Fourier Optics the image field is given as the object field convolved with the pupil function for coherent (phase preserving) imaging, while for incoherent imaging it's the autocorrelation of the exit pupil multiplied by the autocorrelation of the object field (IIRC- my book is at work).

It appears the authors of the article are computing the effects of certain aberrations- something that is not trivial to do.

Does that help?
yes, perhaps it would be better if i tried the convolution form of the equation. The convolution form seems very popular with scientist. I am using this form of the equation for speed (1 FFT Vs. 3 FFT), but it would be nice to see things working to show I'm at least on the right track.

The paper simplifies the eye some what by combining pupil, artifacts in the len, eyelashes and particles in the humors together into one function. This is strictly not correct since the particles in the humors alone will be set in several different plane, but to get the code to run at any speed they have to make this simplification or take an fft per particle.

Another simplification they make is to only calculate the effect on 575nm light and then the other wavelengths can be extracted as being diffracted a little more or a little less based on weather the wavelength is longer or shorter

Thanks :)

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