# Diffraction integral : incident field is FT?

Diffraction theory is formulated with the time-frequency Fourier Transform of the electric field. Namely, if our field is ##u(\mathbf{r},t)## then diffraction theory expresses integrals using the field ##u(\mathbf{r},\omega)##.

When we consider Fresnel and Fraunhofer diffraction for an incident field which is a plane wave, we say the field ##u(\mathbf{r},\omega)=1## in all space.

But this isn't the FT of a plane wave. There is a delta function missing.

My confusion is mainly that... Articles online consider the incident field (the time FT) as all the terms in ##u(\mathbf{r},t)## except ##e^{-i\omega_o t}##.

I have ignored many equations here, since I don't think it is necessary to type in the relevant integrals. If you think this should be necessary tell me.

Thanks!

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Andy Resnick
I don't think that's correct- in time-independent scattering, the far-field diffraction pattern is the *spatial* Fourier transform of the incident field: V(r) -> V(u), where u is the spatial frequency coordinate. And the incident field is not a constant, it's set by the aperture and may have a (complex) phase term.

Does this help?

I don't agree with you because:

1. The incident field is a constant, what sets the aperture is the so-called transmittance function. For instance, a plane wave hitting a square aperture is given by

##u_o=1## in all space, and ##t(x,y)=rect(x/w)rect(y/w)## where ##t## is your transmittance function, and rect functions are some sort of step function that give the relevant geometry.

2. What you get out of diffraction integrals is a position function. The incident field has to be in position as well since diffraction integrals come out of a convolution and not a fourier transform from frequency to space (this will look like a FT in the far-field because you end up ignoring a quadratic phase factor).

To see what I mean http://web.ift.uib.no/AMOS/PHYS261/phys261_Part_II.pdf

Andy Resnick
I don't agree with you because:
<snip>
To see what I mean http://web.ift.uib.no/AMOS/PHYS261/phys261_Part_II.pdf

That reference is fine; if you read it you would see your OP is inconsistent. See, for example, section 0.1.1 and 0.1.2. 0.2.4 covers a circular aperture. And, as a point of fact, the far-field diffraction pattern is specified in terms of angle, not position (the so-called angular frequency spectrum).

Hmm I dont see why it is inconsistent, since in that reference their analysis is from a certain point onwards always with ##u(\mathbf{r},\omega)## (the time FT of the field).

If I recall correctly the circular aperture is specified in terms of angle because of the approximation ##r/z\approx \theta ##. Also the so-called angular spectrum is really a function of spatial frequencies like in 0.1.9.

What am I seeing wrong here?

blue_leaf77
Homework Helper
we say the field u(r,ω)=1u(r,ω)=1u(\mathbf{r},\omega)=1 in all space.
If ##u(\mathbf{r},\omega)## is known to be a monochromatic plane wave of frequency ##\omega_0## and wave vector ##\mathbf{k}_0=\frac{\omega_0}{c}\hat{k}_0##, then I will write it as ##u(\mathbf{r},\omega) = e^{i\mathbf{k}_0\cdot\mathbf{r}} \delta(\omega-\omega_0)##.

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davidbenari
marcusl
Gold Member
I agree with Andy Resnick. All the action for a monochromatic field u(r) is in the spatial transform U(k).

blue_leaf77
Homework Helper
I agree with Andy Resnick. All the action for a monochromatic field u(r) is in the spatial transform U(k).
The expressions of the field in whatever planes are equivalent. One can also analyze the propagation of light beam in position plane ##\mathbf{r}##, but one has to deal with some convolution integrals. However, it's indeed true that working in the spatial frequency-frequency plane ##(\mathbf{k},\omega)## is easier and more direct because working with the transfer functions, which are defined in the spatial frequency-frequency plane, is easier. When written in the spatial frequency-frequency plane, a plane wave of frequency ##\omega_0## and wave vector ##\mathbf{k}_0=\frac{\omega_0}{c}\hat{k}_0## becomes ##u(\mathbf{k},\omega) = \delta(\mathbf{k}-\mathbf{k}_0) \delta(\omega-\omega_0)##.

Blue leaf: I agree with you, however my doubt was precisely why these delta functions are ignored in all sources that demonstrate diffraction by the well-known apertures that are available to analytical study: the circular aperture, rectangular aperture etc.

Is this something we omit since we know how to arrive at the ##u(\mathbf{r},t)## in a simple way?

blue_leaf77