This question is about the Fraunhofer diffraction and the assumptions under which it is valid. My understanding is that given the complex amplitude of an electric field $E_{z=0}(x,y)$ in the $(z=0)$ plane, its value after free space propagation over a distance $z$ will be proportional to its Fourier Transform, if $z_0$ is large enough:

$$E_{z_0}(x,y) \propto \hat{E}_{z=0}\left(\frac{kx}{z_0}, \frac{ky}{z_0}\right)\tag{1}$$

"Large enough" means $W^2/\lambda << z$, where $W$ is the largest dimension in the "aperture", according to wikipedia. Now, aperture seems to suppose a strictly zero field outside of a "hole" which lets the field through.

My first question would be, can Fraunhofer diffraction be applied to a field which does not have a bounded support? Are there conditions on how rapidly decreasing said field must be? What conditions then define the far field?

The questions stem from the following reflection:

If Fraunhofer diffraction could be applied to $E_{z_0}(x,y)$ as given by equation $(1)$, there would exist a $z_{far}$ such that, approximately,

$$E_{z_0+z_{far}}(x,y) \propto \hat{E}_{z_0}\left(\frac{kx}{z_{far}}, \frac{ky}{z_{far}}\right) $$

According to equation $(1)$, $E_{z_0}$ is the function $E_{z_0}: x,y \mapsto \hat{E}_{z=0}\left(\frac{kx}{z}, \frac{ky}{z}\right)$, the Fourier transform of which is given by:

$$\hat{E}_{z_0}: k_x, k_y \mapsto \frac{1}{\left|\frac{k}{z_0}\right|^2}\hat{\hat{E}}_{z=0}\left(\frac{k_x}{\frac{k}{z_0}}, \frac{k_y}{\frac{k}{z_0}}\right)$$


$$E_{z_0+z_{far}}(x,y) \propto \hat{\hat{E}}_{z=0}\left(\frac{x z_0}{z_{far}}, \frac{y z_0}{z_{far}}\right) = E_{z=0}\left(\frac{-x z_0}{z_{far}}, \frac{-y z_0}{z_{far}}\right)$$

But $z_0 + z_{far}$ should still be in $E_{z=0}$'s far field, so $E_{z_0+z_{far}}(x,y) \propto \hat{E}_{z=0}\left(\frac{kx}{z_0 + z_{far}}, \frac{ky}{z_0 + z_{far}}\right)$

So, $E_{z=0}\left(\frac{-x z_0}{z_{far}}, \frac{-y z_0}{z_{far}}\right) \propto \hat{E}_{z=0}\left(\frac{kx}{z_0 + z_{far}}, \frac{ky}{z_0 + z_{far}}\right)$ ? Obviously, something went wrong, but where? I assume that applying Fraunhofer to $E_{z_0}$ is the mistake, thus the initial question.

Looking at the simple example of a square field at $z=0$, the far field is a cardinal sinus, which is not even integrable, so it would be understandable if you could no longer define a far field. Is there a more general result?


1 Answer 1


Sorry for my poor english. My native language is french.

I have no time to look at it in detail but it seems to me that you have to take into account the phase factors which multiply the Fourier transform. They are usually forgotten because they disappear in the calculation of the intensity. But they are not constant and here you have to keep them in the calculation of the second Fourier transform?

In Goodman, "introduction to Fourier optics" : $E(x_1,y_1 )=\exp⁡(jkz)\exp⁡(jk(x_1^2+y_1^2 )/2z) FT(E(x_0,y_0 ))$

  • $\begingroup$ As far as I can tell, the proportionality relationship between the far field amplitude $E_{z}(x,y)$ and the Fourier transform $\hat{E}_{z=0}(x,y)$ is directly between complex amplitudes, which would include phase. Am I misunderstanding something? $\endgroup$
    – liteplane
    Commented Aug 20, 2021 at 11:41
  • $\begingroup$ I have added the formula found In Goodman, "introduction to Fourier optics" $\endgroup$ Commented Aug 20, 2021 at 11:53
  • $\begingroup$ Thanks for your answer. Indeed, most ressources I've seen (including my courses) omit this phase. It solves my problem! $\endgroup$
    – liteplane
    Commented Aug 20, 2021 at 12:19
  • $\begingroup$ Very nice answer. Also mine it is a poor language 😢 $\endgroup$
    – Sebastiano
    Commented Aug 20, 2021 at 13:19

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