How do implement the information that behind the aperture/slit/crystal plane there is a focusing lens in the Fresnel-Kirchhoff integral?

It seems that the solution is understanding the role of the lens as a Fourier transformation on the diffracted waves. But how this is implemented in practice.

I will appreciate if anyone can give some hints and a lead.

  • 1
    $\begingroup$ Could you perhaps rephrase the first sentence? I find it difficult to understand what exactly you are asking. $\endgroup$ Oct 5, 2016 at 4:28
  • $\begingroup$ If you're asking why there is a lens, it's because Fraunhofer's integral requires a plain wave, for which you need a point source on a lens0 focus in order to get straight rays. $\endgroup$
    – FGSUZ
    Jun 26, 2018 at 18:28

2 Answers 2


The electric field a few wavelengths away from the aperture is given by the Fresnel transform (which is an approximation of the Fresnel-Kirchhoff equation).

The Fresnel Transform is the Fraunhofer Transform with a quadratic phase term in the integral:

$$ U(x,y) = \frac{e^{jkz}}{j\lambda z}{e^{k \frac{k}{2z} (x^2 + y^2) }} \int \int_{-\infty}^{\infty} \{ U \left( \xi, \eta \right) e^{j \frac{k}{2z} (\xi^2 + \eta^2 ) } \} e^{-j \frac{2 \pi}{ \lambda z} (x \xi + y \eta ) } d \xi d\eta $$

You see the quadratic phase inside the integral: $ e^{j \frac{k}{2z} (\xi^2 + \eta^2 ) } $

After a certain distance, $z$ is so large that the quadratic phase is approximately $ \approx 1 $. Since it is in the denominator of the argument of the quadratic phase.

Therefore the Fresnel Transform simplifies to the Fraunhofer Transform which is basically the Fourier Transform.

However, in a very special configuration using a lens, it just so happens that the lens imparts a quadratic phase that is equal and opposite to the quadratic phase inside the integral and they cancel out.

So by going through the lens and propagating to the rear focal plane, the quadratic phase term in the Fresnel Transform is canceled out and the Fresnel Transform is simplified to the Fraunhofer transform. And the Fraunhofer transform is the Fourier Transform.

The full answer would require several pages of math and pictures but that's basically the idea of it.


The short (and somewhat cop-out) answer is that in Gaussian optics, one can mathematically show that the intensity distribution at the entrance pupil of a lens undergoes a Fourier transform at the lens' focal plane (which what Fraunhofer diffraction actually is).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.