Acutually this is related to computer simulation, especially MATLAB.

I am very confused, assume a plane wave passes through a convengent lens, where the propagation direction is parallel to paraxial direction of the lens. The light beam would converge to a sigle point in the center (ideal case) in back focal plane, exactly the Fourier plane.

In this situation, is the point equivalent to the central point of the spatial Fourier Transform of the field distribution? If described in fft2 function, is it convenient to the central point of fftshift(fft(field))?


1 Answer 1


In order to get what you expect, you need to put the center of the plane wave field values at index zero. So, the MATLAB command should be analagous to fftshift(fft(ifftshift(field))). This assumes that you have an odd number of values for the field so that the middle index corresponds to $(x,y) = (0,0)$.

Edit after clearing up question in comments:

If a wavefront $u(x,y,z)$, starting at $z = 0$ and propagating in the $z$-direction, encounters a thin lens of focal length $f$ at position $z$, and then propagates a distance equal to $f$, then the final wavefront is equal to

$$u(x,y,z+f) = i\frac{e^{-ik(z+f)+\frac{ik}{2f}\left(\frac{z}{f}-1\right)\left(x^2 + y^2\right)}}{\lambda f}(2\pi)^2F\{u(x,y,0)\}_{p=-kx/f, q = -ky/f}$$

where $\lambda$ is the wavelength of light, $k$ is the wavenumber $2\pi/\lambda$, $F\{\}$ is the Fourier transform, and $p$ and $q$ are horizontal and vertical spatial frequencies (the expressions for $p$ and $q$ after the Fourier transform indicate that these coordiantes are scaled by a factor of $-k/f$). If the lens is at a distance to the initial wavefront equal to the focal lenght ($z=f$), then this simplifies to

$$u(x,y,2f) = i\frac{4\pi^2e^{-ik(2f)}}{\lambda f}F\{u(x,y,0)\}_{p=-kx/f, q = -ky/f}.$$

If you don't care about the intensity of the wavefront, just the shape, then this reduces to

$$u(x,y,2f) = F\{u(x,y,0)\}_{p=-kx/f, q = -ky/f}.$$

So, a drift of length $f$, followed by a thin lens of focal length $f$, followed by another drift of length $f$, results in the wavefront taking the form of a scaled Fourier transform of the original wavefront.

Ref. Optical Electronics in Modern Communications by Amnon Yariv. The question is covered in detail in Appendix D

  • $\begingroup$ This blog post series helped me make sense of discrete Fourier transforms: blogs.mathworks.com/steve/category/fourier-transforms $\endgroup$
    – Mark H
    Jan 20, 2019 at 10:03
  • $\begingroup$ Hi Mark, The 'fft' operator is fast Discrete Fourier Transform in MATLAB. As is known, Fourier Transform is the intergal of the objective function - wave field. So, why the position of plane wave related? $\endgroup$ Jan 20, 2019 at 10:13
  • $\begingroup$ @standerQiu It's been a few years and I don't have the code I wrote to solve the wavefront propagation. I remember fftshift and ifftshift being needed to make the beam profiles come out right. Although, looking at my textbooks, I agree the ifftshift shouldn't make a difference and the center of the wavefront should be in the correct place after fftshift. What specific problem are you having? $\endgroup$
    – Mark H
    Jan 20, 2019 at 10:40
  • $\begingroup$ My problem is that the light passing through an object and a lens would converge to a single point. I want to find out the expression of this single point in MATLAB. Hence, I think is fftshift(fft(object)) enough? $\endgroup$ Jan 20, 2019 at 10:52
  • $\begingroup$ @standerQiu I am unclear about what you mean by "expression of this single point." Do you mean the index of the middle point? $\endgroup$
    – Mark H
    Jan 20, 2019 at 11:14

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