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
