# Fourier Optics: Why a Plane Wave contributes to just a single point

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• Daniel Petka
Daniel Petka
TL;DR Summary
The key of Fourier Optics: Why does a Plane Wave contribute to just a Single Point in the far field? (the plane wave goes through an aperature => actually a superposition of plane waves) I'm afraid my Fourier Optics profs doesn't quite get it either. (source: Fundamentals of Photonics, Saleh/Teich)

If the distance between the input and the output screen d is large enough, then for a plane wave (with some spatial frequencies vx and vy) at the input, the spot at the output will be a point. But if the plane wave is confined (aperature size b in the picture), it's no longer a plane wave... but a sum of plane waves. The proof here mentions the method of stationary phase and a phase factor that goes to zero. Sure, the math works out, but I still don't get intuitively. The only think that makes sense is to treat the plane wave as a gaussian laser beam (and so the dots are also gaussians). But this is not consistent, since a laser beam consists of many plane waves at different angles and here it's clearly mentioned that just one plane wave / spatial frequency contribute to a given point... Thanks a lot for any insight!

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I'm not sure I understand your question. Are there supposed to be images? If so, they didn't attach properly.

Ibix said:
I'm not sure I understand your question. Are there supposed to be images? If so, they didn't attach properly.
Sorry, my bad. Should be fixed now. This is gibberish without images (possibly even with images) The question is: Why do plane waves get transformed into points in the far field?

Ibix
Daniel Petka said:
Sorry, my bad. Should be fixed now. This is gibberish without images (possibly even with images) The question is: Why do plane waves get transformed into points in the far field?
Maybe it would help to think of "the far field" as Fourier transform space (equivalent to 'k-space' in scattering) of the object plane; a single plane wave is of infinite extent and has a single wavevector and thus transforms to a single point.

If that plane wave is instead truncated by an aperture at the object plane, diffraction implies that many planewaves (all of differing wavevectors) are generated, generating an extended pattern in Fourier space (the far-field).

Edit: plane waves and Gaussian beams are not equivalent.

Gleb1964
Andy Resnick said:
Maybe it would help to think of "the far field" as Fourier transform space (equivalent to 'k-space' in scattering) of the object plane; a single plane wave is of infinite extent and has a single wavevector and thus transforms to a single point.

If that plane wave is instead truncated by an aperture at the object plane, diffraction implies that many planewaves (all of differing wavevectors) are generated, generating an extended pattern in Fourier space (the far-field).

Edit: plane waves and Gaussian beams are not equivalent.
I thought about it for a while now.. First of all, the Fourier Transform happens in the far field. People who say that a lens performs a Fourier Transformation are mostly EEs (like me) who think in terms of components. This is fine, but it's not Real understanding. Through propagation alone, it makes zero sense why a plane wave should be transformed into a point. Let's say I shine a plane wave at normal incidence, that is, 0°. Since the plane wave is infinite, you will observe a plane wave at 0° at the output. What kinda makes sense for me rn is ti think in terms of finite size laser beams, so in essence a gaussian aperature (I don't want any frequency spill or anything so gaussian is the way to go). Let's imagine I shine a laser beam at a sinusoidally varying mask. This means that the intensity after the mask is sinusoidal and the Fourier Transform then predicts 3 peaks. You can get this sinusoidal intensity distribution by i nterfering 3 laser beams at each other, coming in at let's say these angles: (-45°,0°,+45°). Obviously, you get the 3 peaks in the far field, then. As far as I understand it, the paraxial approximation just fixes the aberration due perspective: we are shining the light on a plane, so angles won't correspond exactly to points, only if sinx=x. That's all I know right now, hope that something of it makes sense...

Daniel Petka said:
Through propagation alone, it makes zero sense why a plane wave should be transformed into a point.
That doesn't appear to be what it says, though. What it says is that the only plane wave that contributes to the complex amplitude at a point is.... Without bothering to read the proof, I expect what it does is show that plane waves exist in pairs with opposite phase shifts, except for one, do they all cancel out except that one.

Note that we're working in the far field here. Plane waves are all we get, just coming from slightly different angles.

Ibix said:
That doesn't appear to be what it says, though. What it says is that the only plane wave that contributes to the complex amplitude at a point is.... Without bothering to read the proof, I expect what it does is show that plane waves exist in pairs with opposite phase shifts, except for one, do they all cancel out except that one.

Note that we're working in the far field here. Plane waves are all we get, just coming from slightly different angles.
Thanks for the insight! I think I can pin point the bit that I'm struggling with here: If we're dealing with plane waves, then nothing can cancel the DC wave (normal incidence), So the whole plane should be illuminated by the DC. This doesn't get fixed even when there's an aperature...

Daniel Petka said:
I thought about it for a while now.. First of all, the Fourier Transform happens in the far field. People who say that a lens performs a Fourier Transformation are mostly EEs (like me) who think in terms of components. This is fine, but it's not Real understanding. Through propagation alone, it makes zero sense why a plane wave should be transformed into a point. [...]
I don't know what you mean by "real understanding".

A lens doesn't always perform a Fourier transform. What a lens *can* do is bring the far-field a lot closer.

How about this- evaluate the Fourier integral F(k)=∫f(x)eikxdx when f(x) is a plane wave with wave-vector k'. What is the result? How would you interpret this?

Yes, this was my point. The lens just brings the far field to a focal plane. The Fourier Transform happens in the far field.

The Fourier Transform of a plane wave is a Delta peak in k-space whose coordinates are the k-vector itself. For any other plane wave, the wavelengths or the angles dont match, you have just as many positive points as negative and it integrates to zero.
By real understanding, I mean for example answering a simple question: Why is the DC component transform to a point in the far field? Doing the math is necessary, obviously, but understanding is answering simple questions like this, which I can't to be honest.

Now the aperature is finite, which means that there are in fact many plane waves and for a 1mm aperature (let's say you shoot a gaussian laser beam) the far field is at 1m, which leads to a spot size of 1.2mm at the far field. But a gaussian laser beam is of course not a plane wave.

Edit: lol based on your profile picture you might know quite a bit about k-space

Daniel Petka said:
Yes, this was my point. The lens just brings the far field to a focal plane. The Fourier Transform happens in the far field.
Be careful- the FT does not just happen, it is a consequence of Fraunhofer's approximation to the Rayleigh-Sommerfeld diffraction formula.

Daniel Petka said:
The Fourier Transform of a plane wave is a Delta peak in k-space whose coordinates are the k-vector itself. For any other plane wave, the wavelengths or the angles dont match, you have just as many positive points as negative and it integrates to zero.
You had me, then you lost me. The function δ(k-k') does not mean that if k ≠ k', there are equal numbers of positive and negative points. It does mean that there is a single non-zero value at k = k', which corresponds to a single propagation direction.

Daniel Petka said:
Now the aperature is finite, which means that there are in fact many plane waves and for a 1mm aperature (let's say you shoot a gaussian laser beam) the far field is at 1m, which leads to a spot size of 1.2mm at the far field. But a gaussian laser beam is of course not a plane wave.
This is an incorrect calculation. And again, a gaussian beam is not a plane wave. But fine- a 1mm aperture illuminated by green light (0.5 micron wavelength) gives a far-field observation distance (the distance required by the Fraunhofer approximation) of z >> 1.5 meters. Beyond that distance, the diffraction pattern will result in (for a circular aperture) an airy disk, and I will let you work out the angular size of the central peak.

DaveE
Didn't express myself clearly here, sorry. I didn't refer to the delta function when I said that the integral cancel, but to the plane waves. In 1D, you can visualize the integral as area under the curve and if there are equally many positive as negative value, which would be the case if you multipled 2 different frequencies, the integral is zero. This is the same but in 3D. The dirac is not a value, though, you get a value when you integrate over it, but I digress.
I am mentioning gaussian beams to have an aperature that doesn't diffract. If you place a sinusoidal intensity mask, you get 3 dots in the far field. And this is indeed identical to interfering 3 beams that result in a sinusoidal interference pattern. The calculating might be wrong by a few 100s of um, but that doesn't matter. What matters is the order, that the gaussian beam looks like a point on the screen.

Daniel Petka said:
Didn't express myself clearly here, sorry. I didn't refer to the delta function when I said that the integral cancel, but to the plane waves. In 1D, you can visualize the integral as area under the curve and if there are equally many positive as negative value, which would be the case if you multipled 2 different frequencies, the integral is zero. This is the same but in 3D. The dirac is not a value, though, you get a value when you integrate over it, but I digress.
I am mentioning gaussian beams to have an aperature that doesn't diffract. If you place a sinusoidal intensity mask, you get 3 dots in the far field. And this is indeed identical to interfering 3 beams that result in a sinusoidal interference pattern. The calculating might be wrong by a few 100s of um, but that doesn't matter. What matters is the order, that the gaussian beam looks like a point on the screen.
I'm concerned we are not making progress here.

To proceed, I need to know that you can evaluate the Fourier integral F(k)=∫f(x)eikxdx when f(x) is a plane wave with wave-vector k'.

I also need you to provide evidence (for example, a calculation) that verifies your claim "If you place a sinusoidal intensity mask, you get 3 dots in the far field. And this is indeed identical to interfering 3 beams that result in a sinusoidal interference pattern."

Sure, here is the code I used:

from numpy.fft import fft2, ifft2, fftfreq
import numpy as np
from matplotlib import pyplot as plt
x = np.linspace(-2,2,2000)
xv, yv = np.meshgrid(x, x)
fig = plt.figure(figsize=(1,1),frameon=False)
gaussian = np.exp(-200*(xv**2 + yv**2))*(np.cos(80*xv)**2)
plt.figure(figsize=(5,5))
plt.pcolormesh(xv,yv,gaussian, cmap = "hot")
plt.axis('off')
plt.savefig("input_plane.png")
def compute_U(U0, xv, yv, lam, z):
A = fft2(U0)
kx = 2*np.pi * fftfreq(len(x), np.diff(x)[0])
kxv, kyv = np.meshgrid(kx,kx)
k = 2*np.pi/lam
return ifft2(A*np.exp(1j*z*np.sqrt(k ** 2-kxv ** 2-kyv ** 2)))
U = compute_U(gaussian, xv, yv, 0.00000001, z=1000000)
fig = plt.figure(frameon=False)
plt.figure(figsize=(5,5))
plt.pcolormesh(xv,yv,np.abs(U), cmap = "hot")
plt.axis('off')
plt.savefig("output_plane.png")

It's the angular spectrum method. You can tweak the z parameter to see the propagation and the 3 dots appearing. Slight correction, it's the field amplitude, not intensity here. At z=0, you see the interference pattern. If you put a negative value, the beams come from behind the screen. The frequencies get mapped to angles.

Daniel Petka said:
Sure, here is the code I used:

from numpy.fft import fft2, ifft2, fftfreq
import numpy as np
from matplotlib import pyplot as plt
x = np.linspace(-2,2,2000)
xv, yv = np.meshgrid(x, x)
fig = plt.figure(figsize=(1,1),frameon=False)
gaussian = np.exp(-200*(xv**2 + yv**2))*(np.cos(80*xv)**2)
plt.figure(figsize=(5,5))
plt.pcolormesh(xv,yv,gaussian, cmap = "hot")
plt.axis('off')
plt.savefig("input_plane.png")
def compute_U(U0, xv, yv, lam, z):
A = fft2(U0)
kx = 2*np.pi * fftfreq(len(x), np.diff(x)[0])
kxv, kyv = np.meshgrid(kx,kx)
k = 2*np.pi/lam
return ifft2(A*np.exp(1j*z*np.sqrt(k ** 2-kxv ** 2-kyv ** 2)))
U = compute_U(gaussian, xv, yv, 0.00000001, z=1000000)
fig = plt.figure(frameon=False)
plt.figure(figsize=(5,5))
plt.pcolormesh(xv,yv,np.abs(U), cmap = "hot")
plt.axis('off')
plt.savefig("output_plane.png")

It's the angular spectrum method. You can tweak the z parameter to see the propagation and the 3 dots appearing. Slight correction, it's the field amplitude, not intensity here. At z=0, you see the interference pattern. If you put a negative value, the beams come from behind the screen. The frequencies get mapped to angles.
Now I understand why you keep insisting on gaussian beams.

I'm sorry, but this does not demonstrate to me that you can evaluate basic integrals.

sophiecentaur

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