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I don't understand why there is necessarily a diffraction limitation on optical systems. Where does this limitation in focusing light come from?

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That's a good question, and one that looks simple but has a complicated answer. Here's my attempt at an answer with no maths - as usual in physics you'll only really understand it by getting stuck into the mathematics.

It's commonly believed that lenses work by bending the light. You see diagrams like:

lens

showing the light ray bending as it passes through the lens. This is one way of looking at it, but a more fundamental explanation is that the lens changes the phase of the plane wave passing through it. Specifically the phase change produced by the lens varies with distance away from the centre line. So on the left side of the lens we have a plane wave of constant phase, while on the right side we have a plane wave with the phase varying with distance. The result is that on the right side we get an interference pattern - we generally call the interference pattern the image, but it is an interference pattern.

Incidentally, this is why a Fresnel lens can focus light even though it's a completely different shape to your usual convex lens. The Fresnel lens produces the same phase changes as a convex lens, so it focuses light in the same way.

But back to your question: the reason that the image isn't perfect is that it's formed by interference of only a finite portion of light wave i.e. the portion passing through the lens. The bigger the lens the more of the light wave forms the interference pattern and the better the image. But to get a perfect image you need all the light, i.e. an infinitely big lens, to contribute to the interference pattern.

Mathematically, the light intensity at the focal plane is the Fourier transform of the incoming light. The integration limits of a Fourier transform are from $-\infty$ to $+\infty$, but a finite size lens restricts the integration limits and changes the intensity in the focal plane away from the perfect case. It actually convolves the light intensity in the focal plane with the Fourier transform of the aperture through which the light passes. For a round lens this means your image is convolved with an Airy disk, and this smears out the image slightly.

John Rennie
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  • Well you can't have a plane wave with the phase (across the wave front) changing with distance. On the right side you actually have a converging "spherical" wave, with a constant phase across that spherical (assumed aberration free) wave front. I would call it a diffraction pattern, rather than an interference pattern. Interference tends to convey the notion of more than one wave. But you are correct, the truncation of the wavefront by optical apertures, results in the diffraction rings. Gaussian laser beams avoid the rings, by "fading to black" at the perimeter. –  Oct 15 '13 at 19:42
  • Isn't it possible to have a source of light that is fully encapsulated by the lens? Such as a laser, or something of the sort? –  Oct 17 '13 at 01:51
  • I guess that would require the source be a point as well, which a laser clearly is not. So where does the actual number of the size to which you can focus a laser come from? –  Oct 17 '13 at 03:15
  • It comes from the radius of the laser beam. When you attempt to focus the laser you'll find the focus is not a point but is the Fourier transform of the beam intensity profile. For example if the beam has uniform intensity the intensity in the focal plane will be an Airy disk. – John Rennie Oct 17 '13 at 05:54
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Well I suppose by perfectly focus, you mean to a mathematical point. And if we could do that, Heisenberg's principle implies that the momentum uncertainty would be infinite. So by the time we looked where the point was supposed to be, it would have moved to someplace else, in fact it could be anywhere at all, and we would never find it.

The problem is that the wavelength is not zero.

  • Not quite right: it's not the point would be anywhere else because of Heisenberg principle, it's photon which would. The focal point wouldn't move. – Ruslan Oct 16 '13 at 15:07
  • The observer (of the focused beam) would presumably be looking at where ALL of those photons are supposed to have arrived (at a single point); there being no observation of anything, sans a photon from there scattering into the observation instrument. You are asserting that the photon(s); all of them, could be moving to anywhere per Heisenberg. In any case, the observer would never catch up with them, or know where to look. –  Oct 16 '13 at 22:40
  • Practical experience, aka observation, says that the photons are scattered over a theoretically infinite expanse, commonly known as the Airy disc. –  Oct 16 '13 at 22:45
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I'd like to add to George Smith's answer: "The problem is that the wavelength is not zero" because a full answer to this question begins and ends with that statement, altogether independently of lenses and other hardware.

Let's begin with the monochromatic case and think about how an image, i.e. a pattern of light, forms on the focal, or indeed any plane. Any free running (as opposed to evanescent - more below) solution to Maxwell's equations can be decomposed through Fourier analysis into a superposition of plane waves. All these plane waves have the same wavelength and wavenumber - they're just heading in different directions. The interference between these waves is what forms the image at the focal (or any other) plane. So a wave with wave vector (heading direction) pointing at an angle $\theta$ to the focal plane normal varies like $\exp(i\,k\,\sin\theta\,x)$ (or, if you like, $\cos(k\,\sin\theta\,x - \omega\,t+\delta)$) where increasing $x$ is in the plane containing both the focal plane normal and the wavevector. So a wave propagating orthogonal to the focal plane has its phase fronts parallel to the focal plane: it's disturbance at a given time doesn't vary at all in the focal plane. A wave slightly tilted wave gives a low spatial frequency ($k\,\sin\theta$ radians per unit length) variation in the focal plane, $i.e.$ this is a low frequency Fourier component. The highest spatial frequency Fourier component we can get comes from waves running parallel to focal plane, and their spatial frequency is $k$ radians per unit length.

So, simply put, there are no high spatial frequency components to build an arbitrarily-swiftly-varying-with position disturbance from. But if we wanted to have a perfect spot focus, we would need arbitrarily high spatial frequencies. But we don't have these, so we're just stuck with a spot whose characteristic spatial frequencies are in the range $2\pi/\lambda$. This is exactly how an Airy disk, diffraction limited focus looks.

There is a theoretical exception to the above, and that is the evanescent field. Solutions to Helmholtz's equation (which all the Cartesian components of the electric and magnetic field vectors fulfill) fulfill $k^2 = k_x^2 + k_y^2 + k_z^2$, where $(k_x,k_y,k_z)$ is the wavevector. There are solutions to Maxwell's equations where $k_x > k$ and $k_y > k$, i.e. arbitrarily high spatial frequencies in the focal plane, but such waves have imaginary values of $k_z$. These solutions are altogether valid, but their amplitudes dwindle exponentially with distance $z$ away from their source. Theoretically we have solutions growing exponentially away from the source too, but these are unphysical - they would violate energy conservation because evanescent waves, although they are not propagating in the sense of transporting energy, are still an energy store - they represent reactive power shuttling back and forth between neighbouring regions and are thus akin to capacitive and inductive energy stores in an AC circuit.

Therefore, owing to this exponential dwindling with distance from the source, these evanescent waves cannot propagate far from their sources, and so are practically excluded from any focusing field, whose constituent spatial frequencies are thus limited to a maximum of $k$ radians per unit length.

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Selene Routley
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  • Regarding the "evanescent field" situation; those who study "radio-physics", and radio wave propagation from "antennas" (Hams etc) are familiar with the idea of a "near field" region, within a few wavelengths at most of the antenna, and the "far-field" region, which includes the propagating EM waves. The near field is stuck to the antenna, and waxes and wanes with the antenna current, but never disconnects from the antenna, and does not propagate. It is like the soap film, that hasn't disconnected from the wire loop to form a bubble; i.e. The evanescent field (not wave). –  Oct 17 '13 at 00:25
  • @GeorgeE.Smith Good point. I was going to mention antennas (i.e. spontaneously radiating atoms) but thought it might clutter the answer. Evanescent "waves" simply refers to a plane "wave" solution with one or more of the wavevector components imaginary. They sum to an evanescent field. I'm not altogether sure that the near and farfield of an antenna are quite the same thing as the running and evanescent field. Maybe you know something about this? There is certainly a relationship between the two - but all the spherical modes with only spherical Bessels of the first kind ($j_\ell(k r)$) .... – Selene Routley Oct 17 '13 at 00:47
  • ...can be fully and uniquely decomposed into a superposition of running plane waves, but these also have near and farfield behavior. So the nearfield is not only evanescent behavior (I define "evanescent" strictly in terms of the plane "waves" in my answer). I've never seen the singular modes i.e. the ones containing the $y_\ell(k r)$ terms that are unique to antennas, decomposed into plane waves: this would make a good question and I'm not even sure whether it can be done over the "whole domain", i.e. one containing the antenna. It must be possible over regions outside the antenna. – Selene Routley Oct 17 '13 at 00:49
  • @GeorgeE.Smith BTW I seem to have a crisis in understanding of the EM field about once every ten years and I'm overdue for one - the evanescent field really is a concept that bends one's brain and this was the concept that led to my last crisis! I've never fully recovered and can only say that the evanescent field and its relationship to singular solutions of Maxwell's equations is something that has to have a good theoretical treatment somewhere (unless my understanding of the EM field is altogether up the creek)- I've just never found one. – Selene Routley Oct 17 '13 at 00:54
  • Well Rod, My Radio Physics prof was an expert on antenna design; and I find it useful to think of optical emission and absorption as an antenna problem. Remember, "varying AC currents" of radio antennas, are no different from the "accelerated charges" of Maxwell classical EM theory; and the wire antenna, becomes a free space antenna, such as an electric dipole moment in an atom. (and higher order multi-poles as well). By the way Rod, I left you an addendum answer relating to the conditions for laser resonator stability back at that "why do laser beams diverge" question, way back. –  Oct 17 '13 at 05:23
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To give a further wordy answer and extend the discussion of fourier transform I offer the following.

Start with three planes, the source, the image and an aperture plane where our lens sits. The distribution of intensity in the aperture plane will, in the far field limit, be the fourier transform of the intensity in the source plane. Note the 'in the far field' implies a planar wave in the aperture plane.

Taking a lens of finite size we can only intercept a portion of this plane, and therefore the information contained in the portions outside the aperture is lost.

That collected will undergo a reverse fourier transform and an image of the source can be formed. But the information that the image is formed from is limited to that which was collected in the lens. This strongly hints that we cannot 'undo' the loss of information, or image degradation, introduced by the lens. We can make our lens arbitarily large but until it encompasses the entire infinite plane we can't get a perfect image.

Nic
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