What is the diffraction limit of light? I see the following explanation in many papers on plasmonics: Light can be focused to dimensions no smaller than roughly half the wavelength
What is the reason for the above statement being true?
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1$\begingroup$ Diffraction. That is the reason. Let's consider a HeNe laser (632.8nm), and imagine we focus it to 10 times stated the limit: 32 nm. Can you imagine what the wavefront would look like? $\endgroup$– JEBCommented Mar 1, 2018 at 23:40
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$\begingroup$ Short, but a powerful explanation. Thanks. $\endgroup$– kiahCommented Mar 30, 2018 at 20:21
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$\begingroup$ This is known as the Abbe or Rayleigh resolution limit. See en.wikipedia.org/wiki/… $\endgroup$– my2ctsCommented Apr 8, 2020 at 21:14
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1 Answer
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Because propagating radiation comprises only plane waves with the wavelength in question.
Therefore, any propagating light field configuration can be found by using Fourier analysis to add up all the contributions of the field's $\exp(i\,\vec{k}_j\cdot\vec{r}_j)$ basic plane wave constituents.
In any given transverse plane, the variation from each constituent is found by taking a cross section in the plane of the component $\exp(i\,\vec{k}_j\cdot\vec{r}_j)$ in question.
You might have a plane wave propagating normal to the plane. In which case, the plane would be a phase front and that wave can contribute only a zero spatial frequency wigglyness to the Fourier superposition in the plane. You might have one inclined at a shallow angle, in which case it would contribute a very long wavelength wiggliness. The most wiggly any plane wave component can get in the plane is a spatial frequency of $2\pi/\lambda$, arising when the plane wave propagates along a direction in the plane.
So, the Fourier superposition comprises only sinuosoids of spatial frequency $k = 2\,\pi/\lambda$ or less. There are basic known limits in Fourier theory as to how concentrated the superposition can be. But, roughly, they tell us that the high intensity region has to be of the order of $\lambda$ in diameter or more.
Things are a little different when we have conductive metal screens constraining the field through tiny holes. Then the fields are no longer propagating and are evanescent. Thus the resolution of the scanning near field microscope can be much finer than a wavelength. But the catch is one has to bring these screens extremely near (nanometers) of the object being imaged, as evanescent, nonpropagating waves dwindle swiftly with distance, as I discuss further in my answer here.
answered Mar 2, 2018 at 1:34