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I know from the angular spectrum method that given a field $U$ with a wavelength $\lambda$, we can decompose it with Fourier transform.

\begin{equation} U(x, y,0) = \iint{\tilde{U}_0(f_x,f_y)} e^{2 \pi i (f_x x + f_y y)} d f_x d f_y \end{equation}

And propagate it along a distance $z$

\begin{equation} U(x, y, z) = \iint{\tilde{U}_0(f_x,f_y)} e^{2 \pi i (f_x x + f_y y + {z \sqrt{\frac{1}{\lambda^2} - f^2_x - f^2_y}})} d f_x d f_y \end{equation}

When I have $\frac{1}{\lambda^2} - f^2_x - f^2_y <0$ I will have evanescent fields

\begin{equation} e^{ - 2 \pi { z \sqrt{ \bigl| \frac{1}{\lambda^2} - f^2_x - f^2_y}\ \bigr|}} \end{equation}

So the resolution is limited to $\lambda/2$, and this can also be applied for free propagation.

But by studying resolution with an imaging lens (and so the point spread function), I know that the resolution is limited to $ \frac{\lambda}{2 NA}$. The numerical aperture (NA) can be greater than 1, so I can have a better resolution than $\lambda/2$.

How can this happen?

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    $\begingroup$ Consider to spell out acronyms. $\endgroup$ Commented Apr 13, 2023 at 22:08

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Waves become evanescent when the generating grating (associated with a component of the Fourier transform) has periodicity smaller than the wavelength. For NA>1 the medium refractive index (RI) has to be greater than 1. At which case the wavelength in the medium is reduced by a factor of the RI, hence the threshold periodicity for which waves become evanescent is also reduced by RI.

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The angular spectrum method assumes propagation through vacuum. Any objectives with NA>1 are usually oil immersion objectives: they are immersed in a medium with higher refractive index $n$. In a medium where $n\neq 1$ the wavelength obeys $\lambda=\lambda_0/n$. For monochromatic light the frequency always stays constant$^\dagger$, but the wavelength changes with refractive index.

$\dagger$ unless your medium shows nonlinearities, which is quite rare.

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  • $\begingroup$ Immersion lithography uses water. $\endgroup$ Commented Nov 3 at 9:14
  • $\begingroup$ @my2cts Perhaps it would be more accurate to say that the angular spectrum method assumes propagation to a constant refractive index medium. The wavelength that appears in the exponent is then the wavelength of the light in this constant refractive index medium. $\endgroup$ Commented Nov 3 at 9:31
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The diffraction limited resolution is not some fundamental "theoretical number" rather it is an order of magnitude estimate of what one can expect from an optical system when the images of two equal intensity point sources are to be distinguished and detected. The nominal separation of these images are all $k\lambda$ where $k$ is a number that depends on the geometry of the optical system, but for a well-designed system (from RF to optical) this number is $k \approx \frac{1}{2} \cdot \cdot \cdot 1$.

There are several rules of thumb regarding the diffraction limit of which the most commonly used are the so-called 3dB width of the main lobe of the PSF or the separation of the first nulls, etc. Since the propagation wavelength $\lambda=\frac{c}{\tilde n \nu}=\frac{\lambda_0}{\tilde n}$ is inversely proportional to the refractive index $\tilde n$, the smaller the $\lambda$ is the better the resolution. The numerical aperture is a number less than the refractive index so the best resolution you can have is $k\frac{\lambda_0}{\tilde n}$.

Finally, why the resolution is inversely proportional to the refractive index? The answer lies in the reason for the diffraction limit. If you have a circular aperture, say, through which a coherent plane wave passes then the angular width of the central, that is, main lobe is $\theta \approx k_1\frac{\lambda}{D}$ where $D$ is the diameter of the aperture and $\lambda$ is the wavelength in the medium. When projected to a screen this is the size of the first Airy disk. If you change the medium and increase its refractive index $\tilde n$ but keep the aperture the same, the angular spread reduces correspondingly. If you change the aperture shape you will have a different value for $k_1$ that may also depend on polarization but the scaling law between the angular width and the wavelength versus aperture size is the same.

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  • $\begingroup$ thank you. So how the two limits (from angural spectrum and psf) are related? In an imaging system will be relevant the higher one? $\endgroup$ Commented Apr 13, 2023 at 22:09
  • $\begingroup$ I do not understand your last question but it does not seem directly to be related to your original post. Anyhow, if that is the case then ask it in a separate question with some more detail, this forum prefers single questions standing by themselves and answered that way. $\endgroup$ Commented Apr 13, 2023 at 22:16
  • $\begingroup$ yes it is the same question of my original post. I do not understand why he have two limits, the one from angular spectrum method and the other one from point spread function. The first one is for free propagation, the second one is for an imaging system. When we design an optical system, have we to consider both these limits? Or one of them is more general than the other? $\endgroup$ Commented Apr 14, 2023 at 7:16
  • $\begingroup$ It is not clear to me how you got the diffraction limit $\lambda/2$ from the condition $\frac{1}{\lambda^2} - f^2_x - f^2_y <0$ to exclude evanescent waves. The diffraction limit is a resolution condition for two equal strength point sources to estimate the visibility of one as seen in the other's image background that is not a point but is spread out. $\endgroup$ Commented Apr 18, 2023 at 10:07

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