I were studying something about the diffraction limit for a microscope when i found this definition:

For a focusing objective or for lithography applications, the smallest resolvable distance or the so-called critical dimension/size, respectively, is $CD=k_1\lambda/N_A.$

Of course I know the definition $R=0.61\lambda/N_A$ for the optical resolution, which, at least to me, seems to be pretty similar to the "critical dimension". But why to introduce $k_1$? Is it possible to make $k_1\neq0.61$ for an optical objective?


1 Answer 1


I think this formula is probably saying that several different criteria can be used to define resolution, e.g. the Rayleigh criterion gives $k_1=0.61$, and that they all have different values of $k_1$. The formula is saying that, no matter which criterion you use, the dependence on wavelength and numerical aperture is always the same: proportional to the former, inversely proportional to the latter.

Given the context, it would not be surprising if the $k_1$ were a semi-empirical constant that is found to represent the minimum feature size that is readily achieved in current photolithography technology, i.e. it takes into account not only the diffraction limit, but the etching process as well.

Resolution is a vague concept and the resolution one achieves depends heavily on signal to noise. Another criterion used for Gaussian beams measures the resolution and numerical aperture by the $1/e^2$ diameters of the point spread function in both untransformed space and Fourier space, respectively. By these criteria, the formula is $CD = 2 \lambda / (\pi\,NA)$ and $k_1 = 2/\pi$, which is rather close to the Rayleigh criterion value. I discuss these criteria more in my answer here.

  • $\begingroup$ Thank you for the answer, I found it very useful. However, I think $ k_1$ may also be correlated to the light distribution on the aperture, which determine the diffraction figure. I have read that in lithography a ring-shaped illumination could give a smaller $k_1$, as $k_1 \approx 0.36$. I wonder if a similar arrangement could be realized in microscopy... $\endgroup$
    – Alessandro
    Aug 6, 2017 at 10:15

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