1
$\begingroup$

Because the resolution of an objective is limited by the diffraction of its circular aperture, I don't understand why an infinite lens can not have an infinitely good resolution power.

People use Abbe relation to compute the minimum resolvable distance $p_m$ of an objective as a function of its numerical aperture NA:

$p_m = \frac{1.22 \lambda}{NA} = \frac{1.22 \lambda}{n \sin(\theta_m)}$

When the aperture goes to infinity, $\sin \theta_m$ goes to 1 and thus the minimum resolvable distance is : $p_m = \frac{1.22 \lambda}{n}$

Now, the Airy disk created by the diffraction by a circular aperture of diameter $D$ has a radius defined by the zero of the $J_1$ Bessel function : $r_{Airy} = \frac{1.22 \lambda z}{D}$, where $z$ is the distance of the observation plane.

So the radius of the Airy disk goes to zero as the circular aperture goes to infinity. Why isn't it the case for the minimum resolvable distance when the lens diameter goes to infinity ?

Improve this question
$\endgroup$
3
  • $\begingroup$ How would you make an infinite lens? $\endgroup$ – mike stone Nov 4 '20 at 15:07
  • $\begingroup$ more importantly, what are you looking at that needs an infinite lens $\endgroup$ – Topcode Nov 4 '20 at 16:00
  • $\begingroup$ Because I don't understand the derivation of the minimum resolvable distance. $\endgroup$ – magtweezers Nov 5 '20 at 18:12
1
$\begingroup$

Actually, resolution is (classically) limited by the angle of convergence of a collimated beam that is brought to a focus by the lens. That angle of convergence cannot exceed 180 degrees. At 180 degrees, the standing wave spacing is a half wavelength. At any lesser angle, the spacing is larger. The standing wave spacing is the resolution limit.

Improve this answer
$\endgroup$
7
  • $\begingroup$ Thanks for this answer. Any textbook or article that derives this half wavelength limit ? $\endgroup$ – magtweezers Nov 5 '20 at 18:21
  • $\begingroup$ I'll urge you to look up "interference" and learn how to calculate the size of an interference fringe produced by two mutually coherent beams crossing each other at a given angle. Resolution of a microscope is simply the inverse of the same problem. $\endgroup$ – S. McGrew Nov 6 '20 at 1:42
  • $\begingroup$ I don't really see why it's the same problem. That is actually the core of my question When you use a microscope you don't have only two angles like in the case where two counterpropagating planar wave interfere (the 180° case). In the case of an infinite microscope you make interfere all the angles between -180° and 180°. I am looking for this precise computation. $\endgroup$ – magtweezers Nov 6 '20 at 14:34
  • 1
    $\begingroup$ Sure, you don't have just two angles. But if you do the math, you'll see that the most extreme angles produce the smallest interference fringes. It's certainly possible to integrate all the contributions from all the intermediate angles, but it's not necessary because all the smaller angles result in larger fringes; and we can't make smaller features by combining larger features. $\endgroup$ – S. McGrew Nov 6 '20 at 17:44
  • 1
    $\begingroup$ "All frequencies"? That means, in this case, all fringe widths. The point is that the smallest fringe width obtainable by two interfering beams is limited by the wavelength and is 1/2 wavelength. So the Fourier series is truncated. $\endgroup$ – S. McGrew Nov 7 '20 at 14:49
0
$\begingroup$

Besides a lens having $D=\infty$ is a rather difficult engineering task, as @mikestone indicated above, the Airy formula must break down even before we get to that challenging senior management goal...

Think of what the reason is for the finite discriminating ability be proportional to $\lambda/D$. While the formula expresses that the image of a single point source spreads out with an oscillating but decreasing tail it also assumes that when looking at two sources they are independent from each other and this is where the formula breaks down for large enough aperture. When sources are within a wavelength they become electromagnetically coupled and their combined image amplitude becomes strongly dependent on their relative phases. There are so-called super-resolution methods that go below Rayleigh's rule. These methods solve for the underlying interaction between the sources to be discriminated but the result is also strongly signal to noise ratio dependent while Rayleigh's rule ignores noise.

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.