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I'm trying to implement a model for diffraction-limited imaging, following "Microlithography" by Sheats and Smith. You can skip to the bottom for my question, but I'll explain the setup here.

The basic idea is that I have a black and white image $m(x, y)$ (measuring the electric field, with 0 being dark and 1 fully illuminated). The case I'm especially interested in is where $m$ is an array of repeating spaces and lines, but my question applies to general images.

The book says to take the Fourier transform and then multiply by a pupil function, modeling the lens, before taking the inverse Fourier transform to get the final aerial image that I should expect as a result of diffraction and a given numerical aperture (NA) for my lens.

If $M(f_x, f_y) = \mathcal{F}(m(x, y))$ represents the Fourier transform, then the final image should be $\mathcal{F}^{-1}(M(f_x, f_y) P(f_x, f_y))$, where $f_i$ is the spatial frequency and the pupil function $P$ is defined below.

$$P(f_x, f_y) = \begin{cases} 1 & \text{if } \sqrt{f_x^2 + f_y^2} \le \text{NA}/\lambda \\ 0 & \text{if } \sqrt{f_x^2 + f_y^2} > \text{NA}/\lambda \\ \end{cases} $$

My question is about this pupil function. When I coded it up, I had an intuition that the pupil window could be made arbitrarily big by just increasing the numerical aperture. But it seems like $\text{NA} = \sin \alpha = 1$ is the maximum possible numerical aperture (assuming the medium is air with an index of refraction around 1 and $\alpha$ is the maximum half-angle of the diffracted light that still enters the lens), which creates a window of radius $1/\lambda$ in the Fourier domain.

But the book says the following: Does an ideal lens produce a perfect image? No. Because of the finite size of the numerical aperture, only a portion of the diffraction pattern enters the lens. Thus, even an ideal lens cannot produce a perfect image unless the lens is infinitely big. Since in the case of an ideal lens the image is limited only by the diffracted light that does not make it through the lens, we call such an ideal system diffraction limited.

I'm struggling to reconcile these two previous paragraphs. For an infinitely big lens, we should have $\text{NA} = \sin \alpha = 1,$ so the pupil function would have a limited radius of $1/\lambda$, while the quoted passage seems to suggest that the image should be perfect, so the pupil function would be identically 1 everywhere. Which one of these is true? Is the pupil function always limited by $n/\lambda$ ($n$ the index of refraction) or can a large enough lens in theory give a perfect reconstruction of the original image, regardless of the wavelength $\lambda$ of light used?

Any help/clarity you could give would be much appreciated! Thanks for reading.

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To me, even an infinite sized lens is not able to form a perfect image in the sense of image=object. Let's consider an infinite lens that can collect all the diffracted rays by an object illuminated by a plane wave parallel to the optical axis, since 𝜃=𝜋/2 , the numerical aperture NA=1.

As stated by Abbe, a microscope forms an image of an object after interference of the different diffractions orders generated by the object than can be considered as a sum of diffraction gratings of different periodicity. It is equivalent to decompose the object in a Fourier series.

Take a 4𝑓 system composed of two lenses, one that collects the diffracted light and the other that images on a detector. The final image is obtained from the interferences of the diffraction orders. If you look at the interference pattern of the two extreme rays of diffraction angle $𝜃_m=\pm𝜋/2$ , they converge at the detector plane to interfere.

The closest distance between interference fringes in such a case is given by, \begin{equation} 𝑖=\frac{2𝜋}{|\vec{k}_1-\vec{k}_2|}=\frac{\lambda}{2\sin(\alpha/2)} \end{equation} where 𝛼 is the angle between the direction of the two rays $\vec{k}_1$ and $\vec{k}_2$. For the two extreme rays, 𝛼=𝜋 , such that 𝑖=𝜆/2 . So the minimum space between fringes is equal to 𝜆/2

This case represents the most ideal case i.e., the best resolution but it does not provide a "perfect" image in the sense of an image identical to the object. Maybe the author meant the perfect image considering diffraction.

Edit :

The factor 2 arises from the fact that the distance between two bright fringes is only half the period of the interference term in the detector plane. So the period of the interference term is equal to $\lambda$. This is equal to the inverse of the cutoff frequency :) ! In other words, the Fourier series representing the object is truncated. The highest spatial frequencies can't be imaged.

Note that this cut-off frequency of $1/\lambda$ is only valid for a coherent system i.e., illumination of the object by a single plane wave. For an incoherent system, the cut-off frequency is doubled. Check Goodman's book for more information

Edit 2:

The pupil function is equivalent to a low-pass filter i.e., it cuts off the high frequencies. Take a look at the following example,

enter image description here

The original image contains a certain ammount of details, the more detailed is the image, the more high frequencies it contains. The image spectrum is represented in log scale, with the center of the spectrum equivalent to the low frequencies. It is then filtered by a circular function corresponding to the pupil function of the imaging system. After inverse Fourier transform, the image is blurry as the high frequencies have been cutted-off

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  • $\begingroup$ Thank you! I hadn't met your equation for the distance between fringes before, but I'm new to this, so maybe I'll be able to dig deeper soon. I plan on checking out the Goodman Fourier Optics book eventually, and I'm guessing I can find that there? I wonder if the being off by a factor of 2 has anything to do with radius vs. diameter for my pupil function, or maybe the difference between pitch $p$ and the space between fringes $p/2$? Either way, this was helpful, and I like the interpretation in your last paragraph. Thanks again! $\endgroup$ Commented Sep 5 at 20:52
  • $\begingroup$ I edited my answer to explain the factor 2. I'm pretty sure you can find a derivation of the space $i$ online, if not I can provide it here, just let me know ! $\endgroup$ Commented Sep 7 at 14:58
  • $\begingroup$ Thank you so much for your help and kindness! :) $\endgroup$ Commented Sep 9 at 14:07
  • $\begingroup$ I think I might be confused on one point: isn't the pupil function only seeing low spatial frequencies? So the Fourier series would be truncated in such a way that the highest spatial frequencies are lost? I think we're keeping the lows, unless I'm mistaken. $\endgroup$ Commented Sep 12 at 19:56
  • $\begingroup$ Yes ! My bad, there was a typo, I meant the highest frequencies are not imaged... I provided a visual example in the second edit. $\endgroup$ Commented Sep 13 at 7:36

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