This questions concerns three types of microscope systems and the differences between them. Let me lay out my understanding of these three types of systems. I am imagining imaging with a camera so only an intermediate image is needed, no eyepiece. I assume thin optics everywhere so that there is no confusion about different principle planes etc. I will consider the ray tracing for light scattered from the sample but also for hypothetical illumination light (propagating into the microscope from behind the sample) which is perfectly collimated (a laser beam back illumination for example).

Also a brief note about my understanding of lens Fourier transforming. My understanding is that if I have a lens with focal length $f$ then the lens will take whatever is on its front focal plane (distance $f$ before the lens) and output the Fourier transform of that onto its back focal plane (distance $f$ behind the lens). My understanding is that this Fourier transform is particular to these two planes. There are no other two planes which are Fourier transform pairs in this single lens imaging system.

  • Finite conjugate microscope: In this system there will be an objective with focal length $f_O$. The distance between the distance and the sample, $d_S$ will be slightly greater than $f_O$, $d_S>f_O$. The rays from the sample will fill the lens and then refocus down to form an intermediate image at a distance $d_C$ ($C$ for camera) much greater than $f_O$ from the back of the lens. The sensor can be placed at the location of the intermediate image. The magnification of this system is $M = \frac{d_C}{f_O} -1$. Collimated illumination will be focused distance $f_O$ behind the sample and then re-diverge and be diverging as it falls on the sensor.

  • $4f$ imaging system: In this system the distance between the sample and the objective is $d_S = f_O$. The light from the sample is then collimated after the objective. The back focal plane of the objective (located $f_O$ behind the objective) will contain the Fourier transform of the sample pattern. A second lens (I'll call the tube lens) with focal length $f_T$ is placed distance $f_T$ from the Fourier plane. Because the light going into the second lens is collimated it is focussed down by the tube lens distance $d_C = f_T$ behind the tube lens. The camera can be placed at the location of this final image. Also, because the Fourier plane is distance $f_T$ before the tube lens we know that the Fourier transform of the fourier plane (the image again) will appear $f_T$ behind the tube lens. In principle one could put some sort of filter at the location of the Fourier plane and do Fourier filtering which may be of interest for some applications. In this case the distance between the tube lens and the objective is $d_T = f_O + f_T$. The magnification of this system is $M = \frac{f_T}{f_O}$. Collimated illumination light will be focused distance $f_O$ behind the objective (at the location of the Fourier plane for the sample) and then recollimated by the tube lens. It will be collimated on the sensor.

  • Infinite conjugate microscope: Also known as infinity corrected system or objective.. In this system the objective is distance $f_O$ away from the sample, $d_S = f_O$. The light from the sample is thus collimated behind the objective. A tube lens can then be placed (basically any distance from the objective that one likes*) some distance $d_T$ from the objective. Because the light going into the tube lens is collimated it will be focused down at distance $d_C = f_T$. The magnification of this system is $M = \frac{f_T}{f_O}$. collimated illumination light will be focused distance $f_O$ behind the objective. Depending on the location of the tube lens (which may be closer than $f_O$ from the back of the objective?) this light might be converging, diverging, or collimated after the tube lens. It might be focused onto the sensor or it may be diverging, convering, or collimated on the sensor.

It looks to me like an infinite conjugate imaging system is very very similar to a $4f$ imaging system. Here are the only differences that I see.

  • A) For a $4f$ imaging system $d_T$ must equal to $f_O + f_T$ while for an infinite conjugate system $d_T$ is arbitrary

  • B) For a $4f$ imaging system the back focal plane of the objective serves as a Fourier plane which is going to be re-transformed by the second lens to form an image. However, in the infinite conjugate imaging system the second lens can have a different placement. This means that the Fourier plane for the first objective may not be overlapped with the Fourier plane for the second objective. This means that a filter place somewhere between the objective and tube lens (in the "infinity space") will not act as a pure spatial Fourier filter (aperture stopping) but it will do some image clipping or something (field stopping) as well.

  • C) For a $4f$ imaging system the illumination light is always collimated on the sensor while for an infinite conjugate system the light can be collimated or diverging or converging. I don't really know what effect this might have on image properties.

Actual Questions

Now that I've laid out my understanding here are my questions:

  1. Does an infinity corrected system really refer to a system that can have any length $d_T$ between the tube lens and the objective?

If yes then:

  • 2a. Is it correct that a $4f$ system is just a special case of an infinity corrected system?

  • 2b. Is it correct that you cannot do Fourier filtering (or any Fourier masking) in an infinity corrected system unless $d_T = f_O + f_T$?

  • 2c. What goes into the choice for $d_T$? I can see one might want $d_T$ to be short as possible to save space but then why not just put the tube lens right on the back of the objective? In that case you basically just have a new compound objective in a finite conjugate configuration.. Alternatively, one might want the tube length as long as possible so that you can fit many optical components in the path, in this case you would be limited only by vignetting? Why would one not choose $d_T = f_O + f_T$ so that one would always have the ability to do Fourier masking?

  • 2d. Does it matter that the illumination light is not necessary collimated on the sensor? Does this cause any sorts of problems or is there a particular way that it should be? If no then:

  • 3a. I have got the impression from reading resources and from colleagues that the distance between the tube lens and the objective is totally arbitrary in an infinity corrected system. I've also read that this was maybe historically part of the motivation for moving away from fixed length finite conjugate systems**. Are these claims that the length of the infinity space is arbitrary simply wrong?


  1. Is my explanation of Fourier transforming only happening between the front and back focal planes correct?

*The only limit I see to this is that if the tube lens is too far away one will see vignetting in the final image.

**Maybe the story was that for finite conjugate if you want to change out the objective or put any new optics in between the objective and intermediate image you always need to re-adjust the position of the eyepiece or sensor as well, whereas for an infinite conjugate system you can switch out one infinity corrected objective for another while leaving ther rest of the system the same..

  • $\begingroup$ For an infinity corrected system, you CANNOT have arbitrarily long length between the objective and the tube lens, for you will run into the problem of vignetting (less light power collected for objects far away from the optical axis). edmundoptics.com/resources/application-notes/microscopy/… $\endgroup$ – wcc Jul 11 at 19:57
  • $\begingroup$ @wcc thank you for comment. As indicated in my question I am aware of how vignetting sets an upper limit on the distance between the objective and tube lens. Other than that are there any other limitations on that distance? For example, should it always be $d_T = f_O + f_T$ or can it be any distance as long as the distance is shorter than the distance at which vignetting would occur? $\endgroup$ – jgerber Jul 11 at 21:29
  • $\begingroup$ vignetting happens when the light from an object far from the edge of the field of view is traveling at an angle such that by the time it arrives at the plane of the tube lens, it misses the aperture of the tube lens. So if $d_T <= f_O + f_T$ there is no vignetting. $\endgroup$ – wcc Jul 11 at 22:05

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