Infinity objective to tube lens distance question.

[zondar]

Hello All,

For infinity optics, is the distance between the objective and the tube lens substantially irrelevant (within reason, other than leaving room for any other optical components), or what other factors are in play?

Thank you.

[Scarodactyl]

It matters somewhat. Increasing distance shrinks the image circle and can cause vignetting which puts some limits on how many accessories you can add. Depending on the tube lens there can be some relatively minor effect on image quality.

[apochronaut]

The telan lens is placed in a fairly obvious location in the design in order for the accomodation of aux. components and ergonomics to be optimized, plus capture the ray bundle with minimal off axis variance , scatter, random interference , loss of peripheral rays and potential vignetting or diffraction when aux. components are installed. Farther from the objective, the more image quality loss is likely and the closer to it, less likely.
Ray bundles can be canted somewhat and I am sure uniform parallelism of photon axes varies with each objective design , especially at the periphery.

[zondar]

(Not arguing, just trying to figure it out:)

Would increasing the distance necessarily shrink the image circle? If the light is truly "infinite", i.e. completely parallel, then the distance should not matter. Of course, the objective's optics are probably not perfect, and hence not actually completely parallel, but imperfections could go either way - expanding or shrinking, or even "canted" as mentioned (projecting off axis). No?

So assuming no other optics are in the path, is the optimum distance essentially zero, with no down-side to being that close?

[hans]

Would increasing the distance necessarily shrink the image circle? If the light is truly "infinite", i.e. completely parallel, then the distance should not matter.

Yes, increasing the distance necessarily leads to vignetting at some point. Rays from a single point are always parallel with one another but are parallel to the optical axis only for the point at the center of the image. For points off axis the rays exiting the objective are still parallel with one another but are angled relative to the optical axis. See for example figure 2 here:
https://www.microscopyu.com/microscopy- ... al-systems

[zondar]

Yes, I found that link and other discussions while surfing around too. This one in particular presents math to determine the maximum length before vignetting occurs:

https://www.edmundoptics.com/knowledge- ... bjectives/

To answer my own question about what distance is optimal: It's not zero. This link can compute an optimal distance with the goal of matching the optics to the sensor size based on parameters of the objective and tube lens:

https://www.edmundoptics.com/knowledge- ... be-length/

I don't think that calculator is always accurate, though, in that I think it fails in some circumstances, e.g. if the tube lens entrance pupil is small compared to the sensor diagonal. In other words, it doesn't take into account vignetting.

Thank you.

[zondar]

OK, I'm confused about the results of that calculator. It and the formulas (the results match) are showing that a small sensor requires a long objective to tube-lens distance ("L") and a large sensor requires a small one. This feels backwards to me.

To provide the magnification required to fill a large sensor at the sensor's focal plane, one would need a long L, wouldn't one, and vice-versa for a small sensor, right?

If not, what am I missing?

Thank you.

[hans]

To provide the magnification required to fill a large sensor at the sensor's focal plane, one would need a long L, wouldn't one, and vice-versa for a small sensor, right?

L in that calculator is the length of the infinity space, not the focal length of the tube lens which determines magnification.

It and the formulas (the results match) are showing that a small sensor requires a long objective to tube-lens distance ("L") and a large sensor requires a small one.

L is not a "required" distance, just the longest that can be used without vignetting. With a smaller sensor the angular deviation of the rays in the infinity space that hit the corner of the sensor is proportionally smaller, therefore the tube lens can be moved further away before useful ray start hitting beyond the edge of the tube lens leading to vignetting.

[zondar]

L is not a "required" distance, just the longest that can be used without vignetting. With a smaller sensor the angular deviation of the rays in the infinity space that hit the corner of the sensor is proportionally smaller, therefore the tube lens can be moved further away before useful ray start hitting beyond the edge of the tube lens leading to vignetting.

Oh, I see, this is not a calculation fitting to the sensor, it's just another way of calculating the maximum L before vignetting takes place. Makes total sense. Thanks!

But isn't it true that increasing L increases the size of the image at the sensor's focal plane, too? Just simple geometry would suggest so, as does the vignetting effect. And in that case, wouldn't you want to adjust L for a good fit? If not, I suppose I'm still erroneously treating the objective as if it was finite-conjugate.

Thanks again.

[Scarodactyl]

To answer my own question about what distance is optimal: It's not zero.

This varies between objectives and tube lenses based on real-world testing. Some infinity objectives (particularly lower mag ones) can actually perform better with a tube lens not focused at infinity. Sometimes things are optimized and sometimes compromises have to be made for them to work practically in a complete microscope system.

But isn't it true that increasing L increases the size of the image at the sensor's focal plane, too?

No, the image size does not change.

[zondar]

No, the image size does not change.

Well, why then? Doesn't increasing L create a larger image at the tube lens as well as at the sensor plane? Otherwise, vignetting at the tube lens would not occur when L is very large, right? That's the sense in which I mean "magnification." (Not "zooming in" to a narrower field of view.) And if so, then it seems reasonable to try to tune L to match the tube lens' aperture and the sensor size. Or am I all wrong still and in need of an explanation?

Thank you.

[Scarodactyl]

Doesn't increasing L create a larger image at the tube lens as well as at the sensor plane?

No, it does not. Peripheral light is lost but the magnification does not change.

[zondar]

Doesn't increasing L create a larger image at the tube lens as well as at the sensor plane?

No, it does not. Peripheral light is lost but the magnification does not change.

If we look at the geometry of the figure on the tube-lens calculation page, we see that light from a given single point of the subject refracts, etc., to a cone (related to the NA), which enters the objective and is formed into a parallel bundle.

If that point is from the center of the subject, then that parallel bundle exits the objective horizontally and does not diverge from the center-line of the optics. But if the light is from a point at the edge of the subject, then the parallel bundle formed from that point exits at an angle.

The parallel bundle from the center of the subject is not given to vignetting by the tube-lens as long as the tube-lens aperture is equal or larger than the objective's exit pupil.

The parallel bundle from the edge of the subject, exiting at an angle, diverges from the center-line. That bundle is subject to being cut off if L is so long that the diverging bundle never hits the tube lens at all. At the sensor plane, after the parallel bundles are reformed to points, we find that it's the periphery of the subject that is lost (and not, for example, the entire image becoming dimmer). This is classic vignetting.

So far so good? Now, as we make L bigger, more and more of the periphery is cut off. Make L a few meters, even, and the vignetting would be so severe that only the very center of the subject is visible at all. And yet the full image circle projected by the tube-lens is still apparent at the sensor (it hasn't gotten smaller). That full image circle only contains the very center of the subject, hence magnifying it.

Alternately, with the same large L that causes severe vignetting, we now swap in a larger-diameter tube-lens, big enough to catch all diverging bundles without vignetting. Then at the sensor we would find an image of the whole subject, but projected to a larger image circle. Is this not increased magnification?

If this is all wrong, then an explanation of how this all works in reality would be appreciated.

Thank you.

[apochronaut]

No, the image size does not change.

Otherwise, vignetting at the tube lens would not occur when L is very large, right?

Thank you.

Rays can be parallel but not all rays will be parallel with each other, nor parallel with the central axis. Thus the longer the infinity space the more rays escape the optical pathway.

[zondar]

OK, I figured it out.

When L is made larger to the point that vignetting starts to happen, the parallel bundles that still do strike the edge of the tube-lens come in at a shallower angle. This means that the tube lens bends the resulting point of light not straight out parallel to the optics, but inward, towards the center of the image circle. So the image circle does get smaller.

Taken to extremes: Imagine that L is infinite. Then the tube lens sees only light from the central point of the subject - a single point. There is only one parallel bundle left, in a sense - the one from the central point. All the others have diverged off to nowhere. Then the tube lens dutifully focuses that single bundle back to a single point at the center of the nominal image circle. This is "infinite vignetting."

So in the end, I surmise that one mostly just has to avoid vignetting (tube lens aperture > than objective exit pupil, not making L too long, etc.), and otherwise L doesn't matter greatly.

[apochronaut]

Doesn't increasing L create a larger image at the tube lens as well as at the sensor plane?

No, it does not. Peripheral light is lost but the magnification does not change.

If we look at the geometry of the figure on the tube-lens calculation page, we see that light from a given single point of the subject refracts, etc., to a cone (related to the NA), which enters the objective and is formed into a parallel bundle.

If that point is from the center of the subject, then that parallel bundle exits the objective horizontally and does not diverge from the center-line of the optics. But if the light is from a point at the edge of the subject, then the parallel bundle formed from that point exits at an angle.

The parallel bundle from the center of the subject is not given to vignetting by the tube-lens as long as the tube-lens aperture is equal or larger than the objective's exit pupil.

The parallel bundle from the edge of the subject, exiting at an angle, diverges from the center-line. That bundle is subject to being cut off if L is so long that the diverging bundle never hits the tube lens at all. At the sensor plane, after the parallel bundles are reformed to points, we find that it's the periphery of the subject that is lost (and not, for example, the entire image becoming dimmer). This is classic vignetting.

So far so good? Now, as we make L bigger, more and more of the periphery is cut off. Make L a few meters, even, and the vignetting would be so severe that only the very center of the subject is visible at all. And yet the full image circle projected by the tube-lens is still apparent at the sensor (it hasn't gotten smaller). That full image circle only contains the very center of the subject, hence magnifying it.

Alternately, with the same large L that causes severe vignetting, we now swap in a larger-diameter tube-lens, big enough to catch all diverging bundles without vignetting. Then at the sensor we would find an image of the whole subject, but projected to a larger image circle. Is this not increased magnification?

If this is all wrong, then an explanation of how this all works in reality would be appreciated.

Thank you.

If you make the infinity space such that a complete illuminated image created by the tube lens just equals the smallest dimension of the sensor just at the point where vignetting commences, then keeping the sensor to tube lens distance constant, you lengthen the infinity space : the illuminated image circle will shrink on the sensor consistent with any vignetting caused at the tube lens. The image of the tube lens will still be the same diameter as the smallest sensor dimension but the illuminated portion of it will be smaller. Increasing the diameter of the tube lens will just increase the portion of the image that is not illuminated. The increased dimension of the tube lens image will be outside the smallest dimension of the sensor as well.

[zondar]

Yeah, I figured it out in the previous post, including for the case in which L is infinite. The key was to recognize that increasing L results in a decrease in the angle of attack of light at the tube-lens periphery, hence causing the image circle to shrink. If L is infinity, the divergence of the parallel-bundle wavefront becomes zero, and only a single parallel bundle, emitted from a single point of the subject, is received by the tube-lens, which then focuses it down to only a single point again - "infinite vignetting."

I have noted a specific L listed in the specs of some infinity objectives, e.g. 142 mm; probably the maximum L before vignetting occurs when in conjunction with their own tube lens. I'll see if I can verify that.

Thanks all for the discussion.