Author Topic: Fields stops, focal point, and Dawes' limit... where is the "magic" happening.  (Read 332 times)

Damon Brigham

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unless I also moved to a place twice as close to the screen.

I think that's what Jon was going for. "Cut tube in half" = remove the half of the tube closest to you and move forward to where you cut it.

retpoiwerround

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unless I also moved to a place twice as close to the screen.

I think that's what Jon was going for. "Cut tube in half" = remove the half of the tube closest to you and move forward to where you cut it.
Jon

cludertypos

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To obtain a fuller understanding, do not consider just the two triangles resulting from a ray trace for a single image point on axis. That is, the first triangle ahead of the focus and its smaller version between focus and eyepiece. Picture a huge number of such triangles, their common apexes being scattered all over a focal surface of some diameter even larger than your largest field stop.

Now when you insert a variety of eyepieces having different field stop diameters, more or fewer of that multitude of triangles will be clipped by the field stop.

Suppose you have two eyepieces which just happen to have identical field stop diameters, but the focal lengths differ by a factor of 2. The shorter f.l. job will deliver an apparent FoV (AFoV) twice as large, which is 4X more area. Both will deliver the same true field (TFoV), but the shorter one will magnify 2X more.

In terms of the light admitted over the full TFoV, both eyepieces are the same. But the more magnified view will be dimmer in *surface brightness* due to the smaller exit pupil (provided that exit pupil is smaller than your iris, of course.) If the longer f.l. eyepiece delivered an exit pupil no larger than your iris, the shorter f.l. eyepiece's half-diameter exit pupil would dim the image to 1/4 that seen by the lower power ocular. This nicely balanced out; the same *total* light in both images covers 4X or 1/4 the area on your retina, with the surface brightness concomitantly being 1/4 or 4X, respectively.

The resolution of the Fresnel pattern of diffraction is solely dependent on the exit pupil diameter. You can discover your own threshold by looking at any near point-like light source through pinholes in foil. Where you just begin to resolve an Airy disk and first ring, measure the pinhole with a magnifying loupe/reticle, or even with a reversed eyepiece and ruler, to see at what exit pupil you will detect the same.

lehroldwebbdep

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unless I also moved to a place twice as close to the screen.

I think that's what Jon was going for. "Cut tube in half" = remove the half of the tube closest to you and move forward to where you cut it.
I was visualizing a TV more than ten feet away.

Thomas Homer

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Thanks Glenn, I think this concept is along the lines of what I was missing. Simple diagrams I find on the internet are just that, simple. They show you how it works at a high level and not what's really happening.

Greg Fleming

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I'm hoping someone can help me gain a deeper understanding of what's actually happening when an eyepiece is reducing the FOV. I understand the math on FOV, Magnification, Exit Pupil, etc. But when I try to reconcile what is actually happening I end up with a conundrum regardless of how I slice it. As follows:

A short focal length EP is closer to the focal point, and this results in magnification and reducing the FOV, but where does the rest of the light gathered go? Is it cut off by the field stop, or is there some "magic" happening as the light expands from the focal point that results in a more narrow FOV the closer you are to the focal point?

If the latter, is there a word for the "magic" causing the FOV to get wider as you get further from the focal point?

If the former, and the field stop is cutting off the image, why doesn't that affect Dawes' limit (or does it but that's countered by higher magnification?) I know if I were to mask the outer edge on the front of the scope, say making a 5" a 4", then the resolution capability of the scope would be reduced. It seems the same thing should be happening at the back end with a field stop as does happen with the mask on the aperture. But we use high magnification to split double stars and get detail on planets so it appears "masking" at the back end via the field stop has a different effect than masking the aperture. So this is where I run into "magic" again. What happens at the focal point that makes it different on either side, does the light "mix" somehow?

I hope I've explained this well enough so someone can understand my conundrum. A reply with a word to google or link to something that explains this "magic" would be greatly appreciated. And sorry to keep using the word "magic", it's just the word I use for things that happen which I understand the output but don't have a full understanding of the inner workings.Thanks for reading and clear skies,

Bob

The telescope produces an image on a focal plane. The eyepiece is a small magnifier of the image on the focal plane. The shorter the focal length of the eyepiece, the higher its inherent power.

The portion of the telescope's focal plane seen is determined by the "field stop" of the eyepiece. A larger field stop at the same power shows a larger true field.
So lower power eyepieces have larger field stops, which is why they see a larger portion of the telescope's focal plane, and thus, true field.

The longer the focal length of a scope, the larger the image is on its final focal plane, so the larger the image is seen with any eyepiece--also a narrower true field.
The apparent field of the eyepiece doesn't change from scope to scope, but the true field will depending on the image scale on the focal plane of the telescope.

The star in the center of the field has its image produced by the entire objective, whether mirror or lens. If there is no source of vignetting (dimming due to light cut-off) in the system,
this would be true of every star in the entire field. So the maximum resolution of the scope is the same everywhere in the field.

Larger scopes have less diffraction, so their Airy Discs are smaller. Hence, they resolve smaller details.
The light ray from a star is a plane that hits the whole earth and the opening of the scope is a "hole" that causes diffraction after the light passes through it. It's the same as the passage
of a wave through a hole. The bigger that hole, the less diffraction there is.

These are just a few conceptual points to help you grok what's happening a bit better.

mellidonde

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It's misleading to say the Fresnel pattern is *smaller* because "there is less diffraction" for a larger aperture.

A larger aperture creates a smaller diffraction pattern because the ratio of aperture (diameter) to the wavelength of light is larger. But this is modified by the f/ratio. All different apertures at the same f/ratio produce a diffraction pattern of the *same physical size.* Of course, the *angular* size will decrease with increasing aperture (for given f/ratio.)

Irrespective of aperture, the *relative* intensity of diffraction is constant. No matter the aperture, or f/ratio, or actual magnification, at given exit pupil the angular size of diffraction is constant, and its intensity relative to the source is also constant. For example:

You have a 2" and a 20" aperture scope operating at the same 0.7mm exit pupil. Through the 2" you're looking at a star of magnitude 2. Through the 20" you're examining a 7th magnitude star. In either eyepiece the stars will *appear* to be equally bright. (The 10X bigger aperture collects 100X the light, which is equivalent to 5 magnitudes.) And so will the diffraction rings appear to be equally bright, and of the same apparent size.