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Equipment Discussions >> Refractors

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Jon Isaacs
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Re: Reflector/Refractor equivalence formula new [Re: Asbytec]
      #5785454 - 04/08/13 08:54 AM

Quote:

So, how does one reconcile that with the MTF? The image on the focal plane should remain consistent with MTF. It's easier to resolve the Whirlpool's spiral arms than individual stars, the lower spacial frequency arms are set off - contrast wise - by darker space between them and the core. So much so, we can see them visually.




Glenn is pointing out that the MTF of the telescope has little to do with what we see at the eyepiece when looking at faint objects because it's our eye that is the limitation rather than the telescope, any telescope provides sufficient contrast, sufficient resolution. Viewing the moon through a solar filter demonstrates this very nicely. I know the detail is there and clearly the telescope can resolve the detail with little loss of contrast.

But I can't see that detail, that contrast because my eye is a very poor sensor at low light levels. All that is necessary to see that detail is to brighten up the image maybe 10,000 times or even more. Unfortunately, a telescope cannot do this, the brightness of M51 is limited by the 7mm (or so) entrance pupil of the eye. When I look at M51 in my 25 inch with a 6mm exit pupil, it is no brighter than it is when I look at with a 60mm refractor with a 6mm exit pupil, it is just larger so I see more detail.

Jon


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Asbytec
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Re: Reflector/Refractor equivalence formula new [Re: Jon Isaacs]
      #5785739 - 04/08/13 11:31 AM

Jon, again that's interesting. I think the MTF is still at play, but I believe you're correct about brightness. The detail is there, the eye's response is too low. And the MTF does not model brightness. I ma still trying to understand the brightness and contrast relationship, I think it's in the eye. But, does that mean the MTF does nto accurately describe contrast and resolution. Not sure, the relationship between contrast and resolution still exist as evidenced by time exposures.

Still, both a 6" refractor and a C14 will have the same trouble with brightness, so any RoT would hold at a given exit pupil. <CLICK> It just clicked with me the relation between exit pupil and MTF. what we're really taking about is the translation of full contrast of the actual object to the focal plane with the resulting loss of contrast through any telescope. It's just a brightness issue.

But, thinking out loud, its curious a larger scope will brighten the filtered moon and allow some better detail to be resolved with contrast closer to the eye's response. Still trying to figure out what that means.

Thanks, Glenn for bringing that up in the first place and both of you for discussing it.

Edited by Asbytec (04/08/13 01:56 PM)


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Jon Isaacs
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Re: Reflector/Refractor equivalence formula new [Re: Asbytec]
      #5785793 - 04/08/13 11:54 AM

Quote:

I ma still trying to understand the brightness and contrast relationship, I think it's in the eye. But, does that mean the MTF does nto accurately describe contrast and resolution. Not sure, the relationship between contrast and resolution still exist as evidenced by time exposures.





Indeed, it is in the eye, the ability of the eye to detect the contrast differences, to resolve the details is what governs the response of the telescope/eye system. The contrast is still there, the resolution is still there at the focal plane, in the exit pupil.

But it doesn't matter because the eye cannot detect it. The MTF describes the response of the telescope/eyepiece system, it does not describe the response of the Telescope/eyepiece/eye system. That's the critical factor, a poor mirror, a poor objective, a good lens, a good mirror, the MTF describes what is present at the focal plane but it doesn't matter because you just can't see it, it's too dim for the eye to see. All those details on the moon, they are still there, they are clean and sharp, you just cannot see them.

Jon


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Asbytec
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Re: Reflector/Refractor equivalence formula new [Re: Jon Isaacs]
      #5785970 - 04/08/13 01:08 PM

Sure, still a MTF BASED RoT, as plotted by Eddgie above, can be derived because brightness applies to all scopes and apertures pretty much equally and relatively. I would think, anyway.

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azure1961p
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Re: Reflector/Refractor equivalence formula new [Re: Jon Isaacs]
      #5785972 - 04/08/13 01:08 PM

Jon,

A fine testament to the contrast present at the focal plane but undetected is visual versus CCD, with a 10" for example. A decent 10" f/4 system can turn out a pic that you'd swear was from a huge apo of at least that aperture. Seen or not the CCD detects it and the post processing enhances it - even utterly invisible things simply because they were again, at the focal plane.'

Pete


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GlennLeDrew
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Re: Reflector/Refractor equivalence formula new [Re: azure1961p]
      #5786094 - 04/08/13 01:52 PM

For virtually all MTF plots, only the optics ahead of the focus are considered; in other words, the eyepiece is not usually modelled.

The MTF chart is completely independent of target brightness; whether the photon flux be sparse or intense, the optical effect is the same.

The MTF is only fully meaningful when the detector can fully sample down to the lowest resolution of the optic. A CCD with really big pixels (more so when pixels are binned into groups), or visually when the exit pupil is large and/or the target is dim, one is not sampling the regime of the smallest scales (in the lower right of the chart, as usually constructed.)

Nevertheless, whatever extra resolution the optic delivers is there, in spite if the inadequacies of the detector to sample it.


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Jon Isaacs
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Re: Reflector/Refractor equivalence formula new [Re: Asbytec]
      #5786108 - 04/08/13 02:01 PM

Quote:

Sure, still a MTF BASED RoT, as plotted by Eddgie above, can be derived because brightness applies to all scopes and apertures pretty much equally and relatively. I would think, anyway.




I am not sure of the need to continue to discuss the MTF when viewing objects with faint surface brightnesses. I believe Glenn's point is that at these low light levels, remembering that a telescope does not increase the surface brightness of an extended object, the eye simply does not have the ability to take see the contrast, to resolve the differences that the MTF might show.

If the eye could detect 5 arc-minute details of low surface brightness objects, Andromeda would be spectacularly detailed naked eye...

The MTF just doesn't apply to the response of the human eye and it is the that determines what you see. The other night we were looking at the Leo Triplet in my 25 inch and after battling the wind 20mph+ wind we gave up and pulled out the 16 inch. The 25 inch definitely showed it better but neither showed the detail that an 80mm would show were the eye able to detect it.

Leo Triplet in an 80mm

Jon

Edited by Jon Isaacs (04/08/13 02:08 PM)


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Asbytec
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Re: Reflector/Refractor equivalence formula new [Re: GlennLeDrew]
      #5786194 - 04/08/13 03:16 PM

Yes, the MTF applies to the focal plane, our eyes detect what's there somewhat differently.

Image brightness is not modeled. Right, it's important because image brightness appears to affect contrast. But that's not true, it's the detector that is suffering. And we all suffer in nearly similar ways with regards to image brightness. So, the MTF should still be useful as a model. It puts the theoretical contrast and resolution on the focal plane and our nearly similar visual acuity takes it from there.

I think the MTF is still meaningful despite the acuity of the detector,provided the detectors are at least similar: two pairs of eyeballs, or two CCD chips, for example. It might get confusing because dim objects "appear" to the eye to have less contrast, but they don't.

Quote:

The 25 inch definitely showed it better but neither showed the detail that an 80mm would show were the eye able to detect it.




Yes, correct me if I am wrong, but the 25" and 16" still put better contrast and resolution on the focal plan than the 80mm giving our eyes the "opportunity" to get at that detail...despite the wind.

I never thought of it that way - the filtered moon concept, so that paradigm is interesting. Still, the (aperture - CO) modeling we see plotted on the MTF is still meaningful even for faint fuzzes, ya? Its not perfect because our eyes differ slightly, but close enough for a RoT because all scopes and eyes treat brightness relatively. Lemme sleep on it...interesting.

Edited by Asbytec (04/08/13 03:23 PM)


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GOLGO13
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Re: Reflector/Refractor equivalence formula new [Re: Jon Isaacs]
      #5786215 - 04/08/13 03:28 PM

Quote:

remembering that a telescope does not increase the surface brightness of an extended object, the eye simply does not have the ability to take see the contrast




We discuss this a lot...but I wonder if this is somewhat misleading. Or maybe seeing color in the object requires it to be bigger...but that would also explain why DSOs look brighter in the bigger scope (more of the object hitting your eye).

I've seen color in the Ring Nebula with a 30 inch dob. Is that possible with a smaller scope with a matched exit pupil? A better example for me is the Blue Snowball...I've seen color in many scopes with this one...but it's much more obvious in larger scopes. I guess I'd have to match exit pupils to do an exact test. In my friend's 16 inch it's blue at high magnifications as well as a 30 inch. In fact, in the 30 inch it looked like it does in photographs.

Found this source which seems to have a lot of good info: Surface brightness

Edited by GOLGO13 (04/08/13 03:38 PM)


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Jon Isaacs
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Re: Reflector/Refractor equivalence formula new [Re: Asbytec]
      #5786444 - 04/08/13 05:15 PM

Quote:


Yes, correct me if I am wrong, but the 25" and 16" still put better contrast and resolution on the focal plan than the 80mm giving our eyes the "opportunity" to get at that detail...despite the wind.




It was not the contrast or the resolution at the focal plane, the 80mm has plenty of that as witnessed by the photo. Rather, the larger aperture allowed the identical brightness at a greater magnification, the details covered more of the retina.

If I had used a 40% transmission filter on the larger scope the brightnesses would be the same at equal magnifications. The two scopes could operating at the same magnification and image brightness. Two would show essentially identical images.... If the 80mm, with it's contrast and resolution, could be made equally bright, it be right there too.

Try the moon test... when you see how little you can see of the moon when it's surface brightness is reduced by a 12.6 magnitudes, it's instructive.

Jon


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Asbytec
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Re: Reflector/Refractor equivalence formula new [Re: Jon Isaacs]
      #5786662 - 04/08/13 07:15 PM

I have observed the dark side of the moon being earth lit, that may be the same effect as using a strong filter. Yes, no craters were resolved, not that I remember anyway. They were there, but not illuminated.

Still, my 150mm and your 25" deliver more contrast and resolution to the focal plane. I'm trying to imagine it like this. Take an image of Jupiter in a small refractor and a larger obstructed scope. Its bright enough we might agree it will closely follow the MTF in terms of scale and contrast. Now, dim it down to barely detectable on the retina. The information is still on the focal plane following the MTF.

At the same low power magnification, a large exit pupil though different, the MTF effects are not noticeable. This is because, I'm sure you agree if I am saying it correctly, the diffraction effects are most prominent on small scales out to about 2x the Airy disc, give or take. At low power, those effects are not easily seen and the dominant larger scale effects are almost identical. I believe Glenn was alluding to this effect earlier.

At the same exit pupil, the views would be similar in image scale with the larger aperture providing better resolution. Here it gets confusing, but I believe both scopes will show the same contrast at larger scales, then the obstructed instrument will show a bit less at medium scales...and more at higher frequencies.

Ah, my time is gone, gotta run. Thinking about getting this concept right, so I'll leave the above as is even if it's not right. Thanks.


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jhayes_tucson
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Re: Reflector/Refractor equivalence formula new [Re: Jon Isaacs]
      #5786695 - 04/08/13 07:31 PM

Quote:

Quote:


Yes, correct me if I am wrong, but the 25" and 16" still put better contrast and resolution on the focal plan than the 80mm giving our eyes the "opportunity" to get at that detail...despite the wind.




It was not the contrast or the resolution at the focal plane, the 80mm has plenty of that as witnessed by the photo. Rather, the larger aperture allowed the identical brightness at a greater magnification, the details covered more of the retina.

If I had used a 40% transmission filter on the larger scope the brightnesses would be the same at equal magnifications. The two scopes could operating at the same magnification and image brightness. Two would show essentially identical images.... If the 80mm, with it's contrast and resolution, could be made equally bright, it be right there too.

Try the moon test... when you see how little you can see of the moon when it's surface brightness is reduced by a 12.6 magnitudes, it's instructive.

Jon






I agree that if the image gets too dim, you won’t see much; but, you need to separate what the telescope is doing from what you can see. First, the resolution and MTF response of the telescope is unchanged with the solar filter. In fact, the image has exactly the same contrast with and without the filter. You have simply reduced the brightness of the image so that much of image intensity is below the visual threshold--effectively reducing the contrast of the “signal” from the eyeball—not the telescope. In the limiting case, you can make all telescopes equal by putting a card in front of the telescope (ND=infinity) and you won’t see anything—no matter what size objective it has! The issue with faint extended objects is that much of the detail is at a brightness below what you can visually detect and that’s why you don’t see it. That’s the whole reason to use a camera—to integrate photons, which you can’t do with your eye. For a given telescope aperture, the peak intensity of the image of a point object at the focal plane increases with the inverse of the focal ratio. At a given focal ratio, the peak intensity increases with the square of the diameter. So, for stars, big and fast is better. For an extended object with an eyepiece, the surface intensity will vary as the inverse square of the magnification. You will get the maximum surface brightness when the diameter of the telescope exit pupil is matched to the diameter of the iris in your eye, which is the low magnification limit for any telescope. Go lower and your eye becomes the stop, which raises other issues. (As has been pointed out, the sharpest visual acuity doesn’t occur when the iris is completely filled, but that probably doesn’t matter so much at low magnification.) When the object is sufficiently bright, your iris closes and the minimum magnification increases.


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GlennLeDrew
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Re: Reflector/Refractor equivalence formula new [Re: jhayes_tucson]
      #5786704 - 04/08/13 07:38 PM

When a telesco

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GlennLeDrew
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Re: Reflector/Refractor equivalence formula new [Re: jhayes_tucson]
      #5786748 - 04/08/13 08:06 PM

When a telescope is used afocally, with an eyepiece, the objective f/ratio is of no fundamental importance. (In the real world, it's important only insofar as the availability of eyepiece focal lengths, and designs, is concerned.) The exit pupil is key, not objective f/ratio.

The 'rule' which posits an equivalence of aperture - CO is applicable only in the brighter regime. Its purpose is to find an equivalent aperture where the *small scale* diffraction effects are made similar. For dim objects, the eye fails to resolve in this regime, and so one is no longer concerned with *small scale* diffraction. Now the only concern is system transmission.

The central obstruction does remove some light, but it's not of great concern. If one wishes to be rigorous, that light removed by the CO, and and that not transmitted due to the (usually) one or two less-efficient mirror coatings, can be taken into account in order to find the aperture equivalence as regards image brightness. Coincidentally, the aperture ratio as derived from consideration of diffraction could be somewhat similar to that derived from system transmission; in such case, the one rule is applicable in a broader sense.

The important thing in the end is this; for visual application, below some surface brightness threshold, diffraction effects become quite irrelevant. To the point that in this specific regime of the dim, extended object the MTF chart (for otherwise reasonably decent telescopes) becomes largely meaningless in practice. Naturally, for stars, and any other sufficiently bright sources, the MTF is quite relevant.

The interesting thing about the visual system is that it's effectively 'multi-modal.' In the *same* field of view one can have a wide range of object type and brightness. For all of them, the brain simultaneously applies the required processing in order to eak out the best detail. The mon-faint stars will be seen and resolved to the eye's photopic limit (1-2 arcminutes on the retina), whereas a galaxy might have its core resolved to, say 5 arcminutes (again, on the retina), its larger bulge region to, say 20 arcminutes, and the disk to 1-2 degrees. All at the same time, and over a smoother continuum than suggested in this 'quantized' example.


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Jon Isaacs
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Re: Reflector/Refractor equivalence formula new [Re: jhayes_tucson]
      #5786751 - 04/08/13 08:07 PM

Quote:


I agree that if the image gets too dim, you won’t see much; but, you need to separate what the telescope is doing from what you can see. First, the resolution and MTF response of the telescope is unchanged with the solar filter. In fact, the image has exactly the same contrast with and without the filter. You have simply reduced the brightness of the image so that much of image intensity is below the visual threshold--effectively reducing the contrast of the “signal” from the eyeball—not the telescope.




Exactly, I believe I wrote almost exactly that in my first post. The image exists at the focal plane, in the exit pupil with the same contrast, the same resolution. But the eye cannot see it.

However the importance of this is not that I have reduced the brightness of the moon to the point where I can no longer detect the contrast and resolution that exist at the focal plane. Rather, the importance is that most extended deep space objects are even dimmer than the moon viewed through a solar filter, that most of the details that exist at the focal plane, in the exit pupil, are below the "visual threshold."

The importance of viewing the moon through a solar filter is that it provides a first hand example of how poorly the eye does resolve detail at low light levels. It is just not an academic discussion with calculations and discussions of surface brightnesses and visual thresholds and the need for a detail to be degrees in dimension at the retina.

Rather it is a tool to see first hand, the workings of the eye... It's no longer academic, it's a real experience that one can learn from. I definitely recommend it for those who have not tried it.

Jon


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Ziggy943
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Re: Reflector/Refractor equivalence formula new [Re: JimP]
      #5786902 - 04/08/13 09:23 PM

Quote:

Ziggy, if I did not believe that you are completely and absolutely sold on Celestron telescopes I would find your posts to be more objective. As it is, you, me and most others on these forums have a bias toward one type scope or another and find ways to show how the telescope we chose, the one we spent big money on, is the best. The old saying "there are lies and there are statistics" is true.

JimP




Jim,

I'm not sure what I said that made you think that I am "completely and absolutely sold on Celestron telescopes." I'm not. In my experience I have seen some very fine samples and IMHO have been consistently better than Meades but I don't own any SCT and I am not sold on them. They do fill a great niche for amateur astronomy and are one of the great innovations of the last century.

I admit to having a bias toward certain types of telescopes. You have most of them


Edited by Ziggy943 (04/08/13 09:32 PM)


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jhayes_tucson
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Re: Reflector/Refractor equivalence formula new [Re: GlennLeDrew]
      #5787233 - 04/09/13 01:25 AM Attachment (26 downloads)

Quote:

When a telescope is used afocally, with an eyepiece, the objective f/ratio is of no fundamental importance. (In the real world, it's important only insofar as the availability of eyepiece focal lengths, and designs, is concerned.) The exit pupil is key, not objective f/ratio.

The 'rule' which posits an equivalence of aperture - CO is applicable only in the brighter regime. Its purpose is to find an equivalent aperture where the *small scale* diffraction effects are made similar. For dim objects, the eye fails to resolve in this regime, and so one is no longer concerned with *small scale* diffraction. Now the only concern is system transmission.

The central obstruction does remove some light, but it's not of great concern. If one wishes to be rigorous, that light removed by the CO, and and that not transmitted due to the (usually) one or two less-efficient mirror coatings, can be taken into account in order to find the aperture equivalence as regards image brightness. Coincidentally, the aperture ratio as derived from consideration of diffraction could be somewhat similar to that derived from system transmission; in such case, the one rule is applicable in a broader sense.

The important thing in the end is this; for visual application, below some surface brightness threshold, diffraction effects become quite irrelevant. To the point that in this specific regime of the dim, extended object the MTF chart (for otherwise reasonably decent telescopes) becomes largely meaningless in practice. Naturally, for stars, and any other sufficiently bright sources, the MTF is quite relevant.

The interesting thing about the visual system is that it's effectively 'multi-modal.' In the *same* field of view one can have a wide range of object type and brightness. For all of them, the brain simultaneously applies the required processing in order to eak out the best detail. The mon-faint stars will be seen and resolved to the eye's photopic limit (1-2 arcminutes on the retina), whereas a galaxy might have its core resolved to, say 5 arcminutes (again, on the retina), its larger bulge region to, say 20 arcminutes, and the disk to 1-2 degrees. All at the same time, and over a smoother continuum than suggested in this 'quantized' example.







The F/# of an optical system appears in too many performance parameters to list so we'll have to disagree on the importance of F/# as it relates to telescope performance—with or without an eyepiece. Be careful, MTF is relevant ONLY for extended, spatially incoherent sources—not stars (which are spatially coherent)--but that’s an explanation best left for another thread. The discussion here appears to have evolved into a discussion about visual acuity and perception, which I believe has little relevance for comparing the performance of reflectors to refractors. Visual acuity varies greatly between observers and perception issues lead to biased opinions, which are not comparable in any measurable way. In my view, this comparison is possible but it has to be based on physical optics so without including the effects of diffraction, my participation isn't going to be very valuable so I'll wind down active involvement in this discussion.

One thing that I want to mention is that I see a number of confusing references to “the exit pupil" that go beyond simple semantics. So, I want to clarify some first order optics so that everyone is on the same page. The exit pupil is defined as the image of the stop in the "image space." For a simple Newtonian or refractor without an eyepiece, the stop, entrance pupil and exit pupil all lie in the same physical location on top of the mirror or objective lens. With an eyepiece (with positive power,) the exit pupil lies behind the eyepiece at a location where the eyepiece forms an image of the objective. It has a diameter (given by the diameter of the stop divided by the magnification) and a specific axial location. When the eyepiece is focused at infinity, the focus of the eyepiece sits at the focal plane of the objective and the beam of light coming out of the telescope will be collimated (i.e. the marginal rays are parallel) and there will be no "image plane" for objects at infinity (like stars.) Your eye refocuses the collimated beam onto the retina to form the image you see. Placing your eye at the location of the exit pupil allows collimated light from the objective at different field angles to enter the iris all at once so that you see a wide field of view. That’s where eye-relief comes in: It’s the distance of the exit pupil behind the last component in the eyepiece. It needs to be at a comfortable distance or your eye can bang into the lens and that gets uncomfortable. I’ve attached a schematic diagram.

I apologize to all you guys who already know this stuff; but, since we aren't using diagrams, I just want to make sure that the words line up with the meaning or it gets confusing. It screws me up when I see references to images “in the exit pupil” because it sounds like a reference to an image of the primary mirror (or lens) and that makes no sense in this context.


John


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GlennLeDrew
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Re: Reflector/Refractor equivalence formula [Re: jhayes_tucson]
      #5787386 - 04/09/13 05:09 AM

John,
That was a fine summary of the exit pupil!

I take it as given (perhaps too readily?) that all involved here well enough understand at least the basic geometry of the exit pupil. I suspect/hope that its merely idiomatic when we see such as, "in the exit pupil," when it's more correct to say, "through the exit pupil."

I look at the exit pupil this way. It's the coupler between eye and objective. It functionally allows the scope/eye system to behave as though the eye's lens/iris/cornea has been replaced by a lens having the diameter of the objective (as long as the iris is larger than the exit pupil), and a focal length equal to

Obj. diam. * eye f.l. / exit pupil, or

100 * 20 / 2 = 1,000mm f.l., yielding f/10

This makes sense, since the exit pupil is making the eye's own lens work at f/10 (c.f. 20 / 2 = 10.)

For comparison, a 7mm iris makes the eye (at 20mm f.l.) work at f/2.86. In such case, our 100mm scope working at a 7mm exit pupil results in a 'replacement' lens for the eye of 100mm diameter and 286mm f.l.


Please firgive the digression. I tend to become a little apostolic in my zeal to provide perspective. At any rate, it seems the business of diffraction effects and their representation in the MTF chart has about run its course. I'm allowing myself the luxury of exploring other factors which have a bearing on the visual experience.


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Jon Isaacs
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Re: Reflector/Refractor equivalence formula [Re: jhayes_tucson]
      #5787425 - 04/09/13 05:57 AM

Quote:

The F/# of an optical system appears in too many performance parameters to list so we'll have to disagree on the importance of F/# as it relates to telescope performance—with or without an eyepiece.




A telescope system, that being an objective plus an eyepiece, is afocal and so to a first order, it's performance can be described without regard to focal ratio, magnification and exit pupil are sufficient to describe the performance parameters.

It is not necessary to know the focal ratio to understand the brightness of the stars, the surface brightness of an extended object, the resolution possible, this all can be determined from the exit pupil and/or and magnification alone.

Higher order issues, aberrations of all sorts can depend on the focal ratio, design issues, eyepiece choices, there are many places where the focal ratio is important but it is not necessary to know the focal ratio of the telescope to discuss what one sees in the most basic terms because an objective with an eyepiece is afocal, a black box, and its function can be described by the exit pupil and the magnification alone.

Jon


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jhayes_tucson
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Re: Reflector/Refractor equivalence formula [Re: Jon Isaacs]
      #5788288 - 04/09/13 02:26 PM

Quote:

Higher order issues, aberrations of all sorts can depend on the focal ratio, design issues, eyepiece choices, there are many places where the focal ratio is important ...




Ok, on this point, it sounds like we are in violent agreement.


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