

Exit Pupil
#1
Posted 08 March 2005 - 12:43 PM

#3
Posted 08 March 2005 - 01:51 PM
#4
Posted 08 March 2005 - 01:52 PM
Here is a good article in the CN Reports area pertaining to exit pupil.
http://www.cloudynig...al/binoexit.pdf
Patrick
#5
Posted 08 March 2005 - 02:07 PM
It refers to the size of the light cone that "exits" the eyepiece. Think of it as a cylinder of light. The exit pupil is the diameter of this cylinder.
The average human eye has a pupil size of about 7mm when fully dialted. So the view in an eyepiece will get brighter with larger exit pupils until you reach the 7mm size. Then it won't get any brighter.
I've heard that having an exit pupil larger than 7mm wastes light, making it sound like you lose image resolution or something like that. Personally I think that while not all the light enters your eyes, you don't lose any image details. It's just that an 8mm exit pupil is not any brighter than a 7mm exit pupil. I think Al Nagler said something to this effect, too, on his web page.
Hope this helps some

Mark
#6
Posted 08 March 2005 - 02:37 PM
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An exit pupil which is larger than the opening of your own pupil effectively "masks" the telescopes aperture so that in theory you do lose resolution. When your eye is open to 7mm, you are this is not a limiting factor because you eye cannot resolve the detail anyway. This can be an issue during the day when your pupil might only be open a 2 mm or less.
jon
#7
Posted 08 March 2005 - 02:47 PM

Mark
#8
Posted 08 March 2005 - 04:19 PM
#9
Posted 08 March 2005 - 04:25 PM
Jon has already reinforced that any exit puil larger than your eye does indeed cause a loss. The losses are total light and resolution. But as Jon points out and the article clearly explains, these large exit pupils can only be achieved by using low magnifications, and that is not where you are attempting to achieve maximum resolution.
You will find all of that (clearly, I hope) explained in the article. The article uses a lot of references to binoculars, but all the same applies equally to telescopes. After you read my article, you may want to visit the "Discuss this article" forum posts and read what Bill Ferris has to say about Extended Objects which is a better explanation for that specific topic than I gave.
edz
#10
Posted 08 March 2005 - 04:45 PM
Mark
#11
Posted 08 March 2005 - 04:58 PM
#12
Posted 08 March 2005 - 05:30 PM
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Very true...
One issue that has not been mentioned yet:
The reflector, the Central Obstruction and the Exit Pupil...
If you look at the eyepiece from a distance during the day, you will see a circle of light. To first order this is the exit pupil. If the scope in question has a secondary mirror obstructing the light path, you will see a dark spot in the center of the exit pupil, this dark spot is caused by the central obstruction.
The size of the dark "hole" in the exit pupil can easily be estimated. The size of the exit pupil is equal to the focal length of the eyepiece divided by the focal ratio of the telescope.
The central obstruction has its own "focal ratio" which can be found either by dividing the eyepiece focal length by scope's focal ratio or by dividing the focal length of the scope by the diameter of the obstruction. The diameter of the hole is then calculated in the same manner as the exit pupil.
Imagine an F4 scope with a 25% central obstruction viewing during the day or maybe the full moon. The eyepiece is a beautiful 32mm TeleVue Widefield...
The exit pupil is a slightly wasteful 8mm (32mm/F4=8mm) and the exit pupil of the hole in the center is 2mm (8mm/25%= 2mm)
Now it turns out that when the light is bright, your own eye contracts down so that it only open about 2mm.
This means what you see when the light is bright is often not the image but the dark spot in the center. Not a good thing.
This same thing can happen with an SCT that is used with a focal reducer, normally F6.3. SCT's normally have central obstructions larger than Newtonians so the effective focal ratio of the central obstruction can be about the same as a Newtonian with similar consequences.
jon
#13
Posted 08 March 2005 - 05:51 PM
Is there a minimum exit pupil size that you would want to stay away from, or is that scope(apature,focal length,etc.) dependant?What is the minimum useful exit pupil size or is it relevent?
Albie, most folks eyes don't really resolve very well under 0.5mm exit pupil. Also, aside from wide field applications, a good many folks find that eyepieces that produce something in the neighborhood of a 2mm exit pupil are favored for many general situations. The fastest way to figure exit pupil for eyepieces is to divide the focal length of the eyepiece by the focal ratio of the scope. Thus, a 38mm eyepiece used in an f/7 scope gives an exit pupil of ~5.43, whereas a 6mm eyepiece used in the same f/7 scope gives an exit pupil of ~0.86mm. In theory, if your viewing conditions support it, you should be able to go all the way down to a 4mm eyepiece for a 0.58mm exit pupil. But, you may well find you prefer less magnification and a larger exit pupil for most of your viewing.
If you wish to figure exit pupil for binoculars, a more useful formula is objective size in mm divided by magnification. Thus a set of 10x50s have a 5mm exit pupil. This also works for scopes, but is more cumbersome than the first method.
#14
Posted 08 March 2005 - 11:35 PM
#15
Posted 09 March 2005 - 12:57 AM
Keep well in mind that you may find that your local conditions are more limiting than your equipment. I find that's true more often than not. Tonight, nothing beyond a 1mm exit pupil was at all useful. Sometimes it's more restricted than that. So it goes.
Mike Swaim
#16
Posted 09 March 2005 - 07:44 AM
I have a 6" scope, that is f/5. The max recommended magnification is about 300x. If I use a 2.5mm eyepiece, I get, 300x. I also have a .5mm exit pupil.
Mark
#17
Posted 09 March 2005 - 08:11 AM
Could minimum exit pupil be the true source of the 40-50x per inch of aperture rule?
They're simply equivalent.
The source is that a given circular aperture diffracts a point source into an Airy disc (i.e., blurs it), and once this Airy disc becomes large enough to see very clearly (which depends on the resolving power of the human eye, obviously), there's little point in increasing magnification.
Although that isn't true in the *strictest* sense, given that some contrast details have to made larger to be seen than those used to determine the resolving power of the eyes.
But it's a good rule of thumb.
#18
Posted 09 March 2005 - 11:50 AM
See David Knisely's message #364280 in the "Eyepieces" forum for a good summary of the physics and physiology behind the "50X per inch" guideline: Re: Eyepiece question - 233x or 260x?
Regards,
Bill in Flagstaff
#19
Posted 09 March 2005 - 01:35 PM
I had known that resolution limits was the reason for the magnification limit for a given aperture, not counting atmosphere of course.
But I had noticed that Mike's answer of .5mm minimum exit pupil, coincided with the max magnification, at least for my scope; I should see if it's true for other apertures (now where's my calculator..)
This is a case where someone new to this concept asked a question (thanks Ablie!

But to be honest, at least for telescopes, I still can't see why it's much of a factor? Ok, exit pupils larget than 7mm "wastes" aperture, and technically will cause loss of detail, but at the low magnifications we're talking about, it seems to be something the average backyard astronomer is not going to notice. Except on reflectors/SCTs/Maks where the central obstruction becomes a true obstruction

It seems most useful in estimating what your view in a given eyepiece will be like, or if you're comparing different magnifications. Or perhaps, maybe it's just over my head and I'll just forget about it and go outside and look at more stars



Mark
#20
Posted 09 March 2005 - 02:10 PM
#21
Posted 09 March 2005 - 03:50 PM
No need for a calculator.But I had noticed that Mike's answer of .5mm minimum exit pupil, coincided with the max magnification, at least for my scope; I should see if it's true for other apertures (now where's my calculator..)
exit pupil = scope_aperture/magnification (directly, or through FLep/f-ratio = FLep/(FLscope/scope_aperture) ).
i.e. magnification = scope_aperture/exit pupil.
It's only because you express eyepiece focal lengths in mm and scope aperture in inches that you need that 50x deus ex machina. With everything expressed in mm,
exit pupil = 1/2mm => magnification=aperture_mm*2.
exit pupil = 7mm => magnification=aperture_mm/7.
Simple, eh? The joys of SI units...