How much of the mirror is used on a Newt?

[ColHut]

For the technicaly-enabled, Is there a way of working out how much of the mirror is actually used for diferent eyepieces from 5 to 35mm (say)? How does using 2" versus 1.25" affect this???

Somewhat puzzled how this all works.

regards

[Don Pensack]

OK, think of the starlight as a wave that hits the entire Earth. The light from that star fills the entire mirror. This is true for every star in the field of view. For every eyepiece.
And it would be true if your telescope were 100 meters across.
What the eyepiece does is determine how much of the parent focal plane of the telescope is examined.
But the entire mirror is used for every point on that focal plane (excepting the umbra of the secondary shadow).
[In fact, as little as seventy percent of your mirror is illuminating a spot at the edge of the field of view, but that is because the secondary mirror isn't large enough to retrieve the entire light cone at every angle, not because the primary isn't receiving the light.]
Stopping down the mirror will reduce the brightness of the image, but using a different eyepiece does not.
Larger eyepieces see a larger portion of the telescope's focal plane (hence, a larger field of view), but they cannot make the star images brighter and they do not use any more or less of the mirror's surface.
Hope that explanation makes sense.
Don

[Robh]

Don,

I'm not clear on one point. If the entire primary mirror is utilised by any eyepiece, why does the brightness of the field reduce at higher magnifications? If an eyepiece uses a smaller portion of the telescope's focal plane, resulting in a smaller exit pupil, isn't this equivalent to reducing the aperture size?

Regards, Rob.

[AstroJunk]

Break it down into two distinct stages.

First: The mirror, or lens for that matter concentrates the light into an image at the focal plane. Every available bit of aperture is used to form that image. That is a constant.

Second: An eyepiece is used to magnify the image at the focal plane. Different eyepieces will use different amounts of the image at the focal plane.

The resultant effect is that you get an image at whatever magnification and whatever FoV containing all of the light that the telescope collected for the object that you are observing, but not necessarily all the light that the mirror collected (ignoring the issue of exit pupil).

I think I'm going to have to lie down...

[Jone5y]

"I'm not clear on one point. If the entire primary mirror is utilised by any eyepiece, why does the brightness of the field reduce at higher magnifications"

I was always under the impression that the field dims because the available light is being spread over a larger area within the field of view. An analogy being pouring some water (light) over a given area. Pour a litre over a square foot and it looks wet (bright), pour the same litre over an acre and it wont look as wet (bright).

Hope I haven't added to the confusion.

[ColHut]

Ahh thanks -the glimmer of an understanding.

so for different sized mirrors (say 10" and 6") but with eyepieces giving the same magnification in both, will the object viewed have more "intensity" fromm the larger mirror because so much more light was collected?

does it follow that we therefore usually only view a tiny part of the image at the focal plane?

Can this be calculated? It seems that the TFOV through even a 2" eyepiece in a fast dob only uses a tiny part of the actual light collected?:doh:

[AstroJunk]

But remember - it's using ALL the light collected for what ever you are looking at. Increasing the FoV just lets you see other things too. Go look through a toilet roll tube (post your photo here):D

[ColHut]

I can't get my camera to fit on the tube!:)

[Don Pensack]

Yes, if you calculate the field of view per millimeter of focal plane, you'll be able to take the field stop size of the eyepiece and calculate the size of the field of view.
Example:
FL = 1200mm
2" eyepiece inside diameter = 46mm.
True field = 46/1200 * 57.3 = 2.1965 degrees of field=131.8'
So, 46mm = 131.8' of field.
An eyepiece with a 10mm field stop sees 10/46 x 131.8' of field = 28.65' of true field.
Think of the image scale on the telescope's focal plane as being related to its focal length, and an eyepiece acting as a simple magnifier of part of that field.
How much dimmer is the extended image of the background sky in the eyepiece than the apparent illumination at the telescope's focal plane?
Simple.
The True field of that 10mm field stop eyepiece was 28.65'. Let's say the apparent field of the eyepiece was 68 degrees. The "parent" field of the eyepiece is (68x60) 4080'. To get that field width from a 28.65' true field requires 142X. Brightness of the background is 1/(142 squared) = 1/20164 x as bright as the image on the focal plane.
But, stars don't stretch with magnification--they're points. So the background sky is 1/20164 X as bright as the image at the focal plane (about 10.8 magnitudes fainter), but the stars are the same brightness. So the contrast is greatly improved, and we see fainter stars with increased magnification.

Back to the part of the focal plane being viewed:
Assuming two telescopes, one of which has twice the aperture of the other, and having identical focal lengths: the larger scope will put twice the number of photons per square millimeter on the focal plane. The image scale (minutes per millimeter) will be exactly the same, but more photons will be being crammed into that area.
Remember that a star on axis is the result of the entire aperture bringing the image of that star to the focal plane. When you view that star, regardless of what focal length or field width of eyepiece being used, you are using the entire aperture for that on-axis star image. It is true that you are examining only a portion of the focal plane, but the eyepiece does not determine what portion of the primary mirror is used to illuminate anything on the focal plane of the telescope--it merely magnifies that image. So an extended object will get fainter with magnification, but stars do not.
Hope my explanation makes sense. Let me know if it doesn't.

[Don Pensack]

Rob,
The eyepiece magnifies the focal plane to a larger size. The higher the magnification of the eyepiece, the more the small amount of the focal plane being viewed is magnified. In the hypothetical example I listed, a 28.5' field is expanded to 68 degrees by the eyepiece. That is a 142X magnification of the field being viewed, in width. In area, it's 20,164X the area. That means the eyepiece spreads the background sky brightness out from 28.5' to 68 degrees. The same amount of light now covers an area 20,164X as large, meaning the surface brightness of the background went down 10.8 magnitudes.
It's not the equivalent of reducing the aperture. The brightness of each square millimeter on the focal plane is related to the aperture of the scope. The eyepiece has no effect on the brightness AT THE FOCAL PLANE, while reducing the aperture would.

[ColHut]

Okay starting to get the picture.:)
The maximum diameter of the actual focal plane is also limited by the minor axis of the secondary mirror in a newt?


I am struggling with the idea that the image I see looking down the focus tube (unmagnified) seems a wide angle. But if I am only using a part of the mirror when using eyepices what's the point of all the extra mirror? Does this make sense?:help:

[Starkler]

Col when you step back from the eyepiece and look at it you will see a donut shaped circle of light. Thats an image of the primary mirror and the hole in the middle is the image of the secondary. The size of that image is whats called the exit pupil size.

The amount of the primary mirror been seen and used is ALL OF IT regardless of magnification if the secondary mirror is sized correctly and your eye can fully accommodate the exit pupil.

[Robh]

OK. I get all the maths. Now let me see if I have the logic right. The image at the focal plane has a certain diameter (unlike the exit pupil, I take it this is complicated to calculate). Every part of the primary mirror contributes to every part of this image, which will have a constant total brightness dependent basically on the size of the primary mirror. The specs of the eyepiece determine how much of the focal plane image we see. A low power eyepiece will see more of the focal plane image and gather more of its light, a high power eyepiece will see less of this image and gather less of its light. However, every part of the primary mirror still contributes to every part of the eyepiece image. This must imply that the focal plane image must be reinforced with more rays at wider angles than the narrower image of the eyepiece. Low power eyepieces must receive more rays per point of primary mirror than a high power eyepiece.

Regards, Rob.

[Don Pensack]

Actually, it's the other way around.
Let me explain:
100% of the light from the primary is reflected by the secondary ON AXIS, and for a certain distance either side of the axial point. But, at some point, the light cone that is coming to the secondary from an off-axis point starts to miss the edge of the secondary. At the every edge of the focal plane, if the secondary's size is correctly chosen, only about 70% of the light from the primary is reflected and 30% passes by the edge of the secondary.
The secondary could be larger and the percentage of light at the edge would go up, and that is usually the case if the secondary size is chosen for photographic use.
But, visually, we can easily tolerate a 30% drop in illumination at the edge of the field, and it helps keep the secondary mirror smaller and less diffractive.
So, the high power eyepiece, looking at only the center portion of the telescope's focal plane, actually has a HIGHER percentage of illumination at the edge of the field than the low power eyepiece that looks at a larger portion of the telescope's focal plane.

So if we choose the size of the secondary mirror JUST RIGHT, it will illuminate the field of view in our lowest power, largest field stop, eyepiece to 70% at the edge of the field. Illumination on the focal plane of the telescope would be even lower farther off axis, but because we can't see it (or don't see it), it's irrelevant for our viewing.

Now, in order to see a 70% illumination at the edge, more than just a point in the middle of the primary mirror needs to have 100% illumination on the focal plane. That's why, when you look through the focuser without an eyepiece, you see the entire primary plus a little bit outside the primary. That guarantees that a little bit of off-axis light also is fully illuminated. The "standard" amount of 100% illumination is the central 12mm or so of the focal plane. So if your eyepiece has a field stop of that size or smaller, 100% of the field of view is 100% illuminated.

Low power eyepieces see a larger portion of the focal plane of the scope, and since the magnification is lower, the brightness per unit area goes up. The sky background appears brighter, but when the exit pupil exceeds your eye's pupil diameter, your iris is acting like a stop and shaving off some of the edge of the mirror's light. So the image won't continue to get brighter as you lower the magnification from there(the point where exit pupil matches eye pupil diameter).

Now here's the interesting point: the image will never be brighter than the naked eye looking at the sky. As the scope size increases, we can see the same brightness of image with more magnification, though, which makes deep sky objects more visible (ultimate visibility, per the studies, is when the object subtends a size of at least 2 degrees to our vision). Why we see fainter stars with increases of magnification above the point where exit pupil and eye pupil are matched is because of an improvement in contrast.

I don't find this intuitive at all, but it's been demonstrated time and time again.

So when the magnification goes up on a given scope, and the exit pupil size goes down, the contrast on an extended object will not have changed from a low power view(sky and object get fainter at the same rate). However, because the object is larger, we tend to see more in it. But only up to a point, whereafter increased magnification lowers the apparent surface brightness too much.

I still have difficulties with the (obvious) fact that a 30%light drop off at the edge of the field isn't noticeable to me. The plain truth is that our eye+brain combination is much better seeing contrast between objects than absolute brightness.

[ColHut]

oooops!

You mean I was looking at images of the prmary mirror not the light cone coming out... - I think I need a picture with extended objects

sigh:doh:

[AstroJunk]

I could see it quite noticeably with a Nagler on less than perfect nights - when the background is gray rather than black, it was realy easy to spot.

I have since got a bigger secondary and the effect is gone. I think the established wisdom of making diagonals as small as mathematically possible is actually not so wise after all; not for my eyes anyway!

[Robh]

Don,

Many thanks for the extended explanations. I have a better understanding of the concepts, even though at times they seem counter-intuitive.

Regards, Rob

[ColHut]

Thanks Don, Much appreciated. How does your analysis apply to planets and the like.

regards

[Don Pensack]

Small objects, viewed on axis, are always 100% illuminated. And since resolution depends on aperture, bigger aperture will generally show more details, even at the same power.

[multiweb]

Great thread. Although I had to read it many times to understand everything thanks to you Don for sharing all this knowledge. Very instructive. :thumbsup:

[ColHut]

Thanks for your time Don (and others) - It finally makes sense now. If only the clouds would lift.

regards