Confused with aperture, & exit pupil - what will I see?

[Fraunhoffer]

lets imagine I wasn't to see a nice DSO about 15' size and I think it should look good nicely framed with a 1 deg field of view in the EP..

Which would give the better (or higher probability of seeing anything at all ) view from a semi urban light polluted home site (e.g Bortle 6)?

a) an 100mm f/6 refractor (fl 600mm) and a 10mm EP (60 deg afov, gain 60x = fov pf 1 deg)
(and exit pupil of 100mm / 60 = 1.6mm)

or

b) a  200mm SCT with focal reducer to give f/6 (fl 1200mm) and a 20mm EP (60 deg afov, gain 60x = fov of 1 deg)
(and exit pupil of 200 / 60 = 3.3mm)

My gut feeling is that the SCT should give a better view just based upon its 2xaperture - but Im not sure I understand fully the maths why.
Is the larger exit pupil going to result in a better / brighter / more successful view?
Or will the view be 'roughly' the same ?
Or have I got it all wrong.....

Thanks.

[happy-kat]

Exit pupil is eyepiece focal length / telescope focal ratio.

Exit pupil is the size of the light beam not the size of what it sees. Imagine holding binoculars in your hand, exit pupil is the bright little light patch you see on the area youo would look through.

Whether using a focal reducer effects exit pupil calculation I don't know.

Edited May 16, 2019 by happy-kat

[Stu]

50 minutes ago, happy-kat said:

Whether using a focal reducer effects exit pupil calculation I don't know.

Yes it does as it effectively reduces the focal length of the scope so reduces the focal ratio resulting in a larger exit pupil.

[John]

With 2x the aperture a 200mm scope actually has 4x the light gathering area of a 100mm aperture so it's going to show DSO's better.

With SCT's and mak-cassegrains the long focal length restricts the size of the true field of view that is attainable so larger DSO's may not fit into the field of view. But if it can fit in, it should be brighter and more detailed than in a 100mm aperture scope.

 

 

[Fraunhoffer]

The SCT and focal reducer is superfluous to what I was trying to work out.

I was trying to understand why the larger aperture scope would give the better view. Since they both have the same overall F ratio in this example and the fov using the 2 eyepieces is the same.
Im thinking that (assuming the eye pupil remains fully open and is larger than the exit pupil)  the exit pupil is effectively controlling the aperture of the eye. Since size of the rods/cones are fixed in size (and treating them like a camera sensor) - so the larger exit pupil in this case is allowing the eye to have a higher f/ratio of its own and hence brighter and more differentiated light levels between the background and the subject view.
(as well as more detail due to the higher aperture resolution of the scope - and ignoring any sampling at the eye).

[Stu]

12 minutes ago, Fraunhoffer said:

The SCT and focal reducer is superfluous to what I was trying to work out.

Yes, I was just answering the question from happy-kat.

In some ways, the more interesting or relevant question is what happens when you keep the exit pupils the same as it directly illustrates the benefit of a larger scope.

Strangely enough, the even the largest scopes cannot make the surface brightness of extended DSOs brighter than they appear with the naked eye. What they can do is maintain brightness at larger image scales which helps with visibility and allows your eye to detect contrast more easily.

In your example, with both scopes giving an exit pupil of 3.3mm, the 100mm refractor will only be giving a magnification of x30. So, the SCT example can maintain the same surface brightness for the object at twice the image scale.

[happy-kat]

Isn't the exit pupil of both telescopes in this example 1.6mm

10/6

[Fraunhoffer]

7 minutes ago, happy-kat said:

Isn't the exit pupil of both telescopes in this example 1.6mm

10/6

one is 10/6 and the other is 20/6.

exit_pupil = ep.focal_length / (t.focal_ratio)
= ep.focal_lenght / (t.focal_length / t.aperture)
= (ep.focal_length / t.focal_length) * t.aperture
= t.aperture / (t.focal.length/ep.focal_lengh)
= t.aperture / gain

both give the same answer.
sorry for the shorthand.