Hi all, my poor old magnitude 25 brain is struggling with something I read in 'The Backyard Astronomers Guide'. Disussing Focal Ratio it says fast focal ratio scopes are useful for imaging, but when using the scope visually image brightness is only due to aperture and that focal ratio has nothing to do with it. Can anyone explain to my why that is? (please!)
Nothing wrong with having trouble understanding this concept
When imaging the focal ratio is important because it means that you have more photons being concentrated into a smaller area (because of the shorter focal length). It isn't the whole story though.
If I use a KAF-8300 sensor with its 5.4 micron sized pixels on a 10"F/5 Newtonian it will perform virtually as "fast" as a 10" F/8 RC with a KAF-16803 with its 9 micron pixels.
Both telescopes are capturing the same number of photons as they are both 10" in diameter. As both cameras have different pixel sizes, even know the F/8 is technically a slower telescope it has been paired with a camera that has larger pixels.
So, in imaging, the "speed" of a system is to do with the sky coverage of each pixel (expressed in arcseconds/pixel) against the aperture.
To calculate the pixel coverage:
The same principle works with visual astronomy. If I put a 5mm eye piece in the 10" F/5 I'll get a magnification of 254x. If I put an 8mm eye piece in the 10" RC I'll get 254x magnification. As both are the same diameter there is the same amount of photons at the same magnification.
@Finite That's not quite right. There are two ways to look at apparent brightness: surface brightness of an object or the overall brightness of an image (whether it's at the camera sensor or on your retina).
As far as the visually observed surface brightness of an object goes, for a given eyepiece a faster telescope will yield brighter views. Aperture does not matter: only f-ratio and the eyepiece focal length matter. Actually it's a combination of these that determines visual (surface) brightness: the exit pupil diameter = eyepiece focal length divided by f-number of scope = telescope aperture divided by magnification.
However, for the same exit pupil, a larger aperture telescope will yield higher magnification. Hence you will be able to see deeper and (if seeing allows) in more detail. If you are looking at an extended object, like a large gaseous nebula (think Orion or Carina) then it will take up more of your field of view at higher magnifications, so the image overall will appear brighter. But the apparent surface brightness of the nebula itself will still only depend on the exit pupil. It's perhaps a little counter-intuitive at first, till you think about it.
Or you can think of the Moon. Its surface will appear just as bright through 10x50 binoculars as through a 10" telescope (250mm aperture) at 50x magnification. But because the Moon will be 5x bigger in the latter case, the overall perceived brightness will be greater: you have 25x more light entering your eye from that much larger looking Moon. Note that the exit pupil in both these cases is 5mm. (Hope this make sense. )
Useful exit pupils fall in the range of 0.5mm to 7mm (from highest to lowest magnification and dimmest to brightest views). But in practice the range from 1mm to 5mm is more useful. Much below 1mm one tends to see no more detail (the image is just gets bigger, but also dimmer and blurrier). Above 5mm, your start running into the limitations of your own eye: how wide your pupil can open. If you're in your late teens to early 20s, nature gives you 7mm (or perhaps even a bit more). As you age this drops. I think for someone in their 40s, 5mm is average.
I will try and explain the idea as I think about it.
Firstly, imaging uses a camera that accumulates an image. the longer you collect light the brighter the image. Your eye doesn't work that way. It simply responds to the light coming in right now. So imaging systems and visual systems work quite differently.
With an imaging scope, the focal ratio determines the size of the image on the sensor. A long focal length scope, for example an 8" at f10 (typical SCT) will have a focal length of 2000mm and will produce a quite large image compared to a much faster 8" at f5 with a focal length of 1000mm. The amount of light being collected is exactly the same as the aperture is 8" in both cases BUT the amount of light being collected by each pixel on the sensor is more with the faster scope. So it can collect the light in a shorter time (hence it is faster!) This has benefits as guiding is easier, the sensor will stay cooler, so less noise and you can get an image quicker or get a deeper image in the same time.
As your eye cannot accumulate light like a camera, the f ratio no longer matters (as far as the overall brightness of the image is concerned) the only factor is how much light is collected, which is a function of aperture. Having said that, the eyepiece that you use will vary the "magnification" of the object you are looking at. If you increase the magnification with a shorter focal length eyepiece, it spreads out the light over a greater area, in effect reducing the surface brightness of the object, but not the overall brightness. However (just to complicate matters) higher powers also have the effect of darkening the sky background which can make dim objects "pop" into view as the contrast improves.So often times for visual observers, particular objects require an ideal amount of power to be seen best, and you sometimes have to try a few eyepieces to find the one that works the best on that object in your scope.
I have tried not to get to technical with the above, hope it helps!!
I might add that the lower power limit (i.e. upper exit pupil limit) only applies to obstructed optics. Unobstructed, there is no penalty for going as wide TFOV and low power as your eyepieee case permits.
I might add that the lower power limit (i.e. upper exit pupil limit) only applies to obstructed optics. Unobstructed, there is no penalty for going as wide TFOV and low power as your eyepieee case permits.
That's not so. The light from a distant point-like object enters the eye from the eyepiece as a beam of definite width = aperture / magnification. That beam width is what is called exit pupil. If the beam is wider than the pupil of the human eye, it will simply not fit in the eye. You still get an image but if you go wider you are throwing light away (using a smaller effective telescope aperture). That's got nothing to do with obstructions in the optical path. It's easy to show this diagrammatically using basic geometric optics.
Thanks guys, I'm a bit the wiser now. Still trying to assimilate some of that though.
So visually - let's use f/10 sct's as examples - if we have a 8" and a 16" side by side and focussed on the same celestial object, using different eyepieces to produce the same magnification in each scope, is the image going to be brighter in the larger scope than the 8", though the image is the same size in appearance (assuming the viewer has a healthy pupil dilation)?
Love the comment Colin! By the way, it was easier to me to understandthan your first 😀. . Would it be simply a better resolution of the same object then, it would also be brighter to look at? The tract in the book kind of threw me.
I understand how an increase in magnification makes the image dimmer, and how a focal ratio is calculated, how a imaging device builds up the light over time exposure, the very basics etc. Beyond that I'm wading through murky waters, so I thanks guys for the input. I also saw written recently that the way f ratio works in a telescope is different to how it works in a camera. I'll have to look that one up, as I'm pretty sure it's the same principle.? Both are effectively lenses.
Thanks guys, I'm a bit the wiser now. Still trying to assimilate some of that though.
So visually - let's use f/10 sct's as examples - if we have a 8" and a 16" side by side and focussed on the same celestial object, using different eyepieces to produce the same magnification in each scope, is the image going to be brighter in the larger scope than the 8", though the image is the same size in appearance (assuming the viewer has a healthy pupil dilation)?
I hope that makes some sense...?
Yes! That is the advantage of aperture, gathering more photons which gives you a brighter and potentially more detailed image.
An object at 200X looks about the the same in every scope regardless of FR or aperture.
The eyepiece you would use to achieve 200X would differ from scope to scope based on the Focal length of the scope but the image size would be the same.
At the same magnification you will get the same FOV but the image will be brighter. Remember, focal length increases linearly (A*x) where as aperture (light gathering power) increases at A^x. So a 16" will gather 4x the amount of light than an 8" but only have twice the focal length given the same f/ratio.
I also saw written recently that the way f ratio works in a telescope is different to how it works in a camera. I'll have to look that one up, as I'm pretty sure it's the same principle.? Both are effectively lenses.
I could be wrong, but my understanding is that they are exactly the same, focal length divided by aperture. The main difference is that in most cases, the f ratio is fixed in a telescope, while it is variable in a camera lense.
There is a discussion here https://en.wikipedia.org/wiki/F-number
I could be wrong, but my understanding is that they are exactly the same, focal length divided by aperture. The main difference is that in most cases, the f ratio is fixed in a telescope, while it is variable in a camera lense.
There is a discussion here https://en.wikipedia.org/wiki/F-number
Malcolm
the only difference i know of in this area is that photographers refer to the focal ratio as the aperture where astronomers, correctly, refer to the objective diameter as the aperture. this can be a source of confusion but i don't think it has much to do with this.
I could be wrong, but my understanding is that they are exactly the same, focal length divided by aperture. The main difference is that in most cases, the f ratio is fixed in a telescope, while it is variable in a camera lense.
There is a discussion here https://en.wikipedia.org/wiki/F-number
Malcolm
f-number is exactly the same for a telescope or a camera lens: Focal Length divided by Aperture. As you say, camera lenses typically have a variable f-number, by using an adjustable diaphragm to reduce the aperture (and zoom lenses also have an adjustable focal length).
What differs is the common usage of a single measurement to describe a camera lens versus a telescope. A 200 mm camera lens has a focal length of 200 mm; if it is f/4, this implies a maximum wide-open aperture of 50 mm. A 200 mm telescope has an aperture of 200 mm, and will typically have a focal length somewhere between 1000 mm (e.g. an f/5 Dob) to 2000 mm (e.g. an f/10 SCT).
You could always stop down your telscope's aperture to increase the f-ratio in exactly the same way as is done in a camera lens. But you very rarely want to do that because normally you want all the light you can get.
But the same goes really for camera lenses when it comes to astrophotography: you normally want the aperture wide open to collect as much light as possible (unless your lens performs very poorly at full aperture; much like a badly figured telescope mirror that can typically also be improved by masking its edges).
Thanks everyone, I get it I think. I got hung up on thinking f10, no matter what size aperture, meant that the image is going to be the same level of dimness but with an improvement in resolution. All great info, I haven't learnt a lot in my 50 yrs (just ask my wife). Colin, this you submitted resonates with me, thanks.
Quote:
Originally Posted by Atmos
At the same magnification you will get the same FOV but the image will be brighter. Remember, focal length increases linearly (A*x) where as aperture (light gathering power) increases at A^x. So a 16" will gather 4x the amount of light than an 8" but only have twice the focal length given the same f/ratio.
Basically, bigger scopes make brighter images
I am thinking of going SCT again for the goto capability. I am rubbish at star hopping and my eyesight is getting a bit long in the tooth perhaps. Before it goes (or before I go) I would like to sight a galaxy in person. Yep haven't seen one,- own galaxy and its satellites not included that is.
I really appreciate the help from all of you who have contributed.
Your "camera lens" for visual astronomy is not just the f-whatever telescope but the combination of telescope, eyepiece and the lens inside your eye. The way others have explained it probably makes the most intuitive sense: for a given magnification, a larger aperture will yield an intrinsically brighter image (both in terms of surface and overall brightness).
That's true until you hit the limitation of your own eye's aperture. Beyond that you gain nothing upon increasing the telescope's aperture. Roughly around 1m or 40" is the largest useful telescope aperture for visual astronomy (so you still have plenty of room for aperture-fever beyond 23" ).
John
I am seeing a 20" dob in your signature, is that correct?
If you are no good at star hopping you could fit out that dob with an Argo Navis system for a lot less $$$ than a go to SCT and you will be seeing galaxies all night long.
Get in touch with Gary if you are not sure about the fittings required, he will surely be able to help.
"Throwing light away" and "using a smaller effective telescope aperture" by stopping down the exit pupil with your own pupil is of approximately zero consequence in unobstructed optics. In fact, stating it like that is misleading, because it suggests that the image will be darker. Wrong. The image simply stops getting brighter, and that is because it's the same brightness as what you see unaided, only magnified (for extended objects that is - point sources like stars do look brighter because their light is not being spread out over the magnified area). So if you want super widefiled, go for it. Unless of course your scope has a secondary mirror. You see, the exit pupil is just a small image of the entrance pupil - the telescope's objective. So enlarging it enlarges everthing in it as well, such as the secondary' image. And when its size gets close to your own pupil size, while the outside of the exit pupil is being cropped away by your own pupil, well you get the idea. "Lowest practical power" is strictly a problem of obstructed optics.
Apologies to the OP for moving off topic a bit here.
Your "camera lens" for visual astronomy is not just the f-whatever telescope but the combination of telescope, eyepiece and the lens inside your eye. The way others have explained it probably makes the most intuitive sense: for a given magnification, a larger aperture will yield an intrinsically brighter image (both in terms of surface and overall brightness).
That's true until you hit the limitation of your own eye's aperture. Beyond that you gain nothing upon increasing the telescope's aperture. Roughly around 1m or 40" is the largest useful telescope aperture for visual astronomy (so you still have plenty of room for aperture-fever beyond 23" ).
Interesting. I'd not seen a aperture limit put on for visual. Using the Naperville Astronomical Association Telescope calculator gives me max 32mm eyepiece for going wide with the 23" (not usable as of yet) dob, and 28mm with the up-and-running 20" dob. Given my estimated pupil size being 50 yrs age.
)
