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Astronomical f-number (N)

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#1 mmalik

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Posted 06 March 2013 - 07:17 PM

Let's have a talk about astronomical f-number (N) in terms of applied imaging. Here is some background info.


Traditional f-number (f) (also called focal ratio, f-stop, f-ratio) is calculated by altering lens aperture while focal length of a given lens remained the same. That traditional implementation of focal ratio controls depth of field.

Posted Image
l=focal length
D=aperture

Traditional diagram of decreasing apertures ONLY [for a given ‘fixed’ focal length lens]
Posted Image


Astronomical f-number (N) , although derived per below…:

Posted Image
f=focal length
D=aperture


…the traditional implementation of f-number doesn’t hold, in its original spirit, in astronomy because aperture of a given scope remains the same; plus there being no depth of field issue, “the brightness of stellar point sources in terms of total optical power (not divided by area) is a function of absolute aperture area only, independent of focal length”.


To sum up:

•'Traditional f/ratio' is focal length divided by diameter (aperture).

•'Astronomical f/ratio' in essence is diameter (aperture) ONLY.

Focal length controls the scale of the image circle that is presented at the focal plane to a sensor, DSLR, or CCD.

Aperture area (optical power) controls the brightness of stellar point source, independent of focal length.


Some questions to contemplate:

1. F-ratio principles are grounded in lens objective speed; how do they translate to reflective/compound optics?

2. F-ratio principles are grounded in varying aperture, should they even be applied (or referred to as being the same) to fixed aperture systems (scopes).

3. While brightness of stellar point source is independent of focal length (of lens objective), condensing the photons to a smaller unit area of a sensor with devices like HyperStar may reduce exposure times and/or produce bright images in comparison; is that bending the principles of astronomical f-ratio or an illusion of photons per unit area of a sensor?

Thx

#2 Mike Clemens

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Posted 06 March 2013 - 07:25 PM

The f-ratios are quite real even if your aperture is "stuck wide open."

"Sunny 16" rule, "Lunar 11" rule, the guidelines work fine at the same fixed open telescope focal ratios.

#3 Falcon-

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Posted 07 March 2013 - 12:35 AM

•'Traditional f/ratio' is focal length divided by diameter (aperture).


Correct

•'Astronomical f/ratio' in essence is diameter (aperture) ONLY.


NOT correct. Astronomical f/ratio is *still* calculated by focal length divided by diameter!

Focal length controls the scale of the image circle that is presented at the focal plane to a sensor, DSLR, or CCD.


Not quite. Focal Length controls the spatial resolution at the image plane (the arc-second per pixel resolution). Image circle is an artifact of optical design and not entirely tied to focal length or diameter (or f-ratio)

Aperture area (optical power) controls the brightness of stellar point source, independent of focal length.


This is, I believe, correct (to within the limit of the ary-disk resolving power vs pixel size). HOWEVER it only applies to stars - nebulea, galaxies, planets tight-globular clusters and anything else that is not a point-source do not care about diameter. For those non point-source targets focal ratio dictates brightness!

1. F-ratio principles are grounded in lens objective speed; how do they translate to reflective/compound optics?


They apply exactly the same to reflective or compound optics as they do to camera lenses.

2. F-ratio principles are grounded in varying aperture, should they even be applied (or referred to as being the same) to fixed aperture systems (scopes).


F-ratio principles are optical principals, no matter the optic used (camera lens or telescope). The fact that you can vary the aperture in a lens gives the lens some extra flexibility in modifying it's optical properties, but as I previously said the optical principals apply to telescopes.

3. While brightness of stellar point source is independent of focal length (of lens objective), condensing the photons to a smaller unit area of a sensor with devices like HyperStar may reduce exposure times and/or produce bright images in comparison; is that bending the principles of astronomical f-ratio or an illusion of photons per unit area of a sensor?


It is not an illusion because *most* astrophotography is of non-point source targets. For images of anything except single stars hyperstar's f/2 focal ratio certainly DOES give an advantage. Exactly as would comparing a 50mm f/2 lens to a 50mm f/4 lens for daylight photography. :)

#4 mpgxsvcd

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Posted 07 March 2013 - 10:30 AM

Falcon- is right

+1 on everything he said

#5 gavinm

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Posted 07 March 2013 - 04:01 PM

Exactly as would comparing a 50mm f/2 lens to a 50mm f/4 lens for daylight photography. :)


A 50mm f/2 lens produces a brighter image because it has a larger aperture than a 50mm f/4 lens - not the same as hyperstar (although apologies beforehand if that was the point you were trying to make..)

#6 Falcon-

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Posted 07 March 2013 - 09:08 PM

Exactly as would comparing a 50mm f/2 lens to a 50mm f/4 lens for daylight photography. :)


A 50mm f/2 lens produces a brighter image because it has a larger aperture than a 50mm f/4 lens - not the same as hyperstar (although apologies beforehand if that was the point you were trying to make..)



Yes what you say about the lenses is exactly correct, and also no the comparisons still apply with Hyperstar so long as you compair with scopes of similar focal length. ;)

By convention we talk about telescopes in terms of aperture & focal ratio (a 6" f/4 newtonian, an 8" f/10 SCT, etc) and we talk about camera lenses in terms of focal length & focal ratio (a 50mm f/1.8, a 300mm f/4.5). However... that is a convention only, a habit formed by each field, not a requirement.

I previously said 50mm f/2 vs 50mm f/4 camera lenses. I can make exactly the same statement about telescopes like this:

A 425mm focal length f/2.1 telescope will give a significantly brighter image then a 420mm focal length f/6.5 telescope (those happen to be the exact specs of a C8 hyperstar vs a AT65EDQ refractor). Another rather extreme example is AT12IN (1220mm f/4) vs a ETX-90 (1250mm f/13.8).

#7 Falcon-

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Posted 07 March 2013 - 09:11 PM

Ok, here is another way to look at it. Optics are Optics are Optics.

Look at a camera lens and think "oh what a nice compact wide-field refractor telescope!"

Or perhaps more importantly for astrophotography look at your telescope and think "That sure is a bulky camera lens, but wow it has a long focal length!" :grin:

#8 mmalik

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Posted 09 March 2013 - 01:03 PM

While principles of f-ratio are fundamental (being a “calculated” number, NO one is denying that per se), f-number of a scope alone may NOT convey the fastness or slowness of a scope in a meaningful way. I cringe when an imager asks “what’s the f-stop of that scope?”, rather than asking what’s the image scale at the focal plane (i.e., in relation to a particular aperture scope and particular sensor size camera, let alone the size of pixels in the sensor).


In traditional photography f-ratio has more meaning because one can manipulate the aperture and see its effect; in astronomy aperture is fixed so one needs to think ‘/f’ little differently. Yes focal length is still used to calculate “the number” but story doesn’t end there. With varying degrees of image scale (image circle) and vastly varying sizes of sensor chips and pixel sizes, what becomes more important is, understanding and comprehending the density of light (photons) gathered from a fixed aperture scope on to a unit area of a sensor at the focal plane.

Consider these scenarios:

•Image circle C11 EdgeHD: 42mm
•Image circle C11 EdgeHD + HyperStar: 27mm
•Image circle FSQ-106EDXIII: 88mm

Posted Image

•Saying HyperStar is faster [@27mm image circle] is relative; faster for what..., may be APS-C size sensor [in 11" aperture class].

•Conversely, saying FSQ-106 is slower [@88mm image circle] is relative; slower for what..., may be APS-C size sensor [in 106mm aperture class].



In short, for astro imaging, following is more practical way to describe f-ratio than just throwing f-numbers around or asking which scope was faster or slower, and even more annoying astronomy term... “f-stop” [those stops were real and physical stops in traditional imaging, no such stops in astro-imaging with fixed aperture scopes]:


It is how densely a given (fixed) aperture scope converges photons to the unit area of a sensor”.

#9 Hikari

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Posted 09 March 2013 - 01:59 PM

This is the way of calculating the number of photons hitting a pixel. From CCD Arrays, Cameras, and Displays, Second Edition by Gerald C. Holst published by SPIE Press and JCD Publishing, page 34:

n = (pi/4)(L*A/F^2(1+M)^2)To*Ta*t

Where

n is the number of photons striking the pixel
L is the spectral photon sterance
A is the pixel area
F is the f-number
M is the optical magnification
To is the spectrally averaged optical transmittance
Ta is the spectrally averaged atmospheric transmittance
t is the integration time--shutter speed

As you can see, it is simply a matter of the f-number. Thinking a telescope is somehow different because it does not have a variable aperture makes no sense--an aperture stop is an aperture stop is an aperture stop. Light really does not care.

Sorry, but the link between aperture (entrance pupil) and exposure is from a lot of bad thinking on the intertrons. What aperture is doing is impacting object-space resolution.

#10 Peter in Reno

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Posted 09 March 2013 - 02:08 PM

If you are referring to Stan Moore's article about F-ratio myth and you are a firm believer that aperture matters more than F-ratio, you may have been greatly mislead. At the bottom of Stan's article showed comparison photos taken with different focal ratio, it was not clear whether the photos were equal stretched or processed. Also, I personally find the faster f-ratio photo quite a bit better than the slower f-ratio photo.

I posted in another thread that you are aware of about "Busting the F/ ratio myth" about imaging the same DSO object for 10 minutes per sub but at F/10 and F/6.3. I proved that F-ratio DOES matter because the ADU values in F/6.3 image was pretty much doubled than the F/10 image. Don't let auto stretching the images fool you. It did fooled me before and I did thought F-ratio didn't matter until later when I did the actual measurements of F/10 and F/6.3 images. Take a look at my sample images:

I uploaded calibrated and un-processed subs of NGC6964 using F/10 and F/6.3 in FIT format for comparison. They were taken with Celestron 8" SCT, SXVR-M25C OSC camera and each sub was 10 minutes.

C-8 at F/10:
https://www.dropbox....NGC6946_F10.fit

C-8 at F/6.3:
https://www.dropbox....NGC6946_F63.fit

I had an easier time processing F/6.3 image than F/10 image because the ADU values in F/6.3 were bigger so stretching F/6.3 image was less than than F/10 image. The more you stretch the greater the risk for noise to show up.

If you believe aperture is more powerful way than F-ratio to determine exposure times, try image at prime focus, then put a 2X Barlow and image at the EXACT SAME exposure time and make a judgement for yourself.

If you own multiple scopes of different focal ratios, you should try the experiment yourself and prove whether Stan Moore is correct or not.

Peter

#11 Alex McConahay

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Posted 09 March 2013 - 02:26 PM

Mike,

Sorry, Mike, but your both your assumptions and conclusions are faulty.

>>>>>>>1. F-ratio principles are grounded in lens objective speed; how do they translate to reflective/compound optics?

No, F ratio applies equally to all optics. All optics simply bend light. They do this by refracting it with curved optics or reflecting it off curved optics. The different angles of reflection or refraction mean that light originally travelling in parralel rays is made to travel in converging rays. Think light that is travelling along the sides of a cylinder are made to travel to the point of a cone. Or, in two dimensions, think that light is made to travel the sides of a triangle. The ratio of the altitude of that triangle to the base of that triangle is the focal ratio regardless of whether the light is made to bend by refraction or reflection. The Focal ratio is simply a convenient way to express the steepness of the trianle's sides.

>>>>>>>2. F-ratio principles are grounded in varying aperture
No, as above, they are grounded in the steepness of the triangle. In early cameras, there were no aperture masks of "F Stops." Yet, there were still focal ratios.

>>>>>>, should they even be applied (or referred to as being the same) to fixed aperture systems (scopes).
Even assuming the correctness of the premise of the question,(which we cannot do), how does it follow that the F ratio concept does not apply when the F ratio is fixed?

>>>>> While brightness of stellar point source is independent of focal length (of lens objective), condensing the photons to a smaller unit area of a sensor with devices like HyperStar may reduce exposure times and/or produce bright images in comparison

Hyperstar follows the same principles of optics as other optics (Barlows, reducers, etc.....) it is just changing the F ratio. They all change the effective focal length of the optical chain. To continue the earlier analogy, the steepness of the sides of the triangle change after hitting this secondary optic.


>>>>>; is that bending the principles of astronomical f-ratio or an illusion of photons per unit area of a sensor?

It is not an illusion. There are in fact more photons per unit of time hitting that unit of sensor.

And there is no such thing as "Astronomical F-ratio." There is just F ratio.

Mike, you do have it right that a given telescopes primary optics have a given fixed aperture. The rest of the stuff you are saying simply do not follow. YOu have made an assumption that there is a thing called N, and therefore there must not be anything called F ratio. YOu can use the term N if you like, but it really is unnecessary. It is simply your name for an F ratio on an optical device with a fixed aperture.

Alex

#12 Falcon-

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Posted 09 March 2013 - 02:33 PM

The thing is focal-ratio *does* tell you the density of photons at the sensor within the fully illuminated image circle!

This image circle problem is not just a telescope optics problem - there are camera lenses with significant vignetting problems as well. That is because those lenses have an image circle smaller then the camera's sensor in exactly the same way. This is why lens reviews talk about vignetting (such as this example from The Digital Picture)


Be it a camera lens or a telescope, focal ratio does tell you what brightness to expect in the center of the image. I absolutely agree that you must consider vignetting/image circle size when choosing your optics for imaging, but doing so does not require a re-evaluation of focal-ratio numbers as a guide to "speed" of the optic.


I think we could say there are four properties that can be compared between any optics being considered for prime-focus deep-sky-object astrophotography. In camera-centric thinking order those would be:

1) Is the image circle (extent of vignetting) large enough to make good use of the camera's sensor?
2) Is the field flat and aberration free across the image circle? If not is there a corrector for it?
3) Is the image scale produced (the focal length of the scope) correct for the targets I want to image and/or local conditions?
4) Is there another optic with similar answers to 1,2&3 but with a faster focal ratio?

Note that the diameter of the optic does not come into consideration at all. It is just a consequence of the choices for #3 and #4.

#13 Falcon-

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Posted 09 March 2013 - 02:45 PM

BTW - that list I made was camera-centric as stated. Astrophotography is a big game of Choose Your Limiting Factor! ;)

When you are thinking camera-centric focal ratio is last on the list, but for alas that does assume some things like *perfect* mount performance and the like, so quite often it is well worth compromising on #1,2 & 3 in order to get a fast focal ratio that lets you keep your exposure lengths down.

A good example is the AT6RC vs the AT6IN. Same aperture, vastly different focal length and focal ratio. With my CG-5 I had lots of trouble getting accurate enough tracking and guiding to meet the demands of the AT6RC's 1370mm focal length (shows minor tracking/guiding errors) and long exposures required by the f/9 focal ratio. For that mount an AT6IN with it's f/4 focal ratio would be a much easier to use choice - at the expense of imaging smaller targets (due to a shorter focal length).

There are other things that may be limiting factors as well depending on situation - including things as basic as price! I would like a hyperstar system or indeed a FSQ106EDXIII but I can not *afford* either! :p

#14 Nils_Lars

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Posted 09 March 2013 - 06:37 PM

All this math is way over my head but I can tell you that in my own personal experience from imaging and pushing my data very hard (processing) that having a faster lens/scope makes the data so much easier to pull out the faint details.

However as I have found out a lot in this hobby (in life as well) there are always trade offs , resolution , vignetting , cost :bugeyes:

#15 Tonk

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Posted 09 March 2013 - 06:42 PM

+1 for each of the posts refuting the flawed congecture

See - http://en.wikipedia....erical_aperture

I.e. as said above by Alex the speed of your optics is a function of the steepness of the light cone converging on the focus point

#16 Samir Kharusi

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Posted 10 March 2013 - 01:40 AM

What I am still unclear about, and will probably remain unconvinced one way or another unless I sit down and do the maths :bawling: is:

Given a fixed sized scope, say, a C14 and we image something small and dim, say, the Bubble Nebula, at f11 (prime focus) then at f1.9 (Hyperstar) using subs that are long enough to put us equivalently into the Skyfog Statistics Regime (say subs of 45sec each at f1.9, and 25 minutes each at f11) ending with the same Integration Time using the SAME camera (fixed pixel size, no binning), do we end up with the same SNR (as espoused by Stan Moore) or do we end up with a better SNR at f1.9 as my gut feel trends? Problem with Stan Moore's discussion is that with an astroCCD one can bin to match pixel size to the f-ratio, and his example did not span as wide an f-ratio range as is used routinely these days by amateurs. I would love to see such a comparison done by anyone with a Hyperstar and who is willing to torture himself into autoguiding 25-minute subs at 2000mm (C8@f10) to 4000mm (C14@f11) :question:

#17 Hikari

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Posted 10 March 2013 - 11:23 AM

Binning is the slight of hand that Stan is doing. He is simply increasing the sensor response to compensate for the loss of light because of the focal ratio. It is kind of like saying I can drive to Boston and New York from Maine in the same time (because I change my speed), so distance is no longer a factor in travel time.

#18 mmalik

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Posted 11 March 2013 - 04:46 AM

I am curious to know how image circle [in mm] can be calculated from following three data points? Note: I am interested in actual math of it if it can be done. Thx

•Aperture (D)=106mm
•Focal length (F)=530mm
•Focal ratio=f/5

Posted Image

#19 Falcon-

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Posted 11 March 2013 - 10:37 AM

I am curious to know how image circle [in mm] can be calculated from following three data points? Note: I am interested in actual math of it if it can be done. Thx

•Aperture (D)=106mm
•Focal length (F)=530mm
•Focal ratio=f/5


Alas it can not be calculated with only those data points. :( It all depends on the optical design of the system. You could achieve those numbers with an 2 element refractor (archomat), or with a 17 element refractor (telephoto, zoom, image-stabilized camera lens), or with a newtonian (but what size secondary!?) or a cassegrain scope (but what secondary size, what baffles, etc) or a bunch of other potential optical configurations.

To truly *calculate* the image circle you would have to know the details of each element of the optical system that makes up that scope (or camera lens) as well as the physical layout (field stops, baffle tubes, etc) and do the math on exactly what the light paths through the optics will be. This is the domain of optical designers (beyond me!).

Worse once you know the scope's image circle you can still mess it up by using the wrong connections between the camera and the optic. That 88mm image circle you listed for the FSQ-106 is meaningless if you use a 1.25" prime-focus adaptor. In fact I suspect you must use a custom camera adaptor as even a T-ring and 2" adaptor would probably not allow the FULL 88mm image circle to pass.

So we are left with two options:

1) trust the manufacturer's stated specs on image circle
2) do real-world tests to see what the vignetting is (and hopefully report on it here :grin:)

#20 mpgxsvcd

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Posted 11 March 2013 - 12:20 PM

F-number is a ratio we made up to describe a law defined by physics. A telescope is a lens and a telephoto lens is a telescope.

#21 Hikari

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Posted 11 March 2013 - 12:28 PM

I am curious to know how image circle [in mm] can be calculated from following three data points? Note: I am interested in actual math of it if it can be done. Thx

•Aperture (D)=106mm
•Focal length (F)=530mm
•Focal ratio=f/5

Posted Image


+1 Can't be done. However, you can project an image and measure the image circle. You can also use a focusing telescope/microscope and rail as well. You can also measure the illumination cone by looking at the exit pupil and rotating the lens--that will give you an angle and with a focal length, the circle of good illumination can be calculated. All kind of methods, but no real way by simple optical specs.

#22 mmalik

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Posted 11 March 2013 - 01:10 PM

I am curious to know how image circle [in mm] can be calculated from following three data points? Note: I am interested in actual math of it if it can be done. Thx

•Aperture (D)=106mm
•Focal length (F)=530mm
•Focal ratio=f/5


Alas it can not be calculated with only those data points. :( It all depends on the optical design of the system. You could achieve those numbers with an 2 element refractor (archomat), or with a 17 element refractor (telephoto, zoom, image-stabilized camera lens), or with a newtonian (but what size secondary!?) or a cassegrain scope (but what secondary size, what baffles, etc) or a bunch of other potential optical configurations.

To truly *calculate* the image circle you would have to know the details of each element of the optical system that makes up that scope (or camera lens) as well as the physical layout (field stops, baffle tubes, etc) and do the math on exactly what the light paths through the optics will be. This is the domain of optical designers (beyond me!).

Worse once you know the scope's image circle you can still mess it up by using the wrong connections between the camera and the optic. That 88mm image circle you listed for the FSQ-106 is meaningless if you use a 1.25" prime-focus adaptor. In fact I suspect you must use a custom camera adaptor as even a T-ring and 2" adaptor would probably not allow the FULL 88mm image circle to pass.

So we are left with two options:

1) trust the manufacturer's stated specs on image circle
2) do real-world tests to see what the vignetting is (and hopefully report on it here :grin:)


This was not a trick question by the way, but a point is made. While dimensionless, f-ratio can be meaningless (the way it is touted by manufactures and how it is comprehended/misunderstood by amateur astro-imagers).


Short of one becoming an optical engineer, practically speaking, most is left to experimentation (the circles, the reducers, the sensors, the pixels, the focuser, the nose piece, etc.). Your thoughts?

#23 Hikari

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Posted 11 March 2013 - 02:42 PM

Even if you were an optical engineer, why bother trying to calculate that stuff. Optics is not that hard. Just put the system together and go from there. Even putting it together is not that hard--you should be able to get enough product specifications to know what you need.

#24 fco_star

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Posted 11 March 2013 - 03:13 PM


Mates, I was following this post very close, but I lack in ability with the numbers, so I ask a friend of mine (Carlos Ortega) that likes everything related to calculations and this type of stuff to help me with this and this is what he came up with.

To my surprise he use words and no formulas, please read it and let me know what do you think?

I consider Carlos a brilliant mind!

“Instead of stars let’s consider objects with a larger angular size, like the Moon, a planet, a galaxy or a nebula. For this example let’s consider we are imaging M31. The following applies basically the same to refractor and reflector telescopes. These basic concepts can be easily understood doing geometric optics.
For a given focal length, the main lens or mirror produces an image on its focal plane of M31 of a given size in millimeters or fractions of an inch. The amount of photons per unit of time that ‘fall’ on the image increases four times if you double the aperture, or changes as the square of the aperture. Therefore, if you have two main telescope lenses or mirrors with the same focal length but one of them has twice the diameter, both of them will make an image of M31 of the same size, but the one with larger diameter will make an image four times brighter. The exposure time will be four times faster with the larger aperture if you want to get similar results with both of them, from the brightness point of view. We are considering here that you keep the rest of the setup the same in both cases, like using the same eyepiece, or imaging directly to your camera sensor with no eyepiece.
Now, if you have two main telescope lenses or mirrors with the same aperture but one of them has a focal length twice that of the other, the one with the larger focal length will make an image twice as big compared to the other. Since both have the same aperture, the number of photons falling on the image per unit of time is the same, but the photons on the larger image have to spread over a larger area, so you have here four times less photons per unit of area, and as a result your image is four times dimmer. The relationship here is quadratic again.
So in summary, if you have two telescopes, one with an aperture ‘x’ times bigger (gaining a factor of x*x in brightness) and a focal length ‘x’ times larger than the other (losing a factor of x*x in brightness), the net result will be that both telescopes will make an image of about the same brightness and will require about the same exposure time for roughly the same result. The main difference will be that the telescope with the larger focal length will make an image ‘x’ times bigger. In other words, if you keep the focal length to aperture ratio constant, you get a similar result from the point of view of exposure time.”

#25 riverpoet

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Posted 11 March 2013 - 04:08 PM

We can all be friends:

In real life, you want to image & frame and object or types of objects (big nebulas, small galaxies etc). So you have to decide which FOV to use. FOV is a direct function of focal length & sensor size. Say you already have a camera and can only choose focal length. Now you want to gather as much light as possible (=aperture rules), so you want the largest aperture possible. By doing that, while keeping FL fixed because of FOV, you automatically want a faster f-ratio :)






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