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F-number vs "fast" and "slow"

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#1 FlyBD5

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Posted 15 March 2013 - 11:46 AM

I was researching the concept of f-numbers and I now understand what it means (ratio, focal length over aperture).

I think I also understand why my Celestron 6SE has a relatively short tube, but a 1.5m focal length... because the focal length is the distance the light travels inside the telescope (corrector to primary to secondary and back to the focal plane in the case of this SCT)?

Now, is the reason why they call a lens with a low f-number "fast" and one with a high number "slow" is because a fast lens can gather more light in a shorter period of time and produce the same image faster than a lens with a high f-number?

Juan

#2 dan_h

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Posted 15 March 2013 - 12:03 PM

Basically that is it. It is a leftover from the days of film photography.

dan

#3 BillB9430

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Posted 15 March 2013 - 12:06 PM

Juan, I believe your reasoning is correct. Historically, this usage likely came from camera lenses, where a small f/number lens required a "fast" shutter speed and a large f/number lens needed a "slow" shutter speed. Hence "fast" lens and "slow" lens". That is just another way to state the same reasoning you gave, however. - Bill

#4 FlyBD5

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Posted 15 March 2013 - 12:22 PM

Ok, guys, thanks. Slowly but surely I am building up the info required to derive knowledge and some day, wisdom. :)

So, if that's the case, how in the wide, wide world of sports does a focal reducer/corrector turn an f/10 lens into an f/6.3 or f/3.0 lens?

#5 ed_turco

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Posted 15 March 2013 - 01:59 PM

:gramps: This only works for photography of extended objects. The fast lens doesn't gather light more quickly, but packs the light into a smaller area with the concurrent increased intensity.

By the way, for star images alone, the light intensity is the same for an x-magnitude star in either a fast or slow telescope. But put a nebulosity, (an extended object), in the picture and you want the faster system to bring it out on film, digicam, or whatever.:crutch:

#6 Mike I. Jones

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Posted 15 March 2013 - 02:06 PM

A "positive" lens produces a real focal length, meaning it forms a real (erect or inverted) image, one that can be seen on a piece of paper or a camera focal plane.

Adding another positive lens in the path shortens the focal length of the first lens. Thus the "system" focal length is shortened by the additional positive lens.

A 6" f/10 telescope has a (6)(10)=60" focal length. Adding a f/6.3 focal reducer (also called telecompressor) into the path reduces the system focal length to (6)(6.3)=37.8" focal length.

Helpful?
Mike

#7 FlyBD5

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Posted 15 March 2013 - 02:28 PM

:gramps: This only works for photography of extended objects. The fast lens doesn't gather light more quickly, but packs the light into a smaller area with the concurrent increased intensity.

By the way, for star images alone, the light intensity is the same for an x-magnitude star in either a fast or slow telescope. But put a nebulosity, (an extended object), in the picture and you want the faster system to bring it out on film, digicam, or whatever.:crutch:


Hmm. But in a camera, the focal plane has to remain the same, on the sensor or the film. If a fast lens means you can use a faster shutter, wouldn't that mean the image would be smaller?

And did you get someone to design this :gramps: and this :crutch: just for you? LOL! :)

Juan

#8 FlyBD5

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Posted 15 March 2013 - 02:30 PM

A "positive" lens produces a real focal length, meaning it forms a real (erect or inverted) image, one that can be seen on a piece of paper for a camera focal plane.


Ok, so far so good.

Adding another positive lens in the path shortens the focal length of the first lens. Thus the "system" focal length is shortened by the additional positive lens.

A 6" f/10 telescope has a (6)(10)=60" focal length. Adding a f/6.3 focal reducer (also called telecompressor) into the path reduces the system focal length to (6)(6.3)=37.8" focal length.

Helpful?
Mike


That's where you lost me. Is it as simple as multiplying the f-number?

Juan

#9 FlyBD5

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Posted 15 March 2013 - 02:38 PM

Ah, I just looked at a pic showing the workings of the reducer and now I think I understand.

Posted Image

I still don't get how you can lop off almost 23 inches off the focal length and still be able to focus. But that's probably because this image is not to true scale. :)

Juan

#10 Jon Isaacs

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Posted 15 March 2013 - 03:20 PM

I think I also understand why my Celestron 6SE has a relatively short tube, but a 1.5m focal length... because the focal length is the distance the light travels inside the telescope (corrector to primary to secondary and back to the focal plane in the case of this SCT)



Juan:

With a simple telescope like a basic refractor or a Newtonian, the focal length is the distance from the mirror or objective to the focal plane. The focal ratio also represents and angle.

But with a compound telescope like an SCT there are multiple elements and so it is probably better to think in terms of "effective focal length" or "effective focal ratio." Your 6 inch F/10 consists of a corrector plate, a primary mirror and a secondary mirror. The primary mirror is about F/2, the secondary mirror is convex and it magnifies the image about 5X so the result is an effective focal ratio of about F/10.

Unlike the Newtonian or refractor where the distance from objective to the focal plane is essentially the focal length, compound scopes are a combination of two curved mirrors that each make their own contribution. The actual magnification, the actual effective focal length of an SCT depends on the distance between the mirrors. Since SCTs generally focus by moving the primary mirror, the actual effective focal length varies depending on the relationship between those two mirrors.

In terms of the focal reducer, the focal ratio only changes after the reducer so if the reducer is say an 0.6 in an F/10 scope and is placed 100mm inside the focal plane at F/10, the focal plane will be 100mm *.6 = 60mm from the focal reducer and therefore 40mm inside the original focal plane.

With an SCT, this is easily managed because of the ease of moving the F/2 primary mirror. With refractors and Newtonians, this can be a problem.

I hope this helps.

Jon Isaacs

#11 FlyBD5

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Posted 15 March 2013 - 05:15 PM

I think I also understand why my Celestron 6SE has a relatively short tube, but a 1.5m focal length... because the focal length is the distance the light travels inside the telescope (corrector to primary to secondary and back to the focal plane in the case of this SCT)



Juan:

With a simple telescope like a basic refractor or a Newtonian, the focal length is the distance from the mirror or objective to the focal plane. The focal ratio also represents and angle.

But with a compound telescope like an SCT there are multiple elements and so it is probably better to think in terms of "effective focal length" or "effective focal ratio." Your 6 inch F/10 consists of a corrector plate, a primary mirror and a secondary mirror. The primary mirror is about F/2, the secondary mirror is convex and it magnifies the image about 5X so the result is an effective focal ratio of about F/10.

Unlike the Newtonian or refractor where the distance from objective to the focal plane is essentially the focal length, compound scopes are a combination of two curved mirrors that each make their own contribution. The actual magnification, the actual effective focal length of an SCT depends on the distance between the mirrors. Since SCTs generally focus by moving the primary mirror, the actual effective focal length varies depending on the relationship between those two mirrors.

In terms of the focal reducer, the focal ratio only changes after the reducer so if the reducer is say an 0.6 in an F/10 scope and is placed 100mm inside the focal plane at F/10, the focal plane will be 100mm *.6 = 60mm from the focal reducer and therefore 40mm inside the original focal plane.

With an SCT, this is easily managed because of the ease of moving the F/2 primary mirror. With refractors and Newtonians, this can be a problem.

I hope this helps.

Jon Isaacs


That was a GREAT explanation. :) If I got this correctly, the focal reducer only acts on the final path of the light from the secondary to the focal plane, not on the other two (corrector to primary, or primary to secondary). Given the three segments and the relationship between them to come up with the f/10, all the reducer has to do is manipulate the final segment to affect the entire f-number of the scope as a whole and achieve its intended goal. Did I get this right?

Now, that said, what are the tradeoffs of using a focal reducer? There has to be a catch to all this, it can't all be good news. :)

Juan

#12 highfnum

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Posted 15 March 2013 - 05:19 PM

I built 4 inch f/120
Hence my handle name

#13 GlennLeDrew

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Posted 15 March 2013 - 06:30 PM

Juan,
Indeed, there is usually some trade-off when attaching a focal reducer. This can be one or more of; stronger vignetting, stronger field curvature, worse off-axis aberrations or smaller fully-illuminated field.

Note that the faster system has a more steeply converging light cone.

You can have two lenses of the same focal length, and hence identical image scale. But make one of larger diameter and it becomes faster. The larger aperture collects more light for the focal length and so delivers a brighter image.

Note also that when you add an eyepiece, the f/ratio itself does not control image surface brightness; the exit pupil does. That's a whole other topic..

#14 FlyBD5

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Posted 15 March 2013 - 08:40 PM

Juan,
Indeed, there is usually some trade-off when attaching a focal reducer. This can be one or more of; stronger vignetting, stronger field curvature, worse off-axis aberrations or smaller fully-illuminated field.


(dang, here we go again... google v-i-g-n-e-t-t-i-n-g... Oh. Softening around the edges. Ok. Off... axis... aberration... coma. Ok. There goes my english out the door...) Ok, got it.

Note that the faster system has a more steeply converging light cone.

You can have two lenses of the same focal length, and hence identical image scale. But make one of larger diameter and becomes faster. The larger aperture collects more light fir the focal length and so delivers a brighter image.


Ok. I actually got it without having to look up anything in google. :D

Note also that when you add an eyepiece, the f/ratio itself does not control image surface brightness; the exit pupil does. That's a whe other topic..


And I looked that up too! Thanks! :)

Juan

#15 leonard

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Posted 15 March 2013 - 08:56 PM

Hello ,

Interesting subject .

>>>> Now, is the reason why they call a lens with a low f-number "fast" and one with a high number "slow" is because a fast lens can gather more light in a shorter period of time and produce the same image faster than a lens with a high f-number? <<<<<<<<<

Is this thinking correct ? :

If you have two refractors say both 4 inchs in dia. , one at f20 the other at f10 , the f20 is called slower because in photography the image scale will be larger than the f 10 because the f 10 has steeper curves to focus the light closer to it and thus has a smaller image scale .
Is this correct ????????? :question:

Leonard

#16 GlennLeDrew

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Posted 15 March 2013 - 09:09 PM

Leonard,
Image scale has nothing to do with 'speed', only f/ratio. These terms come from the ancient days of film photography. A 'slow' lens is one which builds up an image in a film emulsion more slowly, and a 'fast' lens which builds up an image more quickly.

A 28mm f/2.8 and a 300mm f/2.8, even though they form images of greatly differing scale, are equally fast.

#17 GlennLeDrew

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Posted 15 March 2013 - 09:19 PM

Juan, et al,
Here's another perspective on lens speed, and its function in image surface brightness.

Imagine you're a very tiny bug sitting an a camera's detector. Looking up, you see the aperture as an illuminated circle. The larger this circle, the more light illuminating you. And the size of this light depends *only* on the f/ratio.

More specifically, the illumination received scales directly as the solid angle of the aperture. Solid angle is area, and is much like square degrees, except it's a spherical measure, in steradians. For example, an f/2.8 aperture has a solid angle twice that of an f/4 aperture, and so delivers twice as much light to any given image point.

There are finer points to raise regarding the divergence between geometric f/ratio and aperture solid angle when aperture ratios get *really* fast, but for the purposes of the usual range of lens speed, where faster than f/1 is very rare, we can neglect this.

#18 FlyBD5

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Posted 15 March 2013 - 11:41 PM

Juan, et al,
Here's another perspective on lens speed, and its function in image surface brightness.

Imagine you're a very tiny bug sitting an a camera's detector. Looking up, you see the aperture as an illuminated circle. The larger this circle, the more light illuminating you. And the size of this light depends *only* on the f/ratio.

More specifically, the illumination received scales directly as the solid angle of the aperture. Solid angle is area, and is much like square degrees, except it's a spherical measure, in steradians. For example, an f/2.8 aperture has a solid angle twice that of an f/4 aperture, and so delivers twice as much light to any given image point.

There are finer points to raise regarding the divergence between geometric f/ratio and aperture solid angle when aperture ratios get *really* fast, but for the purposes of the usual range of lens speed, where faster than f/1 is very rare, we can neglect this.


Well, considering that in the grand scale of things we -are- bugs, I guess this pretty much makes sense. :roflmao:

Oh, heck, it's time for some homebrew. 9.5 Belgian Ale or 9.2% IPA? Decisions, decisions... :)

#19 leonard

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Posted 16 March 2013 - 02:06 AM

Hi Glen ,

If I used the 4 inch lenes I posted one F 10 the other F 20 to take a pic. of a tree I'm guessing the F 10 lens would take the image to the desired film exposer faster than the F 20 lens would . If this is true is it because :

1 - the 4 inch F 10 lens lets in more light than the 4 inch F 20 . I don't see how that would work.
2 - Or does the F 10 lens record more of the space in front of it much like if used with an eyepiece of the same FL lets say 20mm and the F 10 lens shows a wider field of view than the F 20 lens would .
So would this mean the F 10 lens lets in more light because it sees more space thus recording faster ?
And if so would the image scale at F 10 be smaller per unit area ( The tree would record smaller ) or would the image scales be the same on each film ?

I ask this because I know little about using a telescope for taking pics.

Thnaks for any clairification , Leonard

#20 Jon Isaacs

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Posted 16 March 2013 - 04:10 AM

1 - the 4 inch F 10 lens lets in more light than the 4 inch F 20 . I don't see how that would work.


Leonard:

Astronomers think in terms of aperture (objective diameter) and focal length because from that we can compute the magnification and the exit pupil, basically all we need to know.

Photographers think in terms of focal length and focal ratio and with good reason.

To a photographer, 100mm F/10 is a lens with a 100mm focal length and 10mm diameter lens.

So, compare the exposure time of a 1000mm F/10 lens to a that of 1000mm F/5 lens. Both have the same image scale but the F/10 lens is 100mm in diameter and the F/5 lens is 200mm in diameter. It captures 4 times the light so it only takes 1/4 the time to expose the film.

The exposure is directly related to the focal ratio and is independent of the focal length. The image scale is directly related to the focal length and independent of the focal ratio.

I hope this helps

Jon

#21 Mike Lockwood

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Posted 16 March 2013 - 01:47 PM

1 - the 4 inch F 10 lens lets in more light than the 4 inch F 20 . I don't see how that would work.
So would this mean the F 10 lens lets in more light because it sees more space thus recording faster ?

You almost answered your own quetsion.

Fast camera lenses are physically larger in diameter. Let's say we have a lens that is f/2 wide open. Let's say the lenses are 4" in diameter. Thus it has an 8" focal length.

At f/2, light from the entire 4" diameter is focused on the film/detector.

Of course the f/2 lens can be used at a slower speed (say f/4). How is that done? Simple - the iris within the lens is closed down, so that less light can get through. This means that only light coming through the central 2" of the lens is allowed through. (This is only 1/4 of the light compared to f/2). But, the focal length is still 8", so it's now operating at f/4.

In telescopes we usually don't stop down the system intentionally, but with camera lenses it is normal to do so.

Camera lenses have stops that are approximately multiples of 1.4, like f/2.0 and f/2.8. (2.8 = 2.0 * 1.4). This is because 1.4 is approximately the square root of two. Since the area of a circle is a function of the radius^2, f/2.0 lets twice the light through the lens compared to f/2.8, so the required exposure time is halved. Then f/1.4 lets through four times as much light as f/2.8.

#22 SteveNH

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Posted 16 March 2013 - 03:38 PM

1 - the 4 inch F 10 lens lets in more light than the 4 inch F 20 . I don't see how that would work.

Another way to look at it:

Aperture determines how much light the lens lets in - in your example, both are 4 inches, so both let in the same light whether f/10 or f/20.

Focal length determines image size. 4" f/20 is twice the image size as 4" f/10, and thus spreads to 4x the area.

Same light, 4x the area, makes the 4" f/20 at least 4x darker than the 4" f/10.

#23 leonard

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Posted 16 March 2013 - 05:18 PM

Hello SteveNH ,

I think we have a winner .............

>>> Another way to look at it:

Aperture determines how much light the lens lets in - in your example, both are 4 inches, so both let in the same light whether f/10 or f/20.

Focal length determines image size. 4" f/20 is twice the image size as 4" f/10, and thus spreads to 4x the area.

Same light, 4x the area, makes the 4" f/20 at least 4x darker than the 4" f/10. <<<<<


Given my example ( two 4 inch refractors with NO diaphragm but different focal lengths ) I beleave your answer is correct at lease it make sense in light of my question . The same amount of light goes to the film since there is NO diaphragm , BUT the image scale IS smaller for the F 10 scope so each point on the exposing film is brighter there fore faster correct exposer .

If this is correct it answers my question , if its NOT correct someone please explain .

Thanks , Leonard

#24 SteveNH

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Posted 16 March 2013 - 06:52 PM

Yes, that's it! Each point is 4x brighter because at f/10, the same light is concentrated in only 1/4 of the area as the f/20's image.

#25 FlyBD5

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Posted 16 March 2013 - 07:14 PM

Yes, that's it! Each point is 4x brighter because at f/10, the same light is concentrated in only 1/4 of the area as the f/20's image.


Okaaaay, I think I'll start the popcorn machine now. :)






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