F-Ratio and brightness

[Tim Armes]

Hi,

I've seen the following sort of commentry stated as fact in many places:

"For photography, faster f/4 of f/6 systems yield shorter exposure times. But when used visually, image brightness depends solely upon the aperture. Focal ratio has nothing to do with it."

Coming from a photographic background I can't understand this at all. If I look through my camera and change from f/8 to f/5.6, thus changing the focal ratio (and thus the aperture), I can phsyically see the difference in brightness through the viewfinder.

For any give focal ratio, the brightness is the same regardless of the focal length/diameter that result in that ratio.

So, is there any techically inclined soul who can please put me out of my misery explain exactly why the image brightness is independent of focal ration for the human eye, but not for a camera??

Thanks,

Tim

[GazOC]

If you altered the focal ratio on a telescope by changing the aperture you'd notice a difference in brightness as well (assuming you kept the magnification the same) with the human eye.

[ollypenrice]

This comes up often! Here's my take on it.

I think the confusion arises simply because in the camera world aperture and f ratio are used synonimously. 'I altered the f-stop' is not wrong but 'I altered the aperture' would be clearer since physically that's what you altered. You (say) reduced the effective diameter of the lens, reduced light throughput so the picture will take longer or be dimmer.

In prime focus telescope imaging with telescope 'X' you can't alter anything. If you want the same image faster you need a telescope of the same focal length (to frame the same obect the same way) and greater aperture. (I'm keeping focal reducers and Barlows out of it for clarity.)

In telescope visual observing you can alter the focal length (the effective focal length) by changing the eyepiece. You can, therefore, achieve the same effective focal length in telescopes of radically different apertures and the larger the aperture the brighter the image, the light grasp rising as the square of the aperture. So when people say that, in visual use, brightness goes with aperture, not focal ratio, they are probably talking about a given focal length, and they are right. I think what they are getting at is that it makes no difference if you use a 100mm f10 scope with a 20mm EP or a hundred mm f5 scope with a 10mm EP.

There are at least two caveats. One, the relationship between brightness and f ratio applies to extended sources not point sources. The brightness of point sources rises with aperture, not f ratio.

Two, the sampling rate in arcseconds per pixel has a bearing on the exposure time required.

Olly

[Tim Armes]

I think the confusion arises simply because in the camera world aperture and f ratio are used synonimously. 'I altered the f-stop' is not wrong but 'I altered the aperture' would be clearer since physically that's what you altered. You (say) reduced the effective diameter of the lens, reduced light throughput so the picture will take longer or be dimmer.

Hmm.

A telescope with a focal length of 1000 and a diameter of 200 would be an f/5.

A telescope with a focal length of 2000 and a diameter of 400 would also be an f/5.

They should both require the same exposure time to take a given picture (of admittedly different FOVs). The amount of light entering the sensor is the same.

According to the statement I quoted earlier, the 400mm will be brighter, even though the f/ratio is the same. I'm not disputing it, but I can't understand it. If the camera sees no difference then nor should I.

What am I missing?

Tim

[Capricorn]

Both f/5 systems have the same brightness of the image.

Just on the smaller diameter and so focal length the image is smaller.

The smaller amount of light gathered by the mirror or lens is matched by the smaller image size. Hence the image remains the same brightness. Standard photographic optics.

When you use a scope visually you are concerned with magnification a lot of the time.

So consider an image magnified 100x by a 4 inch scope and the identical image magnified 100x by a 8 inch scope

In the first you have 16 units of light in the 100x image, in the second you have 64 units in the same 100x image. So it is brighter because of aperture.

[GazOC]

In telescope visual observing you can alter the focal length (the effective focal length) by changing the eyepiece. You can, therefore, achieve the same effective focal length in telescopes of radically different apertures and the larger the aperture the brighter the image, the light grasp rising as the square of the aperture. So when people say that, in visual use, brightness goes with aperture, not focal ratio, they are probably talking about a given focal length, and they are right. I think what they are getting at is that it makes no difference if you use a 100mm f10 scope with a 20mm EP or a hundred mm f5 scope with a 10mm EP.

This is exactly how I understand it from a visual perspective. The eyepiece gives you an extra degree of freedom that a camera doesn't.

For visual use aperture and magnification govern how bright an object is (with Ollys caveats) and you can always alter the magnification with a different eyepiece.

[ollypenrice]

Hmm.

A telescope with a focal length of 1000 and a diameter of 200 would be an f/5.

A telescope with a focal length of 2000 and a diameter of 400 would also be an f/5.

They should both require the same exposure time to take a given picture (of admittedly different FOVs). The amount of light entering the sensor is the same.

According to the statement I quoted earlier, the 400mm will be brighter, even though the f/ratio is the same. I'm not disputing it, but I can't understand it. If the camera sees no difference then nor should I.

What am I missing?

Tim

My view is that they don't take 'a given picture.' They take quite different pictures. One will have four times the field of the other. I know you cover this but then you dismiss it somewhere along the line! You can't do that.

If you do, it comes down to how you choose to present the images. Do you re-scale the 'common denominator' part of the sky seen in the short FL up to the size that it appears in the long FL and then crop out the rest for a direct comparison? Why would you do that? If you do, you find that there is an 'f ratio myth' because the faster image is not brighter (read 'deeper') than the slower one.

To my mind it is pointless to compare images at different focal lengths as if they were the same thing. They simply aren't. Consider the moon with aperture pinned at 'x.' Are you collecting light from the Mare Crisium (long focal length/slow f ratio) or from the whole moon (short focal length/fast f ratio) on a given chip? Surely more light comes off the whole moon than off Crisium? But in both cases the same amount of light comes into the telescope from Crisium. I believe that there you can see the f ratio myth at work. But if you have more aprture at the Crisium focal length you get Crisium faster. No myth.

So if you are imaging a small planetary at f10, 2 metres or f5, 1 metre, on a chip plenty large enough in either case you will fall foul of the f ratio myth because the f5/1 metre image will not be deeper/faster when seen at the same screen size.

In this malarky I always insist on keeping the focal length the same in the discussion because if you don't you are just not taking the same picture.

Olly

[KaStern_Former_Member]

Hi Tim,

But when used visually, image brightness depends solely upon the aperture.

visually image brightness for extended objects depend on exit pupil size.

For Point-like objects aperture counts.

Cheers, Karsten

[Tim Armes]

Everything your all saying about comparing fields of view, taking into account the inverser-quare law, etc. all makes sense taken one at a time, but I'm still missing the big picture.

Let's go back to the quote (taken from the Backyard Astronomer):

"For photography, faster f/4 of f/6 systems yield shorter exposure times. But when used visually, image brightness depends solely upon the aperture. Focal ratio has nothing to do with it."

There seems to be something fundamentally flawed with the concept that a camera receives more light when the ratio is smaller, but that the visual receiver's eye doesn't.

I do realise that there's more going on behind this statement than meets the eye (no pun intended) - and I'm sure that than does have something to do with magnification, but I can't get my head around it. It's frustrating since I deal with optics, exposure values, f-stops etc. on a daily basis.

Perhaps if some kind person could come up with a concrete example using various telescope configurations that demonstrate the above it would click for me?

Tim

[ollypenrice]

I think maybe Karsten is onto your quandary with exit pupil. The light has to get into your eye which has a maximum of 5 to 7mm aperture so the exit pupl mustn't exceed this. You can easily configure a telescope-EP combination which does exceed this.

Olly

[PortableAstronomer]

There seems to be something fundamentally flawed with the concept that a camera receives more light when the ratio is smaller, but that the visual receiver's eye doesn't.

You've almost hit the nail on the head there.

Up to a certain limit, the f/ratio determins the apparent brightness of the image for both visual and photographic use.... until you reach a limit where the low f/ratio (that brightens the image photographically) in an equivalent visual setup at that fast f/ratio then the light cone that exits the eyepiece is greater than the diameter of your pupil. So light often gets wasted visually at very fast f/ratios, in a way that would not happen photograpically.

[Ags]

The telescope concentrates the light from the sky into a focal plane that has a particular intensity that depends on the F ratio. The image in that focal plane is then magnified (stretched and _dimmed_) by the eyepiece.

The role of the eyepiece in stretching and dimming is the key point here. If you double the power of the eyepiece, you stretch the image twice as much, making it four times as faint.

In a camera the focal plane is projected directly onto the chip so the F ratio of the primary lens in the only factor. If you connected your camera to a scope using eyepiece projection, then the dimming effect of the eyepiece would be relevant to the camera too.

As Karsten pointed out, point sources like stars are not subject to stretching by the eyepiece as they are infitintely small to start with - so stars stay at a brightness determined by aperture (not F ratio) regardless of magnification.

[ollypenrice]

This is from Celestron's website;

To calculate the exit pupil, simply divide the aperture of the scope’s objective mirror or lens by the magnification of the eyepiece. Or divide the eyepiece focal length by the f/ratio of the scope. Example: an 8-inch (203.2mm) telescope with a focal length of 2032mm is used with a 20mm eyepiece. The exit pupil of this combination is 2mm.

2032/20 = 102x

203.2/102 = 2mm exit pupil

20/10 = 2mm exit pupil

Something I haven't thought about is the effect of the size of a projected image on the retina with respect to its perceived brightness. How does the eye/brain react to the same amount of light concentrated into a small image as compared with a large? I don't know. Karsten? Anyone?

Olly

[KaStern_Former_Member]

Hello Olly,

How does the eye/brain react to the same amount of light concentrated into a small image as compared with a large?

the eye/brain will see the projected object easier when it is larger on your retina.

So allthough 4" f/6 and an 8" f/6 scope, used with the same eyepiece,

give the same exit pupil, the object can be seen easier with the 8" scope.

For example M101 or so.

In practice there will be some other important things such as straylight suppression.

If it is bad in the 8" and goo in the 4" the 4" scope might win.

Greetings, Karsten

[Tim Armes]

The telescope concentrates the light from the sky into a focal plane that has a particular intensity that depends on the F ratio. The image in that focal plane is then magnified (stretched and _dimmed_) by the eyepiece.

The role of the eyepiece in stretching and dimming is the key point here. If you double the power of the eyepiece, you stretch the image twice as much, making it four times as faint.

In a camera the focal plane is projected directly onto the chip so the F ratio of the primary lens in the only factor. If you connected your camera to a scope using eyepiece projection, then the dimming effect of the eyepiece would be relevant to the camera too.

As Karsten pointed out, point sources like stars are not subject to stretching by the eyepiece as they are infitintely small to start with - so stars stay at a brightness determined by aperture (not F ratio) regardless of magnification.

Hi,

I've not looked into AP at all, so I'd never really considered how the scope would be attached the the camera. Am I right in assuming then that the FOV of the image on the sensor is totally fixed and dependent on the focal length of the scope? There's no other optics involved to change the magnification?

That's certainly a big chuck of information that I hadn't realised, and it goes some way to explaining the reason behind my question (the systems are physically and optically very different).

However, I still don't grasp how the inclusion of the eyepiece is rendering the focal length irrelevant as far as brightness is involved. Is anyone able to compare two fictive systems to give a concrete example of how this works?

Thanks,

Tim

[Bazzaar]

I think the difficulties you are having Tim may be due to the text you have been reading lacking scientific rigour. In an attempt to make them more accessible to the man in the street they have made assumptions about the exact conditions involved. You are wanting the facts beyond the scope of these books.

For instance, this statement;

But when used visually, image brightness depends solely upon the aperture. Focal ratio has nothing to do with it.

That statement works IF it assumed that you would want to do this comparison at the same magnification. So you use a different eyepiece to keep the image the same size to the viewer. There is no equivalent way to change image scale with a camera. So comparisons break down.

The camera in the imaging set up cannot be replaced with the eye in the observing one but needs to be the eyepiece/eye combination.

Barry

[Ags]

Let's try it from a different angle...

The physical quantity of light collected to form an image is determined exclusively by the aperture. That makes complete sense I hope, as the light comes in through the front of the scope, so a wider aperture means more light and it doesn't matter what F-ratio happens inside the scope - the front of a, say, 200mm wide scope intercepts the same amount of light from the sky, regardless of whether its objective is F15 or F5.

The telescope then brings all that light to focus and you can then view the image at different scales by selecting different eyepieces. If you magnify the image by 50x you will get a certain brightness, and then if you magnify by 100x you will have 1/4 the brightness (as image scale is doubled).

So the brightness you see visually through the eyepiece is determined by the aperture (which gives you the total light entering the optical system) and then the magnification, which stretches out the image, making it fainter. As you can see (?) F-ratio is not even mentioned.

[Ags]

Hi,

Am I right in assuming then that the FOV of the image on the sensor is totally fixed and dependent on the focal length of the scope? There's no other optics involved to change the magnification? That's certainly a big chuck of information that I hadn't realised, and it goes some way to explaining the reason behind my question (the systems are physically and optically very different).

Tim

Regarding AP, the typical means of connecting a camera to a scope is by prime focus, in which the telescope acts as a giant telephoto lens and the camera takes the place of the eyepiece (so no eyepiece). So you are right, in prime focus AP to get a different scale you generally use a different scope.

There are other forms of AP that use an eyepiece, like afocal AP or eyepiece projection AP, and I mentioned this only to make the point that the brightness of the image formed by the eyepiece would be the same for a camera connected to the eyepiece, or human eye looking through the eyepiece. Just in case you thought that cameras and eyes were processing the light differently.

[Astro Imp]

Hi Tim,

Thanks for your original post. I have often been confused by this coming from a photography background the same as yourself.

I have read the various replies and am not sure that I totally understand

but will continue to give it thought.

[Tim Armes]

Thanks to all for you answers - I believe you got me onto the right tracks and thinking this through carefully.

Hi Tim,

Thanks for your original post. I have often been confused by this coming from a photography background the same as yourself.

I have read the various replies and am not sure that I totally understand

but will continue to give it thought.

I know how you feel

Anyhow, as alluded to in some of the responses above, I believe the statement comes from the magnification factor. However, without a concrete example I couldn't get my head around it. So, I've been playing with figures, and this is what I now understand.

Important point 1 -

In the case of "prime focus" Astro Photography, the telescope itself acts as the lens, so everything we know about f-numbers and exposure times holds true.

Important point 2 -

When viewing visually, we add an eye piece to magnify the image produced by the telescope. The optics aren't the same as for AP above.

Objective

I'd like to understand this statement:

"For photography, faster f/4 of f/6 systems yield shorter exposure times. But when used visually, image brightness depends solely upon the aperture. Focal ratio has nothing to do with it."

So the objective is to demonstrate that only the diameter, and not the f-number, affects the brightness.

To demonstrate this I'm going to make the assumption that the viewer would like to see a given object at the same size on his retina (i.e. with the same total magnification compared to the naked eye), regardless of the telescope being used.

Proof part 1 - f-number doesn't affect brightness

Let's compare two hypothetical telescopes A and B:

A - Focal length = 800mm, Diameter = 200mm, F-Number = f/4

B - Focal length = 1600mm, Diameter = 200mm, F-Number = f/8

Note that:

1) the diameters are the same, the f-numbers are different

2) the field of view of telescope A is twice as big as telescope B.

3) telescope A collects four times as much light as telescope B (it's 2 stops faster)

With telescope A, the user chooses to magnify the image 100x (using a 8mm eyepiece).

With telescope B, the user only has to magnify by 50x (using a 32mm eyepiece) to see the image at the same size.

Since telescope A requires twice the magnification of B, it requires four times as much light to render the image at the same brightness (inverse square law). Since A collects four times as much light as B this achieved - the image brightness is the same even though the f-number has changed.

Proof part 2 - changing the aperture alone affects brightness

I'm sure the that rest will be obvious now, but we'll continue for completeness.

Let's compare two hypothetical telescopes C and D:

C - Focal length = 800mm, Diameter = 200mm, F-Number = f/4

D - Focal length = 1600mm, Diameter = 400mm, F-Number = f/4

Note that:

1) the diameters are different, the f-numbers are the same

2) the field of view of telescope C is twice as big as telescope D.

3) the two telescopes collect the same amount of light

With telescope C, the user chooses to magnify the image 100x (using a 8mm eyepiece).

With telescope D, the user only has to magnify by 50x (using a 32mm eyepiece) to see the image at the same size.

Since both telescopes have collected the same amount of light, the image created by telescope A will be 4 times dimmer than that created by telescope B since the image is being magnified twice as much.

The large aperture of telescope D results in a brighter image, even though the f-number has stayed the same.

QED.

Conclusion

So, to go back to the original quote:

"For photography, faster f/4 of f/6 systems yield shorter exposure times. But when used visually, image brightness depends solely upon the aperture. Focal ratio has nothing to do with it."

I think that this was what's intended:

"For photography, faster f/4 of f/6 systems yield shorter exposure times. But when used visually, for an object viewed at given final magnification (compared to the naked eye), image brightness depends solely upon the aperture. Focal ratio has nothing to do with it."

Under these conditions, the focal length of the telescope dictates only the field of view, and thus the eyepieces required to achieve a given object size. The diameter dictates brightness (and resolution). F-number holds little other value.

Disclaimer

I've read in several places that the only the aperture controls brightness for visual observing, and not once have I seen this qualified with "for a given magnification" (with the exception of Barry's post above of course), so maybe I'm on completely the wrong tracks here...

For example, here's Wikipedia's comments:

"Even though the principles of focal ratio are always the same, the application to which the principle is put can differ. In photography the focal ratio varies the focal-plane illuminance (or optical power per unit area in the image) and is used to control variables such as depth of field. When using an optical telescope in astronomy, there is no depth of field issue, and 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. The focal length controls the field of view of the instrument and the scale of the image that is presented at the focal plane to an eyepiece, film plate, or CCD."

No mention of why that's the case, or of any assumptions about magnifaction.

[Tim Armes]

Let's try it from a different angle...

The physical quantity of light collected to form an image is determined exclusively by the aperture. That makes complete sense I hope

Actually, no.

The physical amount of light is a function of the aperture AND the focal length.

The focal length changes the FOV, and thus the amount of an object seen, and thus the amount of light entering the imaging system from that object.

For any given exposure, as you increase the focal length you need to open the aperture to increase the amount of light proportionally. The whole point of the f-number is to define a given amount of light ("exposure value") such that, for example, f/4 allows the same amount of light to enter regardless of the focal length (the aperture on the other hand will vary).

Tim

[Tim Armes]

But when used visually, image brightness depends solely upon the aperture. Focal ratio has nothing to do with it.

That statement works IF it assumed that you would want to do this comparison at the same magnification.

Yes, but this lacks rigor. What do you mean by magnification?

If you mean the ratio of the projected image on your retina compared to the size as seen by the naked eye then this holds true (as I've now managed to prove to myself a couple of posts back).

If you mean the specific magnification of the chosen eyepiece, then this statement is false.

Tim

[ollypenrice]

Actually, no.

The physical amount of light is a function of the aperture AND the focal length.

The focal length changes the FOV, and thus the amount of an object seen, and thus the amount of light entering the imaging system from that object.

For any given exposure, as you increase the focal length you need to open the aperture to increase the amount of light proportionally. The whole point of the f-number is to define a given amount of light ("exposure value") such that, for example, f/4 allows the same amount of light to enter regardless of the focal length (the aperture on the other hand will vary).

Tim

I think, 'actually yes.' The front of the telescope, the objective, has no idea what anything behind it is doing. A certain area of light is collected at the front of the telescope. You can put a bung in the back but the same amount of light will go in. How does the aperture know it is supposed to collect less or more light when you change things behind it? How does the incident beam know it is supposed to change?

Changing the EP, and so the effective focal length, cannot possibly alter the amount of light entering the telescope. Newton's inverse square law is predicated upon the idea of imaginary 'spherical shells' of light at certain radii from the point source. Any square metre (or whatever unit) of shell has a certain amount of light intensity (flux) and any square metre of aperture will collect that same flux.

If you don't start with that idea you won't ever crack the issue, I don't think.

The 'physical amount of light' collected is absolutely not connected with focal length. I refer back to my point about imaging the whole moon in short FL or just the Mare Crisium at longer FL, the aperture remaining the same. The same amount of light comes from the mare Crisium in both cases. How can it not?

What alters is what you do with it. Spread it out (with a long focal length) or concentrate it onto a smaller plane (short focal length). In the latter case it will look smaller and brighter, in the former larger and fainter. No? Honestly, I do believe that's it.

Olly

[Ags]

I thought I'd cracked it with my latest explanation... :-) Best of luck figuring it out. I know i'm right though!

[ollypenrice]

I thought I'd cracked it with my latest explanation... :-) Best of luck figuring it out. I know i'm right though!

Yup, but the thng about explanations is, a) getting a clear and accurate one organized - and that's the easy bit. Then there's which is harder. You have to try to find what underlying misconception is lurking in the mind of the person you are talking to. Please, I stress, I say this as a person much afflicted by misconceptions but in this case I do think my mind is clear - for once!

So I want to put my money on the idea that a given aperture collects the same, always the same, precisely the same, incident beam as any other. If the OP can accept this and build from there then I think clarity must follow.

I've been there and know how weird it is when you seem to have the bases covered but cannot accept the official line.

Olly