Does f/4 == f/4? (Explanation of F Stop inside-sbp)

[Vita Rara]

This is something that has been bothering me for a while, and I have not found a sufficient means of performing an experiment to find the answer.

Given a lens such as the 70-200 f/4 and a subject that has an equal level of luminosity throughout would the shutter speed be higher at the 200mm end than at the 70mm end at f/4?

My thoughts on this are that the aperture at the 200mm at f/4 would be 50mm, and at the 70mm end would be 17.5mm. It seems that the 50mm aperture at the 200mm focal length would let in much more light.

So, in the end does f/4 == f/4 or is it a matter of factoring in focal length?

Thanks,

Mark

[Belmondo]

No, the aperture is constant throughout the focal range. It's f/4 at 70mm, and f/4 at 200mm. There are zooms that change aperture through the focal range. For example, the 100-400L is f/4.5 and the wide end, and f/5.6 at the long end.

[ron chappel]

It's all about how the zoom is designed.
Lenses with constant minimum apertures work in such a way that the *apparent* aperture size stays the same due to the way the magnifying/de-magnifying effect of the zoom part are arranged.

Normal zooms have the zoom elements and aperture arranged in a different way ,so that as the lenses zoom parts move towards the long end the aperture *apears* to get smaller (even though it hasn't changed in size)

There are some lenses that do adjust the aperture as they are zoomed but as far as i know they are very rare.

So why not design all lenses to have constant minimum apertures?
I think the answer to that one is that such lenses are far more demanding on the front element groups...and as those are by far the biggest in the lens the price goes up .

[jeric777]

is f4 on one lens the same on other lens? ie f4 on 50 1.8 = to f4 on 28-135.

[robertwgross]

Let me see if I understand the question.

Yes, f/4 = f/4, but what does that mean?

The aperture NUMBER is a constant f/4, but that number is based relative to the focal length. So, the actual SIZE of the aperture at 70mm is different from the actual size of the aperture at 200mm. But, after the math is done, the ability to capture light is f/4.

---Bob Gross---

[kb244]

The aperture numbers are universal way of saying how much light is going to come thru the lens. f/4 on say the 50mm f/1.8 would still be the same ammount of light as f/4 on a very long telephoto. Its a measurement of how many stops of light.

[CyberDyneSystems]

the "f/##" is not measurement of the opening,.. or aperture,. it is a measurement of the "F/Stop"

F/Stop is aperture (or opening) relative to the focal length. (someone will have to offer up the math)

So when I lens like the 70-200mm f/4 has a constant maximum f/stop,. the actual opening where the light goes in is indeed changing (opening more) as you zoom from 70-200mm.. to maintain the constant f/stop

On a lens like the 28-135mm where the f/stop changes,. it changes becuase as the lens zooms from 28 to 135mm the lens opening does NOT increase,. thus forcing the f/stop to change due to the longer focal length.

I think... :rolleyes: ;)

[pradeep1]

f/stop is the ratio between the diameter of the aperture relative to the focal length.

So if you have a 50 mm f/1.8 lens that means that the aperture "hole" in the lens is about 28 mm wide. (50mm/28mm ~ 1.8 )

So a movement in the f/stop equals a doubling or halving of the amount of light entering the camera body, by doubling or halving the area of the hole. Area is calculated (3.14 * (radius)^2) for the aperture hole.

At f/2.8, the aperture diameter is about 18.0 mm or 254 mm^2 (double amount of light)
At f/4.0, the aperture diameter is about 12.5 mm or 123 mm^2.
At f/5.6, the aperture diameter is about 9.0 mm or 64 mm^2 (halving amount)

That explains why we have these weird f/2.8, 5.6, 8, 11, 16 numbers.

So the short answer is that if you have a constant aperture telephoto lens, then the shutter speed will be equivalent at both the wide end and the long end at a constant aperture of f/4.0.

But when testing this out, you results may vary, mainly because of your metering mode. Say for example you take my word and test this theory out by focusing on a bright light that is the sole illumination for a small room. In the first shot, you are at he wide end, say 35 mm and you take a picture at f/8.0. The shutter speed will be maybe 1/60 sec. Then you take the same shot at a longer focal length, say 105 mm. At the same aperture of f/8.0, the shutter speed will be faster...maybe 1/80 or 1/100 sec. How and why is this? My explanation would be because of how the camera meter works. If you are using a center weighted meter, on the wide angle shot, you are including more of the dark background, so the meter reads differently and ups the expsoure to 1/60 to take into account the background. At the zoomed position, you are including only the light in the frame, and the meter reads that directly and cuts down the exposure to say 1/100 to render the picture properly exposed. This phenomena is due to the exposure meter characteristics. I just did this test and got those results so I am telling you. Now to prove that my assertion that f/4=f/4 at all focal lengths and that you should get the same shutter speed for a given f/stop, I took out my Weston light meter and metered the same scene at about the same position I held the camera when metering. Guess what? It gave me an exposure of f/8.0 - 1/60 sec. The light meter just reads how much light is falling on the sensor. It has no understanding of the focal length of the lens you are using. If you moved closer to the source, your shutter speed will change since the amount of light reaching your sensor changes, but if you do not move and meter on an average scene, for example a bright sky, then your shutter speed reading at a constant aperture will not change at the wide or telephoto end.

That's that. Hope I helped. :cool:

[jeric777]

another question, is f4 on 50 1.8 the same with f4 on 50 1.4?

[Belmondo]

another question, is f4 on 50 1.8 the same with f4 on 50 1.4?

In a word, 'Yes.'

[scottbergerphoto]

f/stop is the ratio between the diameter of the aperture relative to the focal length.

So if you have a 50 mm f/1.8 lens that means that the aperture "hole" in the lens is about 28 mm wide. (50mm/28mm ~ 1.8 )

So a movement in the f/stop equals a doubling or halving of the amount of light entering the camera body, by doubling or halving the area of the hole. Area is calculated (3.14 * (radius)^2) for the aperture hole.

At f/2.8, the aperture diameter is about 18.0 mm or 254 mm^2 (double amount of light)
At f/4.0, the aperture diameter is about 12.5 mm or 123 mm^2.
At f/5.6, the aperture diameter is about 9.0 mm or 64 mm^2 (halving amount)

That explains why we have these weird f/2.8, 5.6, 8, 11, 16 numbers.

So the short answer is that if you have a constant aperture telephoto lens, then the shutter speed will be equivalent at both the wide end and the long end at a constant aperture of f/4.0.

But when testing this out, you results may vary, mainly because of your metering mode. Say for example you take my word and test this theory out by focusing on a bright light that is the sole illumination for a small room. In the first shot, you are at he wide end, say 35 mm and you take a picture at f/8.0. The shutter speed will be maybe 1/60 sec. Then you take the same shot at a longer focal length, say 105 mm. At the same aperture of f/8.0, the shutter speed will be faster...maybe 1/80 or 1/100 sec. How and why is this? My explanation would be because of how the camera meter works. If you are using a center weighted meter, on the wide angle shot, you are including more of the dark background, so the meter reads differently and ups the expsoure to 1/60 to take into account the background. At the zoomed position, you are including only the light in the frame, and the meter reads that directly and cuts down the exposure to say 1/100 to render the picture properly exposed. This phenomena is due to the exposure meter characteristics. I just did this test and got those results so I am telling you. Now to prove that my assertion that f/4=f/4 at all focal lengths and that you should get the same shutter speed for a given f/stop, I took out my Weston light meter and metered the same scene at about the same position I held the camera when metering. Guess what? It gave me an exposure of f/8.0 - 1/60 sec. The light meter just reads how much light is falling on the sensor. It has no understanding of the focal length of the lens you are using. If you moved closer to the source, your shutter speed will change since the amount of light reaching your sensor changes, but if you do not move and meter on an average scene, for example a bright sky, then your shutter speed reading at a constant aperture will not change at the wide or telephoto end.

That's that. Hope I helped. :cool:
Great posts. I altered the thread title a little so more people will read it.
Thanks,
Scott

[JZaun]

pradeep1 gave a great explination.

Here is a link to another that some enjoy reading.

A Tedious Explination of F stops by Matthew cole
http://www.uscoles.com/fstop.htm

JZ

[Vita Rara]

So the short answer is that if you have a constant aperture telephoto lens, then the shutter speed will be equivalent at both the wide end and the long end at a constant aperture of f/4.0.

snip...

That's that. Hope I helped. :cool:

I guess it did help, but it's still fairly confusing. Although this doesn't effect my picture taking, it still bothers me.

It would seem that a 70mm lens at f/4 which yields a diaphram opening of 240 mm2 (pi * ( (70/4/2)^2) ) would let in more light than a 14mm lens at f/4 with an opening of 9.6 mm2 (pi * ( (14/4/2)^2) ). Could someone explain to me why it doesn't?

Thanks,

Mark

[Jon]

I guess it did help, but it's still fairly confusing. Although this doesn't effect my picture taking, it still bothers me.

It would seem that a 70mm lens at f/4 which yields a diaphram opening of 240 mm2 (pi * ( (70/4/2)^2) ) would let in more light than a 14mm lens at f/4 with an opening of 9.6 mm2 (pi * ( (14/4/2)^2) ). Could someone explain to me why it doesn't?

Thanks,

Mark
Yes, it will collect more light, but that only really comes into play when you're photographing emission sources. In that case, the absolute aperture is what's important (astrophotography is the prime example; think of the 200 in. reflector at Mt. Palomar, for instance).

If you're dealing with reflected light, where you're trying to form an image, both the aperture area and the focal length of the lens make a difference. This is because the larger physical aperture of the long lens projects a smaller arc of the reflected light from the subject onto any point of the film/sensor, so the sensor catches less of the total light. With a wide-angle lens, there's less magnification, so the available light hitting any given part of the sensor is being collected from a larger part of the subject.

The larger absolute aperture at long focal lengths will also collect more way-off-axis rays (coming from outside the lens' angle of coverage/field of view), which are useless in image formation, and the lens is designed to prevent them reaching the sensor as stray flare. The wide-angle lens is able to use more of the off-axis rays in image formation, because they're not so far off-axis as to be coming from outside the field of view of the sensor.

That's why we use the f/stop, the ratio between aperture and focal length, for most exposure calculations. Because it factors the angle of coverage for the lens into the equation. Also, it's consistent across a wide range of lenses. Without it, exposure meters would need settings for lens f.l. and lens aperture diameter (or area) as well as ISO and and incident/reflected light level. By using f/stops, the first two are combined into one handy setting.

[cactusclay]

Why?

[kfong]

but that only really comes into play when you're photographing emission sources.

Should be revised to POINT emisson source.
The brightness of distributed sources like the sun, moon, planets and nebula depends on f-number like other terrestrial objects.

Ken

[Jon]

Should be revised to POINT emisson source.
The brightness of distributed sources like the sun, moon, planets and nebula depends on f-number like other terrestrial objects.

Ken

No, the moon and planets are reflected sources. Nebulae, strictly speaking, are gas clouds reflecting light. Galaxies, on the other hand . . . And the sun's just another star, albeit a lot greater apparent magnitude than most.

[Vita Rara]

OK, I'm pretty sure I don't get it, but that's ok. It would seem that the larger aperture would result in more light, and a higher shutter speed, but it doesn't. Obviously there is something going on that I don't understand intuitively. Maybe small words and pictures would help me, but I really don't need to know.

Anway, I'll just go back to taking pictures and let the f/stops take care of themselves. I really appreciate the detailed answers though. The knowledge of this community is pretty amazing.

Thanks,

Mark

[pradeep1]

Yeah...you've got a lot of nerds in this community. BTW, OT, are you going to post another one of your montages like you did last year with the Rusted Root Song? I'd love to see your work. I still have the one you did last year and replay it once in a while. That was really cool. :)

[kfong]

No, the moon and planets are reflected sources. Nebulae, strictly speaking, are gas clouds reflecting light. Galaxies, on the other hand . . . And the sun's just another star, albeit a lot greater apparent magnitude than most.


I said "Distributed Source", which include both emission and reflected sources.

Your telescope cannot distinguish whether the photons arriving are emitted from a bunch of hot gas or bounced off some dust particles, so distingushing between emission and reflection is pointless, as far as calculating brightness is concerned. What is important is whether the source is point or distributed.

Theoretically there is no such thing as "Point Reflection source", since a point has no cross section for reflection to occur. So the word "Emission" in the term "Point Emission source" is actually redundant.

The sun is just a star, but it is a distributed source, whereas the other billions of stars are point sources, within the realm of our solar system.

As for galaxies, since the optics are not able to distunguish individual stars, they behave like distributed sources.

Ken

[Adam Hicks]

So this explains why the viewfinder when using the 200mm 1.8 looks so darned good :)

[Vita Rara]

Yeah...you've got a lot of nerds in this community.

Yes, we do, and it's great. I'm a nerd too, when it comes to object oriented software development. I can talk for hours about designs, patterns, and contravariance, and all sorts of esoteric stuff. It's great that we have people who know this stuff and can explain it.

BTW, OT, are you going to post another one of your montages like you did last year with the Rusted Root Song? I'd love to see your work. I still have the one you did last year and replay it once in a while. That was really cool. :)

I probably will. I haven't had as much opportunity to take pictures as I did in the first twelve months though. They really were an incredible twelve months. I rode my bike about 24,000 miles and saw 26 states, met my wonderful girl friend, and did four trips with her in about as many months. Life has not been so exciting since then.

I'm really glad you enjoyed that piece. I find it interesting that someone would save it, and view it from time to time. I do, but I have a personal connection to all of the materials in it. Anyway, I'm really glad you like it. I know I had an amazing time that year, and the photos don't even come close to the reality of it all.

Later,

Mark

[Vita Rara]

So this explains why the viewfinder when using the 200mm 1.8 looks so darned good :)

Actually I think this explains why the viewfinder should look no different from any other f/1.8 lens... I think...

Mark

[pradeep1]

I'm really glad you enjoyed that piece. I find it interesting that someone would save it, and view it from time to time. I do, but I have a personal connection to all of the materials in it. Anyway, I'm really glad you like it. I know I had an amazing time that year, and the photos don't even come close to the reality of it all.


I liked the music very much, and your photos ;) . At that time, I was dreaming of buying a 300D and your post was like an inspiration. Now I am on the market for a 20D or a 350D and I revived your montage to get my motivation up.

[Jon]

I said "Distributed Source", which include both emission and reflected sources.

Your telescope cannot distinguish whether the photons arriving are emitted from a bunch of hot gas or bounced off some dust particles, so distingushing between emission and reflection is pointless, as far as calculating brightness is concerned. What is important is whether the source is point or distributed.

Theoretically there is no such thing as "Point Reflection source", since a point has no cross section for reflection to occur. So the word "Emission" in the term "Point Emission source" is actually redundant.

The sun is just a star, but it is a distributed source, whereas the other billions of stars are point sources, within the realm of our solar system.

As for galaxies, since the optics are not able to distunguish individual stars, they behave like distributed sources.

Ken

Your "correction" was to change "emission source" to "POINT emission source". Your statement here is completely at odds with your earlier one. I'll further submit that "point" emission sources are also theoretically impossible. It's only the size and distance of a light source or other subject that makes itappear as a point. My statement was that the physical size of an aperture, as opposed to the f/ number, only matters when you're dealing with "emission sources". So a 400 mm f/4 is capable of recording more emission sources at the same shutter speed than a 50 mm f/2, even though f/2 is "faster".

[kfong]

I'll further submit that "point" emission sources are also theoretically impossible. It's only the size and distance of a light source or other subject that makes itappear as a point.
The Big Bang was a point source.
Bremstrahlung from an energetic electron is a point source.
My statement was that the physical size of an aperture, as opposed to the f/ number, only matters when you're dealing with "emission sources". So a 400 mm f/4 is capable of recording more emission sources at the same shutter speed than a 50 mm f/2, even though f/2 is "faster".
Try take pictures of a light bulb 10 ft away (distributed emission source) with 400mm f/4 and a 50mm f/2, using the same ISO and shutter time.
You'll find that the 50mm f/2 picture is 4 x brighter than the 400mm f/4.
You'll also find that the 50mm f/2 picture is 8 x smaller than the 400mm f/4.
The light energy collected by the 50mm lens is 4/8/8=1/16 of the 400mm lens.

Repeat the same with the light bulb switched off, but having some outside illumination. You'll get the same results.

Now try take pictures of say, Sirius (a point source) with the same lenses.
You'll find that the 50mm f/2 picture is 16 x less bright than the 400mm f/4.
You'll also find that the 50mm f/2 picture is almost the same size as the 400mm f/4 (slightly bigger diffraction pattern).
The light energy collected by the 50mm lens is 1/16 of the 400mm lens.

So it matters whether the source is distributed or point, but not whether it is emission or reflection.

Ken

[AJSJones]

OK, I'm pretty sure I don't get it, but that's ok. It would seem that the larger aperture would result in more light, and a higher shutter speed, but it doesn't. Obviously there is something going on that I don't understand intuitively. Maybe small words and pictures would help me, but I really don't need to know.
Mark

Mark - It does let in more light but what you are missing is that the lens is focused further away too (200 vs 50 mm if the subject is at "infinity"), so that light is spread out more by the time it hits the sensor/film plane. It's the inverse square law at work. What the same f stop does, regardless of the focal length is let the same light intensity fall *per film/sensor area*. So at 200 you get more photons through the hole than at 50, but they get spread out more - so it all works out , that's why f stop is used to represent "exposure"

Hope that gets you over the intuitive hump :-)

Andy

[Jon]

The Big Bang was a point source.
Bremstrahlung from an energetic electron is a point source.
The exact nature of the big bang in the opening sub-nanosecond interval is still conjectural. It rapidly moved out of any possibility of being considered such, even presuming that some of the newer physical theories haven't superseded it.
And the size of an electron, hence the scale of any internal activity depends on what state you consider it to be in; but considered as a particle it's finite if indeterminate.


Try take pictures of a light bulb 10 ft away (distributed emission source) with 400mm f/4 and a 50mm f/2, using the same ISO and shutter time.
You'll find that the 50mm f/2 picture is 4 x brighter than the 400mm f/4.
You'll also find that the 50mm f/2 picture is 8 x smaller than the 400mm f/4.
The light energy collected by the 50mm lens is 4/8/8=1/16 of the 400mm lens.

Repeat the same with the light bulb switched off, but having some outside illumination. You'll get the same results.

Now try that again with a clear envelope lamp, not a frosted, so you're photographing the light source, not the diffuser (galaxy vs. nebula). I will concede that the image will be smaller, but not dimmer.

Now try take pictures of say, Sirius (a point source) with the same lenses.
You'll find that the 50mm f/2 picture is 16 x less bright than the 400mm f/4.
You'll also find that the 50mm f/2 picture is almost the same size as the 400mm f/4 (slightly bigger diffraction pattern).
The light energy collected by the 50mm lens is 1/16 of the 400mm lens.

So it matters whether the source is distributed or point, but not whether it is emission or reflection.

Ken

Sirius is an emission source - where's your reflection point source by comparison? and how do you justify calling a star twice the mass of the sun a "point source" when you have previously said[QUOTE=kfong] "Theoretically there is no such thing as "Point Reflection source", since a point has no cross section for reflection to occur. So the word "Emission" in the term "Point Emission source" is actually redundant." I submit that if you allow a distributed emission source at sufficient distance to be called as a point source, the same should apply to a reflective source at sufficient distance.

I think that we're arguing semantics as much as optics or physics, though, at this point.