F-stops, angle of view and exposure

couldabin · Nov 3, 2011

Send me straight to the proper thread if this horse has already been beaten to death ...

The standard definition of the f-stop is usually expressed simply as the ratio of the aperture diameter and the focal length. Since the primary utility of f-stop designation is exposure, I'm wondering if the f-stop is actually designated with an intended film format in mind. (Or, more to the point, with a particular circle of illumination in mind.) The illuminance at the film plane of a 210mm set at f16 is going to be a whole lot less (one-fourth) if the lens covers 8x10 as opposed to 4x5. Isn't this what drives the necessity for considering bellows factor? Is it because lenses are so frequently tied to a particular film format that angle of view isn't discussed?

Thanks.

Markster · Nov 3, 2011

I don't think it takes into account format or film.

f/1 would be full light, if I recall. That's what you'd get if you exposed the film with no lense in place. f/1.8 through f/22 are just different fractions of that light (whatever the amount you got from f/1 was), each one half the light amount of the previous.

I think the distance to the film plane and the angle of the light is usually taken into account by the manufacturer, so that on a given mount, on a given camera, the focal plane is generally the same.

On your larger formats there you might get into strange territory, but on the 35mm bodies I'm familiar with all of this is taken into consideration and light metering simply "works"...

Ian C · Nov 3, 2011

The common currency of light intensity at the film plane, the f-stop, is independent of lens view angle and independent of format.

It’s true that the intensity falls off as we move radially outward on the film plane from the central lens axis. But this is significant only for very wide angle lenses. For some (but not all) LF wide-angle lenses there are radially-graduated neutral density filters with maximum density at the center that falls off to match the light falloff characteristics of the lens.

The majority of the central image area is approximately evenly illuminated.

According to the Kodak Professional Photoguide, bellows compensation is only required when the lens-to-subject distance is less than 8f for a lens of focal length f. At lens-to-subject distance of 8f the light falloff is slightly more than 1/3 stop.

With respect to

“The illuminance at the film plane of a 210mm set at f16 is going to be a whole lot less (one-fourth) if the lens covers 8x10 as opposed to 4x5.”

If you mean the total amount of light falling on each area, that’s not how light INTENSITY is reckoned.

It is light intensity that matters in exposure of film. If we have a lens that covers the 8” x 10” format, the intensity of the projected light in the central area is the same whether the image falls on an 8” x 10” sheet or onto a much smaller film (of course the image will be considerably cropped at the same subject distance with the smaller film).

The need for bellows compensation is due to the greater distance the light must travel from the lens to the film when the lens is extended significantly forward of its infinity position for close focusing.

It makes no difference what format the lens is used on. If it’s extended forward 1 focal length so that the total extension is 2f from lens to film, then we must increase the exposure by 2 stops so the film is properly exposed.

That’s true for a lens that covers an 8” x 10” film even if we equip the camera with holder for a smaller film (like a 120 back).

wiltw · Nov 3, 2011

"Aperture is aperture"...no accounting for film format
"FL is FL"...no accounting for film format
'lens distance from rear node to film plane'...often does account for format. For example, a 200mm lens for 135 format is usually a 'telephoto' optical design, where the distance from rear node to film plane is shorter than the true FL of the lens; but for large format, a 200mm lens is not 'telephoto' in design, but a 'long focus' design. For large formats, a 'telephoto' design might be used only when the bellows length would be an issue.
The illuminance at the film plane of a 210mm set at f16 is NOT going to be a whole lot less (one-fourth) if the lens covers 8x10 as opposed to 4x5. The illuminance is the same, the size of the image circle is larger for 8x10 coverage than if the lens were designed only for 4x5 coverage.

couldabin · Nov 3, 2011

I mean the intensity of light per unit area at the film plane. Let's assume for discussion that we have two 210mm lenses, one intended for 4x5, one for 8x10. The one for 4x5 has a circle of illumination of roughly 6.5" (in reality I imagine it is larger than that, but this is just for sake of discussion); the one for 8x10 has a circle of illumination of 13", or twice the area. Using the standard definition of f16, the aperture opening for both lenses will be 13.1. Won't both lenses admit the same amount of light? Won't that quantity of light be spread over four times the area with the 8x10 lens?

As to the bellows compensation -- the light rays themselves don't lose energy over distance; they're being scattered over a larger area. Yes?

Chan Tran · Nov 3, 2011

At the center of the image circle the illumination is the same. Further out lens designed for smaller format would have a lot of light fall off outside of its intended image circle. The one for the larger format would also have some light fall off further out of the center.
One thing is that the lens with small image circle doesn't concentrate light from outer area to the center area.

couldabin · Nov 3, 2011

Do we agree that the diamater of the aperture in both lenses is the same? I'm sure we agree that the area of illumination created by the 8x10 lens is four times that of 4x5. So if the illumination per unit area is the same in both cases (and it has to be if exposure is "correct" in both instances), then there has to be four times as much energy getting through the 8x10 lens even though the aperture opening is no different. Yes? How does that happen?

MattKing · Nov 3, 2011

Think in terms of using a 4x5 reducing back on your 8x10 camera.

Whether or not you use the reducing back or the full 8x10 back, the intensity of light hitting the centre 4x5 of the film is exactly the same.

The difference comes from the fact that, for a given subject distance, the 210mm lens designed to cover 8x10 needs to be able to accept light and render detail from a much wider angle than the 210mm lens designed to cover just 4x5.

If you use the 210mm lens designed for 4x5 on the 8x10 camera, for a given subject distance the centre 4"x5" part of the image on the 8x10 film will be substantially the same as if you had used the 210mm lens designed for 8x10. Outside of that 4"x5" portion though, the "smaller format" lens will most likely vignette and/or lose resolution.

couldabin · Nov 3, 2011

"Whether or not you use the reducing back or the full 8x10 back, the intensity of light hitting the centre 4x5 of the film is exactly the same." -- Agreed!

"If you use the 210mm lens designed for 4x5 on the 8x10 camera, for a given subject distance the centre 4"x5" part of the image on the 8x10 film will be substantially the same as if you had used the 210mm lens designed for 8x10." -- I'd go one step further: the object size on the film would be identical for both lenses. In other words, the image on the 4x5 film using the 4x5 lens would be exactly the same as if we had used the 8x10 lens with the 4x5 reducing back. (The key difference: angle of view.)

But the light coming through the 8x10 iris is being spread over four times the area. If the intensity per unit area is the same with both lenses, the surely must mean four times as much light energy came through the 8x10 iris, even though it's no larger. Yes?

Chan Tran · Nov 3, 2011

We all agree that the center portion of all the lenses the illumination are the same. At the outer area it has to do with vignetting. So I check on the Wiki and found this article.
http://en.wikipedia.org/wiki/Vignetting

MattKing · Nov 3, 2011

couldabin said:
But the light coming through the 8x10 iris is being spread over four times the area. ... Yes?

No

The light coming through the 8x10 iris is coming from a subject that is four times the area, and being spread over film that is four times the area, so it is a wash.

It would only change if you changed the magnification to fill the 8x10 film with the same subject as previously filled the 4x5 centre (and then only if you were working at or near macro magnifications).

couldabin · Nov 3, 2011

If the center (ie, the 4x5 film area) is the same exposure-wise for both lenses (and I would expect it to be if f16 means the same thing exposure-wise on both lenses), then we have to account for the all the extra light energy that falls outside the 6.5" circle when the 8x10 lens is used. The iris opening is the same for both lenses, so I'm not seeing how more light energy gets through the 8x10 lens.

Steve Smith · Nov 3, 2011

couldabin said:
But the light coming through the 8x10 iris is being spread over four times the area. If the intensity per unit area is the same with both lenses, the surely must mean four times as much light energy came through the 8x10 iris, even though it's no larger. Yes?

Yes but the lens is also collecting four times as much light from the subject.

Steve.

jeffreyg · Nov 3, 2011

In my effort to avoid the math and technicality of the optics of this subject I happen to have a 1982 version of Nikkor lenses for large format cameras. According to their chart format coverage shows the arc of the image circle at f22 for each of their lenses. They have lenses for different purposes such as general purpose, telephoto and wide angle. So for an example: their 150mm SW @f22 has an image circle of 10x12 inches and to get the same coverage in the W series all purpose lens you would need a 300mm @f22.

I know this is not the answer you are looking for but the lens design is an influencing factor.

http://www.jeffreyglasser.com/

couldabin · Nov 3, 2011

Ah! But, of course! Don't know why that didn't occur to me at the outset ... but it's clearly the answer ... Thanks much.

holmburgers · Nov 3, 2011

If the aperture was a toll booth and we charged each photon a fare, then indeed, the 8x10" 210mm lens would pay a higher toll than the 4x5" lens, assuming it's area of coverage is much greater. So indeed, it's letting in a larger total quantity of light but not light of a greater intensity.

This extra light is due to lens design and coverage alone, it really has nothing to do with the f/stop.

Right?

holmburgers · Nov 3, 2011

Steve Smith said:
Yes but the lens is also collecting four times as much light from the subject.

Steve.

we have a winner...

couldabin · Nov 3, 2011

Totally. It scares me just a little that I focused on the size of the circle of illumination (and the attendant energy) without thinking about that equally enlarged gathering area in front of the lens ... Duh.

MattKing · Nov 3, 2011

But the mental exercise involved in either:

1) visualizing the reality and understanding the reasons for it; or
2) visualizing the reality and figuring out a way to explain the reasons for it,

is certainly all to the good!

Chan Tran · Nov 3, 2011

couldabin said:
Totally. It scares me just a little that I focused on the size of the circle of illumination (and the attendant energy) without thinking about that equally enlarged gathering area in front of the lens ... Duh.

You got the right answer! Just that one lens with the ability to gather light from a larger area and one doesn't.

couldabin · Nov 3, 2011

It's how I (sometimes) get things through my thick head ...

ic-racer · Nov 3, 2011

couldabin said:
The illuminance at the film plane of a 210mm set at f16 is going to be a whole lot less (one-fourth) if the lens covers 8x10 as opposed to 4x5.

No.
http://en.wikipedia.org/wiki/Illuminance

thuggins · Nov 23, 2011

Another way to think of it is as the amount of light (energy) per unit area. The hole of the aperture doesn't limit the total amount of energy coming thru the lens, only the amount proportional to the full aperture setting. Any given object in your field of view reflects a given amount of light energy (a given number of photons). But the more objects that the lens sees (its field of view) the more total energy (total number of photons) gets reflected to the lens. More photons spread over more film area comes out equal each time. This is why exposure meters don't need a setting for the film format.

I proved this to myself the hard way after spending a bit too much to buy an OM to Pen lens adapter. In an ongoing attempt to photograph the moon, I was hoping to take advantage of the Pen's 1.45X "magnification factor" without the degradation caused by a teleconverter. Of course, for a given lens the image on the film was the same size whether taken with the OM or Pen --> Same moon, same number of photons, same exposure on the same area of film.

F-stops, angle of view and exposure

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F-stops, angle of view and exposure

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