Enlargers and bellows extension

BetterSense · Nov 23, 2009

How are enlarger lens f/stops calculated?

On cameras, lens f/stops are calculated for the focal length that results when the lens is focused on infinity. For any closer focus, in theory the marked f/stops are wrong, and you must calculate a 'bellows factor' to find out the real f/stop.

Since enlargers are NEVER infinity focused, are the marekd f/stops always 'wrong'? Or are enlarger lens f/stops calculated at some intermediate focusing distance? Since enlarger exposures are based on trial-and-error, it's mostly a point of curiosity; it wouldn't matter what the markings were really. But if you used an enlarger lens on a camera, would the f/stops be accurate at infinity?

But, this bellow-factor line of thought brings up another issue.

Generally, when I change print size, I scale the exposure time proportional to the change in area of the print--the same thing as measuring the change in lens height and applying the inverse square law; I just find calculating the change in print area easier. BUT...either method requires you to refocus the enlarger at the new magnification, so in theory the aperture at the new print size is different than that at the old size. A full treatment would require me to

1. Calculate the change in print area (or equivalently, measure the change in lens height and apply the inverse-square law)

2. Measure the before-and-after bellows extension, and adjust my lens so that my aperture is the same at both heights.

I had never considered the aperture 'problem' until I got a view camera, but when making 4x5 enlargements to 8x10 prints, bellows factor on the enlarger must be quite significant.

jeffreyg · Nov 23, 2009

With an enlarger you are focusing on something flat and depth of field is not much of an issue. The lens may be sharper at a particular f/stop but the f/stop will influence the length of exposure.

David A. Goldfarb · Nov 23, 2009

Bellows factor on a camera is an issue whenever you are shooting at a magnification of 1:10 (the image of the subject on film is 1/10 the size of the subject) or greater, and you can calculate bellows factor as a function of magnification alone, without knowing the length of the bellows extension or the focal length of the lens. You just need to know the size of the subject and the size of the image of the subject on the groundglass--works with any camera, any format. That's how devices like the QuikDisk work, but I usually just estimate the size of the object and the size of the groundglass image, or I'll put a ruler in the scene, if I need to be more precise, and convert magnification to exposure factor with a table taped to the camera back, my meter, and my notebook.

So enlarging--you're usually in the same magnification range as the range where you need to think about bellows factor with a camera, but remember, it's not an issue of the bellows swallowing up light. It's purely a function of magnification and how wide an area you're spreading the projected image over compared to the negative size. Now think about the standard paper sizes--5x7, 8x10, 11x14, 16x20, etc. Notice anything familiar about those numbers? 5, 8, 11, 16? They're approximately the standard f:stop series, so when you go from one standard size to the next, the difference in exposure is going to be about one stop.

There may be other factors that change your exposure between formats, like flare from the walls next to the enlarger, or the inherent reduction in contrast when printing at different sizes that might cause you to change paper grade, which might require adjustment in exposure, but if you're just looking for an easy way to determine exposure for prints of different sizes, this works without having to measure bellows extension or calculate bellows factor.

BetterSense · Nov 23, 2009

I'm not interested in what is approximately correct, or what is practical. I'm interested in what is exactly correct, theoretically.

[bellows factor is] purely a function of magnification and how wide an area you're spreading the projected image over compared to the negative size.

Ok, but what function? That's the whole issue here. Let me restate my thinking.

Say I have established my print time T1 for a 4x5 print from a 4x5 negative, and now I want to make an 8x10.

My logic tells me that if I have the same brightness of source, and I spread that brightness across more area, I have to increase my exposure time proportionally, assuming that reciprocity holds. So first I calculate the change in print area--4x--and so to find T2 for my 8x10 print, I multiply T1 by 4. So T2=T1*4.

BUT, if I just raise my enlarger head (and refocus) the image onto an 8x10 area, I have changed my bellows extension--and my 'true' f-stop. Thus, the source is NOT going to be the same brightness at the 8x10 size as at the 4x5 size! I will be spreading the light across 4x the area, AND my brightness will have changed--in this case gotten brighter--because the true f-number of my system decreased compared to what it was at 4x5, when the bellows was racked further out.

It seems that scaling the print area (or using the inverse-square law, which is equivalent) only works if after refocusing at the new size, I adjust my lens aperture ring some amount to account for the new bellows extension at that size, so that I have the same numerical aperture at both sizes. It's a key assumption in the inverse-square method of exposure change that the brightness of the light source doesn't change. Clearly if the brightness changes, because the true aperture of the system has changed, this will effect exposure time. So to accurately estimate the change in exposure time from one print size to another, one has to calculate both the change in image area, and then measure the change in bellows extension and calculate the change in aperture of the enlarger.

David A. Goldfarb · Nov 23, 2009

What you are calling the "true aperture" is usually called the "effective aperture."

One of the great attractions of calculating exposure factor by magnification alone is that it works perfectly, because the extension of the bellows is truly irrelevant. It isn't as if these two things, magnification and bellows extension, are contributing to the difference in exposure. It is just that one way to determine the exposure factor is by knowing the magnification, and another way to determine the magnification without measuring the subject and the image is by knowing the focal length of the lens and the amount of bellows extension.

Exposure factor = (1 + Magnification)^2

That's it. You could figure out the magnification without measuring the subject and the image on the glass by knowing the focal length and the bellows length, and you could express it in terms of effective aperture, if you wanted, but you would be doing the same thing.

BetterSense · Nov 23, 2009

OK. I believe you that the magnification takes the bellows factor into account. It's entirely believeable, but I haven't seen the math yet. When I work it out for myself I will by all means use the magnification...but I need to know the mathematical relationships between the magnification, bellows extension, and effective aperture so that my brain will be satisfied; I can't just use the formula. I suppose I need to find an optics text. In the meantime, I will measure the print area and bellows extension and calculate the exposure change using the method that I understand fully.

BetterSense · Nov 23, 2009

Exposure factor = (1 + Magnification)^2

How is this equation applied to enlarging?
would you find the original exposure factor (1:1 magnification) to be (1+1)^2 = 4
and then the larger factor (1:2) would be (1+2)^2=9

and the difference would be 5? So going from 4x5 to 8x10 print, you would increase exposure time by 5x?

David A. Goldfarb · Nov 23, 2009

Interesting question. I started to compose a response and realized I wasn't applying the formula exactly the right way for enlarging. The answer should be 4x, so you can't exactly subtract the smaller factor from the larger factor, but I'll need to think about that for a bit to see how one does get there, if no one else comes up with an answer first.

BetterSense · Nov 23, 2009

Actually after thinking about it for a bit, I don't believe that magnification alone could be used to predict a new exposure time. It could be used to do so if the effective aperture of the system stayed the same...which it will not due to bellows extension. I think that the common formulas ignore bellows extension and the corresponding change in numerical aperture--which in the neighborhood of 1:1 is not even approximately correct. I think that the formulas that predict the exposure change via magnification would be correct, only if you compensate for bellows factor besides.

David A. Goldfarb · Nov 23, 2009

Here's a quick rundown of the different ways of calculating bellows factor--

http://www.largeformatphotography.info/bellows-factor.html

The methods of computing by magnification, bellows length, and effective aperture should all yield the same result.

In camera terms, if the magnification factor is 1:1, the exposure is two stops greater than the exposure for a subject at infinity, the bellows is twice the length of the bellows at infinity, and the effective aperture is two stops smaller than the marked aperture.

4x= (1+1)^2 by magnification factor

The formula by extension and focal length is

Exposure factor = (bellows extension/focal length) ^2

So for a 150mm lens focused to a magnification of 1:1, say, the bellows extension would be 300mm, so

4x = (300/150)^2

Or alternately, you could say that the effective aperture of a 150mm lens focused at 1:1 is two stops smaller than the marked aperture.

BetterSense · Nov 23, 2009

I'm familiar with how to calculate the "bellows factor" for a camera. But I've never heard bellows factor corrections mentioned when it comes to enlargers. This is probably because print exposures are usually arrived at through trial-and-error, so any bellows factor calculations are just unnecessary because knowing the effective aperture of the system just isn't important.

However, there are formulas to calculate a changed print size that are based on the inverse square law, or on the changes in magnification--these two methods are geometrical and effectively do the same thing. However, I feel those formulas fail to provide the proper warning that they are correct only if the bellows factor changes by only a 'small amount' between the two print sizes. In the vicinity of 1:1 magnification, there could be quite a large difference in "bellows factor" and thus effective aperture between the two enlarger head positions. So in addition to applying the inverse square law, or magnification-based formulas, the user should also calculate the changed in bellows factor between the two print sizes and adjust the aperture ring of the lens so that the enlarger is operating at the same effective aperture at both print sizes. Only then will the magnification-based, or inverse-square-law-based formulas be correct.

David A. Goldfarb · Nov 23, 2009

No. Here's another way of thinking about it. Say you change the focal length of the lens with another lens of the same design (to minimize the effect of transmissive light loss due to factors like the number of glass-air surfaces) and make two prints of the same size from the same negative. The distance from the negative to the paper will be different and the length of the bellows will be different, but the exposure will be exactly the same. "Bellows factor" and magnification factor are two ways of describing the same optical phenomenon.

BetterSense · Nov 23, 2009

The distance from the negative to the paper will be different and the length of the bellows will be different, but the exposure will be exactly the same.

If so, surely, that is the case only if the lenses are set to the same F/number. We agree on this I hope.

What's throwing me off, is that of course the two different lenses will be focused at different bellows draw, and different distance from the baseboard. You are effectively telling me that even though that's true, that both lenses, when set to the same F/number and focused, will suffer then same 'bellows factor' as each other. This is entirely plausible, but until I see the math I don't believe it.

Chuck_P · Nov 23, 2009

David A. Goldfarb said:
It's purely a function of magnification and how wide an area you're spreading the projected image over compared to the negative size. Now think about the standard paper sizes--5x7, 8x10, 11x14, 16x20, etc. Notice anything familiar about those numbers? 5, 8, 11, 16? They're approximately the standard f:stop series, so when you go from one standard size to the next, the difference in exposure is going to be about one stop.

Just like in camera exposure, given the 1:2 relationship that exists between f/stops (and shutter speeds), if the OPs exposure is 15 sec at f/8 for an 8x10 enlargement, then increasing magnifation to 11x14 (an effective one stop reduction in light intensity) would require a doubling of time to 30 sec. While I always understood that increased magnification affects exposure time, it never really clicked in that way with the standard print sizes, cool.

Or, I guess he could keep the same 15 sec enlarging time and double the light intensity by opening up to f/5.6.

cowanw · Nov 23, 2009

I am not sure but I suspect that this has something to do with it. and that is that the camera is a reverse enlarger.
the subject in a camera is to the print in an enlarger.
But the light is coming in from the opposite side.
i.e. light from small negative to bellows to lens to large image for the enlarger. The camera
is light from large image to lens to bellows to small negative. Having the light come from the opposite side of the lens/bellows combination must make a difference? Maybe?

David A. Goldfarb · Nov 23, 2009

BetterSense said:
If so, surely, that is the case only if the lenses are set to the same F/number. We agree on this I hope.

What's throwing me off, is that of course the two different lenses will be focused at different bellows draw, and different distance from the baseboard. You are effectively telling me that even though that's true, that both lenses, when set to the same F/number and focused, will suffer then same 'bellows factor' as each other. This is entirely plausible, but until I see the math I don't believe it.

Yes, this is what I'm saying, but I'll let someone who can explain the math better than I can do that.

Chan Tran · Nov 23, 2009

I am not prepared to do a long post but keep in mind 1 thing. When you shoot with the camera you take the bellow extension (or distance from lens to film) into account but not the subject to lens distance. When you enlarge, the negative is now the subject so you don't care so much for the bellow extension there. What you need to know is the distance between the lens and the paper. So to calculate the effective aperture you use the distance betwen the lens and paper and not the distance between the negative and the lens. Because the effective aperture get smaller with bellow extension the effective aperture of the enlarging lens also gets smaller as you move the lens away from the paper. This is the reason why when you make large print you need to open the lens up or increase the exposure time. If the the effective aperture doesn't change then when you make a 4x6 or when you make an 8x10 you would use the same aperture and exposure time because the negative brightness is the same isn't it? Yes when used in a camera at infinity, enlarging lens transmit the same amount of light as a typical camera lens at the same aperture.

ic-racer · Nov 23, 2009

Your diffraction and the amount of light falling on the baseboard are related to your Effective Aperture which is calculated as follows:

Ne = Ni(1 + m)

Where,
Ne = Effective aperture number (f-stop)
Ni = Indicated aperture number on lens barrel
m = magnification

cowanw · Nov 23, 2009

I will add one further thing, the inverse square law only applies after the light has passed through the lens

dancqu · Nov 24, 2009

Close Ups

cowanw said:
Having the light come from the opposite side of the lens/bellows
combination must make a difference? Maybe?

See the enlarger as a close up camera; a close up camera in
reverse. Camera or enlarger, the same rules apply.

The lens sees as subject a flat lighted film and projects what
it sees upon a sheet of light sensitive paper, or film. Dan

Neil Poulsen · Nov 25, 2009

I can see where the f-stop correction would be affected only by magnification.

But, to what extent does reciprocity failure play a role when exposing photographic paper and adjusting the exposure time for the change in magnification.

dancqu · Nov 26, 2009

Same as Film

Neil Poulsen said:
I can see where the f-stop correction would be affected
only by magnification.

But, to what extent does reciprocity failure play a role
when exposing photographic paper and adjusting the
exposure time for the change in magnification.

Reciprocity failure is a characteristic of the silver
halides at very low levels of illumination. So, to
avoid keep the lens open and the exposures
short. Same as film. Dan

BetterSense · Nov 26, 2009

Or at least, keep exposure times the same. Then reciprocity failures should be no factor (other than for contrast which we assume we are controlling anyway).

Enlargers and bellows extension

BetterSense

jeffreyg

David A. Goldfarb

Moderator

BetterSense

David A. Goldfarb

Moderator

BetterSense

BetterSense

David A. Goldfarb

Moderator

BetterSense

David A. Goldfarb

Moderator

BetterSense

David A. Goldfarb

Moderator

BetterSense

Chuck_P

cowanw

David A. Goldfarb

Moderator

Chan Tran

ic-racer

cowanw

dancqu

Neil Poulsen

dancqu

BetterSense

Enlargers and bellows extension

Moderator

Moderator

Moderator

Moderator

Moderator

Moderator

Privacy & Transparency

Privacy & Transparency