Exposure Calculation

[reinier_de_vlaam]

Hi guys

 

I complex question for me, maybe not for you

 

I set-up my 4x5 with a 210mm lens before a still-life in my studio, I

frame and calculate the required exposure using a spotmeter. I set

the opening to say f8, determine the exposure time and make the shot.

Now I replace the 210 by a 150 and consequently have to re-focus by

moving the backplate backward.

My question now is: Can I use the same exposure time/lens opening? As

I moved the plate backward it is further from the lens and thus there

is a longer travel and thus higher light loss I guess. Is that a

correct assumption?

 

thanks

Reinier

[bruce_pollock1]

Assuming that the lighting conditions have not changed between the two shots, it should make no difference what focal length of lens you are using. The exposure will be the same.

[bill_proud]

Reinier,

Correct assumption except the lens order you stated would be reversed wouldn't they? It takes more bellows for a 210 than a 150.

I'm assuming you are talking extended bellows. Calumet makes an exposure correction gizmo consisting of two pieces. One piece, a square plastic chip is placed on the subject. A second, ruler marked piece is laid on the ground glass and is segmented and indicates exposure correction in 1/2 stops for the magnification and bellows extension used.

[wilbur_wong]

Using different focal length lenses will not require a change in exposure at "normal" distances.

 

If you are photographing something that is close to the camera, your bellows will be extended more than at infinity focus.

 

With extreme bellows extension, such as photographing at a one to one ratio, your lens will be twice as far from the film as when focused at infinity. At this ratio, there is a 2 stop difference in light falling on the film.

 

2 different focal length lenses used "close up" at the same camera (lens) distance from the subject will require different exposures.

At normal distances it will make little difference.

[pete_andrews]

Have you actually used this 5x4 camera yet?<br>In order to refocus from a 210mm lens to a 150mm, you'd have to move the rear standard <i>forward</i> not back, thus <i>reducing</i> the distance between the lens and the filmplane. The exposure would remain exactly the same.<p>It would be a slightly different case if you wanted to keep the magnification the same between the 210mm and the 150mm lens. You would then need to move the whole camera forward, and slightly extend the focus to compensate. This would reduce your effective aperture from f/8(ish) to f/(8 x bellows length in millimetres/150), and you'd need to slightly increase the exposure time to compensate.<p>BTW, f/8 doesn't give you much depth-of-field for a still life, does it?

[wilbur_wong]

At the same subject distance, the longer focal length lens will require more lens extension factor than a short focal length lens

[bruce wilson]

Mr. Pollock is correct, no exposure adjustments are needed. The change in focal length exactly compensates for bellows extension exposure corrections, or you can say that the change in magnification exactly compansates for bellows inverse-square-law losses.

[golliegwillie]

Unless the shot is closeup with long bellows drawn, there shouldn't be an exposure adjustment necessary for either lens. But bracket the hell just to be on the safe side. You can determine the amount of bellow factor necessary with the formula BELLOW FACTOR= (BELLOW EXTENSION)squared / (FOCAL LENGTH)SQUARED. A bellow factor of 1 requires no exposure adjustment; a factor of 2 represents 1 stop; 4 represents 2 stops; and so forth.

[art_haykin]

F/8 is f/8 is f/8 at a given bellows draw on all lenses, but there ARE subtle variables such as the difference in the glass of various lenses: uncoated or coated, low or high dispersion, etc., but these differences are rarely a problem except in highly critical situations. If you are a gearhead, test andcompare all your lenses under highly controlled conditions. And don't forget, mechanical shutters will vary WILDLY! When you take a meter reading, and it says "f/8 @ 1/125th." the meter doesn't know what lens and shutter you are using. If your f/stops and shutter speeds are off, compensate accordingly.

[reinier_de_vlaam]

The attatched picture is what makes me wonder, to me it seems that the 210 gathered more light then the 150 (ofcourse it was reversed sorry)

[bruce wilson]

Perhaps the geometry of the reflection and the flare of each lens resulted in the apparent increase of flare exposure in the 210mm shot. The rest of the shot (bottles and background) don't look too different to me (but then, I'm on a poor monitor right now).

[reinier_de_vlaam]

ps both are polaroids, exposed (10s) and processed (25s) with exactly the same time, both at f8 and only the rear plate moved to focus.

[wilbur_wong]

If your red wine bottle is just under 12 inches tall your bellows is probably extended about 20 to 25% longer than at infinity focus. If I were to be critical in exposure, I would give a small bellows extension factor to give more exposure (working positive should make polaroid darker). However, your 210 actually looks on the overall brighter than the other shot. I am guessing most of this is probably attributable to greater lens flare from your 210 lens reacting to the back-lit light source.

[bill_proud]

Reinier,

 

After reading my earlier response it seems I may have stated it confusingly. We are all on the same page, there is no difference in exposure between lenses at infinity focus, however, at extended bellows draw for macro work there is additional exposure required. Is this as clear as mud or do I need a writing class?

[richard_cochran]

An equivalent formula to the correct ones posted so far

is simply:

<p>

bellows factor = 1 + magnification

<p>

Where magnification is the image size divided by the object size.

At infinity, magnification approaches zero, and the bellows

factor is 1.

<p>

That bellows factor is intended to be used in the formula

<p>

effective f-stop = marked f-stop * bellows factor<br>

or<br>

effective f-stop = marked f-stop * (1 + magnification)

<p>

A nice thing about this formula is that focal length doesn't

appear in it. It wraps up the effects of focal length and measured

bellows extension into the magnification factor.

<p>

So the answer is, if you move your camera so that you fill

the frame with the same sized subject (same magnification),

you can change focal lengths all day and the bellows factor

won't change from one lens to the next. But if you keep the

camera position fixed, and refocus to get a different magnification,

there will be a change in bellows factor. The longer lens will

have a greater magnification, and therefore require a

wider marked f-stop to get the same effective f-stop.

The change will be more significant at close distances (greater magnification).

[reinier_de_vlaam]

So if I understand correctly what I get is this

 

The bottle size on 210 = 10cm and for 150mm 7cm, so the magnification is 1.4 times. This would lead to 1+magnification= 2.4

 

So if I used f8 and 10sec for the 150, I would need f20 and 10sec for the 210, is that correct?

 

That the 210 lens has flare is a fact. I was framed on ebay, I bought the lens marked as top quality and it proved to have cleaning scratches. Actually I'm doing these shots to prove it to the seller, but sofar he didn't respond so I guess I have bad luck. On the other hand, it cost just 250 euro(=dollar) and it has an indeed perfect linhoff shutter which alone is worth 175 I estimate so the loss is not that big. (But if any of you live in Berlin and want to give this guy a visit for me...). That's why I also have started a thread earlier in the vague hope someone dropped/broke the back lens of his symmar and has a spare front element for sale.

[reinier_de_vlaam]

ps sounds like a big difference to me f8->f20. How to determine the first exposure with a light meter? It will give a fixed reading on that distance/lightsource and that is not changed I guess, the lens is still it the same place.

[richard_cochran]

<cite>

 

The bottle size on 210 = 10cm and for 150mm 7cm, so the magnification is 1.4 times. This would lead to 1+magnification= 2.4

</cite>

<p>

No, magnification is defined as the image size divided by the

object size.

<p>

I'll guess that the bottle is 30cm tall in real life.

That gives 10cm/30cm = .33 magnification on the 210.

It gives 7cm/30cm = .23 magnification on the 150.

The bellows factor is 1.33 on the 210, vs. 1.23 on the 150.

<p>

I'd call both bellows factors pretty close to a third of

a stop, and the difference between them is small. To

be precise, the difference is log2 (1.23/1.33) which is

approximately a tenth of a stop.

<p>

I'd guess that coating differences, flare, and shutter

speed calibration issues are more significant.

[david_lazarus]

Just pull some polaroids and see what you get, then make any changes from there. Remember the KISS rule!!! David.

[jorge_gasteazoro4]

I have read all the answers and you have gotten more than enough good advice, the only thing I have not seen mentioned is bellows flare. With the 210 you have more bellows extension and less probability of flare, with the 150 less bellows extension, more flare and a muddy print!

BTW if you are going to be doing these type of photography get the little Calumet gizmo someone mentioned above, it is invaluable for this kind of set up, forget the calculator and all those fomulas, set the little square on the bottles, read the magninfication with the provided ruler and right next to it the bellows adjustement. Piece of pie, easy as cake....uhm how does that go?

[albert_martinez]

Hello Reinier,

 

I have an alternative to Mr. Wiley�s proposal.

Forget squaring the bellows extension and focal length, it requires an extra step that's not required, what your left with is:

 

bellows extension / focal length = exposure compensation

 

A 150mm lens requires about six inches of bellows @ infinity,

it's common to refer to it as a six inch lens.

 

You arrive at the six inches by converting mm to inches.

150mm/25.4mm (1 inch) = 5.905 inches

 

Your 210mm would be 210mm/25.4 = 8.26 inches

 

One caveat, it doesn't apply to telephoto lenses, you'll need to check with the manufacturer or take it outside and focus the lens

at infinity, measure the bellows and use this number instead.

 

What I've found works the easiest for me is to:

 

1. Mark the bellow draw of the lens on the lens board,

so for the 150mm I'd mark it 6"

2. Focus the camera and measure the bellows extension.

3. If it's longer than 6" you'll need to compensate.

4. By how much, that's easy?

5. Say your bellows are now extended to 8"

6. Multiply 8 x 25.4mm = 203.2mm

7. The last step is to divide 203.2mm/150mm (the lens)= 1.3546

8. If you wind up with say a 2, you'll need to compensate by a full

stop. The 1.3546 is somewhere between 1/4 and a 1/2 stop.

 

So in closing all you need to do is measure the bellow draw

and divide it by the know focal length of the lens and your done.

 

I've taken it a step further by marking the rail on my Toyo

with compensation markings for my three most used lenses.

 

With a red Sharpie pen I have my 135mm lens marks at infinity = 0

+1/4, +1/2, and + 1, if I need to compensate even more I measure

and do the simple math. No mess, no targets, no differential

equations ;-).

 

For the 210 it's marked in black, and the 360 in green, use both sides of the rail, it can get a bit busy.

Etching the numbers in another option, a Dremel tool would do a nice

job.

 

I hope this helps simplifies your endeavors, God knows we have enough to consider when were using these great cameras...

 

Regards-Albert

[michael_taylor9]

Try this calculation: M=magnification E=film to lensboard distance

L=lens

Therefore M = E-L/L so if you have a 210 lens streched out to 300mm from the film, subtract 210 from 300 and divide by 210= .42

Now Exposure factor = (M + 1)squared = (.42 + 1)squared = 2.01

so add another stop of exposure. Done.