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Fragomeni
28-Dec-2010, 19:30
I'm working on some macro and micro experiments and a question came up that I've never really had to think about before and it has me stumped. I'm using a 120mm Rodenstock Macro lens and in my experiments I've played with putting some Hoya close up lenses (+1, +2, +4) on for additional magnification. My question is does the use of a close up lens change the actual focal length of the lens being used? This came up because when I go to compensate for bellows length I need to know the focal length of the lens to figure out my compensation. Does the close up lens change my focal length or is it factored in some other way? Thank you.

vinny
28-Dec-2010, 19:37
Just measure your bellows as you would for any other macro shot. At least that's what i do.

Fragomeni
28-Dec-2010, 19:48
Thats what I started to do but to appropriately calculate the compensation necessary for the extended length of the bellows you need to know the focal length of the lens. If I were to compensate using the focal length of the lens (not taking into account the close up lens if it does effectively change the focal length) then the compensation would be off.

I've found some equations that supposedly demonstrate how to calculate the change in focal length when using close up lenses but I'm having a hard time understanding them. Primarily I just need to know if a close up lens does actually change the functional focal length of a lens so I can figure out if its really necessary to even worry about this.

ic-racer
28-Dec-2010, 20:00
Use the magnification (image height comparison) method to get your bellows factor.

Yes, diopter lenses change the focal length. You can see for yourself, just focus at infinity with the diopter in place and with it off.

Fragomeni
28-Dec-2010, 20:08
Can someone simply provide an equation to find the new focal length with the diopter in place and an explanation for using the equation? That will be the most useful thing I can think of right now.

ic-racer
28-Dec-2010, 20:26
Off the top of my head:

+1 diopter is one meter focal length (1000mm)

A 120mm lens is +8.3 diopters

Add them together and you get +9.3 diopters or 107.5 mm focal length (this is a 'thin lens' approximation, best bet is to actually measure it on your standards).

But, you don't need to know this for the bellows factor.

Mark Sawyer
28-Dec-2010, 20:30
First you need to understand what "diopter" means. It's a reciprocal of the focal length of the lens in meters. A +1 diopter lens has a focal length of 1 meter (1000mm), a +2 has a focal length of 1/2 meter (500mm), a +4 has a focal length of 1/4 meter (250mm), a +10 is 1/10 meter (100mm), and so on...

To determine the focal length of the diopter plus your lens, add the two focal lengths together and divide by 4. Example: a 300mm lens and a 500mm (+2) diopter is 300mm + 500mm = 800mm, divided by 4 gives you a 200mm lens.

Don't use the stops indicated on your aperture scale in any calculations. Besides the error from the change in focal length, the aperture must be measured through the front element, and the addition of a diopter will magnify it and change the measurement significantly. Simply calculate the f/stop after you focus: measure the bellows extension and the aperture diameter (through the front element), and divide the former by the latter. No need for calculating the bellows extension factor with this method, as it is already figured in when you measure the bellows extension.

Hope this helps.

Fragomeni
28-Dec-2010, 20:48
K, this is the equation I was looking for. I knew I saw it somewhere before.

To determine what I was looking for i.e. new focal length with diopter lens added:

1) Calculate the dioptric power of a lens -- d=100/f (f=focal length in cm)
2) Add the diopter of your lens to the diopters of the close up lens. Ex. I want to add a +4 diopter to my 120mm lens so 120mm= 8.3 diopters. 8.3+4=12.3 diopters
3) Calculate the new focal length of the lens plus the diopter lens -- f=100/d = 100/12.3= 8.1 = 81mm

Don't try to use the stops as indicated on your aperture scale; besides the error from the change in focal length, the aperture is to be measured through the front element, and the addition of a diopter will magnify it and change the measurement significantly. Simply calculate the f/stop after you focus: measure the bellows extension and the aperture diameter (through the front element), and divide the former by the latter. No need for calculating bellows extension with this method.

This is very interesting, especially because I hadn't yet considered the effect of the diopter lens on the aperture settings. So what you're saying is I can attach my diopter lens, focus getting everything where I want it and then calculate my actual aperture by measuring the bellows extension and the aperture diameter (through the front element), and divide the bellows extension by the aperture diameter? This will give me the true aperture as opposed to what it written on the lens? Just want to make sure I understand that right. And using this method alleviates the need to compensate for bellows extension?

Mark Sawyer
28-Dec-2010, 21:49
...So what you're saying is I can attach my diopter lens, focus getting everything where I want it and then calculate my actual aperture by measuring the bellows extension and the aperture diameter (through the front element), and divide the bellows extension by the aperture diameter? This will give me the true aperture as opposed to what it written on the lens? Just want to make sure I understand that right. And using this method alleviates the need to compensate for bellows extension?

Yes, this is correct. The f/stop is simply the focal length (bellows extension) divided by the aperture (measured through the front element). That's all it ever is, one very simple ratio.

And just to make it clear, when I say "through the front element" that means through the diopter lens too, as it is now your front element, (unless you're putting it on the back of the lens, which would work too).

BTW, if you end up doing longer than one second exposures, as is common with macro photography, don't forget to figure in the reciprocity failure!

Fragomeni
29-Dec-2010, 11:44
Yes, this is correct. The f/stop is simply the focal length (bellows extension) divided by the aperture (measured through the front element). That's all it ever is, one very simple ratio.

And just to make it clear, when I say "through the front element" that means through the diopter lens too, as it is now your front element, (unless you're putting it on the back of the lens, which would work too).

BTW, if you end up doing longer than one second exposures, as is common with macro photography, don't forget to figure in the reciprocity failure!

Thanks for the help! This should be an interesting experiment. Much appreciated!

Fragomeni
29-Dec-2010, 15:35
I just ran some tests to determine the actual effect of using close-up diopter lenses on my LF macro lens and it turns out that the diopter lens does NOT actually change anything that needs to be compensated or corrected for. This is what I did and what I found:

I made test exposures using a 120mm f5.6 macro Rodenstock with no diopter added. I used a Pentax 1 degree spot meter to determine my exposure time and my f-stop. Using Fuji FP-100B (this film allowed me to do the test without needing to expose and then develop film) a test exposure was made at f5.6 @13 sec at ISO 12. These exposure values take into account reciprocity failure and compensation for the length of the bellows and the proportional light fall off that occurs due to the extended distance between the focal plane and film plane in close up photogrpahy. My method for dealing with bellows length in close up photography is to first determine if you are inside of 6X the focal length of the lens. If you are (as you usually will be in close up photography) you take the distance from the focal plane to the film plane (value 1) and the focal length (value 2) and turn both values into f-numbers. Take the difference in stops and then half the iso by that number of stops. Expose at the new iso with everything else unchanged and the image will be exposed perfectly. Subsequent test exposures were made adding +1, +2, and +4 diopters (individually, not stacked) to the 120mm macro, re-focusing, and making exposures at the same exposure values. Aside from magnification the exposures and their densities were identical.

If the diopter lens had changed the functional focal length of the lens then the values used to compensate for the length of the bellows and light fall off would have changed and the densities in the exposures would have consequently been different. By describing this as "functional" I differentiate this from any theoretical change which can be taken into account and calculated with the formulas listed in previous posts and by "functional" I mean a change that will alter the performance of the lens in practice. There was also no change in aperture values as was previously described. Again if there was a change in aperture values due to the diopter lens then the exposures and densities would have differed because the compensation for the bellows length would have changed.

What I believe is happening is the diopter lens simply redirects the angle and projection of light into the front element of the lens. This accounts for the increase in magnification when a diopter lens is used. All other functions of the lens occur as they normally would once the light and image pass through the front element of the lens (not counting the diopter lens). It is effectively like changing the actual size of the subject and photographing it; the size of the image entering the lens is changed but the function of the lens itself is unaffected.

I wanted to share this with anyone who might have run into the same questions. If you experience different results I have no way to explain that. These are simply the results of the tests that I have run in my studio using my equipment and this is what I find to work ideally for me.

Thank you everyone who contributed for your help. Conversation like this (one's that actually help you learn) are why I love this forum.

Dan Fromm
29-Dec-2010, 16:32
Interesting discussion. I've been waiting for someone to pull out a copy of Lester Lefkowitz' book The Manual of Closeup Photography and simply look up the answer. As long as the prime lens is used on its own mount, no exposure compensation is required when a diopter lens is attached to it.

Fragomeni
29-Dec-2010, 16:36
Interesting discussion. I've been waiting for someone to pull out a copy of Lester Lefkowitz' book The Manual of Closeup Photography and simply look up the answer. As long as the prime lens is used on its own mount, no exposure compensation is required when a diopter lens is attached to it.

Hahaha! If I had the book I guess I could have saved myself some trouble here! There are a few cheap used copies on Amazon. I think I might just pick myself up a copy. Thanks for the book recommendation Dan.

ic-racer
29-Dec-2010, 19:35
The diopter lens changes the focal length of your lens. You proved it when you needed to re-focus.

Your method of bellows factor calculation you used is independent of focal length.

To test you focal length change easily, just focus at infinity with and without the diopter. Your change in front or rear standard position after re-focusing at infinity is your change if focal length.

To answer the question of does one NEED an exposure change with a +1 diopter depends on how one is determining bellows factor. If you are not using an equation with focal length in it, then nothing additional needs to be done.

If you are using an equation with focal length in it, the addition of a +1 diopter only changes the focal length 10% so if you ignore it (as stated in the bood reference above) you likely will be OK.

Fragomeni
29-Dec-2010, 19:54
The diopter lens changes the focal length of your lens. You proved it when you needed to re-focus.

Your method of bellows factor calculation you used is independent of focal length.

To test you focal length change easily, just focus at infinity with and without the diopter. Your change in front or rear standard position after re-focusing at infinity is your change if focal length.

To answer the question of does one NEED an exposure change with a +1 diopter depends on how one is determining bellows factor. If you are not using an equation with focal length in it, then nothing additional needs to be done.

If you are using an equation with focal length in it, the addition of a +1 diopter only changes the focal length 10% so if you ignore it (as stated in the bood reference above) you likely will be OK.

We're in agreement of the main point of this. What my question asked was if the diopter lens changes the functional or operational focal length of the lens. This type of change would be like screwing on a different front and rear element (ridiculous I know but it illustrates the point). That type of change would cause all of the factors we're working with such as the true aperture values and how they relate to exposure times and the bellows compensation (depending on how you do it). My point was that the diopter lens does not affect the lens in this way. If it did, an exposure made with a diopter at the same exposure values as the lens without the diopter would result in different exposures with different density levels throughout. This is not the case hence my explanation that the diopter lens simply affects the angle of light hitting the front element of the prime lens. Its like holding up a magnifying glass in front of a dime and photographing it. The magnifying glass does nothing to change the lens, it simply enlarges the subject prior to photographing it. The diopter lens does not affect the lens in any way that requires additional compensation.

ic-racer
29-Dec-2010, 20:51
We're in agreement of the main point of this. What my question asked was if the diopter lens changes the functional or operational focal length of the lens. This type of change would be like screwing on a different front and rear element (ridiculous I know but it illustrates the point). That type of change would cause all of the factors we're working with such as the true aperture values and how they relate to exposure times and the bellows compensation (depending on how you do it). My point was that the diopter lens does not affect the lens in this way. If it did, an exposure made with a diopter at the same exposure values as the lens without the diopter would result in different exposures with different density levels throughout. This is not the case hence my explanation that the diopter lens simply affects the angle of light hitting the front element of the prime lens. Its like holding up a magnifying glass in front of a dime and photographing it. The magnifying glass does nothing to change the lens, it simply enlarges the subject prior to photographing it. The diopter lens does not affect the lens in any way that requires additional compensation.

YES, adding the diopter lens changes the focal length of the lens. There is no question with that.

That you demonstarted no observable difference in exposure is because your test was not sensitive enough to pick up 10% of an f-stop change.

Adding a +1 diopter lens = 10% focal length change = change in aperture scale from f22 to f21

I'd just alter you post to say :"My point was that the diopter lens does not affect the lens in this way [that I can determine from my experiment]" and I'd agree with you.

I commend you on testing this because the short-hand of "diopter addition" is only an approximation. So, the bottom line is one needs to test like you did.

Fragomeni
29-Dec-2010, 21:20
YES the diopter changes the focal length of the lens. There is no question with that.

This is where I think we're missing each other. The diopter lens does not change anything about the prime lens. The aperture f-stops to exposure times do not change and the diameter of the aperture in relation to the focal length does not change. With those facts you can see that the diopter lens has no direct effect on the prime lens. This is why I used the example of the magnifying glass. You could put a very powerful magnifying glass in front of a dime and make it the size of a tea saucer and photograph it and there will be no change in any of the exposure values or mathematical relationships that govern exposure between that image and an exposure made with no magnifying glass using the same prime lens. The magnifying glass is an entirely external factor as far as the prime lens is concerned. The magnifying glass does not literally change the focal length of the prime lens in any way that changes how the lens operates in relation to all other factors. That is my argument. Although the magnifying glass does not literally change the focal length of the prime lens it is a lens itself (as opposed to a piece of glass with no curvature) and you will of course need to re-focus because of the difference in the angle and curvature of light projected through it. This could be described as a change in focal length but would be more accurately described as a combination of two independent focal lengths and we can easily compute the combined focal length of the magnifying glass and the prime lens together but this is different then a literal change in focal length of the prime lens itself that would contribute to differences in the ratios of the diameter of the aperture to focal length and so on i.e. a literal and measurable change to the prime lens itself and its functional operation. Keeping all of that in mind, the additional lens whether it be a diopter lens or a magnifying glass (realistically a diopter lens is nothing but a magnifying glass) does not change any function or any of the relationships that make the prime lens function as it does.
The easiest way to explain it would be that a diopter lens and the prime lens are independent of one another. The focal length of each remains the same whether working in concert with a another lens or not. When working together a combined focal length can be determined but should not be confused with a change in focal length of either lens. A combined focal length of two lenses is different then a direct change in focal length of a lens. Understanding this difference is critical. My question dealt with whether or not this change affected the prime lens in a way that affected how it functioned i.e. caused a literal functioning change of its focal length. I'm fairly sure we're in agreement but we're just thinking about this differently i guess.

Adding a +1 diopter lens = 10% focal length change = change in aperture scale from f22 to f21
My tests covered +1, +2, +4, and later +3, +5, +6, and +7. No change was seen for any of these although I acknowledge that at a high enough magnification there are bound to be some changes somewhere. I need not test any further then I have for my purposes.

I commend you on testing this because the short-hand of "diopter addition" is only an approximation. So, the bottom line is one needs to test like you did.
Exactly. Simply running a test will answer any questions a person might have and when it comes down to it all that matters is what works for you and your equipment regardless of how you explain it or how it's understood.

Let me know if this makes any of what I'm saying more clear. If not I guess we'll just have to agree but disagree all at the same time hahaha but the bottom line is we're talking about the same end result. Going back and forth probably wont clarify anything further.

By the way for anyone reading, this is why this forum is so wonderful. In any other forum this would have turned into a ridiculous flame war of back and forth bickering and name calling. I guess LF photogs just operate on a higher wavelength :D

ic-racer
29-Dec-2010, 23:02
Because the diopter method is just a short hand for optometrists, and the exact optics are somewhat complex, a simple experiment would be interesting to try.

Take your lens and add some diopters to the front of it. Now treat it like an 'unknown' lens and use what you known about lenses to determine the aperture and focal length of the system.

For example measure the entrance pupil (of course looking through the diopter lenses).
Then use any of the popular methods of determining the focal length.

I don't have any lenses in front of me to test right now but I'd predict if you add positive diopters you focal length will shorten (how much?? The 'diopter addition' is only an approximation, so your test will show exactly how much). In terms of the aperture effects on the entrance pupil, I'm predicting the new combined lens-unit actually becomes FASTER! (perhaps explaining everything).

Fragomeni
29-Dec-2010, 23:13
Interesting experiments you just proposed! Inevitably now I'll end up testing each of them since you put them in my mind and knowing myself I'll probably go insane if I don't at least try them! I'm blaming you when my girlfriend gets to yelling at me for playing with the cameras too much and not spending enough time with her!

Emmanuel BIGLER
2-Jan-2011, 05:07
Hello from France and Happy New year to all !

Regarding this question of optics, I am not sure that considering a compound lens made of the association of the close-up lens plus the actual view camera lens is of real interest. But the good way is the one found most convenient, so do what you find most convenient.

The other way to consider how things work is to consider that the view camera lens takes a picture of an image, the image of the original object as transformed by the close-up lens.
Following this approach, there is no need to look for a modified focal length. Suffice to know where the image to be captured is located. This is simple for far-distant objects, since a "+N" dioptres close-up lens places the image at (1/N) metre in front of itself.
Well, in general, if the orginal object is located "closer than infinity" ;) a little calculation might be necessary to find where the image is located.

Regarding bellows correction factors for exposure, again considering that the view camera lens with its nominal f-stop looks at an image yields a very simple approach.
If the close-up lens is large enough so that it does not introduce any vignetting (you should not see the close-up lens mount from the ground glass location), hence the illumination in the film plane is determined by the view camera lens f-stop, the view camera bellows extension and the luminance of the image.
There is small miracle of photometry, namely that the luminance of the image given by the close-up lens is equal to the luminance of the original object, whatever the magnification and position of this image might be !
Hence, when looking at the exposure corrections, simply do as you would do with a real, original object. A spotmeter aimed at the original object will give you an indication which is formally related to the luminance of the object; so, no need to aim at the image of the object through the close-up lens, but if you wish you can place the close-up lens in front of the spot-meter and take a reading through it.
Set your view camera lens according to the nominal f-stop as given by the meter reading, correct for the actual bellows factor and ignore the fact that the image is seen through a close-up lens. If of course the close-up lens does not introduce any vignetting, which is usually the case.
Following this procedure, there is no need to re-compute the focal length of the coupound system. Moreover, as mentioned, the simple addition of inverse focal lengths fails to yield the proper focal length of the compound system, for some technical reasons ... well this is another story for those who love tutorials on geometrical optrics :D

David Casillas
15-Jan-2011, 13:31
Hi all
I´m getting late here, and I wanted to add something.
Per kodak profesional guide you will get a combined focal lenght when using positive or negative supplemantary lenses. The formula was alredy mentioned and is combined focal length = 1000/power of the lens in diopters + the diopters of the supplementary lens.
The required exposure when combination is focused at infinity or at normal distances is Efective f number=indicated f number x (combined focal length/camera lens focal lenght). Of course you need to calculate aperture for combined lens once and the use the same correction if you are working at normal distances. For close up work you need to calculate exposure using the combined focal length and the bellows extension. If you leave the camera lens focused to infinity and then add the supplementary lens and instead of refocusing you move the whole camera to focus the subject, then i believe it is correct to think you dont need any exposure compensation. But at infinity you will get a faster lens with the combination of a lens + a positive supplementary lens and slower lens with the addition of a negative supplementary lens.
You will get better quality if you put the supplementary lens on the back of the camera according to this book.
I´ve been using Nikon 5T and 6T achromats on back of my 210 mm lens with a very good quality. i get combined focal length of 160 mm and 130 mm. If I measure film distance to center of lens this measure longer than calculated combined focal length but if I put supplementary lens on the front of the lens it measure calculated focal length or little less, dont remember very well, but field of view is the same. They cover 4x5 with movements since I use a rodenstock apo Sironar S 210 mm and the angle of coverage keeps the same if there is no physical obstruction.
I also use -2 Kodak Telek suplementary lens. This is the opposite that with positive lens. This work better on the front of the lens and in this way it measures the same than calculated focal length or very close to it. By putting it on the back the distance from the film plane to the center of lens is less and in this case I get a different field of view. Well I hope this helps