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# Thread: Exposure correction with extension tubes

1. I've just bought a 28mm extension tube for my Bronica ETRSi. I know I need to give extra exposure with a tube but how do I calculate it? I know the 'inverse sqare rule' comes into it but how do I apply it - I'm not paticularly mathematically minded! Also, how can I calculate depth of field when using a tube as it's obviosly so shallow?

Rob

2. What you are seeking is a "bellows correction" of which you can find calclators and formula to find the extra aperture or time to add.
The amount varies with the focal length of the lens mounted on the tube.
This is quoted from a posting I made in Exposure Discussions- "Whats your coolest trick"

**Only real "trick" I can think of is figuring the approximate actual aperture with my calculator. Fer'instance 100mm len at f/8. Divide the focal length by aperture to get the aperture factor or 12.5 (which should be the real size of the aperture in mm). Measure the total extension, say its 150 mm from focal plane to lensboard. Divide that by the aperture factor to get 12. That tells me I'm really working with an aperture of f/12 for that magnification.
The math can be swapped around to figure what to set at a particular extension to get the working aperuter desired. Say the 150 mm total extension and the 100 mm lens, you want f/8. Divide 150 by 8 to get the factor of 18.75. Divide 100 by the factor 18.75 and you get 5.33. So you would set the aperture to a smidge wider than f/5.6 to get a working aperture of f/8.
It's worked every time I've used it. Only hard part is finding the damn calculator when I want it, it tends to turn invisible on me at times. **

In your case you will simply add the 28mm to the lenses focal length, and possibly add any amount of extension introduced when using the lenses focusing closer with its helicals and do the math.

3. One operator of large-format cameras says just regard the focal length of your lens as an f stop and add any extension to it to get the next f stop. E.g., if you're putting a 28-mm extension tube behind a 50-mm lens, you're going from f/50 to f/78. These numbers look more like f stops if they're divided by ten: f/5 and f/7.8. So your 28-mm extension tube would call for about a one-stop increase in exposure time or maybe a tad more. This means that f/5.6 on your lens would be about equivalent to f/8. I haven't tried this, but it looks pretty good.

4. Somewhere around here, if it's survived the migration to the new software, I've posted a magnification/exposure table as an MS Word doc file (but such tables can be found in any book on macro photography).

What I do with any format camera is put a ruler in the scene at the subject position or estimate the width of the subject field and compare it to the width of the film, so if I'm shooting a macro scene that's about 4.5" wide with a 2-1/4" square (6x6cm) camera, then my magnification ratio is 1:2. Then I look up the exposure factor on my table and make the correction--1-1/3 stops in this example. No need to think about focal length, lens extension or subject distance with this method--just magnification ratio.

For DOF--you'll usually need all you can get. Just stop down all the way and use multiple flash pops or very long exposures if necessary. There are various DOF calculators on the net. One that I like is called "f/calc" which is a shareware program that can do several photographic calculations.

5. Originally Posted by David A. Goldfarb
. . . put a ruler in the scene at the subject position or estimate the width of the subject field and compare it to the width of the film, so if I'm shooting a macro scene that's about 4.5" wide with a 2-1/4" square (6x6cm) camera, then my magnification ratio is 1:2.
Isn't the width of the image on the film the significant value rather than the width of the film itself?[/i]

6. For the purposes of calculation of magnification ratio, it doesn't really matter whether you compare the size of the image on film to the size of the object in the world being photographed, or the width of the image field (the width of the film frame) to the width of the subject field (the whole area that the lens takes in at the subject distance). These ratios should be identical.

Methods like the "QuikDisk" available for free online and Calumet's device use the former approach, where an object is put into the scene, and a ruler marked in stops is used to measure the image of the object (the QuikDisk or Calumet's square card) on the groundglass, but this is only really practical with large format, where you have easy access to the groundglass. With a 35mm or medium format camera, I find it easier just to use the width of the frame as a known quantity for this purpose and to measure the subject field with a ruler or tape measure.

7. I misinterpreted "scene" to mean the width of the subject being shot, when in fact it includes the subject plus any space on either side that appears on the ground glass along with it.

8. I misinterpreted "scene" to mean the width of the subject being shot, when in fact it includes the subject plus any space on either side that appears on the ground glass along with it.

9. Okay now you all have confused the crud out of me. . .if you put a tube on a lens how can the focal length possibly get larger??? My 760mm lens is wide and shallow whereas my 150mm lens with appropriate cone mount is decided similar to mounting a lens on a tube and the focal length is less. Doesn't focal length mean the greatest possible diagonal which the lens can handle?

Same thing with bellows extensions adjustments. The longer the bellows are extended the larger the aperature needs to be wherein f8 would become f5.6, not the other way around.

10. Oops! I think I misread, particularly Lemaster's remark. Okay. Sorry, jumped in a bit too fast. Analytical dislexia I guess

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