
I need a chalk board....
Basic idea: extension tubes increase the image distance (obvious). The implication is that, for a given focal length, one is able to decrease the object distance (i.e. get closer to the object). This relationship is dictated by the equation:
(1/focal lenth) = (1/image distance) + (1/object distance)
so, for any given lens (i.e. focal length), increasing the distance between the lens and the film (the image distance) requires the the camera to be closer to the subject (object distance decreases) and of course, the closer you are, the bigger the image on the film. Simple...right?
The tricky part is adjusting the exposure for the increased image distance. The relationship that I like to use for this calculation is:
exposure factor = (image distance / focal length) ^2
This will give you a number larger than one that you can apply just like a filter factor (e.g. multiply the shutter speed by this number).
The trouble with SLR lenses is that, it is difficult to accurately measure or estimate the image distance. Technically, the image distance is defined as the distance between the nodal plane of the lens and the film plane. The difficulty arises in locating the nodal plane...luckily, we don't really need to.
We can use the fact that the image distance is equal to the focal length when the lens is focussed at infinity.
Mount the lens on the camera with an extension tube. Set the lens at infinity (notice, I didn't say focus at infinity  cause you probably can't do that with the extension tube in there...I mean set the focussing ring to infinity and leave it there. focus by moving the camera). Now, the image distance is one focal length plus the length of the extension tube.
Let's look at an example, take the 80mm lens and the 16mm extension tube...set focus at infinity. The image distance is now 96mm = 80mm + 16mm so your extension factor is:
factor = (96/80) ^2 = (96 * 96) / (80*80) = 1.44
if your meter reading suggests f/16 at 1/125 second you need to use something like 1/90 second instead because...
1.44 * (1/125) = (1/86)
Alternately, you can open up one half stop. This too can be calculated accurately as follows....
stops = log(factor) / log(2)
Continuing with the example,
stops = log(1.44) / log(2) = 0.53 ==> 1/2 stop
Does it make sense? Any questions?
Last edited by GaussianNoise; 01262005 at 08:27 PM. Click to view previous post history.
Reason: correct dumb spelling mistakes