How to calculate new minimal focus distance with a extension tube? I often see that a 300mm lens will focus 'down-to' XX meters instead of YY meters. Is it a easy formular? if not, is there any available website/software can do this automatically? I just want to know if I can use some telelens in my small room with the extension tube (PN-11, 52mm) on. Thanks a lot. Leon

This doesn't answer your question directly, but you may find it useful. http://www.macrophotography.org/modules.php?name=News&file=print&sid=15 Todd

Calculating this stuff with a simple lens that focuses by extension (moving the optical elements as a unit) is straightforward, but for internally-focused lenses (optical elements move relative to each other) is complex. Try this site. It's mainly aimed at Canon users, in that if you enter (say) 300 mm as the focal length, it will assume the minimum native focus for the Canon 300/4. However, you can make adjustments appropriate to other lenses and the results should be broadly generalizable: Julian Loke's lens calculator The site can be a little slow to respond, so give it a second or two to adjust after you enter a new value.

Suppose you have a lens of focal length F which can focus down to distance X when mounted on your camera. Suppose you use an extension tube of length L. Then you can focus down to the distance given by the formula F(XF +XL - Lf)/(XL -LF +f^2) In words, you do the following. First multiply the shortest distance by the focal length, the multiply it by the tube lenth and then multiply the tube length by the focal length. Add the first two numbers and subtract the third. Then multiply the result by the focal length. That gives you the numerator. Next take the product of the shortest distance by the tube length and the product of the tube length with the focal length, both of which you already calculated, subtract the second from the first, and then add the square of the focal length. That gives you the denominator. Divide the numerator by the denominator, and you are done. Here is an example. Let the focal length be 50 mm, the shortest distance 200 mm and the tube length 25 mm. Then with the tube, you can focus down to 50 x (200 x 50 + 200 x 25 - 25 x 50)/(200 x 25 - 25 x 50 + 50 x 50) which comes out to 110 mm. I wish this were simpler than it is, but I don't see any useful way to simplify it further.

I was recently wondering the same thing as Leon originally asked. Leonard, two questions: 1. What assumptions went into the math? 2. How valid are those assumptions in the range of typical photographic conditions? I ask not to challenge the validitiy but from the point of view of a curious scientist out of his field of expertise. - B

Leonard's equation is derived from the basic equation for focusing a lens. For example, see the Lens Tutorial at http://www.photo.net/learn/optics/lensTutorial. First, take the focusing equation with the two "knowns" of the lens without the extension tube, the focal length and closeest focusing distance, and solve for the maximum image distance. Now add to that image distance the length of the extension tube to obtain the new maximum image distance and use the equation a second time to solve for the closest focusing distance. Some of the assumptions are a lens that focuses as a unit rather than by changing its configuration, as mentioned by Mark -- an internally focusing lens changes its focal length. Also, on a thick lens, which means in practice all photographic lenses, the distances are measured from special points known as the front and rear principal points. If you need to know very exactly the distance from the film to the subject, you need to add up the image distance, the subject distance, and the principal point (or plane) separation. Sometimes lens manufacturers provide the value of the principal point separation.

Michael's response about how I calculated it is correct. On the other hand, I don't believe it usually requires knowing anything about principal points or any other specific information about the lens. The only assumption that is required is that the lens equation makes sense and describes the behvior of the lens when focusing. That equation says that the reciprocal of the the subject distance plus the reciprocal of the image distance adds up to the focal length. Thus, it is necessary that the focal length doesn't change when focusing and also that the points from which the subject and image distances are measured don't change their positions relative to the body of the lens. If those conditions are met, all that happens when you add an extension tube of a fixed length, is that you change the relative position of the film plane, placing it further away by that distance. In response, you have to move the subject an appropriate distance forward, determined by the lens equation, in order for the image to be in focus. If the focal length changes when focusing or if the reference points for measuring subject and image distance change, then the lens equation doesn't really make sense and you can't draw conclusions from it. I don't know how often non-zoom lenses fit into this latter category.

I am confused by this post. I can derive the above equation, but only by using minimum subject distance X in the lens equation, as in: distance=1/(1/f-1/X). However, my understanding of subject distance is that it's the camera-subject distance, and doesn't go directly into the lens equation, which incorporates only camera-lens and lens-subject distances. Of course, subject distance is the sum of these, but correctly using subject distance changes the structure of the lens equation significantly, preventing it from reducing to the equation described above. Let me clarify: Lens equation: 1/f = 1/g + 1/h where f=focal length, g=camera-lens distance, h=lens-subject distance. Therefore, when inserting a minimum lens focal distance X, that is, X = g+h, the equation becomes: 1/f = 1/(X-h) + 1/h Rearranging, this becomes solvable as a quadratic: h^2 - Xh + fX = 0 Using the quadratic formula one then solves for h as follows: h = (X +/- sqrt(X^2 - 4fX))/2 = (X +/- sqrt(X(X-4f))/2 Where the +/- indicates two solutions are possible for h. (BTW, this equation shows why a simple lens cannot focus closer than 4f: the X-4f becomes negative, and the square root of a negative number doesn't make sense in photography.) Using the prior example where X=200 and f=50, the square root becomes zero, and there is only one solution: h = (200 +/- sqrt(200(200-4(50)))/2 = 100 Then, adding the extension tube, h' = h + L = 100 + 25 = 125 Then, solving for g', gives g' = 1/(1/f-1/h') = 1/(1/50-1/125) = 83.3 Finally, the answer to the minimum subject distance for the extension tube is: Min distance = h' + g' = 125 + 83.3 = 208.3 This said, however, I prefer to use the following method. Because modern lenses do typically change their focal length at close distances, calculating the minimum focal length at closest distance will prevent the prior quadratic equation from becoming unsolvable. That is, if a lens is rated at 100mm but focuses down to 300mm, it violates the 4xfocal length minimum distance of 400mm. This means, therefore, that one could say this lens is working at 75mm at the closest distance (300mm/4=75mm), and 75mm should be used as the focal length in the optical equations at this distance. Once you know the working focal length at minimal distance, you could use the maximum magnification (M) at that distance to calculate h', where L is the length of the extension tube: h' = f (1 + M) + L Note that previously I used the lens equation to calculate h', but knowing the maximum magnification and calculating h' from this gives even more accurate information. Then g' is calculated as before from the lens equation, and g' and h' summed to give the subject distance. The results for the previous example turn out to be identical if M=1.0, since h'=f(2)+L=50(2)+25=125, as before, and the example did use a lens with a minimum working distance of exactly four times the focal length.