I wonder if someone out there could help me, please, with the following question. suppose you have a 100mm, a 200mm and a 300mm lens. Suppose these lenses have minimum focusing distances of approximately 1metre, 2 metres and 4 metres apiece. Now you interpose a 50mm extension tube between each lens and the camera. You should be able to get a maximum magnification of 1:2, 1:4 and 1:6, respectively, OK? Well, how does this affect the minimum focusing distance in each case? Thanks.

Wirelessly posted (BlackBerry9000/4.6.0.167 Profile/MIDP-2.0 Configuration/CLDC-1.1 VendorID/102 UP.Link/6.3.0.0.0) I can't tell you those minimum distances, though there is a mathematical formula to determine it. However if your lens focal lengths are a constant, when you change the focal extension from the film plane by a constant distance, the ratio of minimum focal distances between one lens and another will remain constant as well. If you have focal distance of, say, 1 with the 100mm and a min focal dist of 2 with the 200mm lens then your min focal distance with the 100 and a 50mm ext tube will be 1/2 the min focal dist of the 200 with the tube.

Not quite. The maximum magnification will be higher. You only figured in the 50 mm extra extension. Not the extension the lens mount already provides on its own. For instance, the 100 mm lens having a minimum focussing distance of 1 m would mean that it comes with about 12 mm extension in its mount, which has to be added to the extra 50 mm to figure out the maximum (!) magnification. "about" 12 mm, because we don't know the internodal distance of that particular lens, which is part of the (minimum) focusing distance (which is measured from film plane to subject). But supposing infinitely thin lenses (without internodal distance), i.e. assume that the focussing distance is equal to the sum of the object and subject distances, we could use a rearrangement of the classic formula 1/f = 1/u + 1/v. First, to calculate what the subject and image distances (u and v in the formula) are given their respective sums (1 m, 2 m, 4 m) and the focal lengths (f). Then, to calculate for each focal length how the subject distance shrinks when the image distance is increased by 50 mm, and what the resulting sum would be. I don't like maths. Sorry!

OK, thanks for trying, everyone! I have been busy too, and found a graph in a book (SLR Photographers Handbook, HP Books, Tucson AZ, USA, 1977) written by Carl Shipman (a very informative book, by the way and, unlike many books these days, without spelling or grammatical errors) which provides the "Lens to Subject Distance for Magnification by Extension. According to this (non-linear) graph, the answers to my questions are as follows: In the case of the 100mm lens: approximately 45cm. (Prev.:1m) In the case of the 200mm lens: approx. 1m. (Prev.: 2m) In the case of the 300mm lens: approx 2.1m. (Prev.: 4m) Thanks all.

The reason for my curiosity on this matter is that I have been reading some on close focusing techniques and several writers claim to use a variety of medium telephoto lenses with attachments (such as extension tubes) rather than purpose made macro lenses for largeish subjects (say, flowers). They claim to get results indistinguishable (for most purposes) from purpose built macro lenses of long focal length at a very cheap price. One such writer claimed to use a 300 mm lens ( The Olympus OM-Zuiko T4 300mm f/4.5) with an extension tube to obtain the reach necessary to shoot close-ups of marsh plants etc not easily accessible and too far for, say, a 50mm or a 100mm or so macro lens. I simply wondered how much reach such a combination would provide. Actually, now we've looked into it, it does look "dooable". Thanks again.

Wirelessly posted (BlackBerry9000/4.6.0.167 Profile/MIDP-2.0 Configuration/CLDC-1.1 VendorID/102 UP.Link/6.3.0.0.0) Tubes do make a huge difference in reduction of depth of focus and minimum focusing difference. I have the number one tube (45mm) for my RB67. On a regular old Sekor 90mm lens. Really does make a difference.