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I've been researching extension tubes for macro photography and I can't quite get a handle on how the length of the extension tube, combined with the actual lens (say the standard 18mm-55mm) will work.

I may not be able to explain this perfectly (since I'm confused about it) but hang with me...

So basically if you have the normal lens set to 18mm, and add an extension tube that's 21mm. What effect does this have?

I realize there's a point where the camera will not be able to focus if you match an extension tube of x-length with the regular lens of x-length. Can someone summarize this?

I apologize if this seems like a simple question and I have read up on it quite a bit already but I'm still confused about this part.

Any help appreciated.

marked as duplicate by scottbb, xiota, Hueco, Michael C, inkista Aug 27 at 19:06

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.


When we try to work out optical formula stuff, we get approximate answers. This is because optical formulas need precise data as to lens to object and lens to image distances. Our problem is, we don’t know the exact location of two principal points. The lens to object distance is measured from a point called the front nodal and the lens to image is measured from the rear nodal. We can assume these points are centered in the lens barrel but this is not the case. A modern lens is made using several glass lenses some cemented together, some spaced apart. Some have positive power, some negative power. The bottom line, we don’t know where these nodals fall, we would need an optical bench to find them. Also, the forward nodal can fall in the air forward of the lens. In other words its more complicated than you think. In the math of optics: u = object to lens distance v = lens to image distance m = magnification = u ÷ v

Generally, when we take photographs, the camera images the outside world and the image projected by the lens on film or digital sensor is a reduction. In other words, the camera normally yields tiny images of objects.

As we move in closer and closer to the subject, the image of the object becomes larger. If your goal is to make life-size images of objects (magnification 1 sometimes called 1:1) you must position the object twice the focal length forward of the lens. In other words, with a 50mm lens mounted, magnification 1 is achieved when the lens to object distance is 100mm.

Should you achieve focus at magnification 1, the image projected on the surface of the digital sensor (or film) will form 100mm behind the lens.

The math is 100 ÷ 100 = 1 (magnification is at unity).

Table of magnification or reduction for 50mm lens (zoom set to 50mm). u = 100mm v = 100mm m = 1

u = 75mm v = 150mm m = 2

u = 150mm v = 75mm m = 0.5

u = 66.6mm v = 200mm m = 3

u = 200mm v = 66.6mm m = 0.33

u = 62.5mm v = 250mm m = 4

u = 250mm v = 62.5mm m = 0.25

Suppose you mount set the zoom of your lens to 18mm and mount this lens 21mm forward. You focus and achieve a lens to image distance of 40mm. suppose you lens to subject distance is 80mm.

Object to lens u = 80mm Image to lens v = 40mm M = 80 ÷ 40 = 2 written as 2X or magnification 2 (mage is twice life-size).

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