A 16mm objective has a minimum focus distance of 1m. When an extension tube is used to decrease the minimum distance to 0.1m, how long is this extension tube?

If b_1 is the distance between the lens and focal plane without an extension tube and b_2 with one, then b_2 - b_1 is the length of the extension tube. However, I can't calculate it without knowing the focal length. Is it possible to derive the focal length just from the lens' size and minimum focal distance, or is it possible to calculate the length in some other way with the given information?

  • \$\begingroup\$ Is the focal lengths of the lens 16mm, or is the front diameter of the lens 16mm? It's unclear in your question exactly what is 16mm. What is meant by "objective" can be different, depending on what your native language is. In some languages, it means the same thing as "lens" in English, in other languages it means the front optical element in a compound lens system. \$\endgroup\$
    – Michael C
    Commented Apr 19, 2022 at 7:37

2 Answers 2


It doesn't quite work the way you might expect. The one shorthand rule of thumb is that an extension tube length equal to the lens focal length will more or less focus at 1:1 magnification (1:1 means subject object is same size on camera sensor as in real life). However, lens internal focusing can modify that slightly.

And at such close distances, the useful information is normally expressed in terms of magnification instead of focus distance of inches or cm. Magnification is the useful thing to know, since the image has to fit on your camera sensor. Extension tubes generally only pretty much focus at one distance, and then you will have to adjust where to place the camera so the subject is in focus.

There are formulas for computing with extension tubes, some info here:


But it is harder than that, because the distance "in front of lens" in the Thin Lens formula (a simple single element lens) is in a more complex camera lens to a node usually inside the lens body somewhere (except in telephoto lenses, out in front of lens somewhere). Either way, distance is not measurable (likely insignificant for distant subjects, but a major precision loss up close).

So good advice is that what you want to know is magnification.

The focal length marked on the lens is when focused at infinity (and infinity is at zero magnification). There are exceptions, but focal length will often be somewhat longer when focused closer.

So first know that very many lenses normally have a specification of magnification at closest focus distance. B&H Photo Video product pages are good about showing that for current lenses. Using that, and camera sensor size and lens focal length and extension tube length, magnification can be computed (but not focus distance). You will have to determine where it focuses with an extension tube, and the camera can only be located there. Magnification tells you how the subject fits in your camera frame then.

FWIW, my site has such a calculator online at https://www.scantips.com/copycalc.html

That page has other stuff concerned with copying slides, but it does not mention focus distance (subject to lens). However, there are notes down lower on its page, and the calculator might be useful to you.

  • \$\begingroup\$ Good advice -- a 16mm lens operating a unity (magnification 1) will be located 16mm forward of its infinity position. The object distance will be 16mm forward of the front nodal, the object to image distance will be 4x the focal length. \$\endgroup\$ Commented Apr 19, 2022 at 0:15
  • \$\begingroup\$ Ok I see, thanks. \$\endgroup\$ Commented Apr 19, 2022 at 0:23

There are several ways to do this calculation, I like this one:

A 16mm has a power of 1/16 x 1000 = 62.5d (diopter)

Focused to 1 meter the lens racks forward? Convert 1000mm to diopter 1/1000 x 1000 =1d Subtract 62.5 – 1 = 61.5d convert to millimeters thus 1/61.5 x 1000 = 16.2602mm. This is the natural distance lens to senor when focused on an object 1 meter distance. To focus on an object .1 meters distance – .1 meter = 100mm. 1/100 x 1000 = 10 written as 10d (10 diopter)

What is the lens to sensor distance when imaging an object 100mm distant? 62.5 – 10 = 52.5d Covert this power to millimeters thus 1/52.5 x 1000 = 19.0476mm This is distance lens to sensor. Mount a 19.0476 – 16 = 3.0476mm extension A 4mm extension is the one I would select.

  • \$\begingroup\$ Ok, so from your calculations it seems that if an objective is 16mm then the focal length is also 16mm. Is the focal length 16mm because the size of the objective is 16mm or is it just another way of saying that an objective has a focal length of 16mm? \$\endgroup\$ Commented Apr 18, 2022 at 22:17
  • \$\begingroup\$ The lens to sensor distance is measured from a point called the rear nodal. This distance is the focal length when the object being imaged is at infinity. At all closer distances this distance is elongated. If the object is 100mm from the lens, then the lens to sensor distance for this 16mm lens is 19.0476mm from the rear nodal. \$\endgroup\$ Commented Apr 18, 2022 at 23:03
  • \$\begingroup\$ I understand most of what you have written except how you deduced that the power is 62,5d from that the objective is 16mm. I understand the formula you used but I don't understand what a 16mm objective is. Does 16mm refer to the size of the objective, the focal length of the objective, or does it refer to its size but it is so easy to get the focal length from it, it is practically viewed the same? \$\endgroup\$ Commented Apr 19, 2022 at 0:20
  • \$\begingroup\$ A 16mm lens focuses objects that are at super far distances (infinity) when the lens measuring point is 16mm forward from the sensor. This is the definition of focal length. If the object is closer then infinity, the lens position will be greater than 16mm. I have calculated that if the lens is imaging an object 100mm forward of the lens, that object will come to focus 19.9476mm behind the lens. I used a method that convers to diopter power and then back to mm. Other methods can be used. \$\endgroup\$ Commented Apr 19, 2022 at 0:34
  • \$\begingroup\$ The power of a lens in diopters is the reciprocal of the focal length in mm times 1000. The diopter power is an alternative way to express focal length. It is a convenient, eyeglasses are prescribed in diopter power. It allows many problems to be solved by addition or subtraction. \$\endgroup\$ Commented Apr 19, 2022 at 0:43

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