Thread: Extension Tube Math? View Single Post
27th of October 2003 (Mon)   #5
PrimoFelis
Member

Join Date: Oct 2003
Posts: 41
Re: Extension Tube Math?

Quote:
 Littlebike wrote: Is there any formula for extension tubes? i.e. a 12mm extension tube connected to a lens with a minimum focusing distance of 8' will reduce the focusing distance to x.
Hi Littlebike,

Okay, let me take a crack at this too, although it is going to get a bit long.

But first, let me say that I'm not sure if doing the math is always the best way to obtain an accurate or useful answer to your question, for a couple of reasons I am going to mention below. Might I suggest holding a lens of your interest a distance away from the camera's lens mount while looking through the view finder? (Then you can make measurments of your interest?)

But anyway, I aim to please and so I will give the math a try.

---

First, a short quote from John Shaw's excellent book "Closeups in Nature" (ISBN 0-8174-4052-6), that gives a basic formula on the use of extension tubes in macro photography:

Quote:
 John Shaw writes:: The total amount of extension you need in order to get to any given magnification depends on the focal length of the lens you are using. Notice that I say "total" extension; it does not matter how you gain the extension -- some could be added by using extension tubes, some could be in the focusing mount of the lens -- since the entire amount being used is what is important. A basic formula gives approximate magnification rates: Total extension used / Focal length used = Magnification If you want to photograph at 1/2x with a 50mm lens, you can see that you need 25mm of total extension. But on a 100mm lens, the same 25mm total extension yields only 1/4x, and on a 200mm lens it yields 1/8x. Any given amount of extension yields less magnification when used with a longer focal-length lens.
Note that, by "1/2x," etc., he is talking about the magnification of image at the film/sensor plane. So, as an example, at 1/2x magnification and using an EOS 10D, you would be filling the viewfinder with an object of a size 45.4mm x 30.2mm (double the size of the CMOS sensor -- 22.7mm x 15.1mm -- used in 10D). Hope this part is clear.

How to use the formula

In order to use the above basic formula to derive the answer you are seeking, you need to know the amount of (effective) extension built into the lens of your interest.

(I say "effective" because these days many lenses don't use a simple lens extension to achieve a close focus, instead rearranging position of lenses. This often produces a side effect of changing the actual focal length of the lens a bit -- even if it is a prime (i.e., non-zoom) lens -- introducing discrepancies between the theory and reality.)

Usually this "built-in extension" value is not part of a published lens spec.

However, you can derive this value using the lens' maximum image magnification value (or alternatively a minimum coverage area) and the lens' focal length, usually found in the lens spec. (How accurate these published values are remains a question. But let's not get too critical...)

Now, I will refer to such a chart from Canon...

Example: Step 1. Magnification

Let's pick EF50mm f/1.8 II. The table says the closest focus distance for this lens is 0.45 meters (= ~17.7 inches), and the maximam magnification is 0.15x. Using the above basic formula:

x / 50 = 0.15

Solving for x, the derived maximum built-in lens extension is: x = 50 * 0.15 = 7.5mm

If you add Canon's 12mm extension tube to this lens, you will achieve a maximum total extension of 7.5 + 12 = 19.5mm

Therefore, using a 12mm extension tube with EF50 1.8 II lens, you can obtain a maximum magnification of:

19.5 / 50 = 0.39

a magnification of 0.39x [Filling the frame with an object of actual size 58.2mm x 38.7mm]

Example: Step 2. Minimum focusing distance (a rough approximation)

[NOTE: The following was corrected and updated Nov. 7, '03]

To continue with the example and figure out new minimum focusing distance (MFD) from the new magnification, we need the help of a few additional formulas (found in any general physics book):

1) M = b/a
2) 1/a + 1/b = 1/f
3) MFD = a + b

where
M: Magnification
a: lens-to-subject distance
b: lens-to-film/sensor distance
f: focal length

[NOTE: for a and b above, the position of lens being referred to is that of a theoretical equivalent lens of single element, NOT of the actual camera lens which usually have multiple elements.]

So, given the new magnifcation of 0.39x and focal length of 50mm:

M = 0.39 = b/a
1/a + 1/b = 1/50

Solving for a and b, you get:

a=178.2mm
b=69.5mm
MFD = a + b = 247.7mm

That is to say, using EF50mm f/1.8 II lens with a 12mm extension tube, the minimum focus distance should be reduced from 0.45 meters to ~0.25 meters (17.7 inches to ~9.75 inches)

(Would anyone care to run a test to see if this calculated value matches the reality? I don't have an EF50mm 1.8II lens to test with.)