Question About Extension Tubes

Well, I have done some research on this topic, but have found no real answers---so here is the question: Are any of you familiar with a chart, graph or calculation that you can estimate minimum focus distance based on the size of an extension tube and lens? For example, I want to shoot humming birds with my 500mm F4L which has an MFD of 15 ft. What would the MFD be if I used it with a 24mm extension tube? or how much of a reduction in MFD could I expect with a 24 mm Extension tube? I know I can simply go out and do the trial and error thing----But I am looking for some formula, graph or chart that would get me in the ballpark based on the lens/extension tube combination that I am using. If you have any Ideas or refrences please share.


Thanks


Bob

Additional magnification=(extension tube length)/(lens focal length)


Based on this rough rule, the extension tube mounted between the camera body and your 500mm will result in a .05x magnification gain (25mm divided by 500mm).


Hope this helps,


John.

Quote:

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Originally Posted by Fast Glass

Additional magnification=(extension tube length)/(lens focal length)


Based on this rough rule, the extension tube mounted between the camera body and your 500mm will result in a .05x magnification gain (25mm divided by 500mm).


Hope this helps,


John.
<div style="CLEAR: both"]</div>

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Thanks John, I think your getting me pretty close----But how do you apply this formula to MFD? Would this .05 increase in magnification be the same as a .05 decrease in MFD?

Quote:

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Originally Posted by Bob Williams

ut how do you apply this formula to MFD?

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The formulas you need are


1) 1/f = 1/a + 1/b


where a = subject distance, b = distance between the lens and the ccd, and f is the focal length of the lens. and


2) magnification = b / a


From these two formulas it is easy to derive the formula John gave you, and also easy to figure out your new mfd .


Your mfd comes from the fact that travel in your lens only allows it to get so far from the ccd (that is, b can only get so big), thus a can only get so small. Adding a 24mm extension tube essentially adds 24 to your maximum value of b, thus allowing a smaller minimum value for a, thus increasing magnification.


Example: a 100mm macro lens has a maximum magnification of 1. This means that when focusing as close as possible, b / a = 1, so b = a. Thus formula one gives


1 / 100 = 1 / a + 1 / a, so 1/100 = 2/a or a = 200. The mimimum focus distance is 200mm, and when at this mfd, the lens is 200mm from the ccd.


Now imagine adding a 100mm extension tube. Then the furthest we can get the lens from the ccd is b = 300, (we used to be able to get 200mm from the ccd, but we've added 100mm) and we can compute the new a using formula 1), as follows:


1 / 100 = 1 / a + 1 / 300


This gives a = 150mm for our new mfd. (And, of course, our new magnification is 300 / 150 = 2, just as predicted by John's formula).


Having said all of this, it the formula does not really work for some lenses in real life, for two reasons. The first is that focal lengths of some lenses change as they get close to the subject (so the approximation gets worse as you get closer). I guess for your 500mm, the formula will be pretty close, though.


The other problem is that a is the distance of *something* to the subject, but it might not be the front element of the lens. Thus what you really have to do is compute mfd with and without the extension tube, and subtract the difference from your old mfd to get your new mfd.


I hope this helps...

MFD is measured from the image plane too the point of focus, not from the lens element.


John.

Do you mean the image plane to the ccd? (ie, a + b above) Are you sure?


In that case, mfd = a + b above, so according to the lensmakers equation, I get mfd = f[(m+1)^2]/m, where f is the focal length and m is the max magnification. This formula seems to overestimate the mdf reported in lens specs.


If mfd is the distance between the "ideal optic" and the focal plane (a above) I think the formula would be mfd = f(m+1)/m, which agrees much more closely with specs in Bryan's reviews for lenses like the 500 f/4.


Possibly I've made a mistake in my algebra, but I don't think so :)


Anyway, it doesn't really matter what mfd is. Bob- if you want the distance from the front element to the subject, use the second formula. If you want distance from the subject to the ccd, use the first. :)


Just know that real lenses differ from ideal, so neither formula will be exact.

Yes I am sure, just google "how is minimum focus distance is measured" and you can confirm it too. That screws your math a bit[:P]


Cheers,


John.

It doesn

@John and John, Thanks for the detailed explanation and formula's. Sorry I didn't get back to you earlier but this weekend has been a little crazy. It's only been 30 years since I took Algebra 101, but I think I can fight my way through this [:^)]. I do understand that focus distance is measured from the CCD and not the front element on the lens----Canon was even kind enough to put a little mark on my camera so I knew exactly where the ccd is.


I'll wrestle with this tomorrow and let you know what I discover, i.e. Calculation vs reality.


Thanks again,


Bob

Ok Guys----I suppose I should just get out the tape measure and find out for myself what the MFD is with the 500 and 24 mm ext tube---but I am stubborn and hard headed and really want to understand this; I thought I almost had it until I looked at Bryan's Lens Specs which are:
<table align="center" cellpadding="3" cellspacing="0" class="SpecsTable" style="width: 625px;"]
<tbody>
<tr>
<td class="SpecsTableCell"]Specified Min Focus Distance</td>
<td class="SpecsTableCell"]177.2&rdquo;</td>
<td class="SpecsTableCell"]</td>
</tr>
<tr>
<td class="SpecsTableCell"]Tested Min Focus Distance (Wide / Long)</td>
<td class="SpecsTableCell"]163.86&rdquo;</td>
<td class="SpecsTableCell"]</td>
</tr>
<tr>
<td class="SpecsTableCell"]Maximum Magnification (MM)</td>
<td class="SpecsTableCell"]0.12x</td>
<td class="SpecsTableCell"]</td>
</tr>
<tr>
<td class="SpecsTableCell"]MM with 12mm Extension Tube</td>
<td class="SpecsTableCell"]0.15-0.03x</td>
<td class="SpecsTableCell"]</td>
</tr>
<tr>
<td class="SpecsTableCell"]MM with 25mm Extension Tube</td>
<td class="SpecsTableCell"]


0.18-0.05x
</td>
</tr>
</tbody>
</table>





So, as I ponder the last line----where did the .18 come from? If the native magnification is .12 x and the added magnification with the 25 mm ext tube is .05 x---shouldn't the end result be .17x?---Is this just a rounding error or adjustment, or is there something I am missing?


Thanks for your patience, and instructional wizdom,


Bob

Quote:

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Originally Posted by Bob Williams

So, as I ponder the last line----where did the .18 come from? If the native magnification is .12 x and the added magnification with the 25 mm ext tube is .05 x---shouldn't the end result be .17x?---Is this just a rounding error or adjustment, or is there something I am missing?

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The 0.05x addition assumes that the 25mm extension tube is actually adding 25mm of extension. In fact, the EF25 is nominally 25mm in length, but actually provides 27.25mm of extension, meaning 0.1745x final calculated magnifiction (still not quite to 0.18x even with rounding). But, that calculation is based on the specified magnification, which is associated with a specified MFD of 177.2". Most lenses can manually focus closer than the specified MFD, so perhaps Bryan is actually empirically determining the magnifications (his tested MFD is closer than the specified MFD, meaning a slightly higher magnification).

Quote:

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Originally Posted by neuroanatomist

The 0.05x addition assumes that the 25mm extension tube is actually adding 25mm of extension. In fact, the EF25 is nominally 25mm in length, but actually provides 27.25mm of extension, meaning 0.1745x final calculated magnifiction (still not quite to 0.18x even with rounding). But, that calculation is based on the specified magnification, which is associated with a specified MFD of 177.2". Most lenses can manually focus closer than the specified MFD, so perhaps Bryan is actually empirically determining the magnifications (his tested MFD is closer than the specified MFD, meaning a slightly higher magnification).

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Thanks John, That makes sense to me. I just wanted to make sure I wasn't missing something here.


Thanks


Bob