What is the significance of the typical extension tube lengths?

Question

I've been recently looking into getting an extension tube set for some macro photograpy. I've noticed that a lot of these 3 tube sets (for Canon EOS) have the same, very specific lengths: 13mm, 21mm, and 31mm.

I've done some searching on Google as to the significance of these lengths, but have found nothing. Does anyone know why these lengths in particular seem to be so common?

Asked by ltn100

Answer

It might help if you thought of everything in Imperial units rather than metric. A normal lens for 35mm cameras is a 2-inch lens (50mm), and the tubes are (approximately) 1/2-inch (13mm), 3/4-inch (21mm) and 1 1/4-inch (31mm). That makes both magnification calculation and bellows draw (exposure compensation required for the lens extension) relatively easy to calculate, whether the tubes are used alone or in combination, when the lens is focused to its infinity mark. For instance, combining the 21mm and 31mm tubes would give you a 1:1 magnification ratio with a 2-stop exposure compensation.

The calculations are similarly simple for 24mm, 100-105mm and 200mm lenses, at least in a ballpark sense.

Remember that zoom lenses were once far from ubiquitous, and it was anomalous for a photographer to own a 35mm camera without also owning a 50mm lens. It might also help to remember that TTL metering was, once upon a time, and only if your camera actually had it, something you wouldn't really want to rely on most of the time (almost all metering was center-weighted, and rather depended on the condition of your camera's batteries or the age of the selenium cell).

These days we don't need to rely so much on external metering for most photography, and it's as likely as not that you'll be using a zoom lens as a prime, so maybe the tube lengths don't make as much immediate sense as they used to. But that's why millimeter lengths that are around rounded simple fractions (halves and quarters) of an inch are normal.

Answered by Stan Rogers

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How Much Can Lens Magnification Be Improved Without Significantly Lowering Image Quality?

Question

Currently I own only a 1X magnification macro lens (35mm F/2.8) but I am playing with a rented Canon MP-E 65mm lens which can go to 5X. The photography at that magnification is a world apart!

The question is then how much can I increase the magnification of the 35mm Macro through extension tubes or other macro adapters without losing image quality? What would it take to get beyond 2-3X if it possible?

Asked by Itai

Answer

You should be fine stacking on a whole set of extension tubes. You will increase diffraction, however you'll also be magnifying your subject by a greater factor, possible several times more...so fine details will still stand out more than they would at a lower magnification level because the effects of diffraction remain smaller than the magnified details (up to a certain point...diffraction will grow faster than detail magnification, however long before it gets to the point where the airy disc is larger than your original details, other things will limit your ability to keep extending.) The facets of an insects eye become gigantic, and the fine details of EACH FACET could be visible with enough magnification, to the point where they span large clusters of pixels...where the airy disc of diffraction may only span a couple pixels. Extension tubes do not add any optical elements to the light path, so technically speaking, you should be able to extend and gain additional magnification without significantly affecting IQ.

For experiments sake, lets say that hypothetical insect actually is our subject. Lets say we are shooting with an 18mp APS-C camera, at 1:1 magnification. Lets say the facets of our subjects eyes span 8x8 pixel areas (very small!)

If you are shooting 35mm 1:1 @ f/5.6, and slap on a 25mm extension tube. Magnification gain is extension/focalLength, so your adding 25mm/35mm, or 0.714x more magnification. Magnification affects the effective f-stop that you are shooting at. At 1.0x magnification, you are already experiencing some of the effects, and your effective aperture is f/11. With the additional magnification, your effective f-stop is f/5.6 * (1 + 1.714), or f/15. Your subjects eye facets are now about 26x26 pixels in size, and diffraction is affecting about 4 pixel areas.

Similarly, 50mm of extension would be 1.43x additional magnification (50/35), so effective f-stop is f/5.6 * (1 + 2.43), or f/19. Diffraction at that level is visible and will have a moderate impact on IQ, but not anywhere close to as bad as optical aberrations are going to be at f/2.8. It still isn't affecting the ultimate quality of your image, however...as your subject has also grown in detail. Your subject's eye facets are now about 43x43 pixels, and diffraction is affecting about 6 pixel areas.

Lets take the experiment farther...you have to stop down to f/22 to get enough DOF, and your extending by a whole 5x magnification. That gives you an effective aperture of f/22 * (1 + 5), or f/132. At this point, the effects of diffraction would span about a 150 pixel area for an 18mp APS-C sensor (which is VERY high resolution, about 116 lp/mm...line pairs/millimeter.) You might be inclined to think the effects of diffraction are now obliterating all the detail you worked so hard to get. That wouldn't necessarily be the case, though. Your at 5x magnification, almost three orders of magnitude greater than you were at 2.43x magnification before, where those fine details spanned 26x26 pixel areas. The same details should be spanning more than 250x250 pixel areas now. Diffraction has grown, and will likely blur out fine details, but is affecting about 50 pixel areas. You'll still be extracting more detail than you lose to diffraction.

To answer your fundamental question: How much can you magnify before you lose detail? The size of the airy disc will grow slightly faster than the size of the original detail at 1.0x magnification. This is due to the non-uniform nature of diffraction, and the way it will interfere with/amplify itself as its effect grow. Diffraction is also dependent on the wavelength of light...so while I have used the wavelength of yellow-green light (564nm) for my calculations so far, visible light spans the range from about 340nm violet to 790nm deep red. Deep red light will diffract more than other wavelengths, and will produce greater diffraction. You may eventually reach a limit, wherein diffraction affects IQ enough that you don't gain any further benefits. That limit is very far beyond the point where other mechanical limitations prevent you from extending any more.

In normal photography, the more you stop the aperture down, the more the effects of diffraction affect the image. Since the detail in the image is not getting larger as you stop down, the more detail you lose as airy discs grow. When it comes to macro photography, your magnifying the detail as you increase extension...and while your also increasing diffraction, the original details are always larger than the airy disc. You WILL lose some detail as you extend (you'll be bringing finer and finer detail to light, and beyond around 3x magnification, diffraction will start to affect the visibility of finer details than what you started out with at 1.0x.) Eventually the effects of diffraction will prevent you from continuing to make useful gains with additional magnification. But you can push magnification very far. In the general case, you are far more likely to run into the problem where your focal plane ends up too close or actually inside the lens before you actually run into problems with diffraction affecting IQ in a truly detrimental way.

Answered by jrista

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Does maximum aperture change with focus distance with the Canon 60mm f/2.8 macro lens?

Question

This site says

The Canon EF-S 60mm f/2.8 Macro USM Lens loses 1/2 stop at 1:5, 1 stop at 1:3, 1.5 stops at 1:1.5 and 2 stops at 1:1 (lifesize).

Can anyone confirm that at 1:5 magnification, the Canon EF-S 60mm f/2.8 USM lens has a maximum aperture of f/3.4, at 1:3 magnification the maximum aperture is f/4.0, at 1:1.5 it is f/4.8 and anything below that is f/5.6?

Asked by Nick

Answer

My Nikon 105mm drops from f/2.8 to f/4.5 at closest focus, so that sounds right.

A post at betterfamilyphotos has a post where they say (emphasis mine):

You would imagine that using a macro lens is the same as using a normal lens, and you would be right except that with a macro lens when you get close to 1x magnification, you start losing light. My 60mm for example starts losing light at close ranges until it reaches 2 stops of light loss at 1x magnification, this means that the effective aperture is f/5.6 instead of f/2.8 (regarding light quantity entering, not DoF). If you are using auto modes on the camera like aperture priority or using flash in TTL mode then the camera will auto compensate for the light loss, but if you're metering light manually you need to take it into account, Canon has included a table in the user manual with the light loss values at each magnification level.

So if you have access to the manual for this lens, or request one from Canon, it should verify the information. It's expected for a close focusing macro to lose 1-2 stops.

Edit: in response to the comment, the above post is incorrect in saying that the reduction in aperture doesn't affect DOF. It does. Two references for those interested in the physics of it:

Cambridge in Colour has a good explantion here (scroll down to LENS EXTENSION & EFFECTIVE F-STOP.

Also see explanation here: Do normal macro lenses suffer the same light reduction as tubes?

Answered by MikeW

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How can I calculate what the effect of an extension tube will be?

Question

There must be a mathematical description of the difference that an extension tube makes to a lens -- is it something that can be easily described?

(For example, with teleconverters you can say things like "a 2x teleconverter will turn a Y-mm lens into a 2Y-mm lens, and will lose you 2 stops." Is there something similar for extension tubes?)

If there's nothing much you can say about magnification, in general, what about the change in closest focal distance? Is that also lens-dependent?

What about if we factor out the lens: is there any general way to compare the effects of (say) a 12mm and a 24mm extension tube on the same lens?

Answer

I do believe there are some formulas you can use. To Matt Grum's point, I have not tested these with zoom lenses, and to my current knowledge, they apply only to prime (fixed focal length) lenses. You did not specifically specify zoom lenses, so...

The simplest way to calculate the magnification of a lens is via the following formula:

  Magnification = TotalExtension / FocalLength
  M = TE / F

To calculate the magnification with an extension tube, you need to know the total extension...that is, the extension provided by the lens itself, as well as that provided by the extension tube. Most lens statistics these days include the intrinsic magnification. If we take Canon's 50mm f/1.8 lens, the intrinsic magnification is 0.15x. We can solve for the lenses built in extension like so:

   0.15 = TE / 50
   TE = 50 * 0.15
   TE = 7.5mm

The magnification with additional extension can now be computed as follows:

  Magnification = (IntrinsicExtension + TubeExtension) / FocalLength
  M = IE + TE / F

If we assume 25mm of additional extension via an extension tube:

  M = 7.5mm + 25mm / 50mm
  M = 32.5mm / 50mm
  M = 0.65x

A fairly simple formula that allows us to calculate magnification fairly easily, assuming you know the intrinsic magnification of the lens (or its intrinsic extension.) If we assume the wonderful 50mm lens is the lens you are extending, to create a 1:1 macro magnification, you would need 50mm worth of extension. The problem here is that if you add too much extension, the plane of the world that is in focus (the virtual image) might just end up inside the lens itself. Additionally, this assumes a "simple" lens, one with very well-defined and well-known characteristics (i.e. a simple single-element lens.)

In a real-world scenario, having a clear understanding of any particular lenses characteristics is unlikely. With lenses that focus internally, or zoom lenses, the simple formula above is insufficient to allow you to calculate exactly what your minimum focusing distance and magnification can be for any given lens, focal length, and extension. There are too many variables, most of which are likely to be unknown, to calculate a meaningful value.

Here are some resources that I have found that provide some useful information that might help in your endeavor:

• Lens Tutorial
  • Some great mathematics for lenses, including MFD and Mag
• Lens Calculator
  • A javascript based lens calculator that can calculate a wide variety of values for you
  • Not really sure how accurate it is in the case of internally focusing
• Wikipedia: Magnification
• Forum Post about Extension, Magnification, MFD
  • Limited applicability, assumes too much
• Forum post about Extension and MDF
  • Limited applicability, assumes too much

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Does my crop sensor camera actually turn my lenses into a longer focal length?

Question

So, I mount a 200mm lens on my Canon 450D. It effectively becomes a 320mm lens. Is this the equivalent of 320mm on a full frame camera? That is, from what I've figured out I get an equivalent field of view but nothing I've read is indicating that I get the magnification to go along with it.

So as my question says in the title, does my crop sensor camera really turn my lens into a longer one (in terms of magnification), or does it just look like it based on the reduced field of view I get?

Answer

The lens does not actually turn into a different focal length, since that's a real, physical property of the optics that can't be changed without more optics. So from that point of view, the answer is a definitive no.

However, when you get to the question of is it effectively the same in terms of magnification, the answer is "pretty much, given some assumptions."

A key assumption is that you're printing at the same size. That means: you're increasing the magnification of the image from the smaller sensor. If you print at sizes different by the same ratio of the crop factor, you get exactly the same result as if you just took a full-frame photo, printed large, and then cropped out the middle.

So, if you print your full-frame picture at 12×9", and print your crop-factor picture at 7.5×5.6" (for Canon; 8×6" for others, or 6×4.5", or whatever), and then chop down the full-frame print to match, they'll be roughly the same.

"Roughly" comes in because, of course, the actual sensors won't be equivalent in image quality. (The crop-factor print may have larger resolution, but from denser photosites, depending on the technology generation used in each camera.)

Blowing up that cropped image — either from the full-frame cropped print, or from the cropped sensor — has two effects which are very like changing the focal length. And these two things are the most visible effects of changing focal length — field of view, as you've noted; and depth of field, which changes exactly as if you'd adjusted the f-stop by the amount of the crop.

If you've ever used a point and shoot camera with "digital zoom", that's what's actually going on. It's cropping the photo and then expanding it. From a practical point of view, zoom is indistinguishable from cropping. But of course, that raises the spectre of decreased image quality — we all know that digital zoom can be awful. The answer is simply that sensor technology is really very good, and amazing, excellent results can be produced at even large print sizes even with a 1.5 or 1.6× crop — but if you do want to go larger with your prints, eventually you need a larger sensor. And, equivalently, if you want to zoom in more, you can do that with more cropping, but eventually, you need actual higher-focal-length glass.

Note that this doesn't address macro shooting. I don't really do any of that, so I'll let someone else handle that aspect of the question — which I think is well-addressed here: Does a camera's crop factor apply to the magnification of macro shots?

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