My macro set up!

A forum to ask questions, post setups, and generally discuss anything having to do with photomacrography and photomicroscopy.

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rjlittlefield
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Post by rjlittlefield »

Cyclops wrote:I'm intrigued tho how a 50 mm on a digi goes back to being a 50 but with a higher magnification!
Well, I was careful to say "more like". :wink:

If you can stand just a few numbers...

Imagine that the 50 mm is a "thin lens". Then when it's at 1:1 on the small sensor, the working distance is 100 mm and the field width is 22.7 mm.

Suppose we ask the question: what length lens would be required, to give 100 mm working distance and a 22.7 mm field on a full-frame 35 mm camera?

To get 22.7 mm field on a full-frame 35 mm camera requires 1.6:1 (36:22.7 = 1.6:1).

And to get 1.6:1 with 100 mm working distance requires a lens whose focal length is 61 mm.

So, the 50 mm at 1:1 on the small sensor would be equivalent to a 61 mm at 1.6:1 on a full frame, in terms of working distance and field size.

61 mm is a lot closer to 50 mm than it is to 80 mm, hence my comment "more like" a 50 than an 80.

The difference between equivalent focal lengths becomes less and less as the magnification rises. At 5:1 on the digi, the working distance would be 60 mm and the field width would be 4.54 mm. The equivalent magnification on full frame would be 7.9:1, and the equivalent focal length (7.9:1 at 60 mm working distance) would be only 53 mm, not much different from 50 at all!

--Rik

Cyclops
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Post by Cyclops »

Nope,afraid you lost me! The thing is I find numbers fascinating but when presented with a heap of them it becomes a blur. Its kind of like being dyslexic but with numbers! Part of the reason i cant work on checkouts at the shop where i work.
Canon 30D | Canon IXUS 265HS | Cosina 100mm f3.5 macro | EF 75-300 f4.5-5.6 USM III | EF 50 f1.8 II | Slik 88 tripod | Apex Practicioner monocular microscope

rjlittlefield
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Post by rjlittlefield »

Ah yes, the old "eyes glazing over" problem! I have that myself, truth be told, though usually I can get past it by chipping away at the problem.

Ignoring the numbers, then...

The key concept is that there are "equivalent focal lengths" that give the same working distance and field size on the small sensor and on the full-frame camera.

At very small magnifications (long working distances), those equivalent focal lengths differ by 1.6X, hence the popular press.

At very high magnifications (short working distances), the equivalent focal lengths hardly differ at all.

And at intermediate magnifications, they differ by some intermediate factor, depending on the magnification.

Does that work better?

--Rik

Cyclops
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Post by Cyclops »

Yea i see that,but what bothers me is why? Why the difference shrinks as the working distance disappears. Thats whats bothering me.
Canon 30D | Canon IXUS 265HS | Cosina 100mm f3.5 macro | EF 75-300 f4.5-5.6 USM III | EF 50 f1.8 II | Slik 88 tripod | Apex Practicioner monocular microscope

rjlittlefield
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Post by rjlittlefield »

why? Why the difference shrinks as the working distance disappears.
I don't know how to explain that.

I can start with the basic lens formula 1/f = 1/i + 1/o, and the basic magnification formula m = i/o, then slog through the algebra to get a formula for the ratio of equivalent focal lengths as a function of magnification.

That formula ends up being not very complicated. If R is the ratio of equivalent focal lengths, S is the ratio of sensor sizes (larger/smaller), and M is the magnification for the smaller sensor, then I get that

R = (S*(1+M)) / (1+(M*S))

This formula reproduces the numbers that I calculated above for 1:1 and 5:1 on the digi. And by taking limits, it also reproduces the summary about what happens at very small and very large magnifications. (For M=0, it reduces to R = S, and for M = infinity, it reduces to R = 1.) So I'm reasonably convinced that the formula is correct.

But that doesn't help much, because I find the formula itself to be completely opaque -- it means nothing at all to me beyond some rules about how to compute numbers!

So in the end, the best I can say is just "That's the way it works out."

Not very satisfying... :(

--Rik

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Post by rjlittlefield »

On the other hand, here's a geometric explanation that might work.

Once we've agreed to hold the working distance and field size constant, then the lens-to-camera distance scales in proportion to the sensor size. (m = i/o, where o = object-to-lens and i = lens-to-image = lens-to-camera. If o is fixed, then m and i change together, and of course m has to scale in proportion to the sensor size.)

At very low magnifications, the lens-to-camera distance is essentially equal to the lens focal length. So in this regime, the equivalent focal length tracks the sensor size.

At very high magnifications, the required focal length is just a little less than the working distance, and how much less depends only weakly on the lens-to-camera distance. So in this regime, sensor size has very little effect on equivalent focal length.

As usual, when conditions are intermediate, the results are intermediate also. Around 1:1, the equivalent focal length does depend noticeably on sensor size, but not as much as tracking it directly.

Does this version help?

--Rik

Cyclops
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Post by Cyclops »

Thats it,my brain is fried! e=MC2 times 2 of the hypotenuse is equal to the angle of the dangle. reboot,reboot...
Canon 30D | Canon IXUS 265HS | Cosina 100mm f3.5 macro | EF 75-300 f4.5-5.6 USM III | EF 50 f1.8 II | Slik 88 tripod | Apex Practicioner monocular microscope

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Post by rjlittlefield »

:lol: :-# :-# :-# :lol:

--Rik

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