Redneck Macro Photography Math

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I wanna see if I got this right. I am coupling a 150mm w/ lens shade and an 80mm with zippy ties (gotta love redneck photography). Wanna do some macro photography. Now, with the lenses coupled together the overall length from the front element (the back glass of the 80) through the 150 to the film plane is about 8 3/8" (213mm). The maximum aperture on the 150mm is f3.5 (43mm). The maximum aperture on the 80mm is f2.8 (28.5mm).

Assuming that the focal length is 213mm and that the maximum, minimum aperture is 28.5mm, this would make a maximum aperture of f7.5 (213/28.5= 7.47). Since this is less than one quarter stop below f8 (1/4 below would be 7.346) I should expose at f8.

Also, this focal length is with both lenses focused to their closest focusing distance and both apertures opened wide up.

Did I get this right?
 

Ole

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Sounds completely wrong.

The focal length of a combination of two lenses will always be shorter than the shortest of the two!

The real maths is 1/F= 1/f1 + 1/f2 + d, where F is the total focal length, f1 and f2 are the focal lengths of the two lenses, and d is a correction factor due to spacing and such. So in your case, the focal length is something like a bit over 52mm.

There's an easier way to think of this:

With both lenses focused at infinity, and the subject placed where the film plane of the outer lens is supposed to be, you will have sharp focus. Moving the focus rings of either lens introduces so much complications that I would advice against it (at least until I've had three more cups of coffee).

I'll get back to the aperture when I've had another coffee...
 
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Actually I found another equation. F = (F1 x F2) / (F1 + F2) where F = Total Focal Length, F1 = first lens' focal length, F2 = second lens' focal length. But it comes out to 52+ mm. So at least I get warm and fuzzies for figuring part of it out. After an exhaustive 3+ hour search online, I could find nothing that would hint at what I could use for my Maximum relative aperture. AAAAAAAAAUUUUUUUUUUUGGGGGGGGGGHHHHHHHHHH!!!!!!!!!!!!!!!!!!!!!!!!!!!

Thanks, Ole.

Chris
 

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Is knowing the focal length critical? Are you trying to figure out how to set the exposure?

BTW, with the 80mm out front and reversed, changing the focus setting on that lens does nothing except drive the mounting face further out. It doesn't change the distance between lenses. You should also leave the aperture for the 80mm wide open and set exposure with the 150.

Most people stack lenses with TTL metering, in which case you just use whatever your in camera meter reads as correct. I've not seen any method for calculating exposure adjustments with stacked lenses. Essentially your 80mm lens is acting as a +12.5 diopter close up lens (1000/80=12.5), but it's also introducing an smaller aperture at an unknown distance in front of the 150.

At first blush, I'd think that the 80mm aperture of 28.5mm would be your working aperture until the aperture on the 150mm lens became smaller than 28.5mm, giving you a maximum aperture of about f:5.3. So once you get the 150mm stopped down to f:5.6 or smaller (assuming the 80mm lens is left wide open), the 150mm determines your working aperture. I'm basing this thinking on the fact that this is similar to an afocal interface between the lenses, with a parallel light path (neither converging nor diverging ray paths) in the space between lenses.

The formula I've seen for compound lenses is 1/f = 1/f1 + 1/f2 - d/f1 x f2, where f1 and f2 are the lens focal lengths, d is the distance between lenses, and f is the focal length of the compound system. So I'd think you'd need to know the distances between the lens nodes (not sure if it's rear nodes or front nodes in this case) to figure that out. As you likely know, with the 150 set at infinity your magnification is 150/80, or 1.875x.

This is a semi-educated guess on my part, I'm not an optics expert. Likely someone will come along with a better answer.

Lee
 
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Wonderful idea from another site I frequent from time to time but I wanna run it by you all first. Assemble the lenses and hold a piece of white paper to the end that will mount on the camera. Shine a bright light into it and measure the circle of light that results. Divide that into my 52mm effective focal length and arrive at a relative aperture. Now the only kicker I see, will the light circle expand in diameter the further it gets from the aperture? Is this an act in futility or what? This is killin' me and I'm gonna figure it out now. This has become a witch hunt, or, at least snipes, anyway. (Pescky, li'l critters, ain't they?)
 
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Christopher Walrath
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Well, here's my take on the exposure. I tried the paper behind the aperture on each lens seperately and saw that there was not a mathematical constant between the aperture diameter and the light circle's diameter. So I scratched that before continuing. I took the prism from the top of my Mamiya and metered a flashlight through the 150mm lens up through the view screen with a hand held meter and then through both lenses and the reading was the same at f3.5. Then I tried to read the light through the 80 alone and then both together with the 80 on the camera body and there was a one stop drop. So, I have concurred that by metering the subject, I simply adjust the shutter speed to keep my f3.5 aperture or even f4 or f5.6 (the max aperture diameter on the 80 is about f5.6 on the 150) and adjust the 150 accordingly. I mount it directly on the camera body to keep it simple the lens I should adjust. Both focused to infinity and stop wide open. By the by, focusing closer merely moves my hyperfocal distance closer to the front (rear) of the 80mm. And I can see the focus through the viewfinder.

Thank you for all of your help, everybody.
 
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with the lenses coupled together the overall length from the front element (the back glass of the 80) through the 150 to the film plane is about 8 3/8" (213mm).


Also keep in mind that the focal length of the lens has little, if any, correlation to the physical length of the lens. A simple lens can have a focal length of 1000mm, but be only an inch thick.

- Justin
 
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Christopher Walrath
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Thanks, Cafe. Got that already in subsequent posts on this thread. Thank you for the help.
 
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