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Photography Question 
Christopher A. Walrath
BetterPhoto Member Since: 4/25/2006
 

Redneck Macro Photography Question


O. K. Here goes. I have a Mamiya M645j MF and a few lenses. No Macro bellows. So I wanna shoot macro by inverting two lenses. I mount the 150mm on the camera body. It has a lens shade that the 80mm fits into nicely. I don't have a double threaded adapter so this contraption is held together by zippy-ties (that the redneck part of it, it's a beaut).

The 150 has a maximum aperture of f/3.5 (43mm). The eighty has a maximum aperture of f/2.8 (28.5mm).

Now I know that the focal length of two stacked/inverted lenses can be derived from the formula F=(F1*F2)/(F1+F2) where F=Effective Focal Length, F1+ 1st lens Focal Length and F2=2nd lens Focal Length. This would figure to 12,000 divided by 230 equaling 52.3 something, effectively a 52mm focal length.

Now, how do I figure the aperture? I have searching the internet for hours using various search parameters and different search engines. NOTHIN'! What formula do I use to determine the maximum relative aperture so that I can determine exposure correctly without the aid of a TTL meter, of which I have none for this camera.

And how many pancakes does it take to shingle the top of a doghouse?

Thank you all.


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11/18/2007 6:58:44 AM

 
Mark Feldstein
BetterPhoto Member Since: 3/17/2005
  This sounds suspiciously like a trick TRAIN PROBLEM in disguise !!!
Can I take the last part of this question first?

As to the redneck problem, I'd suggest a bandanna, a wide brimmed hat or just staying out of the sun. 0/;>)
M.


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11/18/2007 10:51:34 AM

 
Bob Cammarata
BetterPhoto Member Since: 7/17/2003
cammphoto.com
  This sounds like a job for "Mr. Bracket".

(...unless Jeff Foxworthy logs on with a better solution.)


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11/18/2007 2:05:14 PM

 
Alan N. Marcus   Hi Christopher,

Your math is good, I am proud of you!

Lens makers favor working in a unit called “diopter”.

Convert focal length from millimeters to the diopter unit. Once the power in diopters is known for each lens you simply add the two powers together. This math yields the power in diopters for the total system. To convert to diopters:
d = 1/focal length in mm * 1000
Thus:
d = 1/150 * 1000 = 6.667 d
d = 1/80 * 1000 = 12.5 d

System power is 6.667 + 12.5 = 19.16 d total power in diopter units. Convert back to millimeters 1/19.16 * 1000 = 52.17 mm OK to round to 53mm.

Find working diameter of system:
Dismount from camera. Assemble the pair. Hold a sheet of white paper quite close to the assembled lenses allowing light from a nearby incandescent bulb to shine through the lens set. You will see a bright circle of light shining on the paper. This bright circle of light is known as the exit pupil. Measure the diameter of this circle. This is the working diameter of the system.

Find the working f/number of the system:
Divide the focal length by the exit pupil diameter-- both dimensions to be in the millimeter unit. The answer is the f/number. You can repeat this over and over again with the two lenses at different aperture settings.

Example: Say the exit pupil of the combination is 18.5mm. 52/ 38 = f/2.8

The formula you used is good for thin lens systems. No easy way to accurately calculate the focal length of two complex lenses back-to-back.

However I often use this easy way which is quite accurate Set the dismounted system up and cause it to project an image of a clear plastic clear plastic millimeter ruler onto a nearby screen. Screen can be white paper or ground glass. It might be possible to do this mounted on the camera. Adjust lens-to-screen distance to achieve unity (1:1 life-size). I use another millimeter ruler to measure the projected image size. Now measure; center of the lens system to screen distance. Divide this measurement by 4. You now have an accurate system focal length.

Alan Marcus (marginal technical gobbledygook)
ammarcus@earthlink.net


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11/18/2007 6:48:41 PM

 
Christopher A. Walrath
BetterPhoto Member Since: 4/25/2006
  You rock, Alan. Jeff said to say hi, Bob. And, Mark, I just bought some cinder blocks for a tripod. (This week on the Dukes . . .) Nice to know you guys got my back. Thanks.


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11/18/2007 7:22:38 PM

 
Christopher A. Walrath
BetterPhoto Member Since: 4/25/2006
  Oh, and by the by, the answer is 13 because the ice cream doesn't stick to well.


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11/18/2007 7:24:19 PM

 
Alan N. Marcus   Hi all:

I made a boo boo:

Now measure; center of the lens system to screen distance. Divide this measurement by 4. You now have an accurate system focal length.

Should read:
Now measure ruler (subject) to screen. Divide this measurement by 4. You now have an accurate system focal length.

Alan Marcus (must I stand in the corner again with dunce cap on.

Alan Marcus


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11/18/2007 8:10:42 PM

 
Alan N. Marcus   Hi again Christopher:

I had two additional thoughts. Thoughts are about all a 70 year old can muster.
When you use a burning glass to start a fire, you hold a piece of paper at the focus of a lens looking at the sun. The sun's image on the paper is tiny. Now if you move the paper quite close to the lens you will see a large out-of-focus image of the sun, this is the exit pupil.

Now to the f/number: The exit pupil divided into the focal length is a neat way we calculate the f/numbers. This would be the number engraved on the lens barrel. However this value is accurate only when the camera is focused on infinity. When we close focus the engraved f/number is no longer valid because for close subject the lens is racked further and further away from the focal plane. In other words the working focal lenght much longer at at close focus. The error is 2 f/stops at unity (1:1). Traditionally most camera’s stop the forward rack when the error attain reaches 1/3 of an f/stop. Macro lenses to the rescue. A macro has its aperture gear driven so as you close focus the exit pupil (working diameter) is enlarged to compensate. In a micro this action is transparent to the user.

In your close-up setup you will need to open the aperture or increase exposure time to compensate for this phenomenon called “bellows factor”.

To calculate you need to know the magnification of the set-up. I substitute a millimeter ruler for the subject. I measure the length a 10mm span as seen on the ground glass. If actual size it will measure 10mm thus M=1. If it measures 5mm than M=.5. If it measures 20mm than M = 2 etc.

Bellows Factor formula is (M+1) squared. Thus if at unity M=1 thus BF = (1+1) * (1+1) = 4. If M = 2 than (2+1) * (2+1) = 9 etc.

BF is handled just like a filter factor(any one of the three ways below)
1. Multiply shutter speed by BF to obtain revised shutter speed.
If shutter speed is ½ second and BF = 4 than ½ * 4 = 2 second.
2. Divide ISO by BF to obtain revised ISO for this magnification (set-up).
3. Count on you fingers one finger = 1 f/stop count in powers of 2.
Thus if BF = 8 then count 2 – 4 – 8 = three fingers i.e. open up 3 f/stops.

More gobbledygook from Alan Marcus


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11/18/2007 9:26:34 PM

 
Mark Feldstein
BetterPhoto Member Since: 3/17/2005
  Chris, you put those cinder blocks on a web lawn chair and you'll be one STYLING dude. Don't forget the lovely and stylish Piggly Wiggly camera bags too ! 0/:>). Now in paper AND plastic.
M.


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11/19/2007 9:09:55 AM

 
Mark Feldstein
BetterPhoto Member Since: 3/17/2005
  Alan, you are an incredible wealth of information and we're truly fortunate to have your encyclopedic knowledge available. And no, I'm NOT kidding. You're great !!! Thank you and Happy Thanksgiving to all !!!
Mark


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11/19/2007 9:13:57 AM

 
Christopher A. Walrath
BetterPhoto Member Since: 4/25/2006
  Well, here's my take on the exposure. I tried the paper behind each lens and showed a light through the lenses and measured each light circle 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 on the body up through the viewscreen with a hand held meter and then through both lenses as well and found that the reading was the same for both 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 from the 80mm to the combo. So, I have deduced that by metering the subject, I simply adjust the shutter speed and the aperture remains true. I mount the 150 directly on the camera body to keep it simple which lens I should adjust. Both focused to infinity and stopped 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 so I can play with that.

Thank you for all of your help. Alan, Bob, Mark. I appreciate you guys. More than you can know. Happy Gobble Gobble Day, there, evribuddy.


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11/19/2007 2:20:02 PM

 
Alan N. Marcus   Try the paper behind the lens again using sunlight. I do think this will be helpful.
Likely the light needs to coming from infinity.

Alan Marcus


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11/19/2007 3:14:08 PM

 
Christopher A. Walrath
BetterPhoto Member Since: 4/25/2006
  Thanks, Alan. I'll give it a try. Thank you again.

Chris


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11/19/2007 4:10:19 PM

 
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