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Thread started 23 Sep 2010 (Thursday) 12:50
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the math behind reversing lens

 
ayoyoayoyo
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Sep 23, 2010 12:50 |  #1

hello.

Is there any source on the math/science explaining how reversing lens can allow you to make a macro lens from a standard/wide lens?

i want to know about focal length/focus/[effectiv​e] aperture/magnification etc.

thx!




  
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LordV
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Sep 24, 2010 00:47 |  #2

Assuming we are talking about reversing a lens directly onto the camera body, I've not been able to find any simple maths for this. Just know a reversed 50mm lens gives around 1:1 and a reversed 30mm lens gives around 3:1. Reversing a kit lens gives a rather nice zoom macro lens going from around 0.5:1 to 3:1 magnification.
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ameerat42
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Sep 24, 2010 06:09 |  #3

Think of how the lens works normally: the "subject range" that can be focused onto the sensor/film is from "rather close" to "infinity" , just by moving the lens back and forth. When you reverse the lens onto the camera, your effective "subject range" is much reduced, to a few centimetres (an inch or so) from the back lens element. Just how far you can project the image depends on how much extension there is between, now, the front of the lens, and the sensor/film. Extension tubes and bellows can allow for pretty big magnifications. The cost of big magnifications is light loss and DOF reduction.

The math(s) involved can be simplified by considering the the two ratios:
subject to lens refractive plane (LRP), and LRP to image. (I'm relying on memory that the term is actually LRP.)

The LRP is the position inside the lens barrel where the refraction actually appears to take place. Typically, for my old f=50 mm prime this was about half way along the lens barrel, though somewhat closer to the back. For retro-focus lenses the LRP can be somewhere outside the front of the lens barrel.

Anyway, if the ratio of the two lengths is 4:1 then that's your magnification. As a rule of thumb, your effective aperture is also multiplied by that factor. Old extension tube and bellows came with calculating formulas, but that was essentially it.

Hope this helps a bit. Am.




  
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Lester ­ Wareham
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Oct 03, 2010 11:50 |  #4

To work out the magnification you need to work out how the front of lens principle point is from the sensor. The dof calculation changes also http://www.zen20934.ze​n.co.uk …oF_with_Macro_P​hotography (external link)

The real reason for using a reversed wide angle SLR lens is because the lens will be of a retro focus design the lens rear principle point has to be some way from the rear element to make room for the mirror. When reversed this gives a small improvement in working distance.

When I used to do this in my film days using a 28mm lens on a Canon FD film system I got something like 2.5 to 3X :1 more on bellows, working distance was still fairly small.

The MP-E 65mm appears to effectively be a reversed retro-focus type construction presumably to improve working distance.

I wouldn't want to go back to reversed lenses with manual stop-down, that makes the MP-E look like kids stuff to use. ;)


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ceriltheblade
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Oct 13, 2010 01:46 |  #5

hi there. I found this: http://enticingtheligh​t.com …acros-section-3-the-math/ (external link)

but google has a couple others too about it. I used reversed lens technique math as search parameters


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the math behind reversing lens
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