Convertible Lens

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juan

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I've recently gotten my hands on an old Zeiss Protar lens. One lens group is 41cm - the other is 59cm. But I don't know what they are together. I know there is a formula, but I can't seem to find it. Does anyone know?
Thanks in advance.
Juan
 

SteveGangi

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I don't remember the formula, but you can test it pretty simply. Mount the lens with both sets of elements on your camera. Pick a subject far away (at infinity), and measure the distance from the lensboard to the flim plane. That distance is the focal length.
 

David Vickery

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Hello,
I can't answer your question, but I can add to it. Some time ago I purchased a Series VII Protar (B&L) Cell----without shutter or the additional cell that makes a complete lens. It is marked 600mm. I thought for the longest time that I had wasted my money on this thing, but when I bought it there where other affordable (read cheap) protars on ebay and I thought that I would be able to find something for it to fit in fairly easily. Well, I never did and the cell sat on the shelf for a couple of years. Finally I built a lens board for it and made a box for the front where I can slide in front of the lens an aperture plate. The problem showed up when I put it on my 12x20 camera. It is not 600mm. It looks like about 690mm and it just barely fits on my homemade rails. It would be nice to know the exact focal lenght of this cell so that I could make better calculations for the F#. (But I guess it doesn't really matter as everything else about this camera is an approxamation.)
The really great thing about this contraption is that it has a tremendous amount of coverage. With maximum front and rear swings and front rise(with 12x20!!) I could not get anywhere near the edge of this lens cells' coverage. I have not let it focus on film yet, but will soon.
 

rogein

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Juan,

I don't have the Zeiss specs but the B&L will be close enough. According to the B&L listing what you have is the No. 15 Convertible Protar consisting of a 23 1/4" and a 16 3/16" cells. Together they form a focal length of 10 7/8", max aperture of f/7.7, coverage at small apertures for a 10"x12" 'plate'.

Hope this helps. Have fun!

Cheers,
Roger...
 

Ole

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The simplified formula is:

1/F = 1/f1 + 1/f2

In your case, F = 1/(1/410 + 1/590) mm = 241.9mm

or pretty close to 240mm. There are a couple of corrections for cell spacing and so on, but this should be fairly close.
 
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