# Convertible Lens

• 06-05-2003, 09:53 AM
juan
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?
Juan
• 06-05-2003, 11:06 AM
SteveGangi
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.
• 06-05-2003, 11:11 AM
juan
Thanks, Steve. I was trying to do it the hard way (math) when I should have been thinking of something simple.
j
• 06-05-2003, 11:32 AM
David Vickery
Hello,
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.
• 06-05-2003, 09:02 PM
rogein
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...
• 06-06-2003, 08:29 AM
Ole
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.