# Re: [Rollei] Suplementary lenses, compound system

• Subject: Re: [Rollei] Suplementary lenses, compound system
• From: bigler@ens2m.fr
• Date: Thu, 10 Jan 2002 10:10:00 +0100 (CET)
• References:

``` From Richard K.:

>  The formula for the combined focal length of two lenses is:
>
> F = f1*f2 / f1 + f2 - d
>
>   Where f1 and f2 are the focal lengths of the two lenses and d is the
> distance between them.

Hm... Richard, I am sure that you meant :

F = (f1*f2) / (f1 + f2 - d)

this can also be written as

C= C1 + C2 - d * C1*C2   where C= 1/F ; C1=1/f1 ; C2=1/F2

when d=0 the formula is simply C= C1 + C2, the "powers" of each lenses simply add.

So assuming that f1= 100mm and if we want F=200mm and in the
simplified model of 2 thni lens elements this yields f2= -200 mm i.e.
a negative "close-up" lens of 5 dioptres.

For the most general association of any kind of thick compound lenses
another equivalent formula should be used, apparently very simple also :

F= -f1*f2/D

where D = (F'1F2) is the distance between the image focal point (F'1)
of lens #1 and (F2) the object focal point of lens #2. This may not be
easy apply except if you know exactly where the pricipal planes H1,
H'1 and H2, H'2 of the lenses are actulally located.

Assume that the additional negative lens of unknown focal length f1 is
in fact located at a distance e in front of the principal plane H2 of
the compound photographic lens. For example if f2=10cm, H2 for a
standard lens is located somewere inside the glass so e=+2cm is a
reasonable value. Then a formula identical to Richard's one is found :

F = (f1*f2) / (f1 + f2 - e)
this eventually yields for F=20cm, f2=10cm and e=+2cm

f2=20*(2-10)/(20-10) = -16cm = -6.25 dioptres instead of -20cm when e=0

- --
Emmanuel BIGLER
<bigler

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```