How Close-up Lenses Affect
Prepared 2007-02-02 (169/12614) by Bill Claff
We already learned that
m = v / f - 1
Extension tubes increase m by increasing v.
Close-up lenses increase m by reducing f.
The mathematics is governed by the formula
P = P1 + P2 - dm * P1 * P2
P, P1, and P2 are lens powers in diopters
dm is the distance between lens nodes of P1 and P2 in meters
Diopters are the reciprocal of focal length in meters (we use a capital D to indicate units of diopters).
Since dm is seldom known we will assume dm = 0 so
P = P1 + P2
Note that the powers of lenses are additive.
This is why when you take two close-up lenses, for example a 1D lens and a 2D lens, and put them together you can simply add the powers, to get 3D in this case.
Using P = 1000 / f, P1 = 1000 / f1, and P2 = 1000 / f2 we can restate the combination formula as
f = ( f1 * f2 ) / ( f1 + f2 )
This form is usually easier to use than the power formula.
So, what happens when we put a Canon 500D, a 500mm 2D close-up lens on a 300mm f/4D ED-IF AF-S Nikkor?
(BTW, the D in 500D does not stand for diopter but rather for duplet)
At infinity focus, ( 300mm * 500mm ) / ( 300mm + 500mm ) = 187.5mm
At closest focus, ( 242.9mm * 500mm ) / ( 242.9mm + 500mm ) = 163.5mm
Since m = v / f - 1 that's
At infinity focus, 300mm / 187.5mm - 1 = .60x
At closest focus, 308.5mm / 163.5mm - 1 = .89x
I won't derive the formula but for infinity focus it can be shown that
m = f1 / f2
So note that 300mm / 500mm = .60x, which matches the infinity result above.
Finally, substituting back into
S = ( 1 / m + 1 + 1 + m ) * f
At infinity focus, S = ( 1 / .60 + 1 + 1 + .60 ) * 187.5mm = 800mm
At closest focus, S = ( 1 / .89 + 1 + 1 + .89 ) * 163.5mm = 656mm
Note that in terms of magnification mounting a reversed lens is just a way of attaching a high powered close-up lens.
For example, a 50mm reversed lens acts like a 40D (1000mm/50mm) close-up lens.
I won't cover it here but it's straightforward to determine the effect of extension tubes on a lens that has close-up lens(es) attached.