Is this calculation correct?

[Garimo]

I've swept away the dust from my old optics books. Is this calculation correct?
If the radius of curvature of a lens is 25 mm and the index is 1,67 the lens power is:

F=(n'-1)/r
F=0,67/0,025 = +26,8 D

Then for this lens the focal distance is:

F = n' / f '
26,8 = 1,67 / f'
f' = 16,04 meters.

[shanbaum]

No, the focal length is the reciprocal of the dioptric power, so 1/26.8 = 0.037 meters, or 3.7 centimeters.

[Garimo]

No, the focal length is the reciprocal of the dioptric power, so 1/26.8 = 0.037 meters, or 3.7 centimeters.

Thanks for reply!

Are you sure its not 1/dipters for thin lenses and diopters/n' used for thick lenses?

My books says:
F = (n'-n)/r = n / f = -n'/f'

EDIT: But you are probably right, 16,04 meters doesn't make sense for a 26,8 D lens.

[shanbaum]

Am I sure? As sure as the day is long! I can't make any sense of "F = (n'-n)/r = n / f = -n'/f'", sorry.

[MakeOptics]

Thanks for reply!

Are you sure its not 1/dipters for thin lenses and diopters/n' used for thick lenses?

My books says:
F = (n'-n)/r = n / f = -n'/f'

EDIT: But you are probably right, 16,04 meters doesn't make sense for a 26,8 D lens.

1/D = focal length in air since the index of air is approximately 1, if you were looking for the focal length through the material like for instance the focal length of a rod then yes the 1.67/D would be correct.