of Achromatic Object-Glasses. 15 
Example . — The radius of the two equal surfaces of a flint lens 
(whose index is 1.600) being 10 inches; required its focal 
length. 
Here .600 
2 
1.200)10.000(8.33 focal length. 
The same two rules hold good when both surfaces are con- 
cave, only then the result must be considered as negative. 
3. To determine the same for parallel rays in a plano-convex 
lens ; the radius of the convex side and the index being 
given. 
Rule. — Divide the radius of curvature by the decimal part 
of the index of refraction, and the quotient will be the focal 
length. 
Example .—- Required the focal length of a plano-convex 
crown lens ; the radius of curvature 121 inches, and the index 
of refraction 1 .520. 
124 = 12.5 
.52)12.50(24.04 inches focus. 
4. To determine the focal length of a lens having one concave and 
one convex side, the radii and index of refraction being gi- 
ven, and the rays parallel. 
Rule. — Multiply the two radii together; multiply also their 
difference by the decimal part of the ' refraction : then the for- 
mer product, divided by the latter, will be the focal length ; 
which will be positive when the concave radius is the greater of 
the two, but negative when it is the lesser. 
Example . — Find the focal length of a flint lens, the radius of 
the convex side being 10 inches, and of the concave 16 inches, 
the index 1.600. 
10 
16 
16 
10 
1st product 160 
Difference 6 
.600 
2d product 3.6)160(44.44 focal length 
