Page 65 - Modern Optical Engineering The Design of Optical Systems

48 Chapter Three

The focal length of a thin lens can be derived from Eq. 3.21 by setting
the thickness equal to zero.
1
(n 1) (c c ) (3.25)
1
2
f
1 1 1

(n 1) (3.25a)
f R R
1 2
Since the lens thickness is assumed to be zero, the principal points of
a “thin lens” are coincident with the location of the lens. Thus, in com-
puting object and image positions, the distances s and s′ of Eqs. 2.4, 2.5,
2.7, etc., are measured from the lens itself. The term (c 1 c 2 ) is often
called the total curvature, or simply the curvature of the element.
Note that if the lens index is 1.5, the radii of an equiconvex or equicon-
cave element equals the focal length (R f ) and the radius of a
planoconvex or planoconcave is one-half the focal length (R f/2).

Example 3.2
An object 10-mm high is to be imaged 50-mm high on a screen that is
120-mm distant. What are the radii of an equiconvex lens of index 1.5
which will produce an image of the proper size and location?
The first step in the calculation is the determination of the focal
length of the lens. Since the image is a real one, the magnification will
have a negative sign, and by Eq. 2.7a we have

h′ 50 s′
m ( ) or s′ 5s
h 10 s

For the object and image to be 120 mm apart,

120 s s′ s 5s 6s
s 20 mm

and s′ 5s 100 mm
Substituting into Eq. 2.4 and solving for f, we get

1 1 1

100 f 20

f 16.67 mm