Page 44 - Modern Optical Engineering The Design of Optical Systems

Gaussian Optics: The Cardinal Points 27

and noting from Eq. 2.5 that f/(s f ) is equal to s′/s, we find that

h′ s′
m (2.7a)
h s
Other useful relations are

s′ f (1 m) (2.7b)

1
s f 1 (2.7c)
m
Note that Eqs. 2.3 through 2.7 assume that both object and image are
in air, and also that Figs. 2.3 and 2.4 show a negative magnification.
Longitudinal magnification is the magnification along the optical
axis, i.e., the magnification of the longitudinal thickness of the object
or the magnification of a longitudinal motion along the axis. If s 1 and
s 2 denote the distances to the front and back edges of the object and s′ 1
and s′ 2 denote the distances to the corresponding edges of the image,
then the longitudinal magnification m is, by definition,

s′ 2 s′ 1

m
s s
2 1
Substituting Eqs. 2.7b and 2.7c for the primed distances and manip-
ulating, we get
s′ 1
s′ 2
m m m (2.8)

s s 1 2
1 2
noting that m s′/s. As (s′ 2 s′ 1 ) and (s 2 s 1 ) approach zero, then m 1
approaches m 2 , and

m m 2 (2.9)
This indicates that longitudinal magnification is positive, and that
object and image always move in the same direction.

Example 2.1
Given an optical system with a positive focal length of 10 in, find
the position and size of the image formed of an object 5-in high
which is located 40 in to the left of the first focal point of the system.
See Fig 2.5a.
Using the newtonian equation, we get, by substituting in Eq. 2.3,
f 2 10 2
x′ 2.5 in
x 40