Page 43 - Modern Optical Engineering The Design of Optical Systems

26 Chapter Two

From similar triangles we can write
h ( x) h f ′
and (2.1)
( h′) f ( h′) x′
Setting the right-hand members of each equation equal and clearing
fractions, we get
ff ′ xx′ (2.2)

If we assume the optical system to be in air, then f will be equal to f′ and

f 2
x′ (2.3)
x
This is the “newtonian” form of the image equation and is very useful
for calculations where the locations of the focal points are known.
If we substitute x s f and x′ s′ f in Eq. 2.3, we can derive
another expression for the location of the image, the “gaussian” form.

2
f xx′ (s f) (s′ f )
ss′ sf s′f f 2

2
Canceling out the f terms and dividing through by ss′f, we get
1 1 1
(2.4)
s′ f s
or alternatively,
sf ss′
s′ or f (2.5)
(s f ) (s s′)

Image size
The lateral (or transverse) magnification of an optical system is given
by the ratio of image size to object size, h′/h. By rearranging Eq. 2.1,
we get for the magnification m,
h′ f x′
m (2.6)
h x f

Substituting x s f in this expression to get
h′ f
m
h (s f)