Page 45 - Modern Optical Engineering The Design of Optical Systems

28 Chapter Two

Object h = +5
f 1 f 2
Image h′ = −1.25
40" 10" 10" 2.5
50" 12.5"
Figure 2.5a The formation of a real image. See Example 2.1.

Therefore the image is located 2.5 in to the right of the second focal
point. To find the image height, we use Eq. 2.6.
h′ f 10
m 0.25
h x 40
h′ mh ( 0.25) (5) 1.25 in
Thus if the base of the object were on the optical axis and the top of the
object 5 in above it, the base of the image would also lie on the axis and
the image of the top would lie 1.25 in below the axis.
The gaussian equations can be used for this calculation by noting
that the distance from the first principal plane to the object is given by
s x f 40 10 50; then, by Eq. 2.4,
1 1 1 1 1
0.1 0.02 0.08
s′ f s 10 ( 50)
1
s′ 12.5 in
0.08
and the image is found to lie 12.5 in to the right of the second principal
plane (or 2.5 in to the right of the second focal point, in agreement with
the previous solution).
The height of the image can now be determined from Eq. 2.7a.
h′ s′ 12.5
m 0.25
h s 50
h′ mh ( 0.25) (5) 1.25 in

Example 2.2
If the object of Example 2.1 is located 2 in to the right of the first focal
point, as shown in Fig. 2.5b, where is the image and what is its height?
Using Eq. 2.3,
f 2 10 2
x′ 50 in
x 2