Introductory Physics Volume Two

174 Geometric Optics 8.10
So
y i = − x i
y o = − −8 (9cm) = 3cm
x o 24
The second lens is 20 cm past the first, so the object distance for the
second lens is
x ′ o = 20cm − x i = 20cm − (−8cm) = 28cm.
From this we can compute the final image location
1
x ′ = 1
i f ′ − 1 x ′ = 1
o 16cm − 1
28cm −→ x′ i = 37.33cm
and
y i ′ = − x′ i
x ′ y o = − −37.3 (3cm) = 4cm
o 28
Notice that the second lens has formed a real image of the virtual image
produced by the first lens.
In the following example, the second lens has a virtual object.
Example
Suppose that you have a converging lens, f = 12cm, followed by a
diverging lens, f ′ = −12cm. The lenses are 18 cm apart. You place an
object that is 4cm tall at a distance of 24 cm from the converging lens.
We can construct the final image as follows.
24 cm
18 cm
Image 2
Image 1
Object 1
12cm
12cm
12cm
Object 2
The top ray is not used to construct the first image, since it is not a
principle ray of the first lens. The first image is constructed from the
other two rays, and then the third ray is draw so that it goes to the
first image and also through the center of the second lens, this makes
it a principle ray of the second lens. Similarly the center ray is not a
principle ray of the second lens, it is only used for constructing the first
image. The bottom ray is a principle ray of both lenses.