Introductory Physics Volume Two

8.11 Homework 175
We can also compute the final image from the thin lens equation.
Given f = 12cm and x o = 24cm we can compute
1
= 1 x i f − 1 = 1
x o 12cm − 1
24cm = 1
24cm −→ x i = 24cm
So
y i = − x i
y o = − 24 (4cm) = −4cm
x o 24
The second lens is 18 cm past the first, so the object distance for the
second lens is
x ′ o = 18cm − x i = 18cm − (24cm) = −6cm.
From this we can compute the final image location
1
x ′ = 1
i f ′ − 1 1
x ′ =
o −12cm − 1
−6cm −→ x′ i = 12cm
and
y i ′ = − x′ i
x ′ y o = − 12 (−4cm) = −8cm
o −6
Notice that the diverging lens has formed a real image. Also notice that
the first lens would have formed a real image, but since the diverging
lens enters the optical path before the image is formed, the first image
does not actually get formed.
§ 8.11 Homework
⊲ Problem 8.8
You are designing a movie projector. The film is 8mm wide, and you
wish to project this film onto a screen that is 2.0 meters wide from a
distance of 10 meters.
(a) How far from the lens should the film be?
(b) What should the focal length of the lens be?
⊲ Problem 8.9
An object is located 20cm to the left of a diverging lens having a focal
length f = −32cm. Determine the location and magnification of the
image. Construct a ray diagram for this arrangement.
⊲ Problem 8.10
An object is placed 40 cm in front of a lens with focal length +10 cm.
Describe the image (i.e. location, magnification, etc.).
⊲ Problem 8.11
Two converging lenses, each of focal length 10 cm, are separated by
35 cm. An object is 20 cm to the left of the first lens. (a) Find the