Introductory Physics Volume Two

170 Geometric Optics 8.7
-x i
x o
f f
Notice that we still use the three principle rays. The topmost ray in
the figure above strikes the lens as if it came from the focal point. The
dotted line going back to the focal point is an extension of the actual
path of the light ray. Notice also that even though the ray constructed
in this manner does not hit the lens (since the lens was not tall enough)
we can still use the ray to construct the location of the image.
The thin lens and magnification equations can still be used in this
case, but in order to get the algebra to work out correctly you need to
interpret a negative image distance (x i ) as being on the side of the lens
where the light originates from.
Example
Suppose that in the figure above x 0 = 6cm, y 0 = 2.0cm and f = 9cm.
Putting this into the thin lens equation we find
1
6cm + 1 = 1
x i 9cm −→ 1 = 1
x i 9cm − 1
6cm = − 1
18cm
−→ x i = −18cm
Now we can use the magnification equation to find the size of the image.
y i
= − y o
−→ y i = − x i
y o = − −18 (2cm) = 6.0cm
x i x o
x o 6
⊲ Problem 8.4
You have a lens with a focal length of 80cm. You place an object at
60cm from the lens. The object is 3cm tall.
(a) Construct the image location using the principle rays.
(b) Find the location of the image using the thin lens equation.
(c) Find the height of the image from the magnification equation.
⊲ Problem 8.5
For a converging lens with a focal length of f, over what range of object
distances is the image:
(a) real?