10.3
Applying Inquiry Skills
7. (a) Draw a ray diagram of your own design of a variable-length
pinhole camera made of common household materials, such
as cardboard tubes or shoe boxes.
(b) How would you use your design to determine the relationship between the magnification of the image and the distance between the pinhole and the image?
10.3
Mathematical Relationships
for Thin Lenses
You have studied how a converging lens produces images for light sources placed
at different positions along the principal axis. Now you will study the quantitative relationship between object and image positions.
Investigation 10.3.1
Predicting the Location of Images Produced by a
Converging Lens
The purpose of this investigation is to determine the relationship among the
focal length, the image distance, and the object distance of a converging lens.
INQUIRY SKILLS
Questioning
Hypothesizing
Predicting
Planning
Conducting
Recording
Analyzing
Evaluating
Communicating
Question
What is the relationship among the focal length, the image distance, and the
object distance of a converging lens?
Materials
converging lens (f =10 cm → 25 cm)
small light source (miniature light bulb)
translucent screen (white paper)
optical bench
Prediction
(a) Predict what will happen to the size and distance of an image as an object
is moved closer to a converging lens.
Procedure
Part 1: Real Image
1. Create a table similar to Table 1.
2. In a dark part of the room, hold the lens so that light from a distant object
passes through it and onto the screen, as in Figure 1. Move the screen back
and forth until the image is clearly focused. Measure the focal length, f, of
the lens, the distance between the lens and the screen.
Lenses and the Eye 367
Table 1
Observation
Object
distance (do )
1
2.5f =
2
2.0f =
3
1.5f =
4
f=
5
0.5f =
Image
distance (d i )
size
1
do
Characteristics
attitude
type
object
1
di
1 1
+ do di
lens (at midpoint)
do
Magnification
(M)
image (screen)
di
Figure 1
Setup for Investigation 10.3.1
3. Turn the lens around and repeat step 1. Calculate an average of the two
measurements.
4. Using this average value, calculate the following object distances: 2.5f, 2.0f,
1.5f, f, and 0.5f. Record the information in your table.
5. Place the lens in the exact centre of the optical bench.
6. Using chalk or masking tape, mark on the optical bench the object distances calculated in step 4.
7. Place the object at 2.5f. Move the screen back and forth until the image is
focused clearly on the screen. Record the image distance and the characteristics of the image. Remember to look down along the principal axis to see
if you can observe an image without using a screen. Be careful not to place
your eye at a location where light is focused to a point.
8. Repeat step 7 for the other object distances that result in real images.
1 1
do di
d
1
1
di
9. Complete columns , , and + for the first three observations only.
o
10. Determine the value of the reciprocal of the focal length (f ).
11. Calculate the magnification (M) for the first three observations and enter
the values in the table.
Part 2: Virtual Image
12. Place the object at 0.5f. Hold a pencil above the image that you see in the
lens (you do not want to see the image of the pencil through the lens), then
move the pencil until there is no relative motion between the pencil and
the image when you move your head from side to side.
13. Measure the distance from the pencil to the lens and record the image distance di.
Analysis
(b) Why does the method in step 2 work for finding the focal length of a converging lens?
(c) How do the two focal lengths for both sides of the lens compare?
(d) As the object moves closer to the lens, what regular changes occur in the
size of the image? the distance of the image? the attitude of the image?
368 Chapter 10
10.3
(e) At what object distance was it difficult, if not impossible, to locate a clearly
focused image?
(f) Where would you place an object in relation to the principal focus to form
a real image? a virtual image?
1
1
1
(g) How does the value of relate to the value of + for the cases
f
do di
involving real images?
1
1
1
(h) How does the value of relate to the value of + for the case involving
f
do di
a virtual image?
(i) Answer the Question.
Evaluation
(j) Evaluate your predictions.
(k) Describe the sources of error in the investigation and evaluate their effect on
the results. Suggest one or two improvements to the experimental design.
The Thin Lens Equation
To gain further insight into the mathematical relationships of lenses, we can
derive a quantitative relationship for thin lenses that connects the image distance,
the object distance, and the focal length. We do this by studying the geometry of
two of the special rays that are used in constructing ray diagrams.
In Figure 2, the triangles AOF
and EDF are similar.
ED
AO
DF
OF
DE
BC
Therefore, = or
Also, triangles BCO and EDO
are similar.
DO
CO
Therefore, = di f
h
i = f
ho
or
d
h
i = i 1
f
ho
Therefore, i = i 1
h
d
i = i
ho do
d
f
d
do
d
d
i + 1 = i
f
do
B
A
ho object
O
D
F'
C
F
h i image
E
f
do
di
Figure 2
Using similar triangles to derive the thin
lens equation
Lenses and the Eye 369
thin lens equation: the quantitative
relationship for thin lenses;
1 1 1
+ = do di f
We now divide both sides by di to obtain the thin lens equation:
Thin Lens Equation
1
1
1
+ = f
do di
You saw from triangles BCO and EDO that
DE
DO
BC
CO
DID YOU KNOW ?
The first ratio was defined as the magnification. Using the sign convention
for a real image formed by a converging lens, ho is positive and hi is negative,
whereas do and di are both positive. For this reason, a negative sign must be
added to the second ratio so that it agrees with the sign convention. This gives us
two ways to calculate magnification:
h
M = i
ho
Calculator Exercises
Your calculator has a or x 1 function key
that can be very useful for solving thin lens
equation problems. To ensure that you know
how to use this key, try keying in the following exercises:
1
1. To find the decimal equivalent of 5,
1
enter 5 x .
1
1
2. To find the sum 2 + 4, enter 2x1 + 4x1 = .
3. To solve for f, given do = 20 cm and
di = 5 cm, enter 20x1 + 5x1 = x1 .
Never rely on one method for computing. Try
different methods to check your work. In how
many different ways can you obtain the following answers to the three exercises above?
Answers: 1. 0.2, 2. 0.75, 3. 4 cm
h
d
i i
ho
do
or
1
x
and
d
M = i
do
Sample Problem
(a) At what distance must a postage stamp be placed behind a magnifying
glass with a focal length of 10.0 cm if a virtual image is to be formed
25.0 cm behind the lens?
(b) What are the magnification and attitude of the image?
Solution
(a) f 10.0 cm
(positive, since a magnifying glass is a converging lens)
(negative, since it is a virtual image, on the object side
of the lens)
The diagram for this problem would be similar to Figure 8(e) in section 10.2.
di –25.0 cm
1
1 1
+ = do di f
1 1
1
= do f
di
1
1 1
do = f
di
1
1
= 10.0 cm
25.0 cm
do = 7.1 cm
1
The postage stamp must be placed 7.1 cm behind the magnifying glass.
(b)
d
M = i
do
25.0 cm
= 7.1 cm
M = 3.5
The magnification is 3.5. The image is upright, as indicated by a positive
value for the magnification.
370 Chapter 10
10.3
Practice
Understanding Concepts
1. Positive and negative signs are important when applying the thin
lens and magnification equations. For each of the variables listed,
state one situation in which the value is positive, and one in which it
is negative. (For example, f is positive for a converging lens.) The
variables are f , do, di, ho, hi, and M.
2. Go back to the ray diagrams you drew in Practice questions 3 and 5
in section 10.2. Use the appropriate equations to determine the
image location and magnification in each diagram, then compare the
values with the values you found using the diagrams. If the answers
don’t agree fairly closely, check your diagrams and your use of positive and negative signs in the equations.
SUMMARY
The Thin Lens Equation
• The quantitative relationship for thin lenses is represented by the thin lens
1
do
1
di
1
f
equation, + = .
Section 10.3 Questions
Understanding Concepts
1. A ruler 30.0 cm high is placed 80.0 cm in front of a converging
lens of focal length 25.0 cm.
(a) Using a scale ray diagram, locate the ruler’s image and determine its height.
(b) Using the lens and magnification equations, determine the
image position and its height. Compare your calculations
with your ray diagram.
2. A lamp 20.0 cm high is placed 60.0 cm in front of a diverging lens
of focal length 20.0 cm.
(a) Using a scale ray diagram, locate the image and determine
its height.
(b) Using the appropriate equations, calculate the image position
and the height of the image. Compare your calculations with
your ray diagram.
3. The focal length of a slide projector’s converging lens is 10.0 cm.
(a) If a 35-mm slide is positioned 10.2 cm from the lens, how far
away must the screen be placed to create a clear image?
(b) If the height of a dog on the slide film is 12.5 mm, how tall
will the dog’s image on the screen be?
4. A lens with a focal length of 10.0 cm produces an inverted image
half the size of a 4.0-cm object. How far apart are the object and
image?
Applying Inquiry Skills
5. Do you think Investigation 10.3.1 could be done using a diverging
lens instead of the converging lens suggested? If not, why not? If
so, what method would you use to view the images as the object
distances are changed?
Lenses and the Eye 371