1134
C H A P T E R 3 6 Image Formation
Front
Back
O
I
p
F
C
q
Figure 36.13 Formation of an image by a spherical convex mirror. The image formed
by the real object is virtual and upright.
Convex Mirrors
Front, or
real, side
Back, or
virtual, side
p and q p
S E C T I O N 3 6 . 2 Images Formed by Spherical Mirrors
C
1133
F
L
R
36.2 The Focal Point Is
Not the Focus Point
Henry Leap and Jim Lehman
f
PITFALL PREVENTION
(a)
(b)
Figure 36.12 (a) Light rays from a distant object (p : ) reect from a concave
mirror t
1132
C H A P T E R 3 6 Image Formation
h
O
C
I
h
V
Principal
axis
Figure 36.11 The image
formed by a spherical concave
mirror when the object O lies
outside the center of curvature
C. This geometric construction
is used to derive Equation 36.4.
q
R
p
Figu
Ken Kay/Fundamental Photographs
S E C T I O N 3 6 . 2 Images Formed by Spherical Mirrors
Figure 36.8 Red, blue, and green light rays
are reected by a curved mirror. Note that
the three colored beams meet at a point.
36.2 Images Formed by Spherical Mirrors
1130
C H A P T E R 3 6 Image Formation
Conceptual Example 36.2
The Levitated Professor
Solution This is one of many magicians optical illusions
that make use of a mirror. The box in which the professor
stands is a cubical frame that contains a at vertical
C H A P T E R 3 6 Image Formation
1128
P
h
p
Object
Q
q
P
R
h
Image
Active Figure 36.2 A geometric
construction that is used to locate
the image of an object placed in
front of a at mirror. Because the
triangles PQR and PQR are congruent, p q and h h.
At
S E C T I O N 3 6 . 1 Images Formed by Flat Mirrors
1129
decal can be read from outside the car, you can also read it when looking into your
rearview mirror from inside the car.
We conclude that the image that is formed by a flat mirror has the following
1124
61.
C H A P T E R 3 5 The Nature of Light and the Laws of Geometric Optics
A light ray of wavelength 589 nm is incident at an
angle on the top surface of a block of polystyrene, as
shown in Figure P35.61. (a) Find the maximum value of
for which the
This chapter is concerned with the images that result when light rays encounter at
and curved surfaces. We nd that images can be formed either by reection or by
refraction and that we can design mirrors and lenses to form images with desired characteristi
Answers to Quick Quizzes
illuminated from below. Ray PBB strikes the clear surface at
the critical angle and is totally reected, as are rays such as
PCC. Rays such as PAA emerge from the clear surface. On
the painted surface there appears a dark circle of
Problems
the eastern wall. The light traverses 2.37 m horizontally to
shine perpendicularly on the wall opposite the window. A
young prisoner observes the patch of light moving across
this western wall and for the rst time forms his own
understanding of t
1122
C H A P T E R 3 5 The Nature of Light and the Laws of Geometric Optics
43. In about 1965, engineers at the Toro Company invented a
gasoline gauge for small engines, diagrammed in Figure
P35.43. The gauge has no moving parts. It consists of a at
slab
Problems
Figure 35.25. (b) Find the angle of deviation min for
1 48.6. (c) What If? Find the angle of deviation if the
angle of incidence on the rst surface is 45.6. (d) Find
the angle of deviation if 1 51.6.
33. A triangular glass prism with apex angle
Problems
1119
reected light beam travels before striking mirror 2. (b) In
what direction does the light beam travel after being
reected from mirror 2?
on a smooth surface of water at 25C, at an angle of incidence of 3.50. Determine the angle of refraction
1136
L
C H A P T E R 3 6 Image Formation
PITFALL PREVENTION
36.4 We Are Choosing a
Small Number of
Rays
A huge number of light rays leave
each point on an object (and
pass through each point on an
image). In a principal-ray diagram, which displays the cha
S E C T I O N 3 6 . 2 Images Formed by Spherical Mirrors
Example 36.4
Interactive
The Image formed by a Concave Mirror
Assume that a certain spherical mirror has a focal length of
10.0 cm. Locate and describe the image for object
distances of
1137
image.
1150
C H A P T E R 3 6 Image Formation
where q 1 is the image distance for the rst lens. Treating this image as the object for
the second lens, we see that the object distance for the second lens must be p 2 q 1.
(The distances are the same because the le
S E C T I O N 3 6 . 4 Thin Lenses
this case, the thin lens equation gives
1149
and the magnication of the image is
1
1
1
5.00 cm
q
10.0 cm
M
cm
3.33cm
5.00
0.667
This conrms that the image is virtual, smaller than the
object, and upright.
q 3.33 cm
Inve
1148
C H A P T E R 3 6 Image Formation
Interactive
Example 36.10 The Case of a Diverging Lens
Repeat Example 36.9 for a diverging lens of focal length
10.0 cm.
This result conrms that the image is virtual, smaller than
the object, and upright.
Solution
(A
S E C T I O N 3 6 . 4 Thin Lenses
O
I, F1
F2
O
1147
F2
I
5.00 cm
F1
10.0 cm
10.0 cm
10.0 cm
15.0 cm
30.0 cm
(b)
(a)
Figure 36.30 (Example 36.9) An image is formed by a converging lens. (a) The object
is farther from the lens than the focal point. (b) The
1146
C H A P T E R 3 6 Image Formation
To locate the image of a diverging lens (Fig. 36.28c), the following three rays are
drawn from the top of the object:
Ray 1 is drawn parallel to the principal axis. After being refracted by the lens,
this ray emerge
S E C T I O N 3 6 . 4 Thin Lenses
1145
Various lens shapes are shown in Figure 36.27. Note that a converging lens is
thicker at the center than at the edge, whereas a diverging lens is thinner at the center
than at the edge.
Magnication of Images
Consider
1144
C H A P T E R 3 6 Image Formation
f
F1
f
F2
F1
F2
Henry Leap and Jim Lehman
(a)
F1
F2
F2
F1
f
f
(b)
Front
Back
p positive
q negative
p negative
q positive
Incident light
Refracted light
Figure 36.26 A diagram for
obtaining the signs of p and q for a
S E C T I O N 3 6 . 4 Thin Lenses
n1 > n2
1141
from Table 36.2 that R is negative, we obtain
n1
n
n n1
2 2
p
q
R
3.0 cm
1.50
1
1.00 1.50
2.0 cm
q
3.0 cm
q
n1
2.0 cm
q 1.7 cm
n2
Figure 36.21 (Example 36.7) Light rays from a coin embedded
in a plastic sphe
1142
C H A P T E R 3 6 Image Formation
n1 = 1
R1
I1
Surface 1
R2
Surface 2
n
O
t
p1
C1
q1
p2
(a)
n1 = 1
R1
Surface 1
R2
Surface 2
n
O
I1
p1
t
C1
p2
q1
(b)
Figure 36.23 To locate the image formed
by a lens, we use the virtual image at I 1
formed by surface
S E C T I O N 3 6 . 4 Thin Lenses
R1
O
R2
I
C1
C2
p
q
Figure 36.24 Simplied geometry for a thin lens.
where t is the thickness of the lens. For a thin lens (one whose thickness is small
compared to the radii of curvature), we can neglect t. In this approx
1140
C H A P T E R 3 6 Image Formation
n1 > n2
n1 n2
Flat Refracting Surfaces
If a refracting surface is at, then R is innite and Equation 36.8 reduces to
n1
n
2
p
q
n2
p
q
n1
I
O
(36.9)
q
p
Active Figure 36.20 The image
formed by a at refracting surface
1138
C H A P T E R 3 6 Image Formation
1
1
1
1
p
q
f
0.25 m
1990 Paul Silverman/Fundamental Photographs
1
1
1
q
0.25 m
3.0 m
q 0.23 m
The negative value of q indicates that her image is virtual, or
behind the mirror, as shown in Figure 36.15c.
(B) The mag
S E C T I O N 3 6 . 3 Images Formed by Refraction
n1
P
1
n2
2
d
1139
O
C
I
R
p
q
Figure 36.19 Geometry used to derive Equation 36.8, assuming that n1 n 2.
If we combine all three expressions and eliminate 1 and 2, we nd that
n 1 n 2 (n 2 n 1)
(36.7)
From