Chap 32 SM

Prévia do material em texto

967 
Chapter 32 
Optical Images 
 
Conceptual Problems 
 
1 • Can a virtual image be photographed? If so, give an example. If not, 
explain why. 
 
Determine the Concept Yes. Note that a virtual image is ″seen″ because the eye 
focuses the diverging rays to form a real image on the retina. For example, you 
can photograph the virtual image of yourself in a flat mirror and get a perfectly 
good picture. 
 
2 • Suppose the x, y, and z axes of a 3-D coordinate system are painted a 
different color. One photograph is taken of the coordinate system and another is 
taken of its image in a plane mirror. Is it possible to tell that one of the 
photographs is of a mirror image? Or could both photographs be of the real 
coordinate system from different angles? 
 
Determine the Concept Yes; the coordinate system and its mirror image will 
have opposite handedness. That is, if the coordinate system is a right-handed 
coordinate system, then the mirror image will be a left-handed coordinate system, 
and vice versa. 
 
3 • [SSM] True or False 
 
(a) The virtual image formed by a concave mirror is always smaller than the 
object. 
(b) A concave mirror always forms a virtual image. 
(c) A convex mirror never forms a real image of a real object. 
(d) A concave mirror never forms an enlarged real image of an object. 
 
(a) False. The size of the virtual image formed by a concave mirror when the 
object is between the focal point and the vertex of the mirror depends on the 
distance of the object from the vertex and is always larger than the object. 
 
(b) False. When the object is outside the focal point, the image is real. 
 
(c) True. 
 
(d) False. When the object is between the center of curvature and the focal point, 
the image is enlarged and real. 
 
 
Chapter 32 
 
 
968 
4 • An ant is crawling along the axis of a convex mirror that has a radius 
of curvature R. At what object distances, if any, will the mirror produce (a) an 
upright image, (b) a virtual image, (c) an image smaller than the object, and (d) an 
image larger than the object? 
 
Determine the Concept Let s be the object distance and f the focal length of the 
mirror. 
 
(a) If Rs 21< , the image is virtual, upright, and larger than the object. 
 
(b) If Rs 21< , the image is virtual, upright, and larger than the object. 
 
(c) If Rs > , the image is real, inverted, and smaller than the object. 
 
(d) If RsR <<21 , the image is real, inverted, and larger than the object. 
 
5 • [SSM] An ant is crawling along the axis of a concave mirror that has 
a radius of curvature R. At what object distances, if any, will the mirror produce 
(a) an upright image, (b) a virtual image, (c) an image smaller than the object, and 
(d) an image larger than the object? 
 
Determine the Concept 
(a) The mirror will produce an upright image for all object distances. 
 
(b) The mirror will produce a virtual image for all object distances. 
 
(c) The mirror will produce an image that is that is smaller than the object for all 
object distances. 
 
(d) The mirror will never produce an enlarged image. 
 
6 •• Convex mirrors are often used for rearview mirrors on cars and trucks 
to give a wide-angle view. ″Warning, objects are closer than they appear.″ is 
written below the mirrors. Yet, according to a ray diagram, the image distance for 
distant objects is much shorter than the object distance. Why then do they appear 
more distant? 
 
Determine the Concept They appear more distant because in a convex mirror the 
angular sizes of the images are smaller than they would be in a flat mirror. The 
angular size of the image is 
Ls
y
+'
' , where y′ is the height of the image, s′ is the 
distance between the image and the mirror, and L is the distance between the 
observer and the mirror. For a flat mirror, y′ = y and s′ = −s. 
Optical Images 
 
 
969
We wish to prove that, for a convex 
mirror: Ls
y
Ls
y
+<+'
' (1) 
 
If equation (1) is valid, then: 
y
Ls
y
Ls +>+
'
'
 
or 
y
L
y
s
y
L
y
s +>+
''
'
 (2) 
 
Because the lateral magnification is 
s
s
y
ym '' −== : 
 
s
y
s
y =
'
' (3) 
Combining this with equation (2) 
yields: 
 
y
L
y
L >
'
⇒ yy <' (4) 
It follows that, if y′ < y, then: 
 
ss <' 
To show that ss <' we begin with 
the mirror equation: 
 
fss
11
'
1 =+ 
Solving for s′ yields: 
 fs
fss −=' 
 
Because f is negative and s is 
positive: fs
f
ss +=' ⇒ ss <' 
 
Thus, we have proven that, if equation (1) is valid, then equation (4) is valid. To 
prove that equation 1 is valid, we merely follow this proof in reverse order. 
 
7 • As an ant on the axis of a concave mirror crawls from a great distance 
to the focal point of a concave mirror, the image of the ant moves from (a) a great 
distance toward the focal point and is always real, (b) the focal point to a great 
distance from the mirror and is always real, (c) the focal point to the center of 
curvature of the mirror and is always real, (d) the focal point to a great distance 
from the mirror and changes from a real image to a virtual image. 
 
 
 
 
 
Chapter 32 
 
 
970 
Picture the Problem The ray diagram shows three object positions 1, 2, and 3 as 
the moves from a great distance to the focal point F of a concave mirror. The real 
images corresponding to each of these object positions are labeled with the same 
numeral. )(b is correct. 
 
 
8 • A kingfisher bird that is perched on a branch a few feet above the 
water is viewed by a scuba diver submerged beneath the surface of the water 
directly below the bird. Does the bird appear to the diver to be closer to or farther 
from the surface than the actual bird? Explain your answer using a ray diagram. 
 
Picture the Problem The diagram 
shows two rays (from the bundle of 
rays) of light refracted at the air-water 
interface. Because the index of 
refraction of water is greater than that 
of air, the rays are bent toward the 
normal. The diver will, therefore, think 
that the rays are diverging from a point 
above the bird and so the bird appears 
to be farther from the surface than it 
actually is. 
Air
Water
.
Image
of the
bird
.
The bird
 
 
9 • An object is placed on the axis of a diverging lens whose focal length 
has a magnitude of 10 cm. The distance from the object to the lens is 40 cm. The 
image is (a) real, inverted, and diminished, (b) real, inverted, and enlarged, 
(c) virtual, inverted, and diminished, (d) virtual, upright, and diminished, 
(e) virtual, upright, and enlarged. 
 
Picture the Problem We can use a ray diagram to determine the general features 
of the image. In the diagram shown, the parallel ray and central ray have been 
used to locate the image. 
Optical Images 
 
 
971
cm 40=s
cm 10=f
>
>
 
From the diagram, we see that the image is virtual (only one of the rays from the 
head of the object actually pass through the head of the image), upright, and 
diminished. )(d is correct. 
 
10 •• If an object is placed between the focal point of a converging lens and 
the optical center of the lens, the image is (a) real, inverted, and enlarged, 
(b) virtual, upright, and diminished, (c) virtual, upright, and enlarged, (d) real, 
inverted, and diminished. 
 
Picture the Problem We can use a ray diagram to determine the general features 
of the image. In the diagram shown, the ray parallel to the principle axis and the 
central ray have been used to locate the image. Note that the object has been 
placed farther to the right than suggested in the problem statement. This was done 
so that the image would not be solarge and so far to the left as to make the 
diagram excessively large. 
F F'
>
>
 
From the diagram, we see that the image is virtual (neither ray from the head of 
the object passes through the head of the image), upright, and enlarged. )(c is 
correct. 
 
11 • A converging lens is made of glass that has an index of refraction of 
1.6. When the lens is in air, its focal length is 30 cm. When the lens is immersed 
in water, its focal length (a) is greater than 30 cm, (b) is between zero and 30 cm, 
(c) is equal to 30 cm, (d) has a negative value. 
 
Chapter 32 
 
 
972 
Picture the Problem We can apply the lens maker’s equation to the air-glass lens 
and to the water-glass lens to find the ratio of their focal lengths. 
 
Apply the lens maker’s equation to 
the air-glass interface: 
 
⎟⎟⎠
⎞
⎜⎜⎝
⎛ −−=
21air
11)16.1(1
rrf
 
Apply the lens maker’s equation to 
the water-glass interface: 
 
⎟⎟⎠
⎞
⎜⎜⎝
⎛ −−=
21water
11)33.16.1(1
rrf
 
 
Divide the first of these equations by 
the second to obtain: 
 2.211)33.16.1(
11)16.1(
21
21
air
water =
⎟⎟⎠
⎞
⎜⎜⎝
⎛ −−
⎟⎟⎠
⎞
⎜⎜⎝
⎛ −−
=
rr
rr
f
f 
or airwater 2.2 ff = and )(a is correct. 
 
12 • True or false: 
 
(a) A virtual image cannot be displayed on a screen. 
(b) A negative image distance implies that the image is virtual. 
(c) All rays parallel to the axis of a spherical mirror are reflected through a 
single point. 
(d) A diverging lens cannot form a real image from a real object. 
(e) The image distance for a converging lens is always positive. 
 
(a) True. 
 
(b) True. 
 
(c) False. Where the rays intersect the axis of a spherical mirror depends on how 
far from the axis they are reflected from the mirror. 
 
(d) True. 
 
(e) False. The image distance for a virtual image is negative. 
 
13 • Both the human eye and the digital camera work by forming real 
images on light-sensitive surfaces. The eye forms a real image on the retina and 
the camera forms a real image on a CCD array. Explain the difference between 
the ways in which these two systems accommodate. That is, the difference 
between how an eye adjusts and how a camera adjusts (or can be adjusted) to 
form a focused image for objects at both large and short distances from the 
camera. 
Optical Images 
 
 
973
Determine the Concept The muscles in the eye change the thickness of the lens 
and thereby change the focal length of the lens to accommodate objects at 
different distances. A camera lens, on the other hand, has a fixed focal length so 
that focusing is accomplished by varying the distance between the lens and the 
light-sensitive surface. 
 
14 • If an object is 25 cm in front of the naked eye of a farsighted person, 
an image (a) would be formed behind the retina if it were not for the fact that that 
the light is blocked (by the back of the eyeball) and the corrective contact lens 
should be convex, (b) would be formed behind the retina if it were not for the fact 
that that the light is blocked (by the back of the eyeball) and the corrective 
contact lens should be concave, (c) is formed in front of the retina and the 
corrective contact lens should be convex, (d) is formed in front of the retina and 
the corrective contact lens should be concave. 
 
Determine the Concept The eye muscles of a farsighted person lack the ability to 
shorten the focal length of the lens in the eye sufficiently to form an image on the 
retina of the eye. A convex lens (a lens that is thicker in the middle than at the 
circumference) will bring the image forward onto the retina. )(a is correct. 
 
15 •• Explain the following statement: A microscope is an object magnifier, 
but a telescope is an angle magnifier. Hint: Take a look at the ray diagram for 
each magnifier and use it to explain the difference in adjectives. 
 
Determine the Concept The objective lens of a microscope ordinarily produces 
an image that is larger than the object being viewed, and that image is angularly 
magnified by the eyepiece. The objective lens of a telescope, on the other hand, 
ordinarily produces an image that is smaller than the object being viewed (see 
Figure 32-52), and that image is angularly magnified by the eyepiece. The 
telescope never produces a real image that is larger than the object. 
 
Estimation and Approximation 
 
16 • Estimate the location and size of the image of your face when you hold 
a shiny new tablespoon a foot in front of your face and with the convex side 
toward you. 
 
Picture the Problem We can model the spoon as a convex spherical mirror and 
use the mirror equation and the lateral magnification equation to estimate the 
location and size of the image of your face. 
 
The mirror equation is: 
 fs's
111 =+ 
Chapter 32 
 
 
974 
Because 2rf = , where r is the 
radius of curvature of the mirror: rs's
211 =+ ⇒
sr
rss'
2−= 
 
Let the distance from your face to 
the surface of the spoon be 15 cm. If 
the radius of curvature of the spoon 
is 2 cm, then: 
 
cm 1.1
cm) 15(2cm .02
cm) (15 cm) 0.2( −≈−=s' 
or spoon thebehind cm 1.1about 
The lateral magnification equation 
is: 
 
s
s'
y
y'm −== ⇒ y
s
sy ⎟⎠
⎞⎜⎝
⎛−= '' 
Assume your face is 25 cm long. 
Substitute numerical values and 
evaluate y′: 
( )
cm 2 
cm 25
cm 15
cm 1.1'
≈
⎟⎠
⎞⎜⎝
⎛ −−=y
 
 
17 • Estimate the focal length of the ″mirror″ produced by the surface of 
the water in the reflection pools in front of the Lincoln Memorial on a still night. 
 
Determine the Concept The surface of the ″mirror″ produced by the surface of 
the water in the Lincoln Memorial reflecting pool is approximately spherical with 
a radius of curvature approximately equal to the radius of Earth. The focal length 
of a spherical mirror is half its radius of curvature. Hence Earth21 Rf = 
 
18 •• Estimate the maximum value that could be obtained for the 
magnifying power of a simple magnifier, using Equation 32-20. Hint: Think about 
the smallest focal length lens that could be made from glass and still be used as a 
magnifier. 
 
Picture the Problem Because the focal length of a spherical lens depends on its 
radii of curvature and the magnification depends on the focal length, there is a 
practical upper limit to the magnification. 
 
Use Equation 32-20 to relate the 
magnification M of a simple 
magnifier to its focal length f: 
 
f
x
M np= 
Use the lens-maker’s equation to 
relate the focal length of a lens to its 
radii of curvature and the index of 
refraction of the material from 
which it is constructed: 
 
( ) ⎟⎟⎠
⎞
⎜⎜⎝
⎛ −−=
21
1111
rr
n
f
 
 
Optical Images 
 
 
975
For a plano-convex lens, r2 = ∞. 
Hence: 1
11
r
n
f
−= ⇒ 
1
1
−= n
rf 
 
Substitute in the expression for M 
and simplify to obtain: 
 
( )
1
np1
r
xn
M
−= 
Note that the smallest reasonable value 
for r1 will maximize M. 
 
A reasonable smallest value for the 
radius of a magnifier is 1.0 cm. Use 
this value and n = 1.5 to estimate 
Mmax: 
( )( ) 13
cm0.1
cm2515.1
max =−=M 
 
Plane Mirrors 
 
19 • [SSM] The image of the object point P in Figure 32-57 is viewed by 
an eye, as shown. Draw rays from the object point that reflect from the mirror and 
enter the eye. If the object point and the mirror are fixed in their locations, 
indicate the range of locations where the eye can be positioned and still see the 
image of the object point. 
 
Determine the Concept Rays from the source that are reflected by the mirror are 
shown in the following diagram. The reflected rays appear to diverge from the 
image. The eye can see the image if it is in the region between rays 1 and 2. 
 
 
20 • You are 1.62 m tall andwant to be able to see your full image in a 
vertical plane mirror. (a) What is the minimum height of the mirror that will meet 
your needs? (b) How far above the floor should the bottom of mirror in (a) be 
placed, assuming that the top of your head is 14 cm above your eye level? Use a 
ray diagram to explain your answer. 
 
Chapter 32 
 
 
976 
Determine the Concept A ray diagram 
showing rays from your feet and the top 
of your head reaching your eyes is 
shown to the right. (a) The mirror must 
be half your height, i.e., cm 81 
(b) The top of the minimum-height 
mirror must be 7 cm below the top of 
your head, or 155 cm above the floor. 
The bottom of the mirror should be 
placed cm 47 cm 81 cm 155 =− above 
the floor. 
 
 
21 •• [SSM] (a) Two plane mirrors make an angle of 90º. The light from a 
point object that is arbitrarily positioned in front of the mirrors produces images at 
three locations. For each image location, draw two rays from the object that, after 
one or two reflections, appear to come from the image location. (b) Two plane 
mirrors make an angle of 60º with each other. Draw a sketch to show the location 
of all the images formed of an object on the bisector of the angle between the 
mirrors. (c) Repeat Part (b) for an angle of 120º. 
 
Determine the Concept 
(a) Draw rays of light from the object (P) that satisfy the law of reflection at the 
two mirror surfaces. Three virtual images are formed, as shown in the following 
figure. The eye should be to the right and above the mirrors in order to see these 
images. 
Virtual image Virtual image
PVirtual image
 
(b) The following diagram shows selected rays emanating from a point object (P) 
located on the bisector of the 60° angle between the mirrors. These rays form the 
two virtual images below the horizontal mirror. The construction details for the 
Optical Images 
 
 
977
two virtual images behind the mirror that is at an angle of 60° with the horizontal 
mirror have been omitted due to the confusing detail their inclusion would add to 
the diagram. 
Virtual image
Virtual image
Virtual image
Virtual image
P
 
(c) The following diagram shows selected rays emanating from a point object (P) 
on the bisector of the 120° angle between the mirrors. These rays form the two 
virtual images at the intersection of the dashed lines (extensions of the reflected 
rays): 
Virtual image
Virtual image
P
 
 
22 •• Show that the mirror equation (Equation 32-4 where f = r/2) yields the 
correct image distance and magnification for a plane mirror. 
 
Picture the Problem We can use Equation 32-4 and the definition of the lateral 
magnification of an image to show that the mirror equation yields the correct 
image distance and magnification for a plane mirror. 
 
Chapter 32 
 
 
978 
The mirror equation is: 
 fs's
111 =+ 
 
Because 2rf = : 
 rs's
211 =+ 
 
For a plane mirror, ∞=r . Hence: 
 
011 =+
s's
⇒ ss' −= 
 
The lateral magnification of the 
image is given by: 
1+=−−=−=
s
s
s
s'M 
 
23 •• When two plane mirrors are parallel, such as on opposite walls in a 
barber shop, multiple images arise because each image in one mirror serves as an 
object for the other mirror. An object is placed between parallel mirrors separated 
by 30 cm. The object is 10 cm in front of the left mirror and 20 cm in front of the 
right mirror. (a) Find the distance from the left mirror to the first four images in 
that mirror. (b) Find the distance from the right mirror to the first four images in 
that mirror. (c) Explain why each more distant image becomes fainter and fainter. 
 
Determine the Concept (a) The first image in the mirror on the left is 10 cm 
behind the mirror. The mirror on the right forms an image 20 cm behind that 
mirror or 50 cm from the left mirror. This image will result in a second image 
50 cm behind the left mirror. The first image in the left mirror is 40 cm from the 
right mirror and forms an image 40 cm behind the right mirror or 70 cm from the 
left mirror. That image gives an image 70 cm behind the left mirror. The fourth 
image behind the left mirror is 110 cm behind that mirror. 
 
(a) For the mirror on the left, the images are located 10 cm, 50 cm, 70 cm, and 
110 cm behind the mirror on the left 
 
(b) For the mirror on the right, the images are located 20 cm, 40 cm, 80 cm, and 
100 cm behind the mirror on the right. 
 
(c) The successive images are dimmer because the light travels farther to form 
them. The intensity falls of inversely with the square of the distance the light 
travels. In addition, at each reflection a small percentage of the light intensity is 
lost. Real mirrors are not 100% reflecting. 
 
Spherical Mirrors 
 
24 • A concave mirror has a radius of curvature equal to 24 cm. Use ray 
diagrams to locate the image, if it exists, for an object near the axis at distances of 
Optical Images 
 
 
979
(a) 55 cm, (b) 24 cm, (c) 12 cm, and (d) 8.0 cm from the mirror. For each case, 
state whether the image is real or virtual; upright or inverted; and enlarged, 
reduced, or the same size as the object. 
 
Picture the Problem The easiest rays to use in locating the image are (1) the ray 
parallel to the principal axis and passes through the focal point of the mirror, (2) 
the ray that passes through the center of curvature of the spherical mirror and is 
reflected back on itself, and (3) the ray that passes through the focal point of the 
spherical mirror and is reflected parallel to the principal axis. We can use any two 
of these rays emanating from the top of the object to locate the image of the 
object. 
 
(a) The ray diagram follows. The image is real, inverted, and reduced. 
 
C
F
 
 
(b) The ray diagram is shown to the 
right. The image is real, inverted, 
and the same size as the object. 
C F
Object
Image
 
(c) The ray diagram is shown to the 
right. The object is at the focal plane 
of the mirror. The emerging rays are 
parallel and do not form an image. 
 
 
C F
 
 
Chapter 32 
 
 
980 
(d) The ray diagram is shown to the 
right. The image is virtual, erect, and 
enlarged. 
C F
 
 
25 • [SSM] (a) Use the mirror equation (Equation 32-4 where f = r/2) to 
calculate the image distances for the object distances and mirror of Problem 24. 
(b) Calculate the magnification for each given object distance. 
 
Picture the Problem In describing the images, we must indicate where they are 
located, how large they are in relationship to the object, whether they are real or 
virtual, and whether they are upright or inverted. The object distance s, the image 
distance s′, and the focal length of a mirror are related according to ,111
fs's
=+ 
where rf 21= and r is the radius of curvature of the mirror. In this problem, 
f = 12 cm because r is positive for a concave mirror. 
 
(a) Solve the mirror equation for s′: 
fs
fss −=' where 2
rf = 
 
When s = 55 cm: ( )( )
cm15
cm 35.15
cm12cm55
cm55cm12
=
=−=s' 
 
When s = 24 cm: ( )( ) cm24
cm12cm24
cm24cm12 =−=s' 
 
When s = 12 cm: ( )( ) undefined is 
cm12cm12
cm12cm12' −=s 
 
When s = 8.0 cm: ( )( )
m2.0
cm0.24
cm12cm0.8
cm0.8cm12'
−=
−=−=s 
 
Optical Images 
 
 
981
(b) The lateral magnification of the 
image is: s
s'm −= 
 
When s = 55 cm, the lateral 
magnification of the image is: 
28.0
cm55
cm35.15 −=−=−=
s
s'm 
 
When s = 24 cm, the lateral 
magnification of the image is: 
0.1
cm42
cm24 −=−=−=
s
s'm 
 
When s = 12 cm: undefined. is m 
 
When s = 8.0 cm, the lateral 
magnification of the image is: 
0.3
cm8.0
cm24' =−−=−=
s
sm 
 
Remarks: These results are in excellent agreement with those obtainedgraphically in Problem 24. 
 
26 • A convex mirror has a radius of curvature that has a magnitude equal 
to 24 cm. Use ray diagrams to locate the image, if it exists, for an object near the 
axis at distances of (a) 55 cm, (b) 24 cm, (c) 12 cm, (d) 8.0 cm, (e) 1.0 cm from 
the mirror. For each case, state whether the image is real or virtual; upright or 
inverted; and enlarged, reduced, or the same size as the object. 
 
Picture the Problem The easiest rays to use in locating the image are (1) the ray 
parallel to the principal axis and passes through the focal point of the mirror, (2) 
the ray that passes through the center of curvature of the spherical mirror and is 
reflected back on itself, and (3) the ray that passes through the focal point of the 
spherical mirror and is reflected parallel to the principal axis. We can use any two 
of these rays emanating from the top of the object to locate the image of the 
object. Note that, due to space limitations, the following ray diagrams are not to 
the same scale as those in Problem 24. 
 
(a) The ray diagram follows. The image is virtual, upright, and reduced. 
CF
 
Chapter 32 
 
 
982 
 
(b) The ray diagram follows. The image is virtual, upright, and reduced. 
CF
 
 
(c) The ray diagram follows. The image is virtual, upright and reduced. 
CF
 
 
(d) The ray diagram follows. The image is virtual, upright, and reduced. 
CF
 
 
(e) The ray diagram follows. The image is virtual, upright, and reduced. 
CF
 
 
 
Optical Images 
 
 
983
27 • (a) Use the mirror equation (Equation 32-4 where f = r/2) to calculate 
the image distances for the object distances and mirror of Problem 26. 
(b) Calculate the magnification for each given object distance. 
 
Picture the Problem In describing the images, we must indicate where they are 
located, how large they are in relationship to the object, whether they are real or 
virtual, and whether they are upright or inverted. The object distance s, the image 
distance s′, and the focal length of a mirror are related according to ,111
fs's
=+ 
where rf 21= and r is the radius of curvature of the mirror. In this problem, 
f = −12 cm because r is negative for a convex mirror. 
 
(a) Solve the mirror equation for 
s′: fs
fss' −= 
 
When s = 55 cm: ( )( )( )
cm9.9
cm85.9
cm12cm55
cm55cm12
−=
−=−−
−=s'
 
 
When s = 24 cm: ( )( )( ) cm0.8cm12cm24
cm24cm12 −=−−
−=s' 
 
When s = 12 cm: ( )( )( ) cm0.6cm12cm12
cm12cm12 −=−−
−=s' 
 
When s = 8.0 cm: ( )( )( ) cm8.4cm12cm0.8
cm0.8cm12 −=−−
−=s' 
 
(b) The lateral magnification is given 
by: 
 
s
s'm −= 
When s = 55 cm, the lateral 
magnification of the image is: 
18.0
cm55
cm85.9 =−−=m 
 
When s = 24 cm, the lateral 
magnification of the image is: 
 
33.0
cm42
cm0.8 =−−=m 
When s = 12 cm, the lateral 
magnification of the image is: 
 
50.0
cm21
cm0.6 =−−=m 
Chapter 32 
 
 
984 
When s = 8.0 cm, the lateral 
magnification of the image is: 
60.0
cm8.0
cm80.4 =−−=m 
 
Remarks: These results are in excellent agreement with those obtained 
graphically in Problem 26. 
 
28 •• Use the mirror equation (Equation 32-4 where f = r/2) to prove that a 
convex mirror cannot form a real image of a real object, no matter where the 
object is placed. 
 
Picture the Problem We can solve the mirror equation for 1/s′ and then examine 
the implications of f < 0 and s > 0. 
 
Solve the mirror equation for 
1/s′: 
 
sf
fs
sfs'
−=−= 111 
 
For a convex mirror: 
 
0<f 
For s > 0, the numerator is positive 
and the denominator negative. 
Consequently: 
0 01 <⇒< s'
s'
 
 
29 • [SSM] A dentist wants a small mirror that will produce an upright 
image that has a magnification of 5.5 when the mirror is located 2.1 cm from a 
tooth. (a) Should the mirror be concave or convex? (b) What should the radius of 
curvature of the mirror be? 
 
Picture the Problem We can use the mirror equation and the definition of the 
lateral magnification to find the radius of curvature of the dentist’s mirror. 
 
(a) The mirror must be concave. A convex mirror always produces a diminished 
virtual image. 
 
(b) Express the mirror equation: 
 rfs's
2111 ==+ ⇒
ss
ss'r += '
2 (1) 
 
The lateral magnification of the 
mirror is given by: s
s'm −= ⇒ mss' −= 
 
Substitute for s′ in equation (1) to 
obtain: 
 
m
msr −
−=
1
2 
Optical Images 
 
 
985
Substitute numerical values and 
evaluate r: 
( )( ) cm1.5
5.51
cm1.25.52 =−
−=r 
 
30 •• Convex mirrors are used in many stores to provide a wide angle of 
surveillance for a reasonable mirror size. Your summer job is at a local 
convenience store that uses the mirror shown in Figure 32-58. This setup allows 
you (or the clerk) to survey the entire store when you are 5.0 m from the mirror. 
The mirror has a radius of curvature equal to 1.2 m. Assume all rays are paraxial. 
(a) If a customer is 10 m from the mirror, how far from the mirror is his image? 
(b) Is the image in front of or behind the mirror? (c) If the customer is 2.0 m tall, 
how tall is his image? 
 
Picture the Problem We can use the mirror equation and the relationship 
between the focal length of a mirror and its radius of curvature to find the location 
of the image. We can then use the definition of the lateral magnification of the 
mirror to find the height of the image formed in the mirror. 
 
(a) Solve the mirror equation for s′: 
fs
fss' −= where rf 2
1= 
 
Substitute for f and simplify to 
obtain: rs
rs
rs
rss' −=−= 221
2
1
 
 
Substitute numerical values and 
evaluate s′: 
 
 
 
 
( )( )
( ) ( ) cm6.56m2.1m102
m10m2.1 −=−−
−=s' 
and 
mirror. thefrom cm57 is image the 
 
(b) Because the image distance is 
negative: 
 
mirror. thebehind is image the 
(c) The lateral magnification of the 
mirror is given by Equation 32-5: 
 
s
s'
y
y'm −== ⇒ y
s
s'y' −= 
Substitute numerical values and 
evaluate y′: ( ) cm11m0.2m10
m566.0 =−−=y' 
 
31 •• A certain telescope uses a concave spherical mirror that has a radius 
equal to 8.0 m. Find the location and diameter of the image of the moon formed 
by this mirror. The moon has a diameter of 3.5 × 106 m and is 3.8 × 108 m from 
Earth. 
 
Chapter 32 
 
 
986 
Picture the Problem We can use the mirror equation to locate the image formed 
in this mirror and the expression for the lateral magnification of the mirror to find 
the diameter of the image. 
 
Solve the mirror equation for the 
location of the image of the 
moon: 
 
fs
fss' −= 
Because rf 21= : 
rs
rs
rs
rs
s' −=−= 221
2
1
 
 
Substitute numerical values and 
evaluate s′: 
( )( )( ) m0.4m0.8m108.32 m108.3m0.8 8
8
=−×
×=s' 
 
Express the lateral magnification 
of the mirror: s
s'
y
y'm −== ⇒ y
s
s'y' −= 
 
Substitute numerical values and 
evaluate y′: ( )
cm7.3
m105.3
m108.3
m0.4 6
8
−=
××−=y' 
 
The 3.7-cm-diameter image is 4.0 m in front of the mirror. 
 
32 •• A piece of a thin spherical shell that has a radius of curvature of 
100 cm is silvered on both sides. The concave side of the piece forms a real 
image 75 cm from the piece. The piece is then turned around so that its convex 
side faces the object. The piece is moved so that the image is now 35 cm from the 
piece on the concave side. (a) How far was the piece moved? (b) Was it moved 
toward the object or away from the object? 
 
Picture the Problem We can use the mirror equation to find the focal length of 
the piece of the thin spherical shell and then apply it a second time to find the 
object position after the piece has been moved. 
 
(a) Solve the mirrorequation for f to 
obtain: 
 
ss'
ss'f += 
 
Substitute numerical values and 
evaluate f: 
 
( )( ) cm86.42
cm100cm75
cm75cm100 =+=f 
Optical Images 
 
 
987
Solving the mirror equation for s 
yields: fs'
fs's −= 
 
With the piece turned around, 
 f = −42.86 cm and s′ = − 35 cm. 
Substitute numerical values and 
evaluate s: 
 
( )( )
( ) cm191cm86.42cm35
cm35cm86.42 ≈−−−
−−=s 
 
The distance d the mirror moved 
is: 
 
cm91cm100 cm191 =−=d 
(b) The piece was moved away from the object. 
 
33 •• Two light rays parallel to the optic axis of a concave mirror strike that 
mirror as shown in Figure 32-59. This mirror has a radius of curvature equal to 
5.0 m. They then strike a small spherical mirror that is 2.0 m from the large 
mirror. The light rays finally meet at the vertex of the large mirror. Note: The 
small mirror is shown as planar, so as not to give away the answer, but it is not 
actually planar. (a) What is the radius of curvature of the small mirror? (b) Is that 
mirror convex or concave? Explain your answer. 
 
Picture the Problem Denote the larger mirror on the right using the numeral 1, 
the smaller mirror using the numeral 2, and the distance between the mirrors by d. 
Applying the mirror equation to the two mirrors will lead us to an expression for 
the radius of curvature of the small mirror. 
 
(a) Apply the mirror equation to 
mirror 1: 111
211
r'ss
=+ 
 
Because ∞=1s : 
11
21
r's
= ⇒ 1211 r's = 
 
The image formed by mirror #1 
serves as a virtual object for mirror 
2. Apply the mirror equation to 
mirror 2 to obtain: 
 
222
211
r'ss
=+ 
or, because 12112 rds'ds −=−= , 
2212
1
211
r'srd
=+− 
 
Because the image distance for 
mirror 2 is d: 
 
21
21
2
2
rdrd
=+− 
Chapter 32 
 
 
988 
Substituting numerical values yields: 
 ( ) 2
2
m 0.2
1
m 0.5m 0.22
2
r
=+− 
 
Finally, solve for r2 to obtain: 
 
m3.12 −=r 
 
(b) Because 02212 <= rf , the small mirror is convex . 
 
Images Formed by Refraction 
 
34 •• A very long 1.75-cm-diameter glass rod has one end ground and 
polished to a convex spherical surface that has a 7.20-cm radius. The glass 
material has an index of refraction of 1.68. (a) A point object in air is on the axis 
of the rod and 30.0 cm from the spherical surface. Find the location of the image 
and state whether the image is real or virtual. (b) Repeat Part (a) for a point object 
in air, on the axis, and 5.00 cm from the spherical surface. Draw a ray diagram for 
each case. 
 
Picture the Problem We can use the equation for refraction at a single surface to 
find the images corresponding to these three object positions. The signs of the 
image distances will tell us whether the images are real or virtual and the ray 
diagrams will confirm the correctness of our analytical solutions. 
 
Use the equation for refraction at a 
single surface to relate the image and 
object distances: 
 
r
nn
s'
n
s
n 1221 −=+ (1) 
Solving for s′ yields: 
s
n
r
nn
ns'
112
2
−−
= 
 
(a) Substitute numerical values 
(s = 30.0 cm, n1 = 1.00, n2 = 1.68 
and r = 7.20 cm) and evaluate s′: 
cm 27
0.30
00.1
20.7
00.168.1
68.1 =
−−
=s' 
where the positive distance tells us that 
the image is 27 cm in back of the 
surface and is real. 
 
Optical Images 
 
 
989
 
In the following ray diagram, the object is at P and the image at P′. 
Air Glass
CP P'
n =1.68
 
(b) Substitute numerical values 
(s = 5.00 cm, n1 = 1.00, n2 = 1.68 
and r = 7.20 cm) and evaluate s′: 
cm 16
00.5
00.1
20.7
00.168.1
68.1 −=
−−
=s' 
where the minus sign tells us that the 
image is 16 cm in front of the surface 
and is virtual. 
 
In the following ray diagram, the object is at P and the image at P′. 
Air Glass
CPP'
n =1.68
 
 
35 • [SSM] A fish is 10 cm from the front surface of a spherical fish bowl 
of radius 20 cm. (a) How far behind the surface of the bowl does the fish appear 
to someone viewing the fish from in front of the bowl? (b) By what distance does 
the fish’s apparent location change (relative to the front surface of the bowl) when 
it swims away to 30 cm from the front surface? 
 
Picture the Problem The diagram 
shows two rays (from the bundle of 
rays) of light refracted at the water-air 
interface. Because the index of 
refraction of air is less than that of 
water, the rays are bent away from the 
normal. The fish will, therefore, appear 
to be closer than it actually is. We can 
use the equation for refraction at a 
single surface to find the distance s′. 
We’ll assume that the glass bowl is thin 
enough that we can ignore the 
refraction of the light passing through 
it. 
Fish
Water
Air
Image of
Fish
r
 
 
Chapter 32 
 
 
990 
(a) Use the equation for refraction at 
a single surface to relate the image 
and object distances: 
 
r
nn
s'
n
s
n 1221 −=+ 
Solving for s′ yields: 
s
n
r
nn
ns'
112
2
−−
= 
 
Substitute numerical values 
(s = −10 cm, n1 = 1.33, n2 = 1.00 
and r = 20 cm) and evaluate s′: 
cm .68
cm 10
33.1
cm 20
33.100.1
00.1 =
−−
−=s'
 
and the image is 8.6 cm from the front 
surface of the bowl. 
 
(b) For s = 30 cm: cm 9.35
cm 30
33.1
cm 20
33.100.1
00.1 =
−−
−=s' 
 
The change in the fish’s apparent 
location is: 
cm 27cm 6.8cm 5.93 ≈− 
 
36 •• A very long 1.75-cm-diameter glass rod has one end ground and 
polished to a concave spherical surface that has a 7.20-cm radius. The glass 
material has an index of refraction of 1.68. A point object in air is on the axis of 
the rod and 15.0 cm from the spherical surface. Find the location of the image and 
state whether the image is real or virtual. Draw a ray diagram. 
 
Picture the Problem We can use the equation for refraction at a single surface to 
find the images corresponding to these three object positions. The signs of the 
image distances will tell us whether the images are real or virtual and the ray 
diagrams will confirm the correctness of our analytical solutions. 
 
Use the equation for refraction at a 
single surface to relate the image 
and object distances: 
 
r
nn
s'
n
s
n 1221 −=+ (1) 
Solving for s′ yields: 
s
n
r
nn
ns'
112
2
−−
= 
 
Optical Images 
 
 
991
Substitute numerical values 
(s = 15.0 cm, n1 = 1.00, n2 = 1.68, 
and r = −7.20 cm) and evaluate s′: 
10
cm 0.15
00.1
cm 20.7
00.168.1
68.1 −=
−−
−=s' 
where the minus sign tells us that the 
image is 10 cm in front of the surface 
of the rod and is virtual. 
 
In the following ray diagram, the object is at P and the virtual image is at P′. 
C
Air Glass
P P'
n =1.68
 
 
37 •• [SSM] Repeat Problem 34 for when the glass rod and the object are 
immersed in water and (a) the object is 6.00 cm from the spherical surface, and 
(b) the object is 12.0 cm from the spherical surface. 
 
Picture the Problem We can use the equation for refraction at a single surface to 
find the images corresponding to these three object positions. The signs of the 
image distances will tell us whether the images are real or virtual and the ray 
diagrams will confirm the correctness of our analytical solutions. 
 
Use the equation for refraction at a 
single surface to relate the image 
and object distances: 
 
r
nn
s'
n
s
n 1221 −=+ (1) 
Solving for s′ yields: 
s
n
r
nn
ns'
112
2
−−
= 
 
(a) Substitute numerical values 
(s = 6.00 cm, n1 = 1.33, n2 = 1.68, 
and r = 7.20 cm) and evaluate s′: 
 
cm 7.9
cm 00.6
33.1
cm 20.7
33.168.168.1 −=
−−
=s'
where the negative distance tells us that 
the image is 9.7 cm in front of the 
surface and is virtual. 
 
Chapter 32 
 
 
992 
 
In the following ray diagram, the object is at P and the virtual image is at P′. 
Glass
CPP'
Water
33.11 =n 68.12 =n
 
(b) Substitute numerical values 
(s = 12.0 cm, n1 = 1.33, n2 = 1.68, 
and r = 7.20 cm) and evaluate s′: 
 
cm 27
cm 0.12
33.1
cm 20.7
33.168.1
68.1 −=
−−
=s' 
where the negative distance tells us that 
the image is 27 cm in front of the 
surface and is virtual. 
 
In the following ray diagram, the object is at P and the virtual image is at P′. 
Glass
CPP' Water
33.11 =n 68.12 =n
 
 
38 •• Repeat Problem 36 for when the glass rod and the object are immersed 
in water and the object is 20 cm from the spherical surface. 
 
Picture the Problem We can use the equation for refraction at a single surface to 
find the images corresponding to these three object positions. The signs of the 
image distances will tell us whether the images are real or virtual and the ray 
diagrams will confirm the correctness of our analytical solutions. 
 
Use the equation for refraction at a 
single surface to relate the image 
and object distances: 
 
r
nn
s'
n
s
n 1221 −=+ (1) 
Solving for s′ yields: 
s
n
r
nn
ns'
112
2
−−
= 
 
Optical Images 
 
 
993
Substitute numerical values 
(s = 20 cm, n1 = 1.33, n2 = 1.68, 
and r = −7.20 cm) and evaluate s′: 
15
cm 02
33.1
cm 20.7
33.168.1
68.1 −=
−−
−=s' 
where the minus sign tells us that the 
image is 15 cm in front of the surface 
of the rod and is virtual 
 
In the following ray diagram the object is at P and the virtual image is at P′. 
 
C
Glass
P P'
Water
33.11 =n 68.12 =n
 
 
 
39 •• A rod that is 96.0-cm long is made of glass that has an index of 
refraction equal to 1.60. The rod has its ends ground to convex spherical surfaces 
that have radii equal to 8.00 cm and 16.0 cm. An object is in air on the long axis 
of the rod 20.0 cm from the end that has the 8.00-cm radius. (a) Find the image 
distance due to refraction at the 8.00-cm-radius surface. (b) Find the position of 
the final image due to refraction at both surfaces. (c) Is the final image real or 
virtual? 
 
Picture the Problem (a) We can use the equation for refraction at a single 
surface to find the images due to refraction at the ends of the glass rod. (b) The 
image formed by the refraction at the first surface will serve as the object for the 
second surface. (c) The sign of the final image distance will tell us whether the 
image is real or virtual. 
 
(a) Use the equation for refraction at 
a single surface to relate the image 
and object distances at the surface 
whose radius is 8.00 cm: 
 
r
nn
s'
n
s
n 1221 −=+ 
Solving for s′ yields: 
s
n
r
nn
ns'
112
2
−−
= 
 
Chapter 32 
 
 
994 
Substitute numerical values 
(s = 20.0 cm, n1 = 1.00, n2 = 1.60, 
and r = 8.00 cm) and evaluate s′: 
cm 64
cm 0.02
00.1
cm 00.8
00.160.1
60.1 =
−−
=s' 
 
(b) The object for the second surface 
is 96 cm − 64 cm = 32 cm from the 
surface whose radius is −16.0 cm. 
Substitute numerical values (note 
that now n1 = 1.60 and n2 = 1.00) 
and evaluate s′: 
 
cm 80
cm 23
60.1
cm 0.16
60.100.1
00.1 −=
−−
−=s' 
 
(c) The final image is inside the rod and 96 cm − 80 cm = 16 cm from the surface 
whose radius of curvature is 8.00 cm and is virtual. 
 
40 •• Repeat Problem 39 for an object in air on the axis of the glass rod 
20.0 cm from the end that has the 16.0-cm radius. 
 
Picture the Problem (a) We can use the equation for refraction at a single 
surface to find the images due to refraction at the ends of the glass rod. (b) The 
image formed by the refraction at the first surface will serve as the object for the 
second surface. (c) The sign of the final image distance will tell us whether the 
image is real or virtual. 
 
(a) Use the equation for refraction at 
a single surface to relate the image 
and object distances at the surface 
that has a 16.0 cm radius: 
 
r
nn
s'
n
s
n 1221 −=+ 
Solving for s′ yields: 
s
n
r
nn
ns'
112
2
−−
= 
 
Substitute numerical values 
(s = 20.0 cm, n1 = 1.00, 
n2 = 1.60 and r = 16.0 cm) 
evaluate s′: 
 
cm 103.1cm 128
cm 0.02
00.1
cm 6.01
00.160.1
60.1
2×−=−=
−−
=s'
 
where the minus sign tells us that the 
image is 1.3 × 102 cm to the left of the 
surface whose radius of curvature is 
16.0 cm. 
 
Optical Images 
 
 
995
(b) The object for the second 
surface is 96 cm + 128 cm = 224 cm 
from the surface whose radius of 
curvature is −8.00 cm. Substitute 
numerical values (note that now 
n1 = 1.60 and n2 = 1.00) and 
evaluate s′: 
 
cm15
cm 7.14
cm 242
60.1
cm .008
60.100.1
00.1
=
=
−−
−=s'
 
 
(c) The final image is outside the rod, 15 cm from the end whose radius of 
curvature is 8.00 cm, and is real. 
 
Thin Lenses 
 
41 • [SSM] A double concave lens that has an index of refraction equal to 
1.45 has radii whose magnitudes are equal to 30.0 cm and 25.0 cm. An object is 
located 80.0 cm to the left of the lens. Find (a) the focal length of the lens, (b) the 
location of the image, and (c) the magnification of the image. (d) Is the image real 
or virtual? Is the image upright or inverted? 
 
Picture the Problem We can use the lens-maker’s equation to find the focal 
length of the lens and the thin-lens equation to locate the image. We can use 
s
s'm −= to find the lateral magnification of the image. 
 
(a) The lens-maker’s equation is: 
 ⎟⎟⎠
⎞
⎜⎜⎝
⎛ −⎟⎟⎠
⎞
⎜⎜⎝
⎛ −=
21air
1111
rrn
n
f
 
where the numerals 1 and 2 denote the 
first and second surfaces, respectively. 
 
Substitute numerical values to 
obtain: ⎟⎟⎠
⎞
⎜⎜⎝
⎛ −−⎟⎠
⎞⎜⎝
⎛ −=
cm0.25
1
cm0.30
11
00.1
45.11
f
 
 
Solving for f yields: cm30cm3.30 −=−=f 
 
(b) Use the thin-lens equation to 
relate the image and object 
distances: 
 
fs's
111 =+ ⇒
fs
fss' −= 
Chapter 32 
 
 
996 
 
Substitute numerical values and 
evaluate s′: 
( )( )
( )
cm22
cm 98.21
cm3.30cm0.80
cm0.80cm3.30
−=
−=−−
−=s'
 
The image is 22 cm from the lens and 
on the same side of the lens as the 
object. 
 
(c) The lateral magnification of 
the image is given by: s
s'm −= 
 
Substitute numerical values and 
evaluate m: 
0.27
cm0.80
cm98.21 =−−=m 
 
(d) upright. and virtual is image the0, and 0 Because >< ms' 
 
42 • The following thin lenses are made of glass that has an index of 
refraction equal to 1.60. Make a sketch of each lens and find each focal length in 
air: (a) r1 = 20.0 cm and r2 = 10.0 cm, (b) r1 = 10.0 cm and r2 = 20.0 cm, and 
(c) r1 = –10.0 cm and r2 = –20.0 cm. 
 
Picture the Problem We can use the lens-maker’s equation to find the focal 
length of each of the lenses described in the problem statement. 
 
The lens-maker’s equation is: 
 ⎟⎟⎠
⎞
⎜⎜⎝
⎛ −⎟⎟⎠
⎞
⎜⎜⎝
⎛ −=
21air
1111
rrn
n
f
 
where the numerals 1 and 2 denote the 
first and second surfaces, respectively. 
 
(a) For r1 = 20.0 cm and r2 = 10.0 cm: 
⎟⎟⎠
⎞
⎜⎜⎝
⎛ −⎟⎠
⎞⎜⎝
⎛ −=
cm0.10
1
cm0.20
11
00.1
60.11
f
 
and 
cm33−=f 
 
A sketch of the lens is shown to the 
right: 
 
Optical Images 
 
 
997
(b) For r1 = 10.0 cm and r2 = 20.0 cm: 
 ⎟⎟⎠
⎞
⎜⎜⎝
⎛ −⎟⎠
⎞⎜⎝
⎛ −=
cm0.20
1
cm0.10
11
00.1
60.11
f
 
and 
cm33=f 
 
A sketch of the lens is shown to the 
right: 
 
(c) For r1 = −10.0 cm andr2 = −20.0 cm: 
 
⎟⎟⎠
⎞
⎜⎜⎝
⎛
−−−⎟⎠
⎞⎜⎝
⎛ −=
cm0.20
1
cm0.10
11
00.1
60.11
f
and cm33−=f 
 
A sketch of the lens is shown to 
the right: 
 
Remarks: Note that the lenses that are thicker on their axis than on their 
circumferences are positive (converging) lenses and those that are thinner on 
their axis are negative (diverging) lenses. 
 
43 • The following four thin lenses are made of glass that has an index of 
refraction of 1.5. The radii given are magnitudes. Make a sketch of each lens and 
find each focal length in air: (a) double-convex that has radii of curvature equal to 
15 cm and 26 cm, (b) plano-convex that has a radius of curvature equal to 15 cm, 
(c) double concave that has radii of curvature equal to 15 cm, and (d) plano-
concave that has a radius of curvature equal to 26 cm. 
 
Picture the Problem We can use the lens-maker’s equation to find the focal 
length of each of the lenses. 
 
The lens-maker’s equation is: 
 ⎟⎟⎠
⎞
⎜⎜⎝
⎛ −⎟⎟⎠
⎞
⎜⎜⎝
⎛ −=
21air
1111
rrn
n
f
 
where the numerals 1 and 2 denote the 
first and second surfaces, respectively. 
 
Chapter 32 
 
 
998 
(a) Because the lens is double convex, 
the radius of curvature of the second 
surface is negative. Hence r1 = 15 cm 
and r2 = −26 cm: 
⎟⎟⎠
⎞
⎜⎜⎝
⎛
−−⎟⎠
⎞⎜⎝
⎛ −=
cm26
1
cm15
11
00.1
50.11
f
 
and cm19=f 
 
A double convex lens is shown to 
the right: 
 
 
(b) Because the lens is plano-
convex, the radius of curvature of 
the second surface is negative. 
Hence r1 = ∞ and r2 = −15 cm: 
 
⎟⎟⎠
⎞
⎜⎜⎝
⎛
−−∞⎟⎠
⎞⎜⎝
⎛ −=
cm15
111
00.1
50.11
f
 
and cm30=f 
 
A plano-convex lens is shown to 
the right: 
 
 
(c) Because the lens is double 
concave, the radius of curvature of its 
first surface is negative. Hence 
r1 = −15 cm and r2 = +15 cm: 
⎟⎟⎠
⎞
⎜⎜⎝
⎛ −−⎟⎠
⎞⎜⎝
⎛ −=
cm15
1
cm15
11
00.1
50.11
f
 
and cm15−=f 
 
A double concave lens is shown 
to the right: 
 
 
(d) Because the lens is plano-
concave, the radius of curvature of 
its second surface is positive. Hence 
r1 = ∞ and r2 = +26 cm: 
 
⎟⎟⎠
⎞
⎜⎜⎝
⎛ −∞⎟⎠
⎞⎜⎝
⎛ −=
cm26
111
00.1
50.11
f
 
and cm52−=f 
 
A plano-concave lens is shown to 
the right: 
 
 
Optical Images 
 
 
999
44 • Find the focal length of a glass lens that has an index of refraction 
equal to 1.62, a concave surface that has a radius of curvature of magnitude 
100 cm, and a convex surface that has a radius of curvature of magnitude 
40.0 cm. 
 
Picture the Problem We can use the lens-maker’s equation to find the focal 
length of the lens. 
 
The lens-maker’s equation is: 
 ⎟⎟⎠
⎞
⎜⎜⎝
⎛ −⎟⎟⎠
⎞
⎜⎜⎝
⎛ −=
21air
1111
rrn
n
f
 
where the numerals 1 and 2 denote the 
first and second surfaces, respectively. 
 
Substitute numerical values to 
obtain: ⎟⎟⎠
⎞
⎜⎜⎝
⎛
−−−⎟⎠
⎞⎜⎝
⎛ −=
cm0.40
1
cm100
11
00.1
62.11
f
 
 
Solving for f yields: m 1.1=f 
 
45 •• [SSM] (a) An object that is 3.00 cm high is placed 25.0 cm in front 
of a thin lens that has a power equal to 10.0 D. Draw a ray diagram to find the 
position and the size of the image and check your results using the thin-lens 
equation. (b) Repeat Part (a) if the object is placed 20.0 cm in front of the lens. 
(c) Repeat Part (a) for an object placed 20.0 cm in front of a thin lens that has a 
power equal to –10.0 D. 
 
Picture the Problem We can find the focal length of each lens from the 
definition of the power P, in diopters, of a lens (P = 1/f ). The thin-lens equation 
can be applied to find the image distance for each lens and the size of the image 
can be found from the magnification equation 
s
s'
y
y'm −== . 
 
(a) The parallel and central rays were used to locate the image in the diagram 
shown below. 
 F
 'F
>
>
 
The image is real, inverted, and diminished. 
 
Chapter 32 
 
 
1000 
Solving the thin-lens equation for s′ 
yields: 
 
fs's
111 =+ ⇒ (1) 
fs
fss' −= 
 
Use the definition of the power of 
the lens to find its focal length: 
 
cm 10.0 m100.0
m0.10
11
1 ==== −Pf 
Substitute numerical values and 
evaluate s′: 
 
( )( ) cm7.16
cm0.10cm0.25
cm0.25cm0.10 =−=s' 
Use the lateral magnification 
equation to relate the height of the 
image y′ to the height y of the object 
and the image and object distances: 
 
(2) y
s
s'y'
s
s'
y
y'm −=⇒−==
 
 
 
 
Substitute numerical values and 
evaluate y′: ( ) cm00.2cm00.3cm0.25
cm7.16 −=−=y' 
 
cm7.16=s' , cm00.2−=y' . Because s′ > 0, the image is real, and because 
 y′/y = −0.67 cm, the image is inverted and diminished. These results confirm 
those obtained graphically. 
 
(b) The parallel and central rays were used to locate the image in the following 
diagram. 
 F
 'F
>
>
 
The image is real and inverted and appears to be the same size as the object. 
 
Use the definition of the power of 
the lens to find its focal length: 
 
cm 10.0 m100.0
m0.10
1
1 === −f 
Substitute numerical values in 
equation (1) and evaluate s′: 
 
( )( ) cm0.20
cm0.10cm0.20
cm0.20cm0.10 =−=s' 
Substitute numerical values in 
equation (2) and evaluate y′: ( ) cm00.3cm00.3cm0.20
cm0.20 −=−=y' 
Optical Images 
 
 
1001
 
Because s′ > 0 and y′ = −3.00 cm, the image is real, inverted, and the same size as 
the object. These results confirm those obtained from the ray diagram. 
 
(c) The parallel and central rays were used to locate the image in the following 
diagram. 
 F
 'F
>
>
 
The image is virtual, upright, and diminished. 
 
Use the definition of the power of 
the lens to find its focal length: 
 
cm 0.01 m100.0
m0.10
1
1 −=−=−= −f 
Substitute numerical values in 
equation (1) and evaluate s′: 
 
( )( )
( ) cm67.6cm0.10cm0.20
cm0.20cm0.10 −=−−
−=s'
 
Substitute numerical values in 
equation (2) and evaluate y′: ( ) cm00.1cm00.3cm0.20
cm67.6 =−−=y' 
 
Because s′ < 0 and y′ = 1.00 cm, the image is virtual, erect, and about one-third 
the size of the object. These results are consistent with those obtained graphically. 
 
46 •• The lens-maker’s equation has three design parameters. They consist 
of the index of refraction of the lens and the radii of curvature for its two surfaces. 
Thus, there are many ways to design a lens that has a particular focal length in air. 
Use the lens-maker’s equation to design three different thin converging lenses, 
each having a focal length of 27.0 cm and each made from glass that has an index 
of refraction of 1.60. Sketch each of your designs. 
 
Picture the Problem We can use the lens-maker’s equation to obtain a 
relationship between the two radii of curvature of the lenses we are to design. 
 
Chapter 32 
 
 
1002 
For a thin lens of focal length 
27.0 cm and index of refraction 
of 1.60: 
 
( ) ⎟⎟⎠
⎞
⎜⎜⎝
⎛ −−=
21
11160.1
cm0.27
1
rr
 
or 
cm2.16
111
21
=−
rr
 
 
One solution is a plano-convex lens 
(one with a flat surface and a convex 
surface). Let r2 = ∞. Then 
cm2.161 =r and ∞=2r . 
 
 
 
Another design is a double convex 
lens (one with both surfaces convex 
and radii of curvature that are equal 
in magnitude) obtained by letting 
r2 = −r1. Then cm4.321 =r and 
cm4.322 −=r . 
 
 
 
A third possibility is a double 
convex lens with unequal curvature, 
e.g., let r2 = −12.0 cm. Then 
cm89.61 −=r and 
cm0.122 −=r . 
 
 
 
47 •• Repeat Problem 46, but for a diverging lens that has a focal length in 
air of the same magnitude. 
 
Picture the Problem We can use the lens-maker’s equation to obtain a 
relationshipbetween the two radii of curvature of the lenses we are to design. 
 
For a thin lens of focal length 
−27.0 cm and index of refraction of 
1.60: 
 
( ) ⎟⎟⎠
⎞
⎜⎜⎝
⎛ −−=− 21
11160.1
cm0.27
1
rr
 
or 
cm2.16
111
21 −
=−
rr
 
Optical Images 
 
 
1003
One solution is a plano-concave lens 
(one with a flat surface and a 
concave surface), Let r2 = ∞. Then 
cm2.161 −=r and ∞=2r . 
 
 
Another design is a biconcave 
lens (one with both surfaces 
concave) by letting r2 = −r1. Then 
cm4.321 −=r and 
cm4.322 =r . 
 
 
 
A third possibility is a lens with 
r2 = −8.10 cm. 
Then cm40.51 −=r 
and cm10.82 −=r . 
 
 
48 •• (a) What is meant by a negative object distance? Describe a specific 
situation in which a negative object distance can occur. (b) Find the image 
distance and the magnification for a thin lens in air when the object distance is 
–20 cm and the lens is a converging lens that has a focal length of 20 cm. 
Describe the image—is it virtual or real, upright or inverted? (c) Repeat Part (b) if 
the object distance is, instead, –10 cm, and the lens is diverging and has a focal 
length (magnitude) of 30 cm. 
 
Picture the Problem The parallel and central rays were used to locate the image 
in the diagram shown below. The power P of the lens, in diopters, can be found 
from P = 1/f and the lateral magnification of the image from 
s
s'm −= . 
 
(a) A negative object distance means that the object is virtual; i.e., that extensions 
of light rays, and not the light rays themselves, converge on the object rather than 
diverge from it. A virtual object can occur in a two-lens system when converging 
rays from the first lens are incident on the second lens before they converge to 
form an image. 
 
(b) The thin-lens equation is: 
 fs's
111 =+ ⇒
fs
fss' −= 
 
Chapter 32 
 
 
1004 
Substitute numerical values and 
evaluate s′: 
 
( )( )
( ) cm10cm20cm20
cm20cm20 =−−
−=s' 
The lateral magnification is given 
by: 
 
50.0
cm20
cm 10' =−−=−= s
sm
 
 
The parallel and central rays were used to locate the image in the following ray 
diagram: 
 
 F 'F
>
>
 
s′ = 10 cm, m = 0.50. Because s′ > 0, the image is real, and because m > 0, the 
image is erect. In addition, it is one-half the size of the virtual object. 
 
(c) Proceed as in Part (b) with 
s = −10 cm and f = −30 cm: 
 
( )( )
( ) cm15cm30cm10
cm10cm30 =−−−
−−=s' 
and 
5.1
cm01
cm15 =−−=−= s
s'm 
 
The parallel and central rays were used to locate the image in the following ray 
diagram: 
 F 'F
>
>
Object Image
 
s′ > 0 and m = 1.5. Because s′ > 0, the image is real, and because m > 0, the 
image is erect. In addition, it is one and one-half times the size of the virtual 
object. 
 
Optical Images 
 
 
1005
49 •• [SSM] Two converging lenses, each having a focal length equal to 
10 cm, are separated by 35 cm. An object is 20 cm to the left of the first lens. 
(a) Find the position of the final image using both a ray diagram and the thin-lens 
equation. (b) Is the final image real or virtual? Is the final image upright or 
inverted? (c) What is the overall lateral magnification of the final image? 
 
Picture the Problem We can apply the thin-lens equation to find the image 
formed in the first lens and then use this image as the object for the second lens. 
 
(a) The parallel, central, and focal rays were used to locate the image formed by 
the first lens and the parallel and central rays to locate the image formed by the 
second lens. 
 
Apply the thin-lens equation to 
express the location of the image 
formed by the first lens: 
 
11
11
1 fs
sf's −= (1) 
Substitute numerical values and 
evaluate 's1 : 
( )( ) cm20
cm10cm20
cm20cm10
1 =−='s 
 
Find the lateral magnification of 
the first image: 
0.1
cm20
cm201
1 −=−=−= s
'sm 
 
Because the lenses are separated by 
35 cm, the object distance for the 
second lens is 35 cm − 20 cm = 15 cm. 
Equation (1) applied to the second lens 
is: 
 
22
22
2 fs
sf's −= 
Substitute numerical values and 
evaluate 's2 : 
( )( ) cm30
cm10cm15
cm15cm10
2 =−='s 
and the final image is cm85 to the 
right of the object. 
 
Chapter 32 
 
 
1006 
Find the lateral magnification of 
the second image: 
0.2
cm51
cm302
2 −=−=−= s
'sm 
 
(b) Because 0 2 >s' , the image is real and because m = m1m2 = 2.0, the image is 
erect and twice the size of the object. 
 
(c) The overall lateral magnification 
of the image is the product of the 
magnifications of each image: 
( )( ) 0.20.20.121 =−−== mmm 
 
50 •• Repeat Problem 49 but with the second lens replaced by a diverging 
lens that has a focal length equal to −15 cm. 
 
Picture the Problem We can apply the thin-lens equation to find the image 
formed in the first lens and then use this image as the object for the second lens. 
 
(a) The parallel, central, and focal rays were used to locate the image formed by 
the first lens and the parallel and central rays to locate the image formed by the 
second lens. 
 
Apply the thin-lens equation to 
express the location of the image 
formed by the first lens: 
 
11
11
1 fs
sf's −= (1) 
Substitute numerical values and 
evaluate 's1 : 
( )( ) cm20
cm10cm20
cm20cm10
1 =−='s 
 
Find the lateral magnification of the 
first image: 
 
0.1
cm20
cm201
1 −=−=−= s
s'm 
Because the lenses are separated by 
35 cm, the object distance for the 
second lens is 35 cm − 20 cm = 15 cm. 
Equation (1) applied to the second 
lens is: 
 
22
22
2 fs
sf's −= 
Optical Images 
 
 
1007
 
Substitute numerical values and 
evaluate 's2 : 
( )( )
( ) cm5.7cm15cm15
cm15cm15
2 −=−−
−='s 
and the final image is cm48 to the 
right of the object. 
 
Find the lateral magnification of the 
second image: 
 
50.0
cm51
cm5.72
2 =−−=−= s
'sm 
(b) Because 0 2 <s' and m = m1m2 = −0.50, the image is virtual, inverted, and half 
the height of the object. 
 
(c) The overall lateral magnification 
of the image is the product of the 
magnifications of each image: 
( )( ) 50.050.00.121 −=−== mmm 
 
51 •• (a) Show that to obtain a magnification of magnitude |m| using a 
converging thin lens of focal length f, the object distance must be equal to ( )fm 11 −+ . (b) You want to use a digital camera which has a lens whose focal 
length is 50.0 mm to take a picture of a person 1.75 m tall. How far from the 
camera lens should you have that person stand so that the image size on the light 
receiving sensors of your camera is 24.0 mm? 
 
Picture the Problem We can use the thin-lens equation and the definition of the 
lateral magnification to derive the result. 
 
(a) Start with the thin-lens equation: 
fs's
111 =+ 
 
Express the magnitude of the lateral 
magnification of the image and solve 
for s′: (Note that this ensures that s′ 
is positive for this situation, as it 
must be because the image is real.) 
 
s
s'm −= ⇒ sms' = 
Substitute for s′ to obtain: 
fsms
111 =+ ⇒ ( )fms 11 −+= 
 
(b) The magnitude of the 
magnification m is: 0137.0m75.1
mm24.0 ==−=
y
y'm 
Chapter 32 
 
 
1008 
Substitute numerical values and 
evaluate s: 
( )( ) m70.3
0137.0
mm0.5010137.0 =+=s 
 
52 •• A converging lens has a focal length of 12.0 cm. (a) Using a 
spreadsheet program or graphing calculator, plot the image distance as a function 
of the object distance, for object distances ranging from 1.10f to 10.0f where f is 
the focal length. (b) On the same graph used in Part (a), but using a different y 
axis, plot themagnification of the lens as a function of the object distance. 
(c) What type of image is produced for this range of object distances—real or 
virtual, upright or inverted? (d) Discuss the significance of any asymptotic limits 
your graphs have. 
 
Picture the Problem We can plot the first graph by solving the thin-lens equation 
for the image distance s′ and the second graph by using the definition of the 
magnification of the image. 
 
(a) and (b) Solve the thin-lens 
equation for s′ to obtain: fs
fss' −= 
 
The magnification of the image is 
given by: s
s'm −= 
 
A spreadsheet program to calculate s′as a function of s is shown below. The 
formulas used to calculate the quantities in the columns are as follows: 
 
Cell Content/Formula Algebraic Form 
B1 12 f 
A4 13.2 s 
A5 A4 + 1 s + Δs 
B4 $B$1*A4/(A4 − $B$1) 
fs
fs
− 
C5 −B4/A4 
s
s'− 
 
 A B C 
1 f= 12 cm 
2 
3 s s' m 
4 13.2 132.00 −10.00 
5 14.2 77.45 −5.45 
6 15.2 57.00 −3.75 
7 16.2 46.29 −2.86 
8 17.2 39.69 −2.31 
9 18.2 35.23 −1.94 
Optical Images 
 
 
1009
 
108 117.2 13.37 −0.11 
109 118.2 13.36 −0.11 
110 119.2 13.34 −0.11 
111 120.2 13.33 −0.11 
 
A graph of s′ and m as functions of s follows. 
0
20
40
60
80
100
120
140
0 20 40 60 80 100 120
s (cm)
s '
 (c
m
)
-12.0
-10.0
-8.0
-6.0
-4.0
-2.0
0.0
m
s' 
m
 
(c) The images are real and inverted for this range of object distances. 
 
(d) The equations for the horizontal and vertical asymptotes of the graph of s′ 
versus s are s′ = f and s = f, respectively. These indicate that as the object moves 
away from the lens image distance the image moves toward the first focal point, 
and that as the object moves toward the second focal point the image distance 
becomes large without limit. The equations for the horizontal and vertical 
asymptotes of the graph of m versus s are m = 0 and s = f, respectively. These 
indicate that as the object moves away from the lens, the image size of the image 
approaches zero, and that as the object moves toward the second focal point the 
image becomes large without limit. 
 
53 •• A converging lens has a focal length of 12.0 cm. (a) Using a 
spreadsheet program or graphing calculator, plot the image distance as a function 
of the object distance, for object distances ranging from 0.010f to s = 0.90f, where 
f is the focal length. (b) On the same graph used in Part (a), but using a different y 
axis, plot the magnification of the lens as a function of the object distance. 
(c) What type of image is produced for this range of object distances—real or 
virtual, upright or inverted? (d) Discuss the significance of any asymptotic limits 
your graphs have. 
Chapter 32 
 
 
1010 
Picture the Problem We can plot the first graph by solving the thin-lens equation 
for the image distance s′ and the second graph by using the definition of the 
magnification of the image. 
 
(a) and (b) Solve the thin-lens 
equation for s′ to obtain: fs
fss' −= 
 
The magnification of the image is 
given by: s
s'm −= 
 
A spreadsheet program to calculate s′as a function of s is shown below. The 
formulas used to calculate the quantities in the columns are as follows: 
 
Cell Content/Formula Algebraic Form 
B1 12 f 
A4 0.12 s 
A5 A4 + 0.1 s + Δs 
B4 $B$1*A4/(A4 − $B$1) 
fs
fs
− 
C5 −B4/A4 
s
s'− 
 
 A B C 
1 f= 12 cm 
2 
3 s s' m 
4 0.12 −0.12 1.01 
5 0.22 −0.22 1.02 
6 0.32 −0.33 1.03 
7 0.42 −0.44 1.04 
8 0.52 −0.54 1.05 
9 0.62 −0.65 1.05 
 
108 10.52 −85.30 8.11 
109 10.62 −92.35 8.70 
110 10.72 −100.50 9.37 
111 10.82 −110.03 10.17 
 
Optical Images 
 
 
1011
 
A graph of s′ and m as functions of s follows. 
-110
-90
-70
-50
-30
-10
10
0 2 4 6 8 10 12
s (cm)
s'
 (c
m
)
0.0
2.0
4.0
6.0
8.0
10.0
12.0
m
s'
m
 
(c) The images are virtual and erect for this range of object distances. 
 
(d) The equation for the vertical asymptote of the graph of s′ versus s is f = s. This 
indicates that, as the object moves toward the second focal point, the magnitude 
of the image distance becomes large without limit. The equation for the vertical 
asymptote of the graph of m versus s is s = f. This indicates that, as the object 
moves toward the second focal point, the image becomes large without limit. In 
addition, as s approaches zero, s′ approaches negative infinity and m approaches 
1, so as the object moves toward the lens the image becomes the same size as the 
object and the magnitude of the image distance increases without limit 
 
54 •• An object is 15.0 cm in front of a converging lens that has a focal 
length equal to 15.0 cm. A second converging lens that also has a focal length 
equal to15.0 cm is located 20.0 cm in back of the first. (a) Find the location of the 
final image and describe its properties (for example, real and inverted) and 
(b) draw a ray diagram to corroborate your answers to Part (a). 
 
Picture the Problem We can apply the thin-lens equation to find the image 
formed in the first lens and then use this image as the object for the second lens. 
 
(a) Apply the thin-lens equation to 
express the location of the image 
formed by the first lens: 
 
11
11
1 fs
sf's −= (1) 
Chapter 32 
 
 
1012 
 
Substitute numerical values and 
evaluate 's1 : 
( )( ) ∞=−= cm15.0cm15.0
cm0.15cm0.15
1's 
i.e., no image is formed by the first 
lens. 
 
With 's1 = ∞, the thin-lens equation 
applied to the second lens becomes: 
 
22
11
f's
= ⇒ cm0.1522 == f's 
The overall magnification of the 
two-lens system is the magnification 
of the second lens: 
 
00.1
cm 15.0
cm 0.15
1
2
2 −=−=−== s
'smm 
The final image is 50 cm from the object, real, inverted, and the same size as the 
object. 
 
(b) A corroborating ray diagram is shown below: 
 
Lens 1 Lens 2
 '1F 1F 2F '2F
> >
> > 
 
55 •• [SSM] An object is 15.0 cm in front of a converging lens that has a 
focal length equal to 15.0 cm. A diverging lens that has a focal length whose 
magnitude is equal to 15.0 cm is located 20.0 cm in back of the first. (a) Find the 
location of the final image and describe its properties (for example, real and 
inverted) and (b) draw a ray diagram to corroborate your answers to Part (a). 
 
Picture the Problem We can apply the thin-lens equation to find the image 
formed in the first lens and then use this image as the object for the second lens. 
 
Apply the thin-lens equation to 
express the location of the image 
formed by the first lens: 
 
11
11
1 fs
sf's −= (1) 
Substitute numerical values and 
evaluate 's1 : 
( )( ) ∞=−= cm15.0cm15.0
cm0.15cm0.15
1's 
 
Optical Images 
 
 
1013
 
With 1s' = ∞, the thin-lens 
equation applied to the second 
lens becomes: 
 
22
11
f's
= ⇒ cm0.1522 −== f's 
The overall magnification of the 
two-lens system is the magnification 
of the second lens: 
 
00.1
cm 15.0
cm 0.15
1
2
2 =−−=−== s
'smm 
The final image is 20 cm from the object, virtual, erect, and the same size as the 
object. 
 
A corroborating ray diagram follows: 
 
Lens 1 Lens 2
 '1F 1F 2F '2F
> >
> >
 
 
56 ••• In a convenient form of the thin-lens equation used by Newton, the 
object and image distances x and x′ are measured from the focal points F and F’, 
and not from the center of the lens. (a) Indicate x and x′ on a sketch of a lens and 
show that if x = s – f and x′ = s′ – f, the thin-lens equation (Equation 32-12) can be 
rewritten as 2fxx' = . (b) Show that the lateral magnification is given by 
m = –x′/f = –f/x. 
 
Picture theProblem We can substitute x = s − f and x′ = s′ − f in the thin-lens 
equation and the equation for the lateral magnification of an image to obtain 
Newton’s equations. 
 
(a) The variables x, f, s ,and s′ are shown in the following sketch: 
 'F F 
>
>
x x'ff
s s'
Object Image
 
Chapter 32 
 
 
1014 
Express the thin-lens equation: 
fs's
111 =+ 
 
If x = s − f and x′ = s′ − f, then: 
ffx'fx
111 =+++ 
 
Expand this expression to obtain: ( ) ( )( )
2
2
fx'fxfxx'
fx'fxfxx'f
+++=
++=++
 
or, simplifying, 2fxx' = (1) 
 
(b) The lateral magnification is: 
 s
s'm −= 
or, because x = s − f and x′ = s′ − f, 
fx
fx'm +
+−= 
 
Solve equation (1) for x: 
x'
fx
2
= 
 
Substitute for x and simplify to 
obtain: ( ) f
x'
x'
x'ff
fx'
f
x'
f
fx'm −=+
+−=
+
+−= 2 
 
The lateral magnification is also 
given by: 
 
fx
fxm +
+−= ' 
 
From equation (1) we have: 
 x
fx
2
'= 
 
Substitute to obtain: 
 
x
f
x
fx
x
ff
fx
f
x
f
m −=
⎟⎠
⎞⎜⎝
⎛ +
⎟⎠
⎞⎜⎝
⎛ +
−=+
+
−=
1
1
2
 
 
 
 
 
 
 
 
 
Optical Images 
 
 
1015
57 ••• [SSM] In Bessel’s method for finding the focal length f of a lens, an 
object and a screen are separated by distance L, where L > 4f. It is then possible to 
place the lens at either of two locations, both between the object and the screen, 
so that there is an image of the object on the screen, in one case magnified and in 
the other case reduced. Show that if the distance between these two lens locations 
is D, then the focal length is given by ( ) LDLf 2241 −= . Hint: Refer to Figure 
32-60. 
 
Picture the Problem The ray diagram shows the two lens positions and the 
corresponding image and object distances (denoted by the numerals 1 and 2). We 
can use the thin-lens equation to relate the two sets of image and object distances 
to the focal length of the lens and then use the hint to express the relationships 
between these distances and the distances D and L to eliminate s1, s1′, s2, and s2′ 
and obtain an expression relating f, D, and L. 
 
Relate the image and object 
distances for the two lens positions 
to the focal length of the lens: 
 
f'ss
111
11
=+ and 
f'ss
111
22
=+ 
 
Solve for f to obtain: 
 'ss
'ss
'ss
'ssf
22
22
11
11
+=+= (1) 
 
The distances D and L can be 
expressed in terms of the image and 
object distances: 
 
'ss'ssL 2211 +=+= 
and 
's'sssD 2112 −=−= 
Substitute for the sums of the image 
and object distances in equation (1) 
to obtain: 
 
L
'ss
L
'ssf 2211 == 
 
From the hint: 
 
'ss 21 = and 21 s's = 
Hence L = s1 + s2 and: 12sDL =− and 22sDL =+ 
 
Take the product of L − D and 
L + D to obtain: 
 
( )( )
'ssss
DLDLDL
1121
22
44 ==
−=+−
 
From the thin-lens equation: 
 
fL'srsss 44 1121 == 
Substitute to obtain: 
224 DLfL −= ⇒
L
DLf
4
22 −= 
 
Chapter 32 
 
 
1016 
58 •• You are working for an optician during the summer. The optician 
needs to measure an unknown focal length and you suggest using Bessel’s method 
(see Problem 56) which you used during a physics lab. You set the object-to-
image distance at 1.70 m. The lens position is adjusted to get a sharp image on the 
screen. A second sharp image is found when the lens is moved a distance of 72 cm 
from its first location. (a) Sketch the ray diagram for the two locations. (b) Find 
the focal length of the lens using Bessel’s method. (c) What are the two locations 
of the lens with respect to the object? (d) What are the magnifications of the 
images when the lens is in the two locations? 
 
Picture the Problem We can use results obtained in Problem 57 to find the focal 
length of the lens and the two locations of the lens with respect to the object. Let 1 
refer to the first lens position and 2 to the second lens position. 
 
(a) The distances defined in Problem 57 have been identified in the following ray 
diagram. Note that, for clarity, the two images have been slightly offset from the 
plane of the screen. 
1st lens position 2nd lens position
L
f
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(b) From Problem 57 we have: 
 L
DLf
4
22 −= 
 
Substitute numerical values and 
evaluate f: 
( ) ( )
( ) cm35cm1704
cm72cm170 22 =−=f 
 
(c) Solve the thin-lens equation for 
the image distance to obtain: 
 
sf
fss −=' (1) 
Optical Images 
 
 
1017
In Problem 57 it was established 
that: 
 
12sDL =− and 22sDL =+ 
Solve for s1 and s2: 
21
DLs −= and 
22
DLs += 
 
Substitute numerical values and 
evaluate s1 and s2: 
 
cm49
2
cm72cm170
1 =−=s 
and 
cm121
2
cm72cm170
2 =+=s 
 
(d) The magnification when the lens 
was in its first position is given by: 1
1
1
1
1 s
sD
s
s'm −−=−= 
 
Substituting numerical values yields: 
 
5.2
cm49
cm 49cm 170
1 −=−−=m 
 
The magnification when the lens 
was in its second position is given 
by: 
 
2
2
2
2
2 s
sL
s
s'm −−=−= 
 
Substitute numerical values and 
evaluate m2: 
41.0
cm211
cm 121cm 170
2 −=−−=m 
 
59 ••• An object is 17.5 cm to the left of a lens that has a focal length of 
+8.50 cm. A second lens, which has a focal length of −30.0 cm, is 5.00 cm to the 
right of the first lens. (a) Find the distance between the object and the final image 
formed by the second lens. (b) What is the overall magnification? (c) Is the final 
image real or virtual? Is the final image upright or inverted? 
 
Picture the Problem The ray diagram shows four rays from the head of the 
object that locate images I1 and I2. We can use the thin-lens equation to find the 
location of the image formed in the positive lens and then, knowing the separation 
of the two lenses, determine the object distance for the second lens and apply the 
thin lens a second time to find the location of the final image. 
>
>
>
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 1F
 '
1F
 '
2F 2F
Lens 1 Lens 2
Image 1 Image 2
Chapter 32 
 
 
1018 
(a) Express the object-to-image 
distance d: 
 
'ssd 21 cm5 ++= (1) 
Apply the thin-lens equation to the 
positive lens: 111
111
f'ss
=+ ⇒
11
11
1 fs
sf's −= 
 
Substitute numerical values and 
evaluate 's1 : 
( )( ) cm53.16
cm 50.8cm5.17
cm5.17cm50.8
1 =−='s 
 
Find the object distance for the 
negative lens: 
 
cm11.53
cm16.53cm00.5cm00.5 12
−=
−=−= 'ss
 
The image distance s2′ is given 
by: 22
22
2 fs
sf's −= 
 
Substitute numerical values and 
evaluate s2′: 
( )( )
( ) cm73.18cm0.30cm53.11
cm53.11cm0.30
2 =−−−
−−='s
 
Substitute numerical values in 
equation (1) and evaluate d: cm41
cm73.18cm00.5cm5.17
=
++=d
 
and the final image is 18.7 cm to the 
right of the second lens. 
 
(b) The overall lateral magnification 
is given by: 
 
21mmm = 
Express m1 and m2: 
1
1
1 s
'sm −= and 
2
2
2 s
'sm −= 
 
Substitute for m1 and m2 to obtain: 
 
21
21
2
2
1
1
ss
''ss
s
's
s
'sm =⎟⎟⎠
⎞
⎜⎜⎝
⎛−⎟⎟⎠
⎞
⎜⎜⎝
⎛−= 
 
Substitute numerical values and 
evaluate m: 
 
( )( )
( )( ) 53.1cm53.11cm5.17
cm73.18cm53.16 −=−=m 
(c) Because m < 0, the image is inverted. Because 2's > 0, the image is real. 
 
 
 
 
Optical Images 
 
 
1019
Aberrations 
 
60 • Chromatic aberration is a common defect of (a) concave and convex 
lenses, (b) concave lenses only, (c) concave and convex mirrors, (d) all lenses and 
mirrors. 
 
Determine the Concept Chromatic aberrations are a consequence of the 
differential