Download Chapter 32 Optical Images

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.
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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 s < 12 R , the image is virtual, upright, and larger than the object.
(b) If s < 12 R , the image is virtual, upright, and larger than the object.
(c) If s > R , the image is real, inverted, and smaller than the object.
(d) If
1
2
R < s < R , 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
y'
angular size of the image is
, where y′ is the height of the image, s′ is the
s' + L
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
We wish to prove that, for a convex
mirror:
y'
y
<
s' + L s + L
If equation (1) is valid, then:
s' + L
>
y'
or
s'
y'
+
(1)
s+L
y
L s L
> +
y' y y
(2)
Because the lateral magnification is
y'
s'
m= =− :
y
s
y' y
=
s' s
(3)
Combining this with equation (2)
yields:
L L
> ⇒ y' < y
y' y
(4)
It follows that, if y′ < y, then:
s' < s
To show that s ' < s we begin with
1 1 1
+ =
s' s f
the mirror equation:
Solving for s′ yields:
s' = s
Because f is negative and s is
positive:
s' = s
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f
s− f
f
s+ f
⇒ s' < s
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.
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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.
.
.
Image
of the
bird
The bird
Air
Water
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.
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>
Optical Images
s = 40 cm
f = 10 cm
>
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 so large 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.
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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:
Apply the lens maker’s equation to
the water-glass interface:
Divide the first of these equations by
the second to obtain:
⎛1 1⎞
1
= (1.6 − 1)⎜⎜ − ⎟⎟
f air
⎝ r1 r2 ⎠
1
f water
f water
f air
⎛1 1⎞
= (1.6 − 1.33)⎜⎜ − ⎟⎟
⎝ r1 r2 ⎠
⎛1 1⎞
(1.6 − 1)⎜⎜ − ⎟⎟
⎝ r1 r2 ⎠ = 2.2
=
⎛1 1⎞
(1.6 − 1.33)⎜⎜ − ⎟⎟
⎝ r1 r2 ⎠
or f water = 2.2 f air and (a)
is correct.
True or false:
12
•
(a)
(b)
(c)
A virtual image cannot be displayed on a screen.
A negative image distance implies that the image is virtual.
All rays parallel to the axis of a spherical mirror are reflected through a
single point.
A diverging lens cannot form a real image from a real object.
The image distance for a converging lens is always positive.
(d)
(e)
(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.
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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:
1 1 1
+ =
s s' f
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Because f = r 2 , where r is the
radius of curvature of the mirror:
1 1 2
rs
+ = ⇒ s' =
s s' r
r − 2s
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:
s' =
(2.0 cm) (15 cm)
≈ −1.1 cm
2.0 cm − 2(15 cm)
or about 1.1 cm behind the spoon
y'
s'
⎛ s' ⎞
= − ⇒ y ' = −⎜ ⎟ y
y
s
⎝s⎠
The lateral magnification equation
is:
m=
Assume your face is 25 cm long.
Substitute numerical values and
evaluate y′:
⎛ − 1.1 cm ⎞
y ' = −⎜
⎟ (25 cm )
⎝ 15 cm ⎠
≈
2 cm
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 f = 12 REarth
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:
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:
M=
xnp
f
⎛1 1⎞
1
= (n − 1) ⎜⎜ − ⎟⎟
f
⎝ r1 r2 ⎠
Optical Images
For a plano-convex lens, r2 = ∞.
Hence:
r
1 n −1
=
⇒ f = 1
f
r1
n −1
Substitute in the expression for M
and simplify to obtain:
M=
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(n − 1)xnp
r1
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:
M max =
(1.5 − 1)(25 cm ) =
1.0 cm
13
Plane Mirrors
[SSM] The image of the object point P in Figure 32-57 is viewed by
19 •
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 and want 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.
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Chapter 32
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., 81 cm
(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 155 cm − 81 cm = 74 cm 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
P
Virtual 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
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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
P
Virtual image
Virtual image
(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):
P
Virtual image
Virtual image
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.
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Chapter 32
The mirror equation is:
1 1 1
+ =
s s' f
Because f = r 2 :
1 1 2
+ =
s s' r
For a plane mirror, r = ∞ . Hence:
1 1
+ = 0 ⇒ s' = − s
s s'
The lateral magnification of the
image is given by:
M =−
−s
s'
=−
= +1
s
s
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
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(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.
F
C
(b) The ray diagram is shown to the
right. The image is real, inverted,
and the same size as the object.
Object
C
F
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
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Chapter 32
(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
1 1 1
distance s′, and the focal length of a mirror are related according to + = ,
s s' f
1
where f = 2 r 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′:
s' =
fs
r
where f =
s− f
2
When s = 55 cm:
s' =
(12 cm )(55 cm ) = 15.35 cm
55 cm − 12 cm
= 15 cm
When s = 24 cm:
When s = 12 cm:
When s = 8.0 cm:
s' =
s' =
s' =
(12 cm )(24 cm ) =
24 cm − 12 cm
(12 cm )(12 cm )
12 cm − 12 cm
24 cm
is undefined
(12 cm )(8.0 cm ) = −24.0 cm
8.0 cm − 12 cm
= − 0.2 m
Optical Images
(b) The lateral magnification of the
image is:
m=−
s'
s
When s = 55 cm, the lateral
magnification of the image is:
m=−
s'
15.35 cm
=−
= − 0.28
s
55 cm
When s = 24 cm, the lateral
magnification of the image is:
m=−
s'
24 cm
=−
= − 1.0
s
24 cm
When s = 12 cm:
When s = 8.0 cm, the lateral
magnification of the image is:
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m is undefined.
m=−
− 24 cm
s'
=−
= 3.0
s
8.0 cm
Remarks: These results are in excellent agreement with those obtained
graphically 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.
F
C
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Chapter 32
(b) The ray diagram follows. The image is virtual, upright, and reduced.
F
C
(c) The ray diagram follows. The image is virtual, upright and reduced.
F
C
(d) The ray diagram follows. The image is virtual, upright, and reduced.
F
C
(e) The ray diagram follows. The image is virtual, upright, and reduced.
F
C
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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
1 1 1
distance s′, and the focal length of a mirror are related according to + = ,
s s' f
1
where f = 2 r 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′:
s' =
fs
s− f
When s = 55 cm:
s' =
(− 12 cm )(55 cm ) = −9.85 cm
55 cm − (− 12 cm )
= − 9.9 cm
When s = 24 cm:
s' =
(− 12 cm )(24 cm ) =
24 cm − (− 12 cm )
− 8.0 cm
When s = 12 cm:
s' =
(− 12 cm )(12 cm ) =
12 cm − (− 12 cm )
− 6.0 cm
When s = 8.0 cm:
s' =
(− 12 cm )(8.0 cm ) =
8.0 cm − (− 12 cm )
(b) The lateral magnification is given
by:
m=−
s'
s
When s = 55 cm, the lateral
magnification of the image is:
m=−
− 9.85 cm
= 0.18
55 cm
When s = 24 cm, the lateral
magnification of the image is:
m=−
− 8.0 cm
= 0.33
24 cm
When s = 12 cm, the lateral
magnification of the image is:
m=−
− 6.0 cm
= 0.50
12 cm
− 4.8 cm
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Chapter 32
When s = 8.0 cm, the lateral
magnification of the image is:
m=−
− 4.80 cm
= 0.60
8.0 cm
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′:
1 1 1 s− f
= − =
s' f s
sf
For a convex mirror:
f <0
For s > 0, the numerator is positive
and the denominator negative.
Consequently:
1
< 0 ⇒ s' < 0
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:
1 1 1 2
2ss'
+ = = ⇒r =
s s' f r
s '+ s
The lateral magnification of the
mirror is given by:
m=−
Substitute for s′ in equation (1) to
obtain:
r=
s'
⇒ s' = −ms
s
− 2ms
1− m
(1)
Optical Images
Substitute numerical values and
evaluate r:
r=
985
− 2(5.5)(2.1cm )
= 5.1cm
1 − 5.5
Convex mirrors are used in many stores to provide a wide angle of
30 ••
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′:
s' =
Substitute for f and simplify to
obtain:
s' =
rs
rs
=
1
s − 2 r 2s − r
Substitute numerical values and
evaluate s′:
s' =
(− 1.2 m )(10 m ) = −56.6 cm
2(10 m ) − (− 1.2 m )
fs
where f = 12 r
s− f
1
2
and
the image is 57 cm from the mirror.
(b) Because the image distance is
negative:
the image is behind the mirror.
(c) The lateral magnification of the
mirror is given by Equation 32-5:
m=
Substitute numerical values and
evaluate y′:
y' = −
y'
s'
s'
= − ⇒ y' = − y
y
s
s
− 0.566 m
(2.0 m ) = 11cm
10 m
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.
986
Chapter 32
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:
Because f = 12 r :
s' =
s' =
fs
s− f
1
2
rs
rs
=
1
s − 2 r 2s − r
(8.0 m )(3.8 ×108 m )
Substitute numerical values and
evaluate s′:
s' =
2(3.8 × 10 8 m ) − 8.0 m
Express the lateral magnification
of the mirror:
m=
s'
y'
s'
= − ⇒ y' = − y
s
y
s
Substitute numerical values and
evaluate y′:
y' = −
= 4.0 m
(
4.0 m
3.5 × 10 6 m
3.8 × 10 8 m
= −3.7 cm
)
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 mirror equation for f to
obtain:
f =
ss'
s' + s
Substitute numerical values and
evaluate f:
f =
(100 cm )(75 cm) = 42.86 cm
75 cm + 100 cm
Optical Images
Solving the mirror equation for s
yields:
s=
fs'
s' − f
With the piece turned around,
f = −42.86 cm and s′ = − 35 cm.
Substitute numerical values and
evaluate s:
s=
(− 42.86 cm )(− 35 cm ) ≈ 191cm
− 35 cm − (− 42.86 cm )
The distance d the mirror moved
is:
d = 191 cm − 100 cm = 91 cm
987
(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:
1 1 2
+ =
s1 s1' r1
Because s1 = ∞ :
1 2
= ⇒ s1' = 12 r1
s1' r1
The image formed by mirror #1
serves as a virtual object for mirror
2. Apply the mirror equation to
mirror 2 to obtain:
1
1
2
+
=
s 2 s 2' r2
or, because s 2 = d − s'1 = d − 12 r1 ,
1
1
2
+
=
1
d − 2 r1 s 2' r2
Because the image distance for
mirror 2 is d:
2
1 2
+ =
2d − r1 d r2
988
Chapter 32
Substituting numerical values yields:
2
1
2
+
=
2(2.0 m ) − 5.0 m 2.0 m r2
Finally, solve for r2 to obtain:
r2 = − 1.3 m
(b) Because f 2 = 12 r2 < 0 , 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:
n1 n2 n2 − n1
+
=
s s'
r
Solving for s′ yields:
s' =
(a) Substitute numerical values
(s = 30.0 cm, n1 = 1.00, n2 = 1.68
and r = 7.20 cm) and evaluate s′:
1.68
= 27 cm
1.68 − 1.00 1.00
−
7.20
30.0
where the positive distance tells us that
the image is 27 cm in back of the
surface and is real.
s' =
(1)
n2
n2 − n1 n1
−
r
s
Optical Images
989
In the following ray diagram, the object is at P and the image at P′.
C
P
Air
P'
Glass n =1.68
1.68
= − 16 cm
1.68 − 1.00 1.00
−
7.20
5.00
where the minus sign tells us that the
image is 16 cm in front of the surface
and is virtual.
(b) Substitute numerical values
(s = 5.00 cm, n1 = 1.00, n2 = 1.68
and r = 7.20 cm) and evaluate s′:
s' =
In the following ray diagram, the object is at P and the image at P′.
P'
C
P
Air
Glass 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.
Air
Water
Image of
Fish
Fish
r
990
Chapter 32
(a) Use the equation for refraction at
a single surface to relate the image
and object distances:
n1 n2 n2 − n1
+
=
s s'
r
Solving for s′ yields:
s' =
Substitute numerical values
(s = −10 cm, n1 = 1.33, n2 = 1.00
and r = 20 cm) and evaluate s′:
1.00
= 8.6 cm
1.00 − 1.33
1.33
−
− 10 cm
20 cm
and the image is 8.6 cm from the front
surface of the bowl.
(b) For s = 30 cm:
s' =
The change in the fish’s apparent
location is:
35.9 cm − 8.6 cm ≈ 27 cm
n2
n2 − n1 n1
−
r
s
s' =
1.00
= 35.9 cm
1.00 − 1.33
1.33
−
− 30 cm
20 cm
A very long 1.75-cm-diameter glass rod has one end ground and
36 ••
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:
n1 n2 n2 − n1
+
=
s s'
r
Solving for s′ yields:
s' =
n2
n2 − n1 n1
−
r
s
(1)
Optical Images
Substitute numerical values
(s = 15.0 cm, n1 = 1.00, n2 = 1.68,
and r = −7.20 cm) and evaluate s′:
991
1.68
= − 10
1.68 − 1.00
1.00
−
− 7.20 cm 15.0 cm
where the minus sign tells us that the
image is 10 cm in front of the surface
of the rod and is virtual.
s' =
In the following ray diagram, the object is at P and the virtual image is at P′.
P
P' C
Air
Glass 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:
n1 n2 n2 − n1
+
=
s s'
r
Solving for s′ yields:
s' =
(a) Substitute numerical values
(s = 6.00 cm, n1 = 1.33, n2 = 1.68,
and r = 7.20 cm) and evaluate s′:
1.68
= − 9.7 cm
1.68 − 1.33
1.33
−
7.20 cm
6.00 cm
where the negative distance tells us that
the image is 9.7 cm in front of the
surface and is virtual.
s' =
(1)
n2
n2 − n1 n1
−
r
s
Chapter 32
992
In the following ray diagram, the object is at P and the virtual image is at P′.
P'
P
C
Water
n1 = 1.33
(b) Substitute numerical values
(s = 12.0 cm, n1 = 1.33, n2 = 1.68,
and r = 7.20 cm) and evaluate s′:
Glass
n2 = 1.68
1.68
= − 27 cm
1.68 − 1.33
1.33
−
7.20 cm
12.0 cm
where the negative distance tells us that
the image is 27 cm in front of the
surface and is virtual.
s' =
In the following ray diagram, the object is at P and the virtual image is at P′.
P'
P
Water
n1 = 1.33
C
Glass
n2 = 1.68
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:
n1 n2 n2 − n1
+
=
s s'
r
Solving for s′ yields:
s' =
n2
n2 − n1 n1
−
r
s
(1)
Optical Images
993
1.68
= − 15
1.68 − 1.33 1.33
−
− 7.20 cm 20 cm
where the minus sign tells us that the
image is 15 cm in front of the surface
of the rod and is virtual
Substitute numerical values
(s = 20 cm, n1 = 1.33, n2 = 1.68,
and r = −7.20 cm) and evaluate s′:
s' =
In the following ray diagram the object is at P and the virtual image is at P′.
P
P'
C
Water
n1 = 1.33
Glass
n2 = 1.68
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:
n1 n2 n2 − n1
+
=
s s'
r
Solving for s′ yields:
s' =
n2
n2 − n1 n1
−
r
s
994
Chapter 32
Substitute numerical values
(s = 20.0 cm, n1 = 1.00, n2 = 1.60,
and r = 8.00 cm) and evaluate s′:
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′:
s' =
1.60
= 64 cm
1.60 − 1.00
1.00
−
8.00 cm
20.0 cm
1.00
= − 80 cm
1.00 − 1.60 1.60
−
− 16.0 cm 32 cm
(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:
n1 n2 n2 − n1
+
=
s s'
r
Solving for s′ yields:
s' =
Substitute numerical values
(s = 20.0 cm, n1 = 1.00,
n2 = 1.60 and r = 16.0 cm)
evaluate s′:
s' =
n2
n2 − n1 n1
−
r
s
1.60
1.60 − 1.00
1.00
−
16.0 cm
20.0 cm
= −128 cm = − 1.3 × 10 2 cm
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
(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′:
s' =
995
1.00
= 14.7 cm
1.00 − 1.60
1.60
−
− 8.00 cm 224 cm
= 15 cm
(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'
m = − to find the lateral magnification of the image.
s
(a) The lens-maker’s equation is:
⎞⎛ 1 1 ⎞
1 ⎛ n
= ⎜⎜
− 1⎟⎟ ⎜⎜ − ⎟⎟
f ⎝ nair
⎠ ⎝ r1 r2 ⎠
where the numerals 1 and 2 denote the
first and second surfaces, respectively.
Substitute numerical values to
obtain:
1 ⎛ 1.45 ⎞ ⎛
1
1 ⎞
⎟
=⎜
− 1⎟ ⎜⎜
−
f ⎝ 1.00 ⎠ ⎝ − 30.0 cm 25.0 cm ⎟⎠
Solving for f yields:
f = −30.3 cm = − 30 cm
(b) Use the thin-lens equation to
relate the image and object
distances:
1 1 1
fs
+ = ⇒ s' =
s s' f
s− f
996
Chapter 32
Substitute numerical values and
evaluate s′:
s' =
(− 30.3 cm )(80.0 cm ) = −21.98 cm
80.0 cm − (− 30.3 cm )
= −22 cm
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:
m=−
s'
s
Substitute numerical values and
evaluate m:
m=−
− 21.98 cm
= 0.27
80.0 cm
(d) Because s' < 0 and m > 0, the image is virtual and upright.
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:
(a) For r1 = 20.0 cm and r2 = 10.0 cm:
⎞⎛ 1 1 ⎞
1 ⎛ n
= ⎜⎜
− 1⎟⎟ ⎜⎜ − ⎟⎟
f ⎝ nair
⎠ ⎝ r1 r2 ⎠
where the numerals 1 and 2 denote the
first and second surfaces, respectively.
1 ⎛ 1.60 ⎞ ⎛ 1
1 ⎞
⎟
=⎜
− 1⎟ ⎜⎜
−
f ⎝ 1.00 ⎠ ⎝ 20.0 cm 10.0 cm ⎟⎠
and
f = − 33 cm
A sketch of the lens is shown to the
right:
Optical Images
(b) For r1 = 10.0 cm and r2 = 20.0 cm:
997
1 ⎞
1 ⎛ 1.60 ⎞ ⎛ 1
⎟
=⎜
− 1⎟ ⎜⎜
−
f ⎝ 1.00 ⎠ ⎝ 10.0 cm 20.0 cm ⎟⎠
and
f = 33 cm
A sketch of the lens is shown to the
right:
(c) For r1 = −10.0 cm and
r2 = −20.0 cm:
⎞
1 ⎛ 1.60 ⎞ ⎛
1
1
⎟
=⎜
− 1⎟ ⎜⎜
−
f ⎝ 1.00 ⎠ ⎝ − 10.0 cm − 20.0 cm ⎟⎠
and f = − 33 cm
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) planoconcave 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:
⎞⎛ 1 1 ⎞
1 ⎛ n
= ⎜⎜
− 1⎟⎟ ⎜⎜ − ⎟⎟
f ⎝ nair
⎠ ⎝ r1 r2 ⎠
where the numerals 1 and 2 denote the
first and second surfaces, respectively.
998
Chapter 32
(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:
1 ⎛ 1.50 ⎞ ⎛ 1
1 ⎞
⎟
=⎜
− 1⎟ ⎜⎜
−
f ⎝ 1.00 ⎠ ⎝ 15 cm − 26 cm ⎟⎠
and f = 19 cm
A double convex lens is shown to
the right:
(b) Because the lens is planoconvex, the radius of curvature of
the second surface is negative.
Hence r1 = ∞ and r2 = −15 cm:
1 ⎛ 1.50 ⎞ ⎛ 1
1 ⎞
⎟
=⎜
− 1⎟ ⎜⎜ −
f ⎝ 1.00 ⎠ ⎝ ∞ − 15 cm ⎟⎠
and f = 30 cm
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:
1 ⎛ 1.50 ⎞ ⎛ 1
1 ⎞
⎟
=⎜
− 1⎟ ⎜⎜
−
f ⎝ 1.00 ⎠ ⎝ − 15 cm 15 cm ⎟⎠
and f = − 15 cm
A double concave lens is shown
to the right:
(d) Because the lens is planoconcave, the radius of curvature of
its second surface is positive. Hence
r1 = ∞ and r2 = +26 cm:
A plano-concave lens is shown to
the right:
1 ⎛ 1.50 ⎞ ⎛ 1
1 ⎞
⎟
=⎜
− 1⎟ ⎜⎜ −
f ⎝ 1.00 ⎠ ⎝ ∞ 26 cm ⎟⎠
and f = − 52 cm
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:
⎞⎛ 1 1 ⎞
1 ⎛ n
= ⎜⎜
− 1⎟⎟ ⎜⎜ − ⎟⎟
f ⎝ nair
⎠ ⎝ r1 r2 ⎠
where the numerals 1 and 2 denote the
first and second surfaces, respectively.
Substitute numerical values to
obtain:
⎞
1 ⎛ 1.62 ⎞ ⎛
1
1
⎟
=⎜
− 1⎟ ⎜⎜
−
f ⎝ 1.00 ⎠ ⎝ − 100 cm − 40.0 cm ⎟⎠
Solving for f yields:
f = 1.1 m
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
y'
s'
can be found from the magnification equation m = = − .
y
s
>
(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.
1000 Chapter 32
Solving the thin-lens equation for s′
yields:
fs
1 1 1
+ = ⇒ s' =
s s' f
s− f
Use the definition of the power of
the lens to find its focal length:
f =
1
1
=
= 0.100 m = 10.0 cm
P 10.0 m −1
Substitute numerical values and
evaluate s′:
s' =
(10.0 cm )(25.0 cm ) = 16.7 cm
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:
m=
y'
s'
s'
= − ⇒ y' = − y
y
s
s
Substitute numerical values and
evaluate y′:
y' = −
(1)
25.0 cm − 10.0 cm
(2)
16.7 cm
(3.00 cm ) = −2.00 cm
25.0 cm
s' = 16.7 cm , y' = −2.00 cm . 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:
f =
1
= 0.100 m = 10.0 cm
10.0 m −1
Substitute numerical values in
equation (1) and evaluate s′:
s' =
(10.0 cm )(20.0 cm ) =
Substitute numerical values in
equation (2) and evaluate y′:
y' = −
20.0 cm − 10.0 cm
20.0 cm
20.0 cm
(3.00 cm ) = − 3.00 cm
20.0 cm
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:
f =
1
= −0.100 m = − 10.0 cm
− 10.0 m −1
Substitute numerical values in
equation (1) and evaluate s′:
s' =
(− 10.0 cm)(20.0 cm ) =
20.0 cm − (− 10.0 cm )
Substitute numerical values in
equation (2) and evaluate y′:
y' = −
− 6.67 cm
− 6.67 cm
(3.00 cm ) = 1.00 cm
20.0 cm
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.
1002 Chapter 32
For a thin lens of focal length
27.0 cm and index of refraction
of 1.60:
⎛1 1⎞
1
= (1.60 − 1) ⎜⎜ − ⎟⎟
27.0 cm
⎝ r1 r2 ⎠
or
1 1
1
− =
r1 r2 16.2 cm
One solution is a plano-convex lens
(one with a flat surface and a convex
surface). Let r2 = ∞. Then
r1 = 16.2 cm and r2 = ∞ .
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 r1 = 32.4 cm and
r2 = − 32.4 cm .
A third possibility is a double
convex lens with unequal curvature,
e.g., let r2 = −12.0 cm. Then
r1 = − 6.89 cm and
r2 = − 12.0 cm .
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
relationship between 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:
⎛1 1⎞
1
= (1.60 − 1) ⎜⎜ − ⎟⎟
− 27.0 cm
⎝ r1 r2 ⎠
or
1 1
1
− =
r1 r2 − 16.2 cm
Optical Images 1003
One solution is a plano-concave lens
(one with a flat surface and a
concave surface), Let r2 = ∞. Then
r1 = − 16.2 cm and r2 = ∞ .
Another design is a biconcave
lens (one with both surfaces
concave) by letting r2 = −r1. Then
r1 = − 32.4 cm and
r2 = 32.4 cm .
A third possibility is a lens with
r2 = −8.10 cm.
Then r1 = − 5.40 cm
and r2 = − 8.10 cm .
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
s'
from P = 1/f and the lateral magnification of the image from m = − .
s
(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:
1 1 1
fs
+ = ⇒ s' =
s s' f
s− f
1004 Chapter 32
(20 cm )(− 20 cm ) = 10 cm
− 20 cm − (20 cm )
Substitute numerical values and
evaluate s′:
s' =
The lateral magnification is given
by:
m=−
s'
10 cm
=−
= 0.50
s
− 20 cm
>
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:
s' =
(− 30 cm )(− 10 cm ) = 15 cm
− 10 cm − (− 30 cm )
and
m=−
15 cm
s'
=−
= 1.5
s
− 10 cm
The parallel and central rays were used to locate the image in the following ray
diagram:
>
Object
Image
F'
>
F
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:
s1' =
Substitute numerical values and
evaluate s1' :
s1' =
Find the lateral magnification of
the first image:
m1 = −
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:
s2' =
Substitute numerical values and
evaluate s2' :
s2' =
f1s1
s1 − f1
(1)
(10 cm)(20 cm ) = 20 cm
20 cm − 10 cm
s1'
20 cm
=−
= −1.0
s
20 cm
f 2 s2
s2 − f 2
(10 cm )(15 cm ) = 30 cm
15 cm − 10 cm
and the final image is 85 cm to the
right of the object.
1006 Chapter 32
Find the lateral magnification of
the second image:
m2 = −
s2'
30 cm
=−
= −2.0
s
15 cm
(b) Because s' 2 > 0 , 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:
m = m1m2 = (− 1.0 )(− 2.0 ) = 2.0
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:
s1' =
Substitute numerical values and
evaluate s1' :
s1' =
Find the lateral magnification of the
first image:
m1 = −
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:
s2' =
f1s1
s1 − f1
(10 cm)(20 cm ) = 20 cm
20 cm − 10 cm
s'1
20 cm
=−
= −1.0
s
20 cm
f 2 s2
s2 − f 2
(1)
Optical Images 1007
Substitute numerical values and
evaluate s2' :
s2' =
(− 15 cm )(15 cm ) = −7.5 cm
15 cm − (− 15 cm )
and the final image is 48 cm to the
right of the object.
Find the lateral magnification of the
second image:
m2 = −
− 7.5 cm
s2'
=−
= 0.50
s
15 cm
(b) Because s'2 < 0 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:
m = m1m2 = (− 1.0 )(0.50 ) = − 0.50
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
−1
1 + m f . (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:
1 1 1
+ =
s s'
f
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.)
m=−
Substitute for s′ to obtain:
1
1
1
−1
+
= ⇒ s = 1+ m f
s ms f
(b) The magnitude of the
magnification m is:
s'
⇒ s' = m s
s
(
m= −
)
y' 24.0 mm
=
= 0.0137
y
1.75 m
1008 Chapter 32
Substitute numerical values and
evaluate s:
s=
(0.0137 + 1)(50.0 mm) =
0.0137
3.70 m
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 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.
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:
s' =
The magnification of the image is
given by:
m=−
fs
s− f
s'
s
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
B1
A4
A5
B4
Content/Formula
12
13.2
A4 + 1
$B$1*A4/(A4 − $B$1)
C5
−B4/A4
A
1
2
3
4
5
6
7
8
9
B
f= 12
s
13.2
14.2
15.2
16.2
17.2
18.2
s'
132.00
77.45
57.00
46.29
39.69
35.23
Algebraic Form
f
s
s + Δs
fs
s− f
s'
−
s
C
cm
m
−10.00
−5.45
−3.75
−2.86
−2.31
−1.94
Optical Images 1009
108
109
110
111
117.2
118.2
119.2
120.2
13.37
13.36
13.34
13.33
−0.11
−0.11
−0.11
−0.11
A graph of s′ and m as functions of s follows.
140
0.0
120
-2.0
s'
100
-4.0
80
-6.0
m
s ' (cm)
m
60
-8.0
40
-10.0
20
0
0
20
40
60
80
100
-12.0
120
s (cm)
(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.
1010 Chapter 32
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:
s' =
The magnification of the image is
given by:
m=−
fs
s− f
s'
s
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
B1
A4
A5
B4
Content/Formula
12
0.12
A4 + 0.1
$B$1*A4/(A4 − $B$1)
C5
−B4/A4
A
B
f= 12
Algebraic Form
f
s
s + Δs
fs
s− f
s'
−
s
C
1
2
3
4
5
6
7
8
9
cm
s
0.12
0.22
0.32
0.42
0.52
0.62
s'
−0.12
−0.22
−0.33
−0.44
−0.54
−0.65
m
1.01
1.02
1.03
1.04
1.05
1.05
108
109
110
111
10.52
10.62
10.72
10.82
−85.30
−92.35
−100.50
−110.03
8.11
8.70
9.37
10.17
Optical Images 1011
10
12.0
-10
10.0
-30
8.0
s'
m
-50
6.0
-70
4.0
-90
2.0
-110
m
s ' (cm)
A graph of s′ and m as functions of s follows.
0.0
0
2
4
6
8
10
12
s (cm)
(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:
s1' =
f1s1
s1 − f1
(1)
1012 Chapter 32
Substitute numerical values and
evaluate s1' :
s1' =
(15.0 cm)(15.0 cm) = ∞
15.0 cm − 15.0 cm
i.e., no image is formed by the first
lens.
With s1' = ∞, the thin-lens equation
applied to the second lens becomes:
1
1
=
⇒ s 2' = f 2 = 15.0 cm
s2' f 2
The overall magnification of the
two-lens system is the magnification
of the second lens:
m = m2 = −
s2'
15.0 cm
=−
= − 1.00
s1
15.0 cm
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 2
>
>
Lens 1
F1 '
F2
F1
F2 '
>
>
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:
s1' =
Substitute numerical values and
evaluate s1' :
s1' =
f1s1
s1 − f1
(15.0 cm)(15.0 cm) = ∞
15.0 cm − 15.0 cm
(1)
Optical Images 1013
With s'1 = ∞, the thin-lens
equation applied to the second
lens becomes:
1
1
=
⇒ s2' = f 2 = − 15.0 cm
s2' f 2
The overall magnification of the
two-lens system is the magnification
of the second lens:
m = m2 = −
s2'
− 15.0 cm
=−
= 1.00
s1
15.0 cm
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
>
>
F1 '
F2
F1
>
>
F2 '
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 xx' = f 2 . (b) Show that the lateral magnification is given by
m = –x′/f = –f/x.
Picture the Problem 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:
s
f
x
Object
s'
x'
f
F'
F
Image
>
1014 Chapter 32
Express the thin-lens equation:
1 1 1
+ =
s s' f
If x = s − f and x′ = s′ − f, then:
1
1
1
+
=
x + f x' + f
f
Expand this expression to obtain:
f ( x' + x + 2 f ) = ( x + f )( x' + f )
= xx' + xf + x'f + f 2
or, simplifying, xx' = f
(b) The lateral magnification is:
Solve equation (1) for x:
s'
s
or, because x = s − f and x′ = s′ − f,
x' + f
m=−
x+ f
x=
f2
x'
m=−
The lateral magnification is also
given by:
m=−
Substitute to obtain:
(1)
m=−
Substitute for x and simplify to
obtain:
From equation (1) we have:
2
x' =
x' + f
x'
x' + f
= −
=−
2
(
)
f
f
x'
+
f
f
+f
x'
x'
x'+ f
x+ f
f2
x
f2
+ f
x
m=−
=−
x+ f
⎞
⎛f
f ⎜ + 1⎟
⎠= − f
⎝x
f⎞
x
⎛
x⎜ 1 + ⎟
x⎠
⎝
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 f = 14 L2 − D 2 L . 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:
1 1 1
1
1
1
+ =
and
+
=
s1 s1' f
s2 s2' f
Solve for f to obtain:
f =
The distances D and L can be
expressed in terms of the image and
object distances:
s1s1'
s s'
= 2 2
s1 + s1' s2 + s2'
L = s1 + s1' = s 2 + s 2'
and
D = s 2 − s1 = s1' − s 2'
Substitute for the sums of the image
and object distances in equation (1)
to obtain:
f =
From the hint:
s1 = s2' and s1' = s2
Hence L = s1 + s2 and:
L − D = 2 s1 and L + D = 2 s 2
Take the product of L − D and
L + D to obtain:
(L − D )(L + D ) = L2 − D 2
From the thin-lens equation:
4 s1 s 2 = rs1 s1' = 4 fL
Substitute to obtain:
L2 − D 2
4 fL = L − D ⇒ f =
4L
s1 s1' s 2 s 2'
=
L
L
= 4s1 s 2 = 4s1 s1'
2
2
(1)
1016 Chapter 32
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-toimage 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
Screen
f
>
>
D
L
(b) From Problem 57 we have:
f =
L2 − D 2
4L
2
2
(
170 cm ) − (72 cm )
=
4(170 cm )
Substitute numerical values and
evaluate f:
f
(c) Solve the thin-lens equation for
the image distance to obtain:
s' =
fs
f −s
= 35 cm
(1)
Optical Images 1017
In Problem 57 it was established
that:
L − D = 2 s1 and L + D = 2 s 2
Solve for s1 and s2:
s1 =
L−D
L+D
and s 2 =
2
2
Substitute numerical values and
evaluate s1 and s2:
s1 =
170 cm − 72 cm
= 49 cm
2
and
s2 =
170 cm + 72 cm
= 121cm
2
(d) The magnification when the lens
was in its first position is given by:
m1 = −
s'1
D − s1
=−
s1
s1
Substituting numerical values yields:
m1 = −
170 cm − 49 cm
= − 2.5
49 cm
The magnification when the lens
was in its second position is given
by:
m2 = −
s' 2
L − s2
=−
s2
s2
Substitute numerical values and
evaluate m2:
m2 = −
170 cm − 121 cm
= − 0.41
121 cm
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.
>
Lens 2
>
Lens 1
F1'
Image 1 Image 2
F2'
>
>
F2
F1
1018 Chapter 32
(a) Express the object-to-image
distance d:
d = s1 + 5 cm + s2'
Apply the thin-lens equation to the
positive lens:
1 1 1
fs
+ = ⇒ s1' = 1 1
s1 s1' f1
s1 − f1
Substitute numerical values and
evaluate s1' :
s1' =
Find the object distance for the
negative lens:
s2 = 5.00 cm − s1' = 5.00 cm − 16.53 cm
(1)
(8.50 cm)(17.5 cm) = 16.53 cm
17.5 cm − 8.50 cm
= −11.53 cm
The image distance s2′ is given
by:
s2' =
f 2 s2
s2 − f 2
Substitute numerical values and
evaluate s2′:
s2' =
(− 30.0 cm )(− 11.53 cm) = 18.73 cm
− 11.53 cm − (− 30.0 cm )
Substitute numerical values in
equation (1) and evaluate d:
d = 17.5 cm + 5.00 cm + 18.73 cm
= 41cm
and the final image is 18.7 cm to the
right of the second lens.
(b) The overall lateral magnification
is given by:
m = m1m2
Express m1 and m2:
m1 = −
Substitute for m1 and m2 to obtain:
⎛ s ' ⎞⎛ s ' ⎞ s 's '
m = ⎜⎜ − 1 ⎟⎟⎜⎜ − 2 ⎟⎟ = 1 2
⎝ s1 ⎠⎝ s2 ⎠ s1s2
Substitute numerical values and
evaluate m:
m=
s1'
s'
and m2 = − 2
s1
s2
(16.53 cm )(18.73 cm ) =
(17.5 cm )(− 11.53 cm )
− 1.53
(c) Because m < 0, the image is inverted. Because s2' > 0, the image is real.
Optical Images 1019
Aberrations
Chromatic aberration is a common defect of (a) concave and convex
60 •
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 refraction of light of differing wavelengths by lenses. (a ) is correct.
61 •
Discuss some of the reasons why most telescopes that are used by
astronomers are reflecting rather than refracting telescopes.
Determine the Concept Reasons for the preference for reflectors include: 1) no
chromatic aberrations, 2) cheaper to shape one side of a piece of glass than to
shape both sides, 3) reflectors can be more easily supported from rear instead of
edges, preventing sagging and focal length changes, and 4) support from rear
makes larger sizes easier to handle.
62 •
A symmetric double-convex lens has radii of curvature equal to
10.0 cm. It is made from glass that has an index of refraction equal to 1.530 for
blue light and to 1.470 for red light. Find the focal length of this lens for
(a) red light and (b) blue light.
Picture the Problem We can use the lens-maker’s equation to find the focal
length the this lens for the two colors of light.
The lens-maker’s equation relates
the radii of curvature and the index
of refraction to the focal length of
the lens:
(a) For red light:
⎞⎛ 1 1 ⎞
1 ⎛ n
= ⎜⎜
− 1⎟⎟ ⎜⎜ − ⎟⎟
f ⎝ nair
⎠ ⎝ r1 r2 ⎠
1
f red
⎞
1
⎛ 1.470 ⎞ ⎛ 1
⎟⎟
=⎜
− 1⎟ ⎜⎜
−
⎝ 1.00
⎠ ⎝ 10.0 cm − 10.0 cm ⎠
and f red = 10.6 cm
(b) For blue light:
1
f blue
⎞
1
⎛ 1.530 ⎞ ⎛ 1
⎟⎟
=⎜
− 1⎟ ⎜⎜
−
⎝ 1.00
⎠ ⎝ 10.0 cm − 10.0 cm ⎠
and f blue = 9.43 cm
1020 Chapter 32
The Eye
Find the change in the focal length of the eye when an object originally
63 •
at 3.0 m is brought to 30 cm from the eye.
Picture the Problem We can apply the thin-lens equation for the two values of s
to find Δf.
Express the change Δf in the focal
length:
Δf = f s =0.30 m − f s =3.0 m
Solve the thin-lens equation for s:
f =
Substitute for f to obtain:
ss'
s' + s
s 0.30 m s' 0.30 m
Δf =
s' 0.30 m + s 0.30 m
−
s3.0 m s' 3.0 m
s' 3.0 m + s3.0 m
Because s'3.0 m = s'0.30 m :
⎡
⎤
s 0.30 m
s3.0 m
−
Δf = s' 3.0 m ⎢
⎥
⎢⎣ s' 0.30 m + s 0.30 m s' 3.0 m + s3.0 m ⎥⎦
Substitute numerical values and evaluate Δf:
⎡
⎤
30 cm
300 cm
Δf = (2.5 cm ) ⎢
−
⎥ = −0.172 cm = − 1.7 mm
⎣ 2.5 cm + 30 cm 2.5 cm + 300 cm ⎦
64 •
A farsighted person requires lenses that have powers equal to 1.75 D
to read comfortably from a book that is 25.0 cm from his eyes. What is that
person’s near point without the lenses?
Picture the Problem We can use the thin-lens equation and the definition of the
power of a lens to express the near point distance as a function of P and s.
From the thin-lens equation we
have:
s
1 1 1
+ = = P ⇒ s' =
s s' f
Ps − 1
Substitute numerical values and
evaluate s′:
s' =
25.0 cm
= −44.4 cm
1.75 m (0.250 m ) − 1
(
−1
)
Optical Images 1021
The person’s near point without lenses is 44.4 cm .
65 •
[SSM] If two point objects close together are to be seen as two
distinct objects, the images must fall on the retina on two different cones that are
not adjacent. That is, there must be an unactivated cone between them. The
separation of the cones is about 1.00 μm. Model the eye as a uniform 2.50-cmdiameter sphere that has a refractive index of 1.34. (a) What is the smallest angle
the two points can subtend? (See Figure 32-61.) (b) How close together can two
points be if they are 20.0 m from the eye?
Picture the Problem We can use the relationship between a distance measured
along the arc of a circle and the angle subtended at its center to approximate the
smallest angle the two points can subtend and the separation of the two points
20.0 m from the eye.
(a) Relate θmin to the diameter of the
eye and the distance between the
activated cones:
d eyeθ min ≈ 2.00 μm ⇒ θ min =
2.00 μm
d eye
2.00 μm
= 80.0 μrad
2.50 cm
Substitute numerical values and
evaluate θmin:
θ min =
(b) Let D represent the separation of
the points R = 20.0 m from the eye
to obtain:
D = Rθ min = (20.0 m )(80.0 μrad )
= 1.60 mm
Suppose the eye were designed like a camera that has a lens of fixed
66 •
focal length equal to 2.50 cm that could move toward or away from the retina.
Approximately how far would the lens have to move to focus the image of an
object 25.0 cm from the eye onto the retina? Hint: Find the distance from the
retina to the image behind it for an object at 25.0 cm.
Picture the Problem We can use the thin-lens equation to find the distance from
the lens to the image and then take their difference to find the distance the lens
would have to be moved.
Express the distance d that the lens
would have to move:
d = s' − f
Solve the thin-lens equation for s′:
s' =
fs
s− f
1022 Chapter 32
Substitute for s′ to obtain:
d=
fs
−f
s− f
Substitute numerical values and
evaluate d:
d=
(2.50 cm )(25.0 cm ) − 2.50 cm
25.0 cm − 2.50 cm
= 0.28 cm
That is, the lens would have to move 0.28 cm toward the object.
67 ••
[SSM] The Model Eye I: A simple model for the eye is a lens that
has a variable power P located a fixed distance d in front of a screen, with the
space between the lens and the screen filled by air. This ″eye″ can focus for all
values of object distance s such that xnp ≤ s ≤ xfp where the subscripts on the
variables refer to ″near point″ and ″far point″ respectively. This ″eye″ is said to be
normal if it can focus on very distant objects. (a) Show that for a normal
″eye,″ of this type, the required minimum value of P is given by Pmin = 1 d .
(b) Show that the maximum value of P is given by Pmax = 1 x np + 1 d . (c) The
difference between the maximum and minimum powers, symbolized by A, is
defined as A = Pmax − Pmin and is called the accommodation. Find the minimum
power and accommodation for this model eye that has a screen distance of
2.50 cm, a far point distance of infinity and a near point distance of 25.0 cm.
Picture the Problem The thin-lens equation relates the image and object
distances to the power of a lens.
(a) Use the thin-lens equation to
relate the image and object distances
to the power of the lens:
Because s′ = d and, for a distance
object, s = ∞:
(b) If xnp is the closest distance an
object could be and still remain in
clear focus on the screen, equation
(1) becomes:
(c) Use our result in (a) to obtain:
1 1 1
+ = =P
s s' f
Pmin =
Pmax =
Pmin =
1
1
=
s'
d
1
1
+
xnp d
1
= 40.0 D
2.50 cm
(1)
Optical Images 1023
Use the results of (a) and (b) to
express the accommodation of the
model eye:
A = Pmax − Pmin =
Substitute numerical values and
evaluate A:
A=
1
1 1
1
+ − =
xnp d d xnp
1
= 4.00 D
25.0 cm
68 ••
The Model Eye II: In an eye that exhibits nearsightedness, the eye
cannot focus on distant objects. (a) Show that for a nearsighted model eye
capable of focusing out to a maximum distance xfp, the minimum value of the
power P is greater than that of a normal eye (that has a far point at infinity) and is
given by Pmin = 1 xfp + 1 d . (b) To correct for nearsightedness, a contact lens may
be placed directly in front of the model-eye’s lens. What power contact lens
would be needed to correct the vision of a nearsighted model eye that has a far
point distance of 50.0 cm?
Picture the Problem The thin-lens equation relates the image and object
distances to the power of a lens.
(a) Use the thin-lens equation to
relate the image and object distances
to the power of the lens:
Because s′ = d and s = xfp:
(b) To correct for the nearsightedness
of this eye, we need a lens that will
form an image 25.0 cm in front of the
eye of an object at the eye’s far point:
1 1 1
+ = =P
s s' f
Pmin =
Pmin =
1 1
+
xfp d
1
1
+
50.0 cm − 25.0 cm
= − 2.00 D
69 ••
The Model Eye III: In an eye that exhibits farsightedness, the eye may
be able to focus on distant objects but cannot focus on close objects. (a) Show
that for a farsighted model eye capable of focusing only as close as a distance
'
x np
+ 1 d . (b) Show
′ , the maximum value of the power P is given by Pmax = 1 x np
that, compared to a model eye capable of focusing as close as a distance xnp
(where xnp < x np
′ ), the maximum power of the farsighted lens is too small by
'
1 x np − 1 x np
. (c) What power contact lens would be needed to correct the vision
of a farsighted model eye, with x np
′ = 150 cm, so that the eye may focus on
objects as close as 15 cm?
1024 Chapter 32
Picture the Problem The thin-lens equation relates the image and object
distances to the power of a lens.
(a) Use the thin-lens equation to
relate the image and object distances
to the power of the lens:
Because s′ = d and s = x′np:
(b) For a normal eye:
The amount by which the power of
the lens is too small is the difference
between equations (2) and (1):
1 1 1
+ = =P
s s' f
P'max =
1
1
+
x'np d
1 1
+
xnp d
(2)
Pmax − P' max =
1
1 ⎛ 1
1⎞
+ −⎜
+ ⎟
xnp d ⎜⎝ x' np d ⎟⎠
Pmax =
=
(c) For xnp = 15 cm and
x′np = 150 cm:
(1)
Pmax − P'max =
1
1
−
xnp x' np
1
1
−
15 cm 150 cm
= 6.0 D
70 ••
A person who has a near point of 80 cm needs to read from a computer
screen that is only 45 cm from her eye. (a) Find the focal length of the lenses in
reading glasses that will produce an image of the screen at a distance of 80 cm
from her eye. (b) What is the power of the lenses?
Picture the Problem We can use the thin-lens equation to find f and the
definition of the power of a lens to find P.
(a) Solve the thin-lens equation
for f:
f =
ss'
s' + s
Noting that s′ < 0, substitute
numerical values and evaluate f:
f =
(45 cm)(− 80 cm) = 103 cm
− 80 cm + 45 cm
= 1.0 m
(b) Use the definition of the power
of a lens to obtain:
P=
1
1
=
= 0.97 D
f 1.03 m
Optical Images 1025
71 •• A nearsighted person cannot focus clearly on objects that are more
distant than 2.25 m from her eye. What power lenses are required for her to see
distant objects clearly?
Picture the Problem We can use the thin-lens equation to find f and the
definition of the power of a lens to find P.
1
f
Express the required power of the
lens:
P=
The thin-lens equation is:
1 1 1
+ =
s s' f
For s = ∞:
1 1
=
⇒ f = s'
s' f
Substitute for f to obtain:
P=
1
s'
Substitute for s′ and evaluate P:
P=
1
= 0.444 D
2.25 m
Because the index of refraction of the lens of the eye is not very
72 ••
different from that of the surrounding material, most of the refraction takes place
at the cornea, where the index changes abruptly from 1.00 (air) to approximately
1.38. (a) Modeling the cornea, aqueous humor, lens and vitreous humor is a
transparent homogeneous solid sphere that has an index of refraction of 1.38,
calculate the sphere’s radius if it focuses parallel light on the retina a distance
2.50 cm away. (b) Do you expect your result to be larger or smaller than the
actual radius of the cornea? Explain your answer.
Picture the Problem We can use the equation for refraction at a single surface
(Equation 32-6) with s = ∞ to find the radius of the cornea modeled as a
homogeneous sphere with an index of refraction of 1.38.
(a) Use Equation 32-6 to relate the
radius of the cornea to its index of
refraction and that of air:
n1 n2 n2 − n1
+ =
s s'
r
Because n2 = n, n1 = 1, and s = ∞:
n n −1
s' (n − 1) ⎛ 1 ⎞
=
⇒r =
= ⎜1 − ⎟ s'
n
s'
r
⎝ n⎠
1026 Chapter 32
Substitute numerical values and
evaluate r:
1 ⎞
⎛
r = ⎜1 −
⎟ (2.50 cm ) = 0.688 cm
⎝ 1.38 ⎠
(b) The calculated value for the distance to the retina is too large. The eye is not a
homogeneous sphere, and in a real eye additional refraction occurs at the lens. By
modeling the eye as a homogeneous solid sphere we are ignoring the refraction
that takes place at the lens.
73 •• The near point of a certain person’s eyes is 80 cm. Reading glasses are
prescribed so that he can read a book at 25 cm from his eye. The glasses are
2.0 cm from the eye. What diopter lenses should be used in the glasses?
Picture the Problem We can use the definition of the power of a lens and the
thin-lens equation to find the power of the lens that should be used in the glasses.
1
1
+
f eye f glasses
Express the power of the lens that
should be used in the glasses:
P = Peye + Plens =
Because the glasses are 2.0 cm from
the eye:
s' = −80 cm + 2.0 cm = −78 cm
and
s = 25 cm − 2.0 cm = 23 cm
Apply the thin-lens equation to the
eye with s′ = ∞:
1
1
⇒ f eye = s
=
s f eye
Apply the thin-lens equation to the
glasses with s = ∞:
1
1
⇒ f glasses = s'
=
s' f glasses
Substitute for f eye and f glasses in
P=
1 1
+
s s'
P=
1
1
+
= 3.1 D
0.23 m − 0.78 m
equation (1) to obtain:
Substitute numerical values and
evaluate P:
(1)
74 ••• At age 45, a person is fitted for reading glasses that have a power
equal to 2.10 D in order to read at 25 cm. By the time she reaches the age of 55,
she discovers herself holding her newspaper at a distance of 40 cm in order to see
it clearly with her glasses on. (a) Where was her near point at age 45? (b) Where
is her near point at age 55? (c) What power is now required for the lenses of her
reading glasses so that she can again read at 25 cm? Assume the glasses are
placed 2.2 cm from her eyes.
Optical Images 1027
Picture the Problem We can use the thin-lens equation and the distance from her
eyes to her glasses to derive an expression for the location of her near point.
(a) Express her near point, xnp, at
age 45 in terms of the location of her
glasses:
xnp = s' + 2.2 cm
Because the glasses are 2.2 cm from
her eye:
s = 25 cm − 2.2 cm = 22.8 cm
Apply the thin-lens equation to the
glasses:
1 1
1
+ =
=P
s s' f glasses
Solving for s′ yields:
s' =
(1)
s
1
=
Ps − 1 P − 1
s
Substitute in equation (1) to obtain:
xnp =
Substitute numerical values and
evaluate xnp:
x np =
1
1
P−
s
+ 2.2 cm
1
1
2.10 m −
0.228 m
(2)
+ 2.2 cm
−1
= 46 cm
(b) At age 55:
Substitute numerical values in
equation (2) and evaluate s′:
s = 40 cm − 2.2 cm = 37.8 cm
x np =
1
1
2.10 m −
0.378 m
= 1.9 m
(c) Solve the thin-lens equation for f:
f =
ss'
s' + s
−1
+ 2.2 cm
1028 Chapter 32
From the definition of P:
P=
1 s' + s
=
f
s's
For s = 22.8 cm and s′ = 183.3 cm:
P=
183.3 cm + 22.8 cm
= 4.9 D
(183.3 cm)(22.8 cm)
The Simple Magnifier
75 •
[SSM] What is the magnifying power of a lens that has a focal
length equal to 7.0 cm when the image is viewed at infinity by a person whose
near point is at 35 cm?
Picture the Problem We can use the definition of the magnifying power of a lens
to find the magnifying power of this lens.
The magnifying power of the lens
is given by:
M =
Substitute numerical values and
evaluate M:
M=
xnp
f
35 cm
= 5.0
7.0 cm
76 ••
A lens that has a focal length equal to 6.0 cm is used as a simple
magnifier by one person whose near point is 25 cm and by another person whose
near point is 40 cm. (a) What is the effective magnifying power of the lens for
each person? (b) Compare the sizes of the images on the retinas when each person
looks at the same object with the magnifier.
Picture the Problem Let the numerals 1 and 2 denote the 1st and 2nd persons,
respectively. We can use the definition of magnifying power to find the effective
magnifying power of the lens for each person. The relative sizes of the images on
the retinas of the two persons is given by the ratio of the effective magnifying
powers.
(a) The magnifying power of the lens
is given by:
M =
Substitute numerical values and
evaluate M1 and M2:
M1 =
xnp
f
25 cm
= 4.17 = 4.2
6.0 cm
and
M2 =
40 cm
= 6.67 = 6.7
6.0 cm
Optical Images 1029
If xnp = 25 cm, then M = 4.2, and if xnp = 40 cm, then M = 6.7.
(b) From the definition of
magnifying power we have:
y1
y
M1
f
=
= 1
M 2 y2 y2
f
Substitute for M1 and M2 and
evaluate the ratio of y1 to y2:
y1 4.17
=
= 0.63
y2 6.67
For a person with xnp = 40 cm, the image on the retina is 1.6 times larger than it is
for a person with xnp = 25 cm.
77 ••
[SSM] In your botany class, you examine a leaf using a convex
12-D lens as a simple magnifier. What is the angular magnification of the leaf if
image formed by the lens (a) is at infinity and (b) is at 25 cm?
Picture the Problem We can use the definition of angular magnification to find
the expected angular magnification if the final image is at infinity and the thinlens equation and the expression for the magnification of a thin lens to find the
angular magnification when the final image is at 25 cm.
(a) Express the angular
magnification when the final
image is at infinity:
M =
xnp
= xnp P
f
where P is the power of the lens.
(
)
Substitute numerical values and
evaluate M:
M = (25 cm ) 12 m −1 = 3.0
(b) Express the magnification of
the lens when the final image is
at 25 cm:
m=−
Solve the thin-lens equation for
s:
s=
Substitute for s and simplify to
obtain:
m=−
s'
s
fs'
s' − f
s' − f
s'
s'
=−
= − +1
fs'
f
f
s' − f
= 1 − s'P
1030 Chapter 32
Substitute numerical values and
evaluate m:
(
)
m = 1 − (− 0.25 m ) 12 m −1 = 4.0
The Microscope
78 ••
Your laboratory microscope objective has a focal length of 17.0 mm. It
forms an image of a tiny specimen at 16.0 cm from its second focal point.
(a) How far from the objective is the specimen located? (b) What is the
magnifying power for you if your near point distance is 25.0 cm and the focal
length of the eyepiece is 51.0 mm?
Picture the Problem We can use the thin-lens equation to find the location of the
object and the expression for the magnifying power of a microscope to find the
magnifying power of the given microscope for a person whose near point is at
25.0 cm.
(a) Using the thin-lens equation,
relate the object distance s to the
focal length of the objective lens f0:
f s'
1 1
1
+ = ⇒s = 0
s s' f 0
s' − f 0
The image distance for the image
formed by the objective lens is:
s' = f 0 + L = 1.70 cm + 16.0 cm = 17.7 cm
Substitute numerical values and
evaluate s:
s=
(b) Express the magnifying power
of a microscope:
M =−
Substitute numerical values and
evaluate M:
⎛ 16.0 cm ⎞ ⎛ 25.0 cm ⎞
⎟⎟ ⎜⎜
⎟⎟ = − 46.1
M = −⎜⎜
1.70
cm
5
.
10
cm
⎝
⎠⎝
⎠
(1.70 cm )(17.7 cm ) =
17.7 cm − 1.70 cm
18.8 mm
L xnp
f0 fe
79 ••
[SSM] Your laboratory microscope objective has a focal length of
17.0 mm. It forms an image of a tiny specimen at 16.0 cm from its second focal
point. (a) How far from the objective is the specimen located? (b) What is the
magnifying power for you if your near point distance is 25.0 cm and the focal
length of the eyepiece is 51.0 mm?
Picture the Problem The lateral magnification of the objective is
mo = − L f o and the magnifying power of the microscope is M = mo M e .
Optical Images 1031
(a) The lateral magnification of the
objective is given by:
mo = −
Substitute numerical values and
evaluate mo:
mo = −
(b) The magnifying power of the
microscope is given by:
M = mo M e
Substitute numerical values and
evaluate M:
L
fo
16 cm
= −18.8 = − 19
8.5 mm
where Me is the angular magnification
of the lens.
M = (− 18.8)(10) = − 1.9 × 10 2
80 ••
A crude, symmetric handheld microscope consists of two 20-D lenses
fastened at the ends of a tube 30 cm long. (a) What is the tube length of this
microscope? (b) What is the lateral magnification of the objective? (c) What is
the magnifying power of the microscope? (d) How far from the objective should
the object be placed?
Picture the Problem We can find the tube length from the length of the tube to
which the lenses are fastened and the focal lengths of the objective and eyepiece.
We can use their definitions to find the lateral magnification of the objective and
the magnifying power of the microscope. The distance of the object from the
objective can be found using the thin-lens equation.
(a) The tube length L is given by:
L = D − fo − fe
= 0.30 m −
2
= 20 cm
20 m −1
L
20 cm
=−
= − 4.0
fo
5.0 cm
(b) The lateral magnification of the
objective mo is given by:
mo = −
(c) The magnifying power of the
microscope is given by:
M = mo M e = mo
Substitute numerical values and
evaluate M:
M = (− 4 )
(d) From the thin-lens equation we
have:
1 1
1
+
=
so so' f o
where so' = f o + L
xnp
fe
25 cm
= − 20
5.0 cm
1032 Chapter 32
Substitute for so' to obtain:
f ( f + L)
1
1
1
+
=
⇒ so = o o
so f o + L f o
L
Substitute numerical values and
evaluate so:
so =
(5.0 cm )(5.0 cm + 20 cm )
20 cm
= 6.3 cm
81 ••
A compound microscope has an objective lens that has a power of
45 D and an eyepiece that has a power of 80 D. The lenses are separated by
28 cm. Assuming that the final image is formed by the microscopy is 25 cm from
the eye, what is the magnifying power?
Picture the Problem The magnifying power of a compound microscope is the
product of the magnifying powers of the objective and the eyepiece.
Express the magnifying power of the
microscope in terms of the
magnifying powers of the objective
and eyepiece:
The magnification of the eyepiece is
given by:
M = mo me
me =
xnp
fe
(1)
+ 1 = Pe xnp + 1
The magnification of the objective is
given by:
L
fo
where L = D − f o − f e
Substitute for L to obtain:
mo = −
Substitute for me and mo in equation
(1) to obtain:
⎛ D − fo − fe ⎞
⎟⎟
M = (Pe xnp + 1)⎜⎜ −
fo
⎝
⎠
mo = −
D − fo − fe
fo
Substitute numerical values and evaluate M:
⎛ 28 cm − 2.22 cm − 1.25 cm ⎞
⎟⎟ = − 230
M = [(80 D )(0.25 m ) + 1] ⎜⎜ −
2.22 cm
⎝
⎠
Optical Images 1033
82 ••• A microscope has a magnifying power of 600. The eyepiece has an
angular magnification of 15.0. The objective lens is 22.0 cm from the eyepiece.
Calculate (a) the focal length of the eyepiece, (b) the location of the object so that
it is in focus for a normal relaxed eye, and (c) the focal length of the objective
lens.
Picture the Problem We can find the focal length of the eyepiece from its
angular magnification and the near point of a normal eye. The location of the
object such that it is in focus for a normal relaxed eye can be found from the
lateral magnification of the eyepiece and the magnifying power of the
microscope. Finally, we can use the thin-lens equation to find the focal length of
the objective lens.
(a) The focal length of the eyepiece
is related to its angular magnifying
power:
Me =
xnp
fe
⇒ fe =
xnp
Me
25.0 cm
= 1.67 cm
15.0
Substitute numerical values and
evaluate fe:
fe =
(b) Relate s to s′ through the lateral
magnification of the objective:
mo = −
Relate the magnifying power of the
microscope M to the lateral
magnification of its objective m0 and
the angular magnification of its
eyepiece Me:
M = mo M e ⇒ mo =
Substitute for mo to obtain:
s=−
Evaluate s′:
s' = 22 cm − f e
s'
s'
⇒ s=−
s
mo
M
Me
s'M e
M
= 22 cm − 1.67 cm = 20.33 cm
Substitute numerical values and
evaluate s:
s=−
(20.33 cm )(15.0) =
(c) Solving the thin-lens equation
for fo yields:
fo =
ss'
s' + s
− 600
0.508 cm
1034 Chapter 32
Substitute numerical values and
evaluate fo:
fo =
(0.508 cm)(20.33 cm)
20.33 cm + 0.508 cm
= 0.496 cm
The Telescope
83 •
[SSM] You have a simple telescope that has an objective which has
a focal length of 100 cm and an eyepiece which has a focal length of 5.00 cm.
You are using it to look at the moon, which subtends an angle of about 9.00 mrad.
(a) What is the diameter of the image formed by the objective? (b) What angle is
subtended by the image formed at infinity by the eyepiece? (c) What is the
magnifying power of your telescope?
Picture the Problem Because of the great distance to the moon, its image formed
by the objective lens is at the focal point of the objective lens and we can use
D = f oθ to find the diameter D of the image of the moon. Because angle
subtended by the final image at infinity is given by θ e = Mθ o = Mθ, we can solve
(b) and (c) together by first using M = −fo/fe to find the magnifying power of the
telescope.
(a) Relate the diameter D of the
image of the moon to the image
distance and the angle subtended by
the moon:
D = so'θ
Because the image of the moon is at
the focal point of the objective lens:
so' = f o and D = f oθ
Substitute numerical values and
evaluate D:
D = (100 cm )(9.00 mrad ) = 9.00 mm
(b) and (c) Relate the angle
subtended by the final image at
infinity to the magnification of the
telescope and the angle subtended at
the objective:
θ e = Mθ o = Mθ
Express the magnifying power of
the telescope:
M =−
fo
fe
Optical Images 1035
Substitute numerical values and
evaluate M and θe:
M =−
100 cm
= − 20.0
5.00 cm
and
θ e = − 20.0 (9.00 mrad ) = 180 mrad
The objective lens of the refracting telescope at the Yerkes
84 •
Observatory has a focal length of 19.5 m. The moon subtends an angle of about
9.00 mrad. When the telescope is used to look at the moon, what is the diameter
of the image of the moon formed by the objective?
Picture the Problem Because of the great distance to the moon, its image formed
by the objective lens is at the focal point of the objective lens and we can use
D = f oθ to find the diameter D of the image of the moon.
Relate the diameter D of the image
of the moon to the image distance
and the angle subtended by the
moon:
D = so'θ
Because the image of the moon is at
the focal point of the objective lens:
so' = f o and D = f oθ
Substitute numerical values and
evaluate D:
D = (19.5 m )(9.00 mrad ) = 17.6 cm
The 200-in (5.08-m) diameter mirror of the reflecting telescope at Mt.
85 ••
Palomar has a focal length of 16.8 m. (a) By what factor is the light-gathering
power increased over the 40.0-in (1.02-m) diameter refracting lens of the Yerkes
Observatory telescope? (b) If the focal length of the eyepiece is 1.25 cm, what is
the magnifying power of the 200-in telescope?
Picture the Problem Because the light-gathering power of a mirror is
proportional to its area, we can compare the light-gathering powers of these
mirrors by finding the ratio of their areas. We can use the ratio of the focal lengths
of the objective and eyepiece lenses to find the magnifying power of the Palomar
telescope.
1036 Chapter 32
(a) Express the ratio of the lightgathering powers of the Palomar
and Yerkes mirrors:
π
2
d Palomar
PPalomar
4 mirror
mirror
=
=
π 2
PYerkes
AYerkes
d Yerkes
mirror
4 mirror
APalomar
=
2
d Palomar
mirror
2
Yerkes
mirror
d
Substitute numerical values and
evaluate PPalomar/PYerkes:
PPalomar (200 in )2
=
= 25.0
PYerkes (40.0 in )2
(b) Express the magnifying power of
the Palomar telescope:
M =−
fo
fe
Substitute numerical values and
evaluate M:
M =−
16.8 m
= − 1340
1.25 cm
86 ••
An astronomical telescope has a magnifying power of 7.0. The two
lenses are 32 cm apart. Find the focal length of each lens.
Picture the Problem We can use the expression for the magnifying power of a
telescope and the fact that the length of a telescope is the sum of focal lengths of
its objective and eyepiece lenses to obtain simultaneous equations in fo and fe.
The magnifying power of the
telescope is given by:
M =−
The length of the telescope is the
sum of the focal lengths of the
objective and eyepiece lenses:
L = f o + f e = 32 cm
Solve these equations simultaneously
to obtain:
f o = 28 cm and f e = 4 cm
fo
= −7
fe
87 ••
[SSM]
A disadvantage of the astronomical telescope for terrestrial
use (for example, at a football game) is that the image is inverted. A Galilean
telescope uses a converging lens as its objective, but a diverging lens as its
eyepiece. The image formed by the objective is at the second focal point of the
eyepiece (the focal point on the refracted side of the eyepiece), so that the final
image is virtual, upright, and at infinity. (a) Show that the magnifying power is
given by M = –fo/fe, where fo is the focal length of the objective and fe is that of the
Optical Images 1037
eyepiece (which is negative). (b) Draw a ray diagram to show that the final image
is indeed virtual, upright, and at infinity.
Picture the Problem The magnification of a telescope is the ratio of the angle
subtended at the eyepiece lens to the angle subtended at the objective lens. We
can use the geometry of the ray diagram to express both θe and θo.
Lens 1
Lens 2
>
(b) The ray diagram is shown below:
>
F1' F2
θo
h θ
e
>
>
F1
(a) Express the magnifying power M
of the telescope:
M =
θe
θo
Because the image formed by the
objective lens is at the focal point,
F'1 :
θo =
h
fo
where we have assumed that θo << 1 so
that tan θo ≈ θo.
Express the angle subtended by
the eyepiece:
θe =
Substitute to obtain:
h
f
f
M = e = o
h
fe
fo
h
where fe is negative.
fe
and M = −
fo
is positive.
fe
Remarks: Because the object for the eyepiece is at its focal point, the image is
at infinity. As is also evident from the ray diagram, the image is virtual and
upright.
1038 Chapter 32
88 ••
A Galilean telescope (see Problem 87) is designed so that the final
image is at the near point, which is 25 cm (rather than at infinity). The focal
length of the objective is 100 cm and the focal length of the eyepiece is –5.0 cm.
(a) If the object distance is 30.0 m, where is the image of the objective? (b) What
is the object distance for the eyepiece so that the final image is at the near point?
(c) How far apart are the lenses? (d) If the object height is 1.5 m, what is the
height of the final image? What is the angular magnification?
Picture the Problem We can use the thin-lens equation to find the image distance
for the objective lens and the object distance for the eyepiece lens. The separation
of the lenses is the sum of these distances. We can use the definition of the
angular magnification and the angles subtended at the objective and eyepiece
lenses to find the height of the final image.
(a) Solve the thin-lens equation
for so' :
so' =
Substitute numerical values and
evaluate so' :
so' =
(b) Solve the thin-lens equation for
se:
se =
f e se'
se' − f e
Noting that se′ = −25 cm, substitute
numerical values and evaluate se:
se =
(− 5.0 cm )(− 25 cm )
− 25 cm − (− 5.0 cm )e
f o so
so − f o
(1.00 m )(30.0 m ) =
30.0 m − 1.00 m
103 cm
= −6.25 cm
− 6.3 cm
where the minus sign tells us that the
object of the eyepiece is virtual.
(c) Express the separation D of the
lenses:
D = so' + se
Substitute numerical values and
evaluate D:
D = 103 cm − 6 cm ≈ 97 cm
(d) Express the height h′ of the final
image in terms of the magnification
M of the telescope:
h' = Mh
Optical Images 1039
The magnification of the telescope is
the product of the magnifications of
the objective and eyepiece lenses:
M = mo me =
Substitute for M to obtain:
h' =
Substitute numerical values and
evaluate h′:
⎛ 103.45 cm ⎞ ⎛ − 25 cm ⎞
⎟⎟ ⎜⎜
⎟⎟(1.5 m )
h' = ⎜⎜
⎝ 3000 cm ⎠ ⎝ − 6.25 cm ⎠
so' se'
so se
so' se'
h
so se
= 21 cm
Express the angular magnification of
the telescope:
M =
θe
θo
The angle subtended by the object is:
θo =
h
so
The angle subtended by the image is:
Substitute for θe and θ o and simplify
to obtain:
Substitute numerical values and
evaluate M:
⎛ h' ⎞
⎟⎟
⎝ se ⎠
θ e = tan −1 ⎜⎜
⎛ h' ⎞
tan −1 ⎜⎜ ⎟⎟
⎝ se ⎠ = so tan −1 ⎛⎜ h' ⎞⎟
M =
⎜s ⎟
h
h
⎝ e⎠
so
M=
⎛ 20.7 cm ⎞
30 m
⎟⎟ = 26
tan −1 ⎜⎜
1.5 m
⎝ 6.25 cm ⎠
89 ••• If you look into the wrong end of a telescope, that is, into the
objective, you will see distant objects reduced in angular size. For a refracting
telescope that has an objective with a focal length equal to 2.25 m and an eyepiece
with a focal length equal to 1.50 cm, by what factor is the angular size of the
object changed?
Picture the Problem The roles of the objective and eyepiece lenses are reversed.
Express the magnifying power of
the ″wrong end″ telescope:
M =−
fe
fo
1040 Chapter 32
Substitute numerical values and
evaluate M:
M =−
1.50 cm
= − 6.67 × 10 −3
2.25 m
General Problems
90 •
To focus a camera, the distance between the lens and the imagesensing surface is varied. A wide-angle lens of a camera has a focal length of
28 mm. By how much must the lens move to change from focusing on an object
at infinity to an object at a distance of 5.00 m in front of the camera?
Picture the Problem We can express the distance Δs that the lens must move as
the difference between the image distances when the object is at 30 m and when it
is at infinity and then express these image distances using the thin-lens equation.
Express the distance Δs that the lens
must move to change from focusing
on an object at infinity to one at a
distance of 5 m:
Δs = s' 5 − s' ∞
Solve the thin-lens equation for s′ to
obtain:
s' =
fs
s− f
Δs =
fs5
fs∞
−
s5 − f s ∞ − f
Substitute s'5 and s'∞ and simplify to
obtain:
=
fs5
f
−
s5 − f 1 − f s ∞
⎡ s
⎤
= f ⎢ 5 − 1⎥
⎣ s5 − f
⎦
Substitute numerical values and
evaluate Δs:
⎡
⎤
5.0 m
Δs = (28 mm ) ⎢
− 1⎥
⎣ 5.00 m − 0.028 m ⎦
= 0.16 mm
A converging lens that has a focal length equal to 10 cm is used to
91 •
obtain an image that is twice the height of the object. Find the object and image
distances if (a) the image is to be upright and (b) the image is to be inverted.
Draw a ray diagram for each case.
Optical Images 1041
Picture the Problem We can use the thin-lens and magnification equations to
obtain simultaneous equations that we can solve to find the image and object
distances for the two situations described in the problem statement.
(a) Use the thin-lens equation to
relate the image and object distances
to the focal length of the lens:
1 1 1
+ =
s s' f
Because the image is twice as large
as the object:
m=−
Substitute for s' to obtain:
1
1
1
+
= ⇒s =
s − 2s f
Substitute numerical values and
evaluate s and s′:
s=
1
2
s'
⇒ s' = −2 s
s
(10 cm ) =
1
2
f
5.0 cm
and
s' = −2(5.0 cm ) = − 10 cm
(b) If the image is inverted, then:
s' = 2 s and
Substitute numerical values and
evaluate s and s′:
s=
1
2
1 1
1
+
= ⇒s =
s 2s f
[3(10 cm)] =
3
2
f
15 cm
and
s' = 2(15 cm ) = 30 cm
The ray diagrams for (a) (left) and (b) (right) follow:
92 ••
You are given two converging lenses that have focal lengths of 75 mm
and 25 mm. (a) Show how the lenses should be arranged to form a telescope.
State which lens to use as the objective, which lens to use as the eyepiece, how
far apart to place the lenses and, what angular magnification you expect. (b) Draw
a ray diagram to show how rays from a distant object are refracted by the two
lenses.
1042 Chapter 32
Picture the Problem In an astronomical telescope the eyepiece (short focal
length) and objective (long focal length) lenses are separated by the sum of their
focal lengths. Given these two lenses, we’ll use the 25 mm lens as the eyepiece
lens and the 75 mm lens as the objective lens and mount them 100 mm apart. The
f
75 mm
= −3 .
angular magnification is then M = − 0 = −
25 mm
fe
(a) The lens with a focal length of 75 mm should be the objective. The two lenses
should be separated by 100 mm. The angular magnification is –3.
(b) A ray diagram showing how rays from a distant object are refracted by the two
lenses follows. A real and inverted image of the distant object is formed by the
objective lens near its second focal point. The eyepiece lens forms an enlarged
and inverted image of the image formed by the objective lens.
Objective
Light from
distant object
Image formed by
eyepiece lens is
at infinity
Eyepiece
Image formed by
objective lens
fo
fe
93 ••
[SSM] (a) Show how the same two lenses in Problem 92 should be
arranged to form a compound microscope that has a tube length of 160 mm. State
which lens to use as the objective, which lens to use as the eyepiece, how far
apart to place the lenses, and what overall magnification you expect to get,
assuming the user has a near point of 25 cm. (b) Draw a ray diagram to show how
rays from a close object are refracted by the lenses.
Picture the Problem
Because the focal lengths appear in the magnification formula as a product, it
would appear that it does not matter in which order we use them. The usual
arrangement would be to use the shorter focal length lens as the objective but we
get the same magnification in the reverse order. What difference does it make
then? None in this problem. However, it is generally true that the smaller the
focal length of a lens, the smaller its diameter. This condition makes it harder to
use the shorter focal length lens, with its smaller diameter, as the eyepiece lens.
Optical Images 1043
In a compound microscope, the
lenses are separated by:
δ = L + fe + f0
Substitute numerical values and
evaluate δ:
δ = 16 cm + 7.5 cm + 2.5 cm = 26 cm
The overall magnification of a
compound microscope is given
by:
Substitute numerical values and
evaluate M:
M = m0 M e = −
L xnp
f0 fe
⎛ 16 cm ⎞ ⎛ 25 cm ⎞
⎟⎟ ⎜⎜
⎟⎟ = −21
M = −⎜⎜
⎝ 7.5 cm ⎠ ⎝ 2.5 cm ⎠
The lens with a focal length of 25 mm should be the objective. The two lenses
should be separated by 210 mm. The angular magnification is –21.
(b) A ray diagram showing how rays from a near-by object are refracted by the
lenses follows. A real and inverted image of the near-by object is formed by the
objective lens at the first focal point of the eyepiece lens. The eyepiece lens forms
an inverted and virtual image of this image at infinity.
Eyepiece
Objective
fo
L
fe
94 ••
On a vacation, you are scuba-diving and using a diving mask that has a
face plate and that bulges outward with a radius of curvature of 0.50 m. As a
result, a convex spherical surface exists between the water and the air in the
mask. A fish swims by you 2.5 m in front of your mask. (a) How far in front of
the mask does the fish appear to be? (b) What is the lateral magnification of the
image of the fish?
1044 Chapter 32
Picture the Problem We can use the equation for refraction at a single surface to
locate the image of the fish and the expression for the magnification due to
refraction at a spherical surface to find the magnification of the image.
(a) Use the equation describing
refraction at a single surface to relate
the image and object distances:
n1 n2 n2 − n1
+ =
s s'
r
Solving for s′ yields:
s' =
n2 rs
(n2 − n1 )s − n1r
Substitute numerical values and
evaluate s′:
s' =
(1)(0.50 m )(2.5 m )
(1 − 1.33)(2.5 m ) − (1.33)(0.50 m )
= −0.839 m = − 84 cm
Note that the fish appears to be only
84 cm in front of your mask.
(b) Express the magnification
due to refraction at a spherical
surface:
m=−
n1s'
n2 s
Substitute numerical values and
evaluate m:
m=−
(1.33)(− 0.839 m ) =
(1)(2.5 m )
0.45
Note that the fish appears to be smaller
than it actually is.
95 •• [SSM] A 35-mm digital camera has a rectangular array of CCDs
(light sensors) that is 24 mm by 36 mm. It is used to take a picture of a person
175-cm tall so that the image just fills the height (24 mm) of the CCD array. How
far should the person stand from the camera if the focal length of the lens is 50
mm?
Picture the Problem We can use the thin-lens equation and the definition of the
magnification of an image to determine where the person should stand.
Use the thin-lens equation to relate
s and s′:
1 1 1
+ =
s s' f
Optical Images 1045
The magnification of the image
is given by:
m=−
s'
2.4 cm
=−
= −1.37 ×10 −2
s
175 cm
and
s' = − ms
Substitute for s' to obtain:
1 1
1
⎛ 1⎞
−
= ⇒ s = ⎜1 − ⎟ f
s ms f
⎝ m⎠
Substitute numerical values and
evaluate s:
1
⎛
⎞
(50 mm) = 3.7 m
s = ⎜1 −
−2 ⎟
⎝ − 1.37 ×10 ⎠
96 ••
A 35-mm camera that has interchangeable lenses is used to take a
picture of a hawk that has a wing span of 2.0 m. The hawk is 30 m away. What
would be the ideal focal length of the lens used so that the image of the wings just
fills the width of the light sensitive area of the camera, which is 36 mm?
Picture the Problem We can use the thin-lens equation and the definition of the
magnification of an image to determine the ideal focal length of the lens.
Use the thin-lens equation to
relate s and s′:
1 1 1
+ =
s s' f
The magnification of the image
is given by:
m=−
s'
3.6 cm
=−
= −1.80 × 10−2
s
200 cm
and
s' = − ms
Substitute for s' to obtain:
1 1
1
s
−
= ⇒f =
1
s ms f
1−
m
Substitute numerical values and
evaluate f:
f =
30 m
= 53 cm
1
1−
− 1.80 × 10 − 2
97 ••
An object is placed 12.0 cm in front of a lens that has a focal length
equal to 10.0 cm. A second lens that has a focal length equal to 12.5 cm is placed
20.0 cm in back of the first lens. (a) Find the position of the final image. (b) What
is the magnification of the image? (c) Sketch a ray diagram showing the final
image.
1046 Chapter 32
Picture the Problem Let the numeral 1 refer to the first lens and the numeral 2 to
the second lens. We apply the thin-lens equation twice; once to locate the image
formed by the first lens and a second time to find the image formed by the second
lens. The magnification of the image is the product of the magnifications
produced by the two lenses.
(a) Solve the thin-lens equation for
the location of the image formed by
the first lens:
s1' =
Substitute numerical values and
evaluate s1' :
s1' =
Because the second lens is 20 cm to
the right of the first lens:
s2 = 20 cm − 60 cm = −40 cm
Solve the thin-lens equation for the
location of the image formed by the
second lens:
s2' =
Substitute numerical values and
evaluate s2' :
s2' =
f1s1
s1 − f1
(10 cm )(12 cm ) = 60 cm
12 cm − 10 cm
f 2 s2
s2 − f 2
(12.5 cm)(− 40 cm) =
− 40 cm − 12.5 cm
9.5 cm
i.e., the final image is 9.5 cm in back of
the second lens.
(b) Express the magnification of the
final image:
⎛ s ' ⎞⎛ s ' ⎞ s 's '
m = m1m2 = ⎜⎜ − 1 ⎟⎟⎜⎜ − 2 ⎟⎟ = 1 2
⎝ s1 ⎠⎝ s2 ⎠ s1s2
Substitute numerical values and
evaluate m:
m=
(60 cm)(9.52 cm) =
(12 cm)(− 40 cm)
− 1.2
i.e., the final image is about 20% larger
than the object and is inverted.
(c) The enlarged, inverted image formed by the first lens serves as a virtual object
for the second lens. The image formed from this virtual object is the real, inverted
image shown in the diagram.
Optical Images 1047
Lens 1
Lens 2
F1'
F1
F2'
F2
98 ••
(a) Show that if fa is the focal length of a thin lens in air, its focal
length in water is given by f w = (n w na )(n − na ) (n − n w ) f a where nw is the index
of refraction of water, n is that of the lens material and na is that of air.
(b) Calculate the focal length in air and in water of a double concave lens that has
an index of refraction of 1.50 and radii of magnitude 30 cm and 35 cm.
Picture the Problem We can apply the equation for refraction at a surface to both
surfaces of the lens and add the resulting equations to obtain an equation relating
the image and object distances to the indices of refraction. We can then use the
lens maker’s equation to complete the derivation of the given relationship
between f ′ and f.
(a) Relate s and s′ at the waterlens interface:
nw n n − nw
+ =
s s1'
r1
Relate s and s′ at the lens-water
interface:
n
n n −n
+ = w
r2
− s1' s'
Add these equations to obtain:
⎛1 1⎞
⎛1 1 ⎞
nw ⎜ + ⎟ = (n − nw )⎜⎜ − ⎟⎟
⎝ s s' ⎠
⎝ r1 r2 ⎠
Let
1
1 1
= + to obtain:
f w s s'
⎛1 1⎞
nw
= (n − n w )⎜⎜ − ⎟⎟
fw
⎝ r1 r2 ⎠
(1)
1048 Chapter 32
The lens-maker’s equation is:
Substitute for
1 1
− in equation
r1 r2
(1) to obtain:
Solving for f ′ yields:
(b) Use the lens-maker’s equation to
find the focal length of the lens in
air:
Use the result derived in Part (a) to
find fw:
⎞⎛ 1 1 ⎞
1 ⎛ n
= ⎜⎜ − 1⎟⎟ ⎜⎜ − ⎟⎟
f a ⎝ na
⎠ ⎝ r1 r2 ⎠
and
na
1 1
1
=
− =
(n − na ) f a
r1 r2 ⎛ n
⎞
⎜⎜ − 1⎟⎟ f a
⎝ na
⎠
⎛
nw
na
= (n − n w )⎜⎜
fw
⎝ (n − na ) f a
fw =
⎞
⎟⎟
⎠
n w (n − na )
fa
na (n − n w )
1 ⎛ 1.50 ⎞ ⎛ 1
1 ⎞
⎟
=⎜
− 1⎟ ⎜⎜
−
f a ⎝ 1.00 ⎠ ⎝ − 30 cm 35 cm ⎟⎠
and
f = −32.3 cm = − 32 cm
fw =
(1.33)(1.50 − 1) (− 32.3 cm )
1.50 − 1.33
= − 1.3 m
99 •• While parked in your car, you see a jogger in your rear view mirror,
which is convex and has a radius of curvature whose magnitude is equal to
2.00 m. The jogger is 5.00 m from the mirror and is approaching at 3.50 m/s.
How fast is the image of the jogger moving relative to the mirror?
Picture the Problem The speed of the jogger as seen in the mirror is
v' = ds' dt. We can use the mirror equation to derive an expression for v′ in terms
of f and ds/dt.
Solve the mirror equation for s′:
⎛ 1 1⎞
s' = ⎜⎜ − ⎟⎟
⎝ f s⎠
−1
(1)
Optical Images 1049
Differentiate s′ with respect to time
to obtain:
ds' d ⎛ 1 1 ⎞
= ⎜ − ⎟
v' =
dt dt ⎜⎝ f s ⎟⎠
−1
−2
⎛ 1 1 ⎞ ⎛ 1 ⎞ ds
= −⎜⎜ − ⎟⎟ ⎜ 2 ⎟
⎝ f s ⎠ ⎝ s ⎠ dt
2
Simplify this result to obtain:
⎛ s' ⎞
v' = −⎜ ⎟ v
⎝s⎠
Rewrite equation (1) in terms of r:
⎛ 2 1⎞
s' = ⎜ − ⎟
⎝r s⎠
Find s′ when s = 5.00 m:
⎛
2
1 ⎞
⎟⎟ = −0.8333 m
s' = ⎜⎜
−
−
2
.
00
m
5
.
00
m
⎝
⎠
Use equation (2) to find v' when
⎛ − 0.8333 m ⎞
⎟⎟ (3.50 m/s )
v' = − ⎜⎜
⎝ 5.00 m ⎠
v = 3.50 m/s:
(2)
−1
−1
2
= 9.72 cm/s
100 ••
A 2.00-cm-thick layer of water (n = 1.33) floats on top of a 4.00-cmthick layer of carbon tetrachloride (n = 1.46) in a tank. How far below the top
surface of the water does the bottom of the tank appear, according to an observer
looking down from above at normal incidence?
Picture the Problem Let the numeral 1
denote the CCl4-H2O interface and the
numeral 2 the H2O-air interface. We
can locate the final image by applying
the equation for refraction at a single
surface to both interfaces. The ray
diagram shown below shows a spot at
the bottom of the tank and the rays of
light emanating from it that form the
intermediate and final images.
Air
nair = 1
H2O
2.00 cm
nH 2 O = 1.33
s' 2
s2
CCl4
nCCl 4 = 1.46
s'1
s1 = 4.00 cm
1050 Chapter 32
Use the equation for refraction at a
single surface to relate s and s′ at the
CCl4-H2O interface:
nCCl4
s1
+
nH2O
s1'
=
nH2O − nCCl4
r
or, because r = ∞,
nH O s1
nCCl4 nH2O
+
= 0 ⇒ s1' = − 2
(1)
nCCl4
s1
s1'
(1.33)(4.00 cm) = −3.64 cm
Substitute numerical values and
evaluate s1′:
s1' = −
The depth of this image, as viewed
from the H2O-air interface is:
s2 = 2.00 cm − s1'
At the H2O-air interface equation (1)
becomes:
s2' = −
Substitute numerical values and
evaluate s2′:
s2' = −
1.46
= 2.00 cm − (− 3.64 cm ) = 5.64 cm
nair s2
n H 2O
(1)(5.64 cm ) = −4.24 cm
1.33
The apparent depth is 4.24 cm.
101 ••• [SSM] An object is 15.0 cm in front of a thin converging lens that
has a focal length equal to 10.0 cm. A concave mirror that has a radius equal to
10.0 cm is 25.0 cm in back of the lens. (a) Find the position of the final image
formed by the mirror-lens combination. (b) Is the image real or virtual? Is the
image upright or inverted? (c) On a diagram, show where your eye must be to see
this image.
Picture the Problem We can use the thin-lens equation to locate the first image
formed by the converging lens, the mirror equation to locate the image formed in
the mirror, and the thin-lens equation a second time to locate the final image
formed by the converging lens as the rays pass back through it.
(a) Solve the thin-lens equation for
s1′:
s1' =
Substitute numerical values and
evaluate s1′:
s1' =
Because the image formed by the
converging lens is behind the
mirror:
s2 = 25 cm − 30 cm = −5 cm
fs1
s1 − f
(10.0 cm )(15.0 cm ) = 30 cm
15.0 cm − 10.0 cm
Optical Images 1051
Solve the mirror equation for s2′:
Substitute numerical values and
evaluate s2′:
s2' =
s2' =
fs2
s2 − f
(5 cm)(− 5 cm) = 2.50 cm
− 5 cm − 5 cm
and the
image is 22.5 cm from the lens; i.e.,
s3 = 22.5 cm.
Solve the thin-lens equation for s3′:
Substitute numerical values and
evaluate s3′:
s3' =
s 3' =
fs3
s3 − f
(10.0 cm )(22.5 cm ) = 18 cm
22.5 cm − 10.0 cm
The final position of the image is 18 cm from the lens, on the same side as the
original object.
(b) The ray diagram is shown below. The numeral 1 represents the object. The
parallel and central rays from 1 are shown; one passes through the center of the
converging lens, the other is paraxial and then passes through the focal point F′.
The two rays intersect behind the mirror, and the image formed there, identified
by the numeral 2, serves as a virtual object for the mirror. Two rays are shown
emanating from this virtual image, one through the center of the mirror, the other
passing through its focal point (halfway between C and the mirror surface) and
then continuing as a paraxial ray. These two rays intersect in front of the mirror,
forming a real image, identified by the numeral 3. Finally, the image 3 serves as a
real object for the lens; again we show two rays, a paraxial ray that then passes
through the focal point F and a ray through the center of the lens. These two rays
intersect to form the final real, upright, and diminished image, identified as 4.
Real object
for the final
image
F'
4
1
C
3
F
Final Object
image
(c) To see this image the eye must be to the left of the final image.
Image formed
by the lens
2
1052 Chapter 32
102 ••• When a bright light source is placed 30 cm in front of a lens, there is
an upright image 7.5 cm from the lens. There is also a faint inverted image 6.0 cm
from the lens on the incident-light side due to reflection from the front surface of
the lens. When the lens is turned around, this weaker, inverted image is 10 cm in
front of the lens. Find the index of refraction of the lens.
Picture the Problem The mirror surfaces must be concave to create inverted
images on reflection. Therefore, the lens is a diverging lens. Let the numeral 1
denote the lens in its initial orientation and the numeral 2 the lens in its second
orientation. We can use the mirror equation to find the magnitudes of the radii of
the lens’ surfaces, the thin-lens equation to find its focal length, and the lens
maker’s equation to find its index of refraction.
Solve the mirror equation for r1 :
r1 =
2s1s1'
s1' + s1
Substitute numerical values and
evaluate r1 :
r1 =
2(30 cm )(6.0 cm )
= 10.0 cm
6.0 cm + 30 cm
Solve the mirror equation for r2 :
r2 =
2s2 s2'
s2' + s2
Substitute numerical values and
evaluate r2 :
r2 =
2(30 cm )(10 cm )
= 15.0 cm
10 cm + 30 cm
Solve the thin-lens equation for f:
f =
ss'
s' + s
Substitute numerical values and
evaluate f:
f =
(30 cm )(− 7.5 cm) = −10.0 cm
Solve the lens-maker’s equation for
n to obtain:
n=
Because the lens is a diverging lens,
r1 = −10 cm and r2 = 15 cm.
Substitute numerical values and
evaluate n:
n=
− 7.5 cm + 30 cm
na
⎛1 1⎞
f ⎜⎜ − ⎟⎟
⎝ r1 r2 ⎠
+ nair
1.00
⎛
⎞
(− 10 cm )⎜⎜ 1 − 1 ⎟⎟
⎝ − 10 cm 15 cm ⎠
= 1.6
+ 1.00
Optical Images 1053
103 ••• A concave mirror that has a radius of curvature equal to 50.0 cm is
oriented with its axis vertical. The mirror is filled with water that has an index of
refraction equal to 1.33 and a maximum depth of 1.00 cm. At what height above
the vertex of the mirror must an object be placed so that its image is at the same
position as the object?
Picture the Problem Assume that the object is very small compared to r so that
all incident and reflected rays traverse 1 cm of water. The problem involves two
refractions at the air−water interface and one reflection at the mirror. Let the
numeral 1 refer to the first refraction at the air−water interface, the numeral 2 to
the reflection in the mirror surface, and the numeral 3 to the second refraction at
the water−air interface.
Use the equation for refraction at a
single surface to relate s1 and s1′:
n1 n2 n2 − n1
+ =
s1 s1'
r
or, because r = ∞,
ns
n1 n2
+ = 0 ⇒ s1' = − 2 1
n1
s1 s1'
Let n2 = n. Because n1 = 1:
s1' = −ns1
Find the object distance for the
mirror:
s2 = 1 − s1' = 1 + ns1
where 1 has units of cm.
Solve the mirror equation for s2′:
⎛2 1 ⎞
s2' = ⎜⎜ − ⎟⎟
⎝ r s2 ⎠
Substitute for s2:
⎛2
1 ⎞
⎟⎟
s2' = ⎜⎜ −
⎝ r 1 + ns1 ⎠
Find the object distance s3 for the
water−air interface:
⎛2
1 ⎞
⎟⎟
s3 = 1 − s2' = 1 − ⎜⎜ −
⎝ r 1 + ns1 ⎠
Use the equation for refraction at
a single surface to relate s3 and
s3′:
n1 n2 n2 − n1
+ =
s3 s3'
r
−1
−1
or, because r = ∞,
ns
n1 n2
+ = 0 ⇒ s3' = − 2 3
n1
s3 s3'
−1
1054 Chapter 32
Because n2 = 1 and n1 = n:
⎛2
1 ⎞
⎟⎟
1 − ⎜⎜ −
+
1
r
ns
s3
1⎠
s3' = − = − ⎝
n
n
Equate s3′ and s1:
⎛2
1 ⎞
⎟
1 − ⎜⎜ −
r 1 + ns1 ⎟⎠
⎝
s1 = −
n
Simplifying yields:
s12 +
Substitute numerical values and
simplify:
s12 +
−1
−1
2−r
1− r
s1 + 2 = 0
n
n
2.00 cm − 50.0 cm
s1
1.33
1.00 cm − 50.0 cm
+
=0
(1.33)2
or
s12 − 36.09 s1 − 27.70 = 0
where s1 is in cm.
Solve for the positive value of s1
to obtain:
s1 = 37 cm
104 ••• A concave side of a lens has a radius of curvature of magnitude equal
to 17.0 cm, and the convex side of the lens that has a radius of curvature of
magnitude equal to 8.00 cm. The focal length of the lens in air is 27.5 cm. When
the lens is placed in a liquid that has an unknown index of refraction, the focal
length increases to 109 cm. What is the index of refraction of the liquid?
Picture the Problem We can use the lens maker’s equation, in conjunction with
the result given in Problem 108, to find the index of refraction of the liquid.
Solve the lens-maker’s equation
for n:
n=
nair
⎛1 1⎞
f ⎜⎜ − ⎟⎟
⎝ r1 r2 ⎠
+ nair
Substitute numerical values and evaluate n:
n=
1.00
⎛
⎞
(27.5 cm )⎜⎜ 1 − 1 ⎟⎟
⎝ − 17.0 cm − 8.00 cm ⎠
+ 1.00 = 1.55
Optical Images 1055
From Problem 98, the focal length of
the lens in the liquid, fL, is related to
the focal length of the lens in air, f,
according to:
fL =
n L (n − nair )
f
nair (n − nL )
Solving for nL yields:
nL =
nair nf L
(n − nair ) f + nair f L
Substitute numerical values and evaluate nL:
nL =
(1.55)(1.00)(109 cm )
=
(1.55 − 1.00)(27.5 cm ) + (1.00)(109 cm )
1.4
105 ••• A solid glass ball of radius 10.0 cm has an index of refraction equal to
1.500. The right half of the ball is silvered so that it acts as a concave mirror
(Figure 32-63). Find the position of the final image formed for an object located
at (a) 40.0 cm, and (b) 30.0 cm to the left of the center of the ball.
Picture the Problem The problem involves two refractions and one reflection.
We can use the refraction at spherical surface equation and the mirror equation to
find the images formed in the two refractions and one reflection. Let the numeral
1 refer to the first refraction at the air-glass interface, the numeral 2 to the
reflection from the silvered surface, and the numeral 3 refer to the refraction at the
glass-air interface.
(a) The image and object distances
for the refraction at the surface of the
ball are related according to:
Solve for s1′ to obtain:
Substitute numerical values and
evaluate s1′:
n1 n2 n2 − n1
+ =
s1 s1'
r
s1' =
n2 rs1
(n2 − n1 )s1 − n1r
s1' =
(1.500)(10.0 cm )(30.0 cm )
(1.500 − 1)(30.0 cm ) − (1)(10.0 cm)
= 90.0 cm
The object for the mirror surface is
behind the mirror and its distance
from the surface of the mirror is:
s 2 = 20.0 cm − 90.0 cm = −70.0 cm
1056 Chapter 32
Use the mirror equation to relate
s2 and s2′:
rs 2
1 1 2
+
= ⇒ s 2' =
s2 s2' r
2s 2 − r
Substitute numerical values and
evaluate s2′:
s 2' =
The object for the second refraction
at the glass-air interface is in front of
the mirrored surface and its distance
from the glass-air interface is:
s3 = 20.0 cm − 4.67 cm = 15.3 cm
The image and object distances for
the second refraction are related
according to:
n1 n2 n2 − n1
+ =
s3 s3'
r
Solve for s3′ to obtain:
s3' =
(10.0 cm )(− 70.0 cm ) = 4.67 cm
2(− 70.0 cm ) − 10.0 cm
n2 rs3
(n2 − n1 )s3 − n1r
Noting that r = −10.0 cm, substitute numerical values and evaluate s3′:
s3' =
(1)(− 10.0 cm )(15.3 cm )
= −6.76 cm
(1.500 − 1)(15.3 cm) − (1.500)(− 10.0 cm )
The final image is 3.2 cm to the left of the center of the ball.
(b) With s1 = 20.0 cm:
s1' =
(1.500)(10.0 cm )(20.0 cm )
(1.500 − 1)(20.0 cm ) − (1)(10.0 cm )
= undefined
Because s1' is undefined, no image is formed when the object is 20.0 cm to the left
of the glass ball. (Alternatively, an image is formed an infinite distance to the left
of the ball.)
106 ••• (a) Show that a small change dn in the index of refraction of a lens
material produces a small change in the focal length df given approximately by
df/f = –dn/(n – nair). (b) Use this result to estimate the focal length of a thin lens
for blue light, for which n = 1.530, if the focal length for red light, for which
n = 1.470, is 20.0 cm.
Picture the Problem We can solve the lens maker’s equation for f and then
differentiate with respect to n and simplify to obtain df/f = −dn/(n − nair).
Optical Images 1057
(a) The lens maker’s equation is:
Differentiating both sides gives:
⎞⎛ 1 1 ⎞
1 ⎛ n
= ⎜⎜
− 1⎟⎟ ⎜⎜ − ⎟⎟
f ⎝ nair
⎠ ⎝ r1 r2 ⎠
−
df
dn
=
2
nair
f
(1)
⎛1 1⎞
⎜⎜ − ⎟⎟
⎝ r1 r2 ⎠
(2)
Diving equation (2) by equation (1) and simplifying yields:
−
dn ⎛ 1 1 ⎞
dn
dn
df
⎜⎜ − ⎟⎟
2
nair ⎝ r1 r2 ⎠
nair
n
f
=
= air =
1
n − nair
n
⎛ n
⎞⎛ 1 1 ⎞
−1
⎜⎜
− 1⎟⎟ ⎜⎜ − ⎟⎟
f
nair
⎝ nair
⎠ ⎝ r1 r2 ⎠ nair
Simplifying both sides yields:
df
dn
= −
f
n − nair
(b) Express the focal length for blue
light in terms of the focal length for
red light:
f blue = f red + Δf
Approximate df/f by Δf/f and dn by
Δn to obtain:
Δf
Δn
≈−
f
n − nair
Solving for Δf yields :
Substitute for Δf in equation (1) to
obtain:
Δf = −
(1)
f Δn
n − nair
f blue = f red −
f red Δn
n − nair
⎛
Δn
= f red ⎜⎜1 −
⎝ nred − nair
Substitute numerical values and evaluate f blue :
⎛ 1.530 − 1.470 ⎞
f blue = (20.0 cm )⎜1 −
⎟ = 17 cm
1.470 − 1.00 ⎠
⎝
⎞
⎟⎟
⎠
1058 Chapter 32
107 ••• [SSM] The lateral magnification of a spherical mirror or a thin lens
is given by m = –s′/s. Show that for objects of small horizontal extent, the
longitudinal magnification is approximately –m2. Hint: Show that ds′/ds = – s′2/s2.
Picture the Problem Examine the amount by which the image distance s′
changes due to a change in s.
Solve the thin-lens equation for s′:
⎛ 1 1⎞
s' = ⎜⎜ − ⎟⎟
⎝ f s⎠
−1
Differentiate s′ with respect to s:
−1
ds' d ⎡⎛ 1 1 ⎞ ⎤
s' 2
1
1
=
−
= −m 2
= ⎢⎜⎜ − ⎟⎟ ⎥ = −
2 2
2
ds ds ⎢⎝ f s ⎠ ⎥
s
⎛ 1 1⎞ s
⎦
⎣
⎜⎜ − ⎟⎟
⎝ f s⎠
The image of an object of length Δs will have a length − m 2 Δs.