# Ch 26 CQ 6e.DOC - Ch 26

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```					CHAPTER 26 THE REFRACTION OF LIGHT:
LENSES AND
OPTICAL INSTRUMENTS
CONCEPTUAL QUESTIONS
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1.   REASONING AND SOLUTION Since the index of refraction of water is greater than that
of air, the ray in Figure 26.2a is bent toward the normal at the angle w1 when it enters the
water. According to Snell's law (Equation 26.2), the sine of w1 is given by

nair sin1         sin1
sin  w1                                                (1)
nwater          nwater

where we have taken nair 1.000 . When a layer of oil is added on top of the water, the
angle of refraction at the air/oil interface is oil and, according to Snell's law, we have

nair sin1       sin1
sin  oil                                           (2)
noil           noil

But oil is also the angle of incidence at the oil/water interface. At this interface the angle of
refraction is w2 and is given by Snell's law as follows:

noil sin oil        noil sin 1            sin 1
sin  w2                                                        (3)
nwater           nwater noil             nwater

where we have substituted Equation (2) for sin oil . According to Equation (1), this result is
equal to sin  w1 . Therefore, we can conclude that the angle of refraction as the ray enters
the water does not change due to the presence of the oil.
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2.   REASONING AND SOLUTION When light travels from a material with refractive index
n1 into a material with refractive index n2, the angle of refraction 2 is related to the angle of
1
incidence 1 by Equation 26.2: n1 sin 1  n2 sin  2 or  2  sin (n1 / n2 )sin 1  . When
              
n1 < n2, the angle of refraction will be less than the angle of incidence. The larger the value
of n2, the smaller the angle of refraction for the same angle of incidence. The angle of
refraction is smallest for slab B; therefore, slab B has the greater index of refraction.
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156    THE REFRACTION OF LIGHT: LENSES AND OPTICAL INSTRUMENTS

3.    REASONING AND SOLUTION When an observer peers
over the edge of a deep empty bowl, he does not see the entire
bottom surface, so a small object lying on the bottom is
hidden from view. However, when the bowl is filled with
water, the object can be seen.                                                        Image
When the object is viewed from the edge of the bowl, light
rays from the object pass upward through the water. Since
nair < nwater, the light rays from the object refract away from
the normal when they enter air. The refracted rays travel to                           Object
the observer, as shown in the figure at the right. When the
rays entering the air are extended back into the water, they
show that the observer sees a virtual image of the object at an apparent depth that is less than
the actual depth, as indicated in the drawing. Therefore, the apparent position of the object
in the water is in the line of sight of the observer, even though the object could not be seen
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4.    REASONING AND SOLUTION Two identical containers, one filled with water (n = 1.33)
and the other filled with ethyl alcohol (n = 1.36) are viewed from directly above. According
to Equation 26.3, when viewed from directly above in a medium of refractive index n2, the
apparent depth d in a medium of refractive index n1 is related to the actual depth d by the
relation d   d (n2 /n1 ) . Assuming that the observer is in air, n2 = 1.00. Since n1 refers to
the refractive index of the liquid in the containers, we see that the apparent depth in each
liquid is inversely proportional to the refractive index of the liquid. The index of refraction
of water is smaller than that of ethyl alcohol; therefore, the container filled with water
appears to have the greater depth of fluid.
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5.    REASONING AND SOLUTION When you look through an aquarium window at a fish,
the fish appears to be closer than it actually is. When light from the fish leaves the water
and enters the air, it is bent away from the normal as shown below. Therefore, the apparent
location of the image is closer to the observer than the actual location of the fish.

observer
fish          image

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6.    REASONING AND SOLUTION At night, when it is dark outside and you are standing in a
brightly lit room, it is easy to see your reflection in a window. During the day it is not so
easy. If we assume that the room is brightly lit by the same amount in both cases, then the
light reflected from the window is the same during the day as it is at night. However, during
Chapter 26 Conceptual Questions    157

the day, light is coming through the window from the outside. In addition to the reflection,
the observer also sees the light that is refracted through the window from the outside. The
light from the outside is so intense that it obscures the reflection in the glass.
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7.   REASONING AND SOLUTION
a. The man is using a bow and arrow to shoot a fish. The light from the fish is refracted
away from the normal when it enters the air; therefore, the apparent depth of the image of
the fish is less than the actual depth of the fish. When the arrow enters the water, it will
continue along the same straight line path from the bow. Therefore, in order to strike the
fish, the man must aim below the image of the fish. The situation is similar to that shown in
Figure 26.5a; we can imagine replacing the boat by a dock and the chest by a fish.

b. Now the man is using a laser gun to shoot the fish. When the laser beam enters the water
it will be refracted. From the principle of reversibility, we know that if the laser beam
travels along one of the rays of light emerging from the water that originates on the fish, it
will follow exactly the same path in the water as that of the ray that originates on the fish.
Therefore, in order to hit the fish, the man must aim directly at the image of the fish.
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8.   REASONING AND SOLUTION Two rays of light converge to a point on a screen, as
shown below.

poi nt of
c onv ergenc e

s creen

A plane-parallel plate of glass is placed in the path of this converging light, and the glass
plate is parallel to the screen, as shown below. As discussed in the text, when a ray of light
passes through a pane of glass that has parallel surfaces, and is surrounded by air, the
emergent ray is parallel to the incident ray, but is laterally displaced from it. The extent of
the displacement depends on the angle of incidence, on the thickness, and on the refractive
index of the glass.

glas s pl ate
poi nt of
c onv ergenc e

s creen
As shown in the scale drawing above, the point of convergence does not remain on the
screen. It will move away from the glass as shown.
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158    THE REFRACTION OF LIGHT: LENSES AND OPTICAL INSTRUMENTS

9.    REASONING AND SOLUTION Light from the sun is unpolarized; however, when the
sunlight is reflected from horizontal surfaces, such as the surface of an ocean, the reflected
light is partially polarized in the horizontal direction. Polaroid sunglasses are constructed
with lenses made of Polaroid (a polarizing material) with the transmission axis oriented
vertically. Thus, the horizontally polarized light that is reflected from horizontal surfaces is
blocked from the eyes.
Suppose you are sitting on the beach near a lake on a sunny day, wearing Polaroid
sunglasses. When a person is sitting upright, the horizontally polarized light that is reflected
from the water is blocked from her eyes, as discussed above, and she notices little
discomfort due to the glare from the water. When she lies on her side, the transmission axis
of the Polaroid sunglasses is now oriented in a nearly horizontal direction. Most of the
horizontally polarized light that is reflected from the water is transmitted through the
sunglasses and reaches her eyes. Therefore, when the person lies on her side, she will notice
that the glare increases.
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10. REASONING AND SOLUTION Light from the sun is unpolarized; however, when the
sunlight is reflected from horizontal surfaces such as the surface of a swimming pool, lake,
or ocean, the reflected light is partially polarized in the horizontal direction. Polaroid
sunglasses are constructed with lenses made of Polaroid (a polarizing material) with the
transmission axis oriented vertically. Thus, the horizontally polarized light that is reflected
from horizontal surfaces is blocked from the eyes.
If you are sitting by the shore of a lake on a sunny and windless day, you will notice that
the effectiveness of your Polaroid sunglasses in reducing the glare of the sunlight reflected
from the lake varies depending on the time of the day. As the angle of incidence of the sun's
rays increases from 0 , the degree of polarization of the rays in the horizontal direction
increases. Since Polaroid sunglasses are designed so that the transmission axes are aligned
in the vertical direction when they are worn normally, they become more effective as the sun
gets lower in the sky. When the angle of incidence is equal to Brewster's angle, the reflected
light is completely polarized parallel to the surface, and the sunglasses are most effective.
For angles of incidence greater than Brewster's angle, the glasses again become less
effective.
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11. REASONING AND SOLUTION According to the principle of reversibility (see
Section 25.5), if the direction of a light ray is reversed, the light retraces its original path.
While the principle of reversibility was discussed in Section 25.5 in connection with the
reflection of light rays, it is equally valid when the light rays are refracted. Imagine
constructing a mixture of colored rays by passing a beam of sunlight through a prism in the
usual fashion. By orienting a second prism so that the rays of colored light are incident on
the second prism with angles of incidence that are equal to their respective angles of
refraction as they emerge from the first prism, we have a perfectly symmetric situation. The
rays through the second prism will follow the reverse paths of the rays through the first
prism, and the light emerging from the second prism will be sunlight.
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Chapter 26 Conceptual Questions    159

12. REASONING AND SOLUTION For glass (refractive index ng), the critical angle for the
glass/air interface can be determined from Equation 26.4:

1.0
sin  c                                       (1)
ng
In Figure 26.7 the angle of incidence at the upper glass/air interface is 2. Total internal
reflection will occur there only if 2  c. But 2 is also the angle of refraction at the lower
air/glass interface and can be obtained using Snell's law as given in Equation 26.2:

1.0
(1.0)sin 1  ng sin  2   or     sin  2        sin 1
ng
Using Equation (1) for 1.0/ng, we obtain

sin  2  sin  c sin 1                            (2)

For all incident angles 1 that are less than 90 , Equation (2) indicates that sin 2 < sin c,
since sin 1 < 1. Therefore, 2 < c and total internal reflection can not occur at the upper
glass/air interface.
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13. REASONING AND SOLUTION
a. When a rainbow is formed, light from the sun enters a spherical water droplet and is
refracted by an amount that depends on the refractive index of water for that wavelength.
Light that is reflected from the back of the droplet is again refracted at it reenters the air, as
suggested in Figure 26.21. Although all colors are refracted for any given droplet, the
observer sees only one color, because only one color travels at the proper angle to reach the
observer. The observer sees the full spectrum in the rainbow because each color originates
from water droplets that lie at different elevation angles.
As shown in Figure 26.21, the sun must be located behind the observer, if the observer is
to see the rainbow. Therefore, if you want to make a rainbow by spraying water from a
garden hose into the air, you must stand with the sun behind you, and adjust the hose so that
it sprays a fine mist of water in front of you. The distance between the observer and the
droplets is not crucial. The important factor is the angle formed by the intersection of the
line that extends from the sun to the droplet with the line that extends from the droplet to the
observer. Remark: When the distance is only a few meters, as it would be in the case of a
"garden-hose rainbow", each eye would receive rays from different parts of the mist.
Therefore, the observer could see two rainbows that cross over each other.

b. Each color of light that leaves a given droplet travels in a specific direction that is
governed by Snell's law. You can't ever walk under a rainbow, because each color that
originates from a single droplet travels in a unique direction. To walk under a rainbow, all
the colors would have to be refracted vertically downward, which is not the case. Therefore,
you can't walk under a rainbow, because the rays are traveling in the wrong directions to
reach the observer's eyes.
160    THE REFRACTION OF LIGHT: LENSES AND OPTICAL INSTRUMENTS

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14. REASONING AND SOLUTION A person is floating on an air mattress in the middle of a
swimming pool. His friend is sitting on the side of the pool. The person on the air mattress
claims that there is a light shining up from the bottom of the pool directly beneath him. His
friend insists, however, that she cannot see any light from where she sits on the side.
Rays from a light source on the bottom of the pool will radiate outward from the source in
all directions. However, only rays for which the angle of incidence is less than the critical
angle will emerge from the water. Rays with an angle of incidence equal to, or greater than,
the critical angle will undergo total internal reflection back into the water, as shown in the
following figure.

illuminat ed circle on                                  pool-side
wat er' s surf ace                                      observer
air mat tress

c          c

light source

Because of the geometry, the rays that leave the water lie within a cone whose apex lies at
the light source. Thus, rays of light that leave the water emerge from within an illuminated
circle just above the source. If the mattress is just over the source, it could cover the area
through which the light would emerge. A person sitting on the side of the pool would not
see any light emerging. Therefore, the statements made by both individuals are correct.
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15. REASONING AND SOLUTION Total internal reflection occurs only when light travels
from a higher-index medium (refractive index = n1) toward a lower-index medium
(refractive index = n2). Total internal reflection does not occur when light propagates from a
lower-index to a higher-index medium. The smallest angle of incidence for which total
internal reflection will occur at the higher-index/lower-index interface is called the critical
angle and is given by Equation 26.4: sin  c  n2 / n1 where n1  n2 .
A beam of blue light is propagating in glass. When the light reaches the boundary
between the glass and the surrounding air, the beam is totally reflected back into the glass.
However, red light with the same angle of incidence is not totally reflected and some of the
light is refracted into the air. According to Table 26.2, the index of refraction of glass is
greater for blue light than it is for red light. From Snell's law, therefore, we can conclude
that the critical angle is greater for red light than it is for blue light. Therefore, if the angle
of incidence is equal to or greater than the critical angle for blue light, but less than the
critical angle for red light, blue light will be totally reflected back into the glass, while some
of the red light will be refracted into the air.
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Chapter 26 Conceptual Questions        161

16. REASONING AND SOLUTION A beacon light in a lighthouse is to produce a parallel
beam of light. The beacon consists of a bulb and a converging lens. As shown in
Figure 26.22b, paraxial rays that are parallel to the principal axis converge to the focal point
after passing through the lens. From the principle of reversibility, we can deduce that if a
point source of light were placed at the focal point, the emitted light would travel in parallel
rays after passing through the lens. Therefore, in the construction of the beacon light, the
bulb should be placed at the focal point of the lens.
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17. REASONING AND SOLUTION The figure at the right shows
a converging lens (in air). The normal to the surface of the lens
is shown at five locations on each side of the lens.
A ray of light bends toward the normal when it travels from a
medium with a lower refractive index into a medium with a
higher refractive index. Likewise, a ray of light bends away from
the normal when it travels from a medium with a higher
refractive index into a medium with a lower refractive index. When rays of light traveling in
air enter a converging lens, they are bent toward the normal. When these rays leave the right
side of the lens, they are bent away from the normal; however, since the normals diverge on
the right side of the lens, the rays again converge.
If this lens is surrounded by a medium which has a higher index of refraction than the
lens, then when rays of light enter the lens, the rays are bent away from the normal, and,
therefore, they diverge. When the rays leave the right side of the lens, they are bent toward
the normal; however, since the normals diverge on the right side of the lens, the rays diverge
further. Therefore, a converging lens (in air) will behave as a diverging lens when it is
surrounded by a medium that has a higher index of refraction than the lens.
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18. REASONING AND SOLUTION A spherical mirror and a lens are immersed in water. The
effect of the mirror on rays of light is governed by the law of reflection; namely  r   i . The
effect of the lens on rays of light is governed by Snell's law; namely, n1 sin 1  n2 sin  2 .
The law of reflection, as it applies to the mirror, does not depend on the index of refraction
of the material in which it is immersed. Snell's law, however, as it applies to the lens,
depends on both the index of refraction of the lens and the index of refraction of the material
in which it is immersed. Therefore, compared to the way they work in air, the lens will be
more affected by the water.
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19. REASONING AND SOLUTION A converging lens is used to project a real image onto a
screen, as in Figure 26.27b. A piece of black tape is then placed on the upper half of the
lens. The following ray diagram shows the rays from two points on the object, one point at
the top of the object and one point on the lower half of the object. As shown in the diagram,
rays from both points converge to form the image on the right side of the lens. Therefore,
the entire image will be formed. However, since fewer rays reach the image when the tape is
present, the intensity of the image will be less than it would be without the tape.
162    THE REFRACTION OF LIGHT: LENSES AND OPTICAL INSTRUMENTS

blacktape

F                      F

lens
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20. REASONING AND SOLUTION When light travels from a material with refractive index
n1 into a material with refractive index n2, the angle of refraction 2 is related to the angle of
incidence 1 by Snell's law (Equation 26.2): n1 sin 1  n2 sin  2 .
A converging lens is made from glass whose index of refraction is n. The lens is
surrounded by a fluid whose index of refraction is also n. This situation is known as index
matching and is discussed in Conceptual Example 7. Since the refractive index of the
surrounding fluid is the same as that of the lens, n1 = n2, and Snell's law reduces to
sin 1  sin  2 . The angle of refraction is equal to the angle of incidence at both surfaces
of the lens; the path of light rays is unaffected as the rays travel through the lens. Therefore,
this lens cannot form an image, either real or virtual, of an object.
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21. REASONING AND SOLUTION The expert claims that the height of the window can be
calculated from only two pieces of information : (1) the measured height on the film, and
(2) the focal length of the camera. The expert is not correct. According to the thin-lens
equation (Equation 26.6), 1/ do 1/ di 1/ f , where do is the object distance, di is the image
distance, and f is the focal length of the lens. The magnification equation (Equation 26.7),
relates the image and object heights to the image and object distances: ho / hi  do / di .
These two equations contain five unknowns. To determine any one of the unknowns, three
of the other unknowns must be known. In this case, all that we know is the height of the
image, hi, and the focal length of the camera, f. Therefore, we do not have enough
information given to determine the distance from the ground to the window (the height of
the object in this case), ho. We still need to know either the distance from the photographer
to the house (the object distance, do), or the distance from the center of the lens to the film
(the image distance, di). We can conclude, therefore, that the expert is incorrect.
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22. REASONING AND SOLUTION Suppose two people who wear glasses are camping. One
is nearsighted, and the other is farsighted. It is desired to start a fire with the sun's rays. A
converging lens can be used to focus the nearly parallel rays of the sun on a sheet of paper.
Chapter 26 Conceptual Questions       163

If the paper is placed at the focal point of the lens, the sun's rays are concentrated to give a
large intensity, so that the paper heats up rapidly and ignites. As shown in Figures 26.36 and
26.37, nearsightedness can be corrected with diverging lenses, and farsightedness can be
corrected using converging lenses. Therefore, the glasses of the farsighted person would be
useful in starting a fire, while the glasses of the nearsighted person would not be useful.
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23. REASONING AND SOLUTION A 21-year-old with normal vision (near point = 25 cm) is
standing in front of a plane mirror. The near point is the point nearest the eye at which an
object can be placed and still produce a sharp image on the retina. Therefore, if the 21-year
old wants to see himself in focus, he can stand no closer to the mirror than 25 cm from his
image. As discussed in Chapter 25, the image in a plane mirror is located as far behind the
mirror as the observer is in front of the mirror. If the 21-year-old is 25 cm from his image,
he must be 25 cm/2 = 12.5 cm in front of the mirror's surface. Therefore, he can stand no
closer than 12.5 cm in front of the mirror and still see himself in focus.
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24. REASONING AND SOLUTION The distance between the lens of the eye and the retina is
constant; therefore, the eye has a fixed image distance. The only way for images to be
produced on the retina for objects located at different distances is for the focal length of the
lens to be adjusted. This is accomplished through with the ciliary muscles. If we read for a
long time, our eyes become "tired," because the ciliary muscle must be tensed so that the
focal length is shortened enough to bring the print into focus. When the eye looks at a
distant object, the ciliary muscle is fully relaxed. Therefore, when your eyes are "tired" from
reading, it helps to stop and relax the ciliary muscle by looking at a distant object.
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25. REASONING AND SOLUTION As discussed in the text, for light from an object in air to
reach the retina of the eye, it must travel through five different media, each with a different
index of refraction. About 70 % of the refraction occurs at the air/cornea interface where the
refractive index of air is taken to be unity and the refractive index of the cornea is 1.38.
To a swimmer under water, objects look blurred and out of focus. However, when the
swimmer wears goggles that keep the water away from the eyes, the objects appear sharp
and in focus. Without the goggles, light from objects must undergo the first refraction at a
water/cornea interface. Since the index of refraction of water is 1.33 while that of the cornea
is 1.38, the amount of refraction is smaller than it is when the person is in air, and the
presence of the water prevents the image from being formed on the retina. Consequently,
objects look blurred and out of focus. When the swimmer wears goggles, incoming light
passes through the volume of air contained in the goggles before it reaches the eyes of the
swimmer. The first refraction of the light in the eye occurs at an air/cornea interface. The
refraction occurs to the proper extent, so that the image is formed on the retina. Therefore,
when the swimmer wears the goggles, objects appear to be sharp and in focus.
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26. REASONING AND SOLUTION The refractive power of the lens of the eye is 15 diopters
when surrounded by the aqueous and vitreous humors. If this lens is removed from the eye
and surrounded by air, its refractive power increases to about 150 diopters. From Snell's
164    THE REFRACTION OF LIGHT: LENSES AND OPTICAL INSTRUMENTS

law, we know that the effect of the lens on incoming light depends not only on the refractive
index of the lens, but also on the refractive index of the materials on either side of the lens.
The refractive index of the lens is 1.40, while that of the aqueous humor is 1.33, and that of
the vitreous humor is 1.34. Light that leaves the lens has been refracted twice, once when it
enters the lens and again when it leaves the lens. Since the refractive indices of these three
media are not very different, the amount of refraction at each interface is small. When the
lens is surrounded by air, the light is again doubly refracted. In this case, however, the
refractive indices at each interface differ substantially, so the amount of refraction at each
interface is much larger. Therefore, when the lens is in air, its focal length is much smaller
than it is when the lens is in place in the eye. According to Equation 26.8, the refractive
power of a lens is equal to 1/ f , where the refractive power is expressed in diopters when the
focal length is in meters. The smaller the focal length of the lens, the larger its refractive
power. Consequently, the refractive power of the lens is much greater when the lens is
surrounded by air.
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27. REASONING AND SOLUTION A full glass of wine acts, approximately, as a converging
lens and focuses the light to a spot on the table. An empty glass consists only of thin glass
layers on opposite sides, which do not refract the light enough to act as a lens and produce a
focused image.
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28. REASONING AND SOLUTION The angle  subtended by the image as measured from
the principal axis of the lens of the eye is equal to the angle subtended by the object. This
angle is called the angular size of both the image and the object and is given by   ho / do ,
where  is expressed in radians.
Jupiter is the largest planet in our solar system. Yet to the naked eye, it looks smaller
than Venus. This occurs because the distance from Earth to Jupiter is about 15 times greater
than the distance from Earth to Venus, while the diameter of Jupiter is only about 12 times
larger than that of Venus. Consequently, the angular size of Jupiter is about 12/15 or 0.80
times as large as that of Venus. Therefore, Jupiter looks smaller than Venus.
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29. REASONING AND SOLUTON
a. The figure below is a ray diagram that shows that the eyes of a person wearing glasses
appear to be smaller when the glasses use diverging lenses.

Diverging lens

F

Eye

Image of
eye
Chapter 26 Conceptual Questions        165

F                             F

b. The figure below is a ray diagram that shows that the eyes of a person wearing glasses
appear to be larger when the glasses use converging lenses.

F

Eye
Image
of eye

Notice that in both cases, the eye lies between the focal length of the lens and the lens, and
that both images are virtual images.
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30. REASONING AND SOLUTION As discussed in the text, regardless of the position of a
real object, a diverging lens always forms a virtual image that is upright and smaller relative
to the object. The figures below show this for two cases: one in which the object is within
the focal point, and the other in which the object is beyond the focal point. In each case, the
image is smaller than the object. Therefore, a diverging lens cannot be used as a magnifying
glass.
166    THE REFRACTION OF LIGHT: LENSES AND OPTICAL INSTRUMENTS

Diverging lens                                                   Diverging lens

F                               F                              F                                 F

Object                                           Object
Image                                                    Image

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31. REASONING AND SOLUTION A person whose near point is 75 cm from the eyes, must
hold a printed page at least 75 cm from his eyes in order to see the print without blurring,
while a person whose near point is 25 cm can hold the page as close as 25 cm and still find
the print in focus. If the size of the print is small, it will be more difficult to see the print at
75 cm than at 25 cm, even though the print is in focus. Therefore, the person whose near
point is located 75 cm from the eyes will benefit more by using a magnifying glass.
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32. REASONING AND SOLUTION The angular magnification of a telescope is given by
Equation 26.12: M   fo / fe , where fo is the focal length of the objective, and fe is the
focal length of the eyepiece. In order to produce a final image that is magnified, fo must be
greater than fe. Therefore if two lenses, whose focal lengths are 3.0 and 45 cm are to be
used to build a telescope, the lens with the 45 cm focal length should be used for the
objective, and the lens with the 3.0 cm focal length should be used for the eyepiece.
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33. REASONING AND SOLUTION A telescope consists of an objective and an eyepiece. The
objective focuses nearly parallel rays of light that enter the telescope from a distant object to
form an image just beyond its focal point. The image is real, inverted, and reduced in size
relative to the object. The eyepiece acts like a magnifying glass. It is positioned so that the
image formed by the objective lies just within the focal point of the eyepiece. The final
image formed by the eyepiece is virtual, upright and enlarged.
Two refracting telescopes have identical eyepieces, although one is twice as long as the
other. Since the eyepiece is positioned so that the image formed by the objective lies just
within the focal point of the eyepiece, the longer telescope has an objective with a longer
focal length. The angular magnification of a telescope is given by Equation 26.12:
M   fo / fe , where fo is the focal length of the objective, and fe is the focal length of the
eyepiece. Both telescopes have the same value for fe. The longer telescope has the larger
value of fo; therefore, the longer telescope has the greater angular magnification.
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Chapter 26 Conceptual Questions       167

34. REASONING AND SOLUTION In a telescope the objective forms a first image just
beyond the focal point of the objective and just within the focal point of the eyepiece. Thus,
as Figure 26.42 shows, the distance between the two converging lenses is L  f o  f e . For
the two lenses specified, this would mean that L  4.5 cm + 0.60 cm = 5.1 cm. But L is
given as L = 14 cm, which means that there is a relatively large separation between the focal
points of the objective and the eyepiece. This arrangement is like that for a microscope
shown in Figure 26.33. Thus, the instrument described in the question is a microscope.
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35. REASONING AND SOLUTION
a. A projector produces a real image at the location of the screen.

b. A camera produces a real image at the location of the film.

c. A magnifying glass produces a virtual image behind the lens.

d. Eyeglasses produce virtual images that the eye then sees in focus.

e. A compound microscope produces a virtual image.

f. A telescope produces a virtual image.
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36. REASONING AND SOLUTION Chromatic aberration occurs when the index of refraction
of the material from which a lens is made varies with wavelength. Lenses obey Snell's law.
If the index of refraction of a lens varies with wavelength, then different colors of light that
pass through the lens refract by different amounts. Therefore, different colors come to a
focus at different points. Mirrors obey the law of reflection. The angle of reflection depends
only on the angle of incidence, regardless of the wavelength of the incident light; therefore,
chromatic aberration occurs in lenses, but not in mirrors.
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s. Therefore, different colors come to a focus at different points.
Mirrors obey the law of reflection. The angle of reflection depends only on the angle of incidence,
regardless of the wavelength of the incident light; therefore, chromatic aberration occurs in lenses,
but not in mirrors.
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