Download CHAPTER 26 THE REFRACTION OF LIGHT: LENSES AND

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 θ w1 when it enters the water. According
to Snell's law (Equation 26.2), the sine of θ w1 is given by
sinθ w1 =
n air sinθ 1
n water
=
sinθ 1
(1)
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 θ oil and, according to Snell's law, we have
sinθ oil =
n air sinθ 1
n oil
=
sinθ 1
(2)
n oil
But θ oil is also the angle of incidence at the oil/water interface. At this interface the angle of
refraction is θ w2 and is given by Snell's law as follows:
sinθ w2 =
n oil sinθ oil
n water
=
noil sinθ 1
n water noil
=
sinθ 1
n water
(3)
where we have substituted Equation (2) for sinθ oil . According to Equation (1), this result is equal to
sin θw1 . 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 θ2 is related to the angle of incidence θ1 by
Equation 26.2: n1 sin θ 1 = n2 sin θ 2 or θ 2 = sin −1 [(n1 / n 2 )sin θ 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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3.
REASONING AND SOLUTION When light travels from a material with refractive index n1 into a
material with refractive index n2, the angle of refraction θ2 is related to the angle of incidence θ1 by
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THE REFRACTION OF LIGHT: LENSES AND OPTICAL INSTRUMENTS
Equation 26.2: n1 sin θ 1 = n2 sin θ 2 . Therefore, n1 = n 2 sin θ 2 /(sin θ 1 ) . Both blocks are made from the
same material; therefore, n2 is the same for each system. Furthermore, the angle of incidence is the
same for each system; therefore, sin θ1 is the same in both liquids. Thus, the liquid with the greater
index of refraction will be the one for which sin θ2, and therefore θ2, is the largest. This occurs in
liquid A. Therefore, the index of refraction of liquid A is greater than that of liquid B.
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4.
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.
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 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
image
apparent position of the object in the
water is in the line of sight of the
observer, even though the object could not
object
be seen before the water was added.
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5.
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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6.
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.
Chapter 26 Conceptual Questions
147
observer
fish
image
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7.
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 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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8.
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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9.
REASONING AND SOLUTION
below.
Two rays of light converge to a point on a screen, as shown
point of
convergence
screen
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
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THE REFRACTION OF LIGHT: LENSES AND OPTICAL INSTRUMENTS
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.
glass plate
point of
convergence
screen
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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10. 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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11. 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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12. 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
Chapter 26 Conceptual Questions
149
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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13. REASONING AND SOLUTION For glass (refractive index ng), the critical angle for the glass/air
interface can be determined from Equation 26.4:
sinθ c =
1.0
ng
(1)
In Figure 26.7 the angle of incidence at the upper glass/air interface is θ 2 . Total internal reflection
will occur there only if θ 2 ≥ θ c . But θ 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)sin θ 1 = n g sinθ 2
or
sin θ 2 =
1.0
sinθ 1
ng
Using Equation (1) for 1.0/ng, we obtain
sinθ 2 = sinθ c sin θ 1
(2)
For all incident angles θ 1 that are less than 90° , Equation (2) indicates that sinθ 2 ≥ sinθ c , since
Therefore, θ 2 < θ c and total internal reflection can not occur at the upper glass/air interface.
sinθ 1 < 1 .
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14. 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 26.22.
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 26.22, 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.
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THE REFRACTION OF LIGHT: LENSES AND OPTICAL INSTRUMENTS
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.
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15. 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.
illuminated circle on
water's surface
θc
air mattress
pool-side
observer
θc
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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16. 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 higherindex/lower-index interface is called the critical angle and is given by Equation 26.4: sinθ c = n 2 / n1
where n1 > n 2 .
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.
Chapter 26 Conceptual Questions
151
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17. 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 26.23b, 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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18. 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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19. 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 θ r = θ i . The effect of the
lens on rays of light is governed by Snell's law; namely, n1 sin θ 1 = n2 sin θ 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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20. REASONING AND SOLUTION A converging lens is used to project a real image onto a screen, as
in Figure 26.28b. A piece of black tape is then placed on the upper half of the lens.
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THE REFRACTION OF LIGHT: LENSES AND OPTICAL INSTRUMENTS
black tape
F
F
lens
The ray diagram above 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.
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21. REASONING AND SOLUTION When light travels from a material with refractive index n1 into a
material with refractive index n2, the angle of refraction θ2 is related to the angle of incidence θ1 by
Snell's law (Equation 26.2): n1 sin θ 1 = n2 sin θ 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 θ 1 = sin θ 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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22. 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/ d o + 1/ d i = 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: h o / h i = − d o / d i . 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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Chapter 26 Conceptual Questions
153
23. 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. 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 26.37 and 26.38, 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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24. 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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25. 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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26. 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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THE REFRACTION OF LIGHT: LENSES AND OPTICAL INSTRUMENTS
27. 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 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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28. 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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29. REASONING AND SOLUTION The angle θ 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 θ ≈ h /d , where θ is expressed in
o
o
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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30. REASONING AND SOLUTION
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
F
eye
image of
eye
Chapter 26 Conceptual Questions
155
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.
converging lens
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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31. 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.
diverging lens
F
diverging lens
F
F
object
F
object
image
image
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32. 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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33. REASONING AND SOLUTION The angular magnification of a telescope is given by Equation
26.12: M ≈ − f o / f e , 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
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THE REFRACTION OF LIGHT: LENSES AND OPTICAL INSTRUMENTS
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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34. 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 / f e , 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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35. 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 26.43
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 26.34. Thus, the instrument described in
the question is a microscope.
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Chapter 26 Conceptual Questions
157
36. 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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37. 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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