Physics 106 - How Things Work - Spring, 1998
Problem Set 3 - Answers
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Due Monday, April 13, 1998, in class (note that this is 1 week
later than appears in the course syllabus)
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Please answer each problem as concisely as possible
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You may discuss the problems with one another but you must write them
up separately and in your own words.
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There are 5 problems containing a total of 25 parts. Each part will be
worth 4 points.
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Write your name legibly on the problem set before turning
it in.
Problem 1: Chapter 15, Case 1a-e (Pg. 553)
The color of the sky depends on the weather and the air.
a. On a very humid day, when the air contains many water droplets
that are about 1 micron (0.000001 m) in diameter, the air appears hazy
and white. Why?
Answer: Some light (about 4%) reflects each time it enters or
leaves a water droplet, so sunlight is scattered in all directions.
Why: The water droplets are large enough--larger than the wavelength
of visible light--to reflect about 4% of light on entry and exit from the
drop. Since this reflection occurs at a variety of angles and involves
many separate droplets of water, the reflected light is sent in all directions
evenly and the mist of water droplets appears white.
b. On a very clear, dry day when there is no dust or water in the
air, the sky is very dark blue. Why?
Answer: There are few particles in the air to Rayleigh scatter
sunlight, although what little sunlight does scatter is mostly blue. This
dim blue light is all that you see from the sky.
Why: Without many particles in the air, there is little Rayleigh
scattering and the sky is relatively dark. Since blue light Rayleigh scatters
more effectively than redder light, the only light that you see is bluish
and the sky appears dark blue.
c. Why is the sky black on the airless moon?
Answer: There are no particles in the sky to scatter sunlight
at all.
Why: Without any scattered light coming from the sky, the sky
appears black. The sun and stars are sharp and clear, but the sky around
them sends no light toward your eyes.
d. When you look at the surface of a calm lake, you see a reflection
of the sky. The water is not a metal, yet you see a reflection. What change
occurs as light passes from air to water that causes some of the light
to be reflected?
Answer: The light changes speed (it slows down) and part of
the light is reflected.
Why: When light changes speeds abruptly in going through the
boundary surface between two materials, some of that light reflects from
the boundary.
e. Mirages appear on sunny days in the desert when very hot air
from the earth’s surface bends light from the horizon upward so that you
see it as though it were coming from the ground in the distance. Hot air
has fewer air molecules in it than colder air at the same pressure. Why
does light bend slightly as it moves from colder air to hotter air?
Answer: Light travels faster in hotter air than in colder air
so there is refraction at the interface between the two airs.
Why: When they are at the same pressures--as they are in this
question--hot air is less dense than cold air. With fewer molecules in
any given volume, the hotter air slows light less than colder air would.
Since light bends when it passes at an angle from one region to another
and changes speeds, light bends when it passes from colder air to hotter
air at an angle.
Problem 2: Chapter 15, Case 5a-e (Pg. 554)
While fluorescent lamps can be dimmed, it’s not as easy as dimming
an incandescent light bulb. Fluorescent lamps
tend to turn off when they don’t run at full power, so a dimmed
fluorescent fixture must make sure that the discharge continues to operate,
even at low power.
a. Why can a fluorescent fixture be dimmed only if it continuously
heats the filaments of its fluorescent tubes?
Answer: As you dim the discharge, the discharge won't be able
to keep the filaments hot on its own. The fixture must actively heat the
filaments to allow them to emit electrons for the discharge.
Why: The filaments at the ends of the fluorescent tube must
remain hot enough to emit electrons or the discharge will stop. While a
full-brightness discharge can heat those filaments on its own via electron
impact hitting, a dimmed discharge cannot. That's why a dimmable fixture
must run currents through the filaments to keep them hot.
b. While an incandescent light bulb can be dimmed by reducing the
voltage drop across its filament, reducing the voltage drop across a fluorescent
tube will spoil its discharge. With the voltage drop across the tube reduced,
the average energy of its electrons would be substantially lower than normal
and its mercury atoms would emit almost no ultraviolet light at all. Why
is there a minimum energy that an electron must have in order to cause
a mercury atom to emit ultraviolet light?
Answer: It takes a certain amount of energy to excite a mercury
atom from its ground state to its lowest-energy excited state and the electron
must have at least that much energy.
Why: For the mercury atom to emit a photon of ultraviolet light,
it must be excited by an electron impact. That electron must give it at
least enough energy to promote it from its ground state to its first excited
state or the excitation won't occur and the ultraviolet photon won't be
produced.
c. Reducing the voltage drop across the fluorescent tube would virtually
stop the production of positive mercury ions in the vapor. That would make
the tube’s filaments last longer, but the discharge wouldn’t operate. With
no positively charged particles inside the tube, what would happen to the
electrons as they flowed from one electrode to the other?
Answer: The electrons would push one another outward to the
walls of the tube.
Why: If the only electrically charged particles in the tube
were electrons, they would repel strongly enough to push one another to
the wall of the tube and the discharge would stop working. The presence
of positively charged mercury ions in the tube balance this repulsion and
keep the electrons flying straight through the tube.
d. Rather than reducing the voltage drop across its fluorescent
tubes, a dimmed fluorescent fixture shortens the time that the discharge
is on following each reversal of current in the power line. An electronic
switch allows current to flow through the tube and its ballast for only
part of each half cycle of the alternating current. The switch senses the
voltage reversal in the power line and then waits a certain amount of time
before closing to allow current to flow. The switch remains closed until
the current stops on its own at the next reversal of the power line, at
which time the switch opens and begins waiting again. The longer into each
half cycle the switch waits, the dimmer the fixture becomes. This arrangement
allows the ballast to store energy as the current through it increases
and use that stored energy to create light as the current through it decreases.
Why shouldn’t the switch try to open the circuit while there is a large
current flowing through the fluorescent tube and the ballast?
Answer: The ballast will try to keep current flowing by adding
energy to the current (increasing the current's voltage) and this high
voltage current will damage the switch.
Why: The ballast stores energy in its magnetic field. If you
try to stop the current flow through the ballast, this magnetic field will
begin to decrease and its stored energy will leave the ballast in the current--the
current's voltage will increase. The ballast will push the current so hard,
and add so much energy to it, that the current will flow through the switch
even as that switch tries to open. The switch may be destroyed by this
"unstoppable" current.
e. Current passing through the electronic switching system of the
dimmed fluorescent fixture experiences a small voltage drop. Why does the
switching system get warm?
Answer: If a current experiences a voltage drop as it passes
through the switching system, that current must be giving up power to the
switching system. This lost energy becomes thermal energy in the switch
and makes it warmer.
Why: Whenever charge experiences a voltage drop as it flows
through a circuit, its energy decreases. That's what a voltage drop means.
In this case, the lost energy becomes thermal energy in the switch.
Problem 3: Chapter 16, Case 1a-e (Pg. 591)
A fiber optic light guide consists of a narrow glass fiber core coated
by a second layer of glass with a lower refractive index. The core is made
of an extremely pure and homogeneous glass so that it absorbs very little
light.
a. Light in the fiber core hits the core’s surface at a very
shallow angle and is reflected. Why can’t the light propagate into the
surrounding layer of glass?
Answer: The light experiences total internal reflection as it
hits the boundary between the glass fiber core and the outer glass layer.
Why: Light travels faster in the outer layer of glass than in
the glass core. That means that when light strikes the boundary between
those two glasses from the core side at an angle, this light bends away
from the normal to the boundary (the line that is perpendicular to the
boundary surface). If the incidence angle is shallow enough, the light
will bend so far that it won't enter the outer glass layer at all. Instead,
that light will be perfectly reflected.
b. The glass is free of bubbles. What would happen to the light
if it had to pass through a series of air bubbles on its way through the
fiber?
Answer: The light would be scattered in all directions.
Why: As light entered or exited one of the air bubbles, about
4% of it would be reflected. The light that passed through would be refracted
and wouldn't leave the bubble in the same direction it arrived. Overall,
light would be sent in all directions and the amount continuing down the
fiber would diminish substantially.
c. Light passing through an optical fiber must travel through an
enormous amount of glass. If impurities in the glass halved the light intensity
each time it traveled 1 km, how much light would remain after 10 km of
fiber?
Answer: 1 part in 1,024 (about 0.1%).
Why: If the fiber loses 1/2 of the light with each kilometer
of travel, it will be down to 1/4 after 2 km, 1/8 after 3 km, 1/16 after
4 km, ... 1/1024 after 10 km. More generally, the light intensity will
be (1/2)n where n is the number of kilometers of fiber. Note
that (1/2)10 is 1/1024.
d. Information is usually sent through a fiber as brief pulses of
light. Those pulses are formed by effectively blocking and unblocking the
light from a laser. Since that technique is equivalent to amplitude modulating
a carrier light wave, the light pulses contain sideband waves at slightly
different frequencies and wavelengths from the carrier wave. Because each
pulse of light is formed by the combination of many waves with different
frequencies and wavelengths, the fiber must exhibit very little dispersion.
What would happen to a single pulse of light if it passed through a long
fiber with strong dispersion?
Answer: It would become longer as its various different waves
drifted apart.
Why: Since the different waves parts of a single pulse travel
at different speeds through the fiber, they soon become separated and arrive
at different times. Instead of being a well defined and brief pulse of
light, the pulse turns into a long spread out package of light.
e. Since even the best glass fibers absorb some light, light passing
through a long fiber must be amplified by short segments of laser amplifier
fiber that are spliced into the main fiber. As light passes through a segment
of laser amplifier fiber, each photon is duplicated hundreds of times.
Explain why this laser amplifier fiber needs power to operate.
Answer: Duplicating photons requires energy because those photons
carry away energy from the amplifier. Since this duplication goes on steadily,
something must supply power to the laser amplifier.
Why: Laser amplifiers use stored energy to duplicate passing
photons. To do this duplication over and over again, something must replace
the stored energy and so power must flow into the laser amplifier from
somewhere else.
Problem 4: Chapter 16, Case 5a-e (Pg. 592)
A video camera uses a converging lens to form a real image of a scene
on an electronic imaging device. This device forms a video signal out of
the pattern of light on its surface.
a. Why must the lens of the video camera be a converging lens,
rather than a diverging lens?
Answer: Only a converging lens can bring light rays together
to form a real image.
Why: When the light rays from one spot on an object pass through
a diverging lens, those rays diverge even more and never form a real image.
Only by passing those rays through a converging lens can a real image be
formed.
b. The camera focuses automatically by analyzing the contrast in
the video image. As you shift the camera from a distant object to one that’s
closer, the lens moves. Does it move toward the imaging device or away
from it? Explain.
Answer: The lens moves away from the imaging device because
the rays from the closer object diverge more and it is harder for the lens
to bring them together. They focus farther from the lens.
Why: As you move an object nearer to a converging lens, the
real image that lens forms moves away from the lens. That's because it
is harder to focus light rays from a nearer object--those rays diverge
more as they enter the lens so the lens is less able to focus them. They
may still form a real image, but that image is farther from the lens than
it would be for a more distant object.
c. The camera has an automatic iris—a diaphragm that controls the
lens’s aperture. When you turn the camera toward a bright scene, the iris
reduces the lens’s aperture. What effect does this change have on the camera’s
depth of focus?
Answer: The camera's depth of focus increases.
Why: When the iris shrinks, the lens effectively becomes smaller
in diameter. Since the light rays that can now pass through this smaller
lens are already pretty close together, they are fairly close together
both in front of and behind the actual real image. As a result, there isn't
as much need to focus the lens--just about everything is nearly in focus.
While the camera will focus on a particular object in its view, objects
that are closer or more far away will appear relatively sharp because of
this small-lens effect.
d. The camera has a zoom lens that changes its focal length at the
touch of a button. When you zoom in on a person for a close-up, does the
focal length increase or decrease?
Answer: The focal length increases.
Why: To form a larger real image (which is what you want for
a close up), you need to give the light more room behind the lens to spread
out away from the middle of the image. That is accomplished by delaying
the focus of the light by using a longer focal length lens. With a longer
focal length lens, the real image forms farther from the lens and is larger
as a result.
e. As the lens changes its focal length, it also moves toward or
away from the electronic image device. As you zoom out to view more of
the scene, does the lens move toward or away from the imaging device?
Answer: The lens moves toward the imaging device.
Why: Since the zoom lens is becoming a shorter focal length
lens, it needs to be closer to the imaging device. The light focuses more
quickly so the lens must be nearer to the imaging device.
Problem 5: Chapter 17, Case 4a-e (Pg. 634)
Ancient people had to make do with simple metals that either occurred
naturally in metallic form or were easy to extract from their ores. Among
these were copper and tin.
a. While early metal smiths could melt copper and cast it into finished
shapes, they found that this technique produced objects that were too soft
to be useful. Why were these objects so easy to bend or dent?
Answer: They contained large crystals of pure copper and these
crystals could deform easily via the slip process.
Why: The sheets of atoms in the large copper crystals slip easily
across one another so that pure cast copper deforms easily when subjected
to stress.
b. To make copper objects that were significantly stiffer than those
produced by casting, the metal smiths would pound and hammer the copper
into shape. Why was beaten copper so much harder than cast copper?
Answer: The crystals in beaten copper are much smaller than
in cast copper. With many small crystals at various angles within the metal,
slip becomes difficult and the copper is much harder to deform permanently.
Why: Work hardening of this sort impedes slip so that plastic
deformation is more difficult to achieve.
c. Tin is a very soft metal that melts at only 232° C (450°
F). But when a small amount of tin is added to copper, the two metals form
bronze, an alloy that is much harder than either copper or tin. How might
the presence of two different atoms in the metal crystals make this metal
so hard?
Answer: Building the crystals from two different atoms roughens
the crystalline planes so that they can't slip as easily across one another.
Why: The tin atoms disrupt the copper crystal enough to make
slip difficult in that crystal. The bronze is more resistant to plastic
deformation as a result.
d. One important use of bronze was in bells. Bell metal is a copper–tin
alloy containing about 25% tin. It melts quite easily, a useful feature
for casting, and is extremely hard. Why is hardness important in a bell?
Answer: It takes energy to deform a metal permanently. Since
a hard metal doesn't experience permanent deformation easily, it doesn't
waste its energy. A hard bell vibrates a long time without wasting energy
to deformation.
Why: A vibrating bell must keep its energy for many vibrations.
If the metal were able to deform easily, it would experience internal frictional
effects that would quickly sap its energy and it would not ring well. A
hard metal doesn't experience this internal energy loss and rings nicely.
e. Unfortunately, bronze bells were also susceptible to cracking.
Why did they crack rather than dent when they were struck too hard?
Answer: Because a hard metal doesn't experience slip easily,
it tends to crack rather than undergo plastic (permanent) deformation.
Why: For a metal to dent, it must be able to experience slip.
Many hard metals can't undergo appreciable slip and simply crack when struck.