Physics 106 - How Things Work - Spring, 1998

Problem Set 3 - Answers

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?

b. On a very clear, dry day when there is no dust or water in the air, the sky is very dark blue. Why? c. Why is the sky black on the airless moon? 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? 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? 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?

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? 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? 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? 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? 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?

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? 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? 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? 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. 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?

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. 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? 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? 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? 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? 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? 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? 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? e. Unfortunately, bronze bells were also susceptible to cracking. Why did they crack rather than dent when they were struck too hard?