Mastering Physics Solutions Chapter 25 Electromagnetic Waves
Chapter 25 Electromagnetic Waves Q.1CQ
Explain why the “invisible man” would be unable to see
Solution:
672-25-1CQ RID: 514 I 04/03/2013
RIDI: 2791 31/08/2015
An invisible person allows light to pass through himself/herself uninterruptedly. That makes himself/herself invisible, if incase. there happens any kind of phenomena of light that makes light little bit hindered or unease to pass through. human brain detects that, and tells that there is some medium that hinders light. This way the existence of the invisible person is exposed. and he/she no longer stays invisible. Now for an invisible person to see, the incident light has to be absorbed by the retina, though very small. This absorption can readily be detected if not by naked eye. by sensitive instrument. So the invisibility of the person is exposed. In order for a person to maintain invisibility, the eye of himlher absorbs zero amount of light, which makes himlher blind, completely unable to see.
Chapter 25 Electromagnetic Waves Q.1P
If the electric field in an electromagnetic wave is increasing in magnitude at a particular time, is the magnitude of the magnetic field at the same time increasing or decreasing? Explain.
Solution:
lithe electric field in an electromagnetic wave increases in magnitude at a particular time then the magnitude of the magnetic field at the same time increases because the electric and magnetic fields in an electromagnetic wave vary in phase with one another According to the equation E = cB we can say that if Electric field increases then magnetic field also increases
Chapter 25 Electromagnetic Waves Q.2CQ
The magnitude of the Dopplcr effect tells how rapidly a weather system is moving. What determines whether the system is approaching or receding?
Solution:
Chapter 25 Electromagnetic Waves Q.2P
The electric field of an electromagnetic wave points in the positive y direction. At the same time, the magnetic field of this wave points in the positive z direction. In what direction is the wave traveling?
Solution:
Chapter 25 Electromagnetic Waves Q.3CQ
Explain why radiation pressure is more significant on a grain of dust in interplanetary space when the grain is very small.
Solution:
As the grain of dust becomes smaller, its volume and mass decreases more rapidly than does its area. The radiation pressure which acts on the surface of the grain, becomes increasingly important as the size of the grain is decreased. So, the gravity which acts on the mass of the grain becomes less important.
Chapter 25 Electromagnetic Waves Q.3P
An electric charge on the x axis oscillates sinusoidally about the origin. A distant observer is located at a point on the +y axis. (a) In what direction will the electric field oscillate at the observer’s location? (b) In what direction will the magnetic field oscillate at the observer’s location? (c) In what direction will the electromagnetic wave propagate at the observer’s location?
Solution:
(a)
The electric field propagates parallel to the oscillations.
In this case, electric charge on the x axis oscillates simultaneously about the origin.
Therefore, electric field oscillates in x direction.
(b)
The direction of electric field and magnetic field perpendicular to each other, they are also perpendicular to the direction of propagation.
In this case, electric field direction is x and direction of propagation is +y. Thus, the only direction perpendicular to x and y directions is the z direction.
Therefore, the magnetic field oscillates in the z direction.
(c)
The observer is located at a point on the +y axis. From the rectilinear propagation of light, light travels in the straight line. Thus, the electromagnetic wave must traveled in the +y direction because of it have traveled from origin to a point on the +y axis.
Therefore, the electromagnetic wave propagates in the +y direction.
Chapter 25 Electromagnetic Waves Q.4CQ
While wearing your Polaroid sunglasses at the beach, you notice that they reduce the glare from the water better when you are sitting upright than when you are lying on your side. Explain.
Solution:
Light reflected from a horizontal surface has a polarization in the horizontal direction. It follows that when you sit upright , with the transmission axis of your glasses in the vertical direction , they will block most of the reflected light when you lie on your side , however , the transmission axis is horizontal. This allows most of the reflected light to enter your eyes.
Chapter 25 Electromagnetic Waves Q.4P
An electric charge on the z axis oscillates sinusoidally about the origin. A distant observer is located at a point on the +y axis. (a) In what direction will the electric field oscillate at the observer’s location? (b) In what direction will the magnetic field oscillate at the observer’s location? (c) In what direction will the electromagnetic wave propagate at the observer’s location?
Solution:
Electric charge is oscillating on z-axis about origin and location of distant observer is
+y-axis. Electric field and magnetic field are mutually perpendicular to each other in electromagnetic wave.
The oscillation of electromagnetic wave is shown in figure.
(a)
Since electric field propagates in the direction of oscillation of charge, hence the direction oscillations of electric field will be z-axis.
(b)
In electromagnetic wave the electric and magnetic field are mutually perpendicular to each other in all condition. And magnetic field is also perpendicular to the direction of propagation. So the magnetic field oscillates in x-axis direction.
(c)
The electromagnetic wave propagates in +y direction, in straight line from origin to observer’s location and also it propagates +y direction after it reaches observer’s location.
Chapter 25 Electromagnetic Waves Q.5CQ
You want to check the time while wearing your Polaroid sunglasses. If you hold your forearm horizontally, you can read the time easily. If you hold your forearm vertically, however, so that you are looking at your watch sideways, you notice that the display is black. Explain.
Solution:
This is because, the light coming from the display is linearly polarized. If the polarization direction of the display and sunglasses align you can read the time. If these directions are at 90 to one another no light will pass through the sunglasses and the display will appear black.
When the direction of the linearly polarized light ray is parallel to the axis of the Polaroid then we can see the light .If it is perpendicular to the axis of the Polaroid we can not see light ray.
Here the watch is a LCD display watch that means the light coming from it will be a linearly polarized light .Now if our four arm is horizontal then the light coming from the watch is parallel to the Polaroid sun glass. Therefore we can see the display of the watch. Now if our four arms are vertical then the light coming from the watch is perpendicular to the Polaroid sun glass. Therefore we can not see the display of the watch .Therefore now the display becomes black.
Chapter 25 Electromagnetic Waves Q.5P
Give the direction (N, S, E, W, up, or down) of the missing quantity for each of the four electromagnetic waves listed in Table 25-1.
Solution:
Chapter 25 Electromagnetic Waves Q.6CQ
Polarization and the Ground Spider The ground spider Drassodes cupreus, like many spiders, has several pairs of eyes. It has been discovered that one of these pairs of eyes acts as a set of polarization filters, with one eye’s polarization direction oriented at 90° to the other eye’s polarization direction. In addition,
experiments show that the spider uses these eyes to aid in navigating to and from its burrow. Explain how such eyes might aid navigation.
Solution:
We know that the amount of light received by the polarizer is depend on the orientation of the polarizer with respect to the axis of the source .Here light from the sky is polarized perpendicular to the sun .Thus the amount of light received by the two polarizing eyes depend on the orientation of spider with respect to the sun. By monitoring the amount of light received by each eye, the spider can navigate to and from its burrow.
Chapter 25 Electromagnetic Waves Q.6P
Give the direction (±x, ±y, ±z) of the missing quantity for each of the four electromagnetic waves listed in Table 25-2.
Solution:
Chapter 25 Electromagnetic Waves Q.7CQ
The electromagnetic waves we pick up on our radios are typically polarized. In contrast, the indoor light we see every day is typically unpolarized. Explain.
Solution:
The indoor light we see generally from the sun and from light bulbs is unpolarized light because the atoms which are emitting the light can have any orientation relative to one another. Hence, even if individual atoms emit polarized light, the net result from a group of atoms is light with no preferred direction of polarization.
The radio stations generate their electromagnetic waves with the large vertical antennas. As a result, these waves are polarized in the vertical direction.
Chapter 25 Electromagnetic Waves Q.7P
Solution:
Chapter 25 Electromagnetic Waves Q.8CQ
You are given a sheet of Polaroid material. Describe how to determine the direction of its transmission axis if none is indicated on the sheet.
Solution:
We can determine the direction of its transmission axis by viewing a linearly polarized light whose direction of polarization is known.
For example, the light reflected from the horizontal has a polarization in the horizontal direction. If we view this light through the sheet and rotate it until we get the maximum intensity, at maximum intensity the transmission axis is horizontal.
Chapter 25 Electromagnetic Waves Q.8P
Solution:
Chapter 25 Electromagnetic Waves Q.9CQ
Can sound waves be polarized? Explain.
Solution:
Sound waves are longitudinal waves. So, they cannot be polarized. In case of longitudinal propagation of the wave the particles can moves in only one direction i.e. back and forth along the direction of propagation. For a wave to be polarized there must be two directions of propagations. Hence the sound waves are not polarized.
Chapter 25 Electromagnetic Waves Q.9P
Three electromagnetic waves have electric and magnetic fields pointing in the directions shown in Figure 25-23. For each of the three cases, state whether the wave propagates in the +x, −x, +y, −y, +z, or −z direction.
Solution:
From the figure (1) using right hand rule
The propagation of the wave is along x-direction because the electric field is along the y-direction and magnetic field is along z-direction.
From the figure (2) using right hand rule
The propagation of the wave is along the +z-direction because the electric field is along the x-direction and magnetic field is along y-direction.
From the figure (3) using the right hand rule
The propagation of the wave is along the -x – direction because the electric field is along the z-direction and magnetic field is along y-direction.
Chapter 25 Electromagnetic Waves Q.10CQ
At a garage sale you find a pair of “Polaroid” sunglasses priced to sell. You are not sure, however, if the glasses are truly Polaroid, or if they simply have tinted fenses. How can you tell which is the case? Explain.
Solution:
We observe the light from the pair of Polaroid sunglasses and rotate one of the glass with respective to the other. If the intensity of the light is varying then they are Polaroid glasses. If they are tinted lenses then the intensity of the light does not change.
Chapter 25 Electromagnetic Waves Q.10P
The light-year (ly) is a unit of distance commonly used in astronomy. It is defined as the distance traveled by light in a vacuum in one year. (a) Express 1 ly in km. (b) Express the speed of light, c, in units of ly per year. (c) Express the speed of light in feet per nanosecond.
Solution:
Chapter 25 Electromagnetic Waves Q.11CQ
3-D Movies Modern-day 3-D movies are produced by projecting two different images onto the screen, with polarization directions that are at 90° relative to one another. Viewers must wear headsets with polarizing filters to experience the 3-D effect. Explain how this works.
Solution:
The two projected images give a view of a scene from slightly different angles, just as our eyes view a three dimensional object from different angles. Without headsets, the screen is confusing superposition of the two images. With the headsets our right eye sees one view of the scene and our left eye sees the other view. As these views are combined in our brain, then we can experience a realistic three dimensional effect.
Chapter 25 Electromagnetic Waves Q.11P
Alpha Centauri, the closest star to the sun, is 4.3 ly away. How far is this in meters?
Solution:
Chapter 25 Electromagnetic Waves Q.12P
Mars Rover When the Mars rover was deployed on the surface of Mars in July 1997, radio signals took about 12 min to travel from Earth to the rover. How far was Mars from Earth at that time?
Solution:
Chapter 25 Electromagnetic Waves Q.13P
A distant star is traveling directly away from Earth with a speed of 37,500 km/s. By what factor are the wavelengths in this star’s spectrum changed?
Solution:
Chapter 25 Electromagnetic Waves Q.14P
A distant star is traveling directly toward Earth with a speed of 37,500 km/s. (a) When the wavelengths in this star’s spectrum are measured on Earth, are they greater or less than the wavelengths we would find if the star were at rest relative to us? Explain. (b) By what fraction are the wavelengths in this star’s spectrum shifted?
Solution:
Given data:
Velocity of the star u=37,500 km/s
(a) ) When the star is moving towards the Earth, the frequencies of the electromagnetic wave increases. But, since the wave length are inversely proportional to the frequencies. So the measured wavelengths are less than what they would be if the stars are at rest relative to us.
Chapter 25 Electromagnetic Waves Q.15P
The frequency of light reaching Earth from a particular galaxy is 12% lower than the frequency the light had when it was emitted. (a) Is this galaxy moving toward or away from Earth? Explain. (b) What is the speed of this galaxy relative to the Earth? Give your answer as a fraction of the speed of light.
Solution:
Chapter 25 Electromagnetic Waves Q.16P
Measuring the Speed of Light Galileo attempted to measure the speed of light by measuring the time elapsed between his opening a lantern and his seeing the light return from his assistant’s lantern. The experiment is illustrated in Figure 25–24.
What distance, d, must separate Galileo and his assistant in order for the human reaction time, ∆t = 0.2 s, to introduce no more than a 15% error in the speed of light?
Solution:
Chapter 25 Electromagnetic Waves Q.17P
Measuring the Speed of Light: Michelson In 1926, Albert Michelson measured the speed of light with a technique similar to that used by Fizeau. Michelson used an eight-sided mirror rotating at 528 rev/s in place of the toothed wheel, as illustrated in Figure 25-25. The distance from the rotating mirror to a distant reflector was 35.5 km. If the light completed the 71.0-km round trip in the time it took the mirror to complete one-eighth of a revolution, what is the speed of light?
Solution:
Chapter 25 Electromagnetic Waves Q.18P
Communicating with the Voyager Spacecraft When the Voyager I and Voyager II spacecraft were exploring the outer planets, NASA flight controllers had to plan the crafts’ moves well in advance. How many seconds elapse between the time a command is sent from Earth and the time the command is received by Voyager at Neptune? Assume the distance from. Earth to Neptune is 4.5 × 1012 m.
Solution:
Chapter 25 Electromagnetic Waves Q.19P
A father and his daughter are interested in the same baseball game. The father sits next to his radio at home and listens to the game; his daughter attends the game and sits in the outfield bleachers. In the bottom of the ninth inning a home run is hit. If the father’s radio is 132 km from the radio station, and the daughter is 115 m from home plate, who hears the home run first? (Assume that there is no time delay between the baseball being hit and its sound being broadcast by the radio station. In addition, let the speed of sound in the stadium be 343 m/s.)
Solution:
Chapter 25 Electromagnetic Waves Q.20P
(a) How fast would a motorist have to be traveling for a yellow (λ = 590 nm) traffic light to appear green (λ = 550 nm) because of the Doppler shift? (b) Should the motoristbe traveling toward or away from the traffic light to see this effect? Explain.
Solution:
Chapter 25 Electromagnetic Waves Q.21P
Most of the galaxies in the universe are observed to be moving away from Earth. Suppose a particular galaxy emits orange light with a frequency of 5.000 × 1014 Hz. If the galaxy is receding from Earth with a speed of 3325 km/s, what is the frequency of the light when it reaches Earth?
Solution:
Chapter 25 Electromagnetic Waves Q.22P
Two starships, the Enterprise and the Constitution, are approaching each other head-on from a great distance. The separation between them is decreasing at a rate of 782.5 km/s. The Enterprise sends a laser signal toward the Constitution. If the Constitution observes a wavelength λ = 670.3 nm, what wavelength was emitted by the Enterprise?
Solution:
Chapter 25 Electromagnetic Waves Q.23P
Baseball scouts often use a radar gun to measure the speed of a pitch. One particular model of radar gun emits a microwave signal at a frequency of 10.525 GHz. What will be the increase in frequency if these waves are reflected from a 90.0-mi/h fastball headed straight toward the gun? (Note: 1 mi/h = 0.447 m/s)
Solution:
Chapter 25 Electromagnetic Waves Q.24P
A state highway patrol car radar unit uses a frequency of 8.00 × 109 Hz. What frequency difference will the unit detect from a car receding at a speed of 44.5 m/s from a stationary patrol car?
Solution:
Chapter 25 Electromagnetic Waves Q.25P
Consider a spiral galaxy that is moving directly away from Earth with a speed V = 3.600 × 105 m/s at its center, as shown in Figure 25-26. The galaxy is also rotating about its center, so that points in its spiral arms are moving with a speed v = 6.400 × 105 m/s relative to the center. If light with a frequency of 8.230 × 1014 Hz is emitted in both arms of the galaxy, what frequency is detected by astronomers observing the arm that is moving (a) toward and (b) away from Earth? (Measurements of this type are used to map out the speed of various regions in distant, rotating galaxies.)
Solution:
Chapter 25 Electromagnetic Waves Q.26P
A highway patrolman sends a 24.150-GHz radar beam toward a speeding car. The reflected wave is lower in frequency by 4.04 kHz. (a) Is the car moving toward or away from the radar gun? Explain. (b) What is the speed of the car? [Hint:For small values of x, the following approximation may be used: (1 + x)2≈ 1 + 2x.]
Solution:
Chapter 25 Electromagnetic Waves Q.27P
Dental X-rays X-rays produced in the dentist’s office typically have a wavelength of 0.30 nm. What is the frequency of these rays?
Solution:
Chapter 25 Electromagnetic Waves Q.28P
Find the frequency of blue light with a wavelength of 460 nm.
Solution:
Chapter 25 Electromagnetic Waves Q.29P
Yellow light has a wavelength λ = 590 nm. How many of these waves would span the 1.0-mm thickness of a dime?
Solution:
Chapter 25 Electromagnetic Waves Q.30P
How many red wavelengths (λ = 705 nm) tall are you?
Solution:
Chapter 25 Electromagnetic Waves Q.31P
A cell phone transmite at a frequency of 1.75 × 108 Hz. What is the wavelength of the electromagnetic wave used by this phone?
Solution:
Chapter 25 Electromagnetic Waves Q.32P
Human Radiation Under normal conditions, humans radiate electromagnetic waves with a wavelength of about 9.0 microns. (a) What is the frequency of these waves? (b) To what portion of the electromagnetic spectrum do these waves belong?
Solution:
Chapter 25 Electromagnetic Waves Q.33P
UV Radiation. Ultraviolet light is typically divided into three categories. UV-A, with wavelengths between 400 nm and 320 nm, has been linked with malignant melanomas. UV-B radiation, which is the primary cause of sunburn and other skin cancers, has wavelengths between 320 nm. and 280 nm. Finally, the region known as UV-C extends to wavelengths of 100 nm. (a) Find the range of frequencies for UV-B radiation. (b) In which of these three categories does radiation with a frequency of 7.9 × 1014 Hz belong?
Solution:
Chapter 25 Electromagnetic Waves Q.34P
Communicating with a Submarine Normal radiofrequency waves cannot penetrate more than a few meters below the surface of the ocean. One method of communicating with submerged submarines uses very low frequency (VLF) radio waves. What is the wavelength (in air) of a 10.0-kHz VLF radio wave?
Solution:
Chapter 25 Electromagnetic Waves Q.35P
Solution:
Chapter 25 Electromagnetic Waves Q.36P
(a) Whichcolor of light has the higher frequency, red or violet? (b) Calculate the frequency of blue light with a wavelength of 470 nm, and red light with a wavelength of 680 nm.
Solution:
Chapter 25 Electromagnetic Waves Q.37P
ULF (ultra low frequency) electromagnetic waves, produced in the depths of outer space, have been observed with wavelengths inexcess of 29 million kilometers. What is the period of such a wave?
Solution:
Chapter 25 Electromagnetic Waves Q.38P
A television is tuned to a station broadcasting at a frequency of 6.60 × 107 Hz. For best reception, the rabbit-ear antenna used by the TV should be adjusted to have a tip-to-tip length equal to half a wavelength of the broadcast signal. Find the optimum length of the antenna.
Solution:
Chapter 25 Electromagnetic Waves Q.39P
An AM radio station’s antenna is constructed to be λ/4 tall, where λ is the wavelength of the radio waves. How tall should the antenna be for a station broadcasting at a frequency of 810 kHz?
Solution:
Chapter 25 Electromagnetic Waves Q.40P
As you drive by an AM radio station, you notice a sign saying that its antenna is 112 m high. If this height represents one quarter-wavelength of its signal, what is the frequency of the station?
Solution:
Chapter 25 Electromagnetic Waves Q.41P
Find the difference in wavelength (λ1 − λ2) for each of the following pairs of radio waves: (a) f1 = 50 kHz and f2 = 52 kHz, (b) f1 = 500 kHz and f2 = 502 kHz.
Solution:
Chapter 25 Electromagnetic Waves Q.42P
Find the difference in frequency (f1 − f2) for each of the following pairs of radio waves: (a) λ1 = 300.0 m and λ2 = 300.5 m, (b) λ1 = 30.0 m and λ2 = 30.5 m.
Solution:
Chapter 25 Electromagnetic Waves Q.43P
If the rms value of the electric field in an electromagnetic wave is doubled, (a) by what factor does the rms value of the magnetic field change? (b) By what factor does the average intensity of the wave change?
Solution:
a) If the rms value of the electric field doubled then the rms value of the magnetic field also get doubled since the electric and magnetic fields in an electromagnetic field are proportional to one another according to the relation.E = cB
b) The intensity of an electromagnetic wave depends on the square of the electric and magnetic fields. As a result, doubling the fields quadruples the intensity of the wave.
Chapter 25 Electromagnetic Waves Q.44P
The radiation pressure exerted by beam of light 1 is half the radiation pressure of beam of light 2. If the rms electric field of beam 1 has the value E0, what is the rms electric field in beam 2?
Solution:
Chapter 25 Electromagnetic Waves Q.45P
The maximum magnitude of the electric field in an electromagnetic wave is 0.0400 V/m. What is the maximum magnitude of the magnetic field in this wave?
Solution:
Chapter 25 Electromagnetic Waves Q.46P
What is the rms value of the electric field in a sinusoidal electromagnetic wave that has a maximum electric field of 88 V/m?
Solution:
Chapter 25 Electromagnetic Waves Q.47P
The magnetic field in an electromagnetic wave has a peak value given by B = 3.7 μ T. For this wave, find (a) the peak electric field strength, (b) the peak intensity, and (c) the average intensity.
Solution:
Chapter 25 Electromagnetic Waves Q.48P
What is the maximum value of the electric field in an electromagnetic wave whose maximum intensity is 5.00 W/m2?
Solution:
Chapter 25 Electromagnetic Waves Q.49P
What is the maximum value of the electric field in an ele tromagnetic wave whose average intensityis 5.00 W/m2?
Solution:
Chapter 25 Electromagnetic Waves Q.50P
Electromagnetic wave 1 has a maximum electric field of E0 = 52 V/m, and electromagnetic wave 2 has a maximum magnetic field of B0 = 1.5 μ T.(a) Which wave has the greater intensity? (b) Calculate the intensity of each wave.
Solution:
Chapter 25 Electromagnetic Waves Q.51P
A 65-kW radio station broadcasts its signal uniformly in all directions. (a) What is the average intensity of its signal at a distance of 250 m from the antenna? (b) What is the average intensity of its signal at a distance of 2500 m from the antenna?
Solution:
Chapter 25 Electromagnetic Waves Q.52P
At what distance will a 45-W lightbulb have the same apparent brightness as a 120-W bulb viewed from a distance of 25 m? (Assume that both bulbs convert electrical power to light with the same efficiency, and radiate light uniformly in all directions.)
Solution:
Chapter 25 Electromagnetic Waves Q.53P
What is the ratio of the sunlight intensity reaching Pluto compared with the sunlight intensity reaching Earth? (On average, Pluto is 39 times as far from the Sun as is Earth.)
Solution:
Chapter 25 Electromagnetic Waves Q.54P
In the following, assume that lightbulbs radiate uniformly in all directions and that 5.0% of their power is converted to light. (a) Find the average intensity of light at a point 2.0 m from a 120-W red lightbulb (λ = 710 nm). (b) Is the average intensity 2.0 m from a 120-W blue lightbulb (λ = 480 nm) greater than, less than, or the same as the intensity found in part (a)? Explain. (c) Calculate the average intensity for part (b).
Solution:
Chapter 25 Electromagnetic Waves Q.55P
A 5.0-mW laser produces a narrow beam of light. How much energy is contained in a 1.0-m length of its beam?
Solution:
Chapter 25 Electromagnetic Waves Q.56P
What length of a 5.0-mW laser’s beam will contain 9.5 mJ of energy?
Solution:
Chapter 25 Electromagnetic Waves Q.57P
Sunlight Intensity After filtering through the atmosphere, the Sun’s radiation illuminates Earth’s surface with an average intensity of 1.0 kW/m2. Assuming this radiation strikes the 15-m × 45-m black, flat roof of a building at normal incidence, calculate the average force the radiation exerts on the roof.
Solution:
Chapter 25 Electromagnetic Waves Q.58P
(a) Find the electric and magnetic field amplitudes inan electromagnetic wave that has an average energy density of 1.0 J/m3. (b) By what factor must the field amplitudes be increased if the average energy density is to be doubled to 2.0 J/m3?
Solution:
Chapter 25 Electromagnetic Waves Q.59P
Lasers for Fusion Two of the most powerful lasers in the world are used in nuclear fusion experiments. The NOVA laser produces 40.0 kJ of energy in a pulse that lasts 2.50 ns, and the N1F laser (under construction) will produce a 10.0-ns pulse with 3.00 MJ of energy. (a) Which laser produces more energy in each pulse? (b) Which laser produces the greater average power during each pulse? (c) If the beam diameters are the same, which laser produces the greater average intensity?
Solution:
Chapter 25 Electromagnetic Waves Q.60P
You are standing 2.5 m from a 150-W lightbulb. (a) If the pupil of your eye is a circle 5.0 mm in diameter, how much energy enters your eye per second? (Assume that 5.0% of the lightbulb’s power is converted to light.) (b) Repeat part (a) for the case of a 1.0-mm-diameter laser beam with a power of 0.50 mW.
Solution:
Chapter 25 Electromagnetic Waves Q.61P
Laser Safety A 0.75-mW laser emits a narrow beam of light that enters the eye, as shown in Figure 25−27. (a) How much energy is absorbed by the eye in 0.2 s? (b) The eye focuses this beam to a tiny spot on the retina, perhaps 5.0 μ m in diameter. What is the average intensity of light (in W/cm2) at this spot? (c) Damage to the retina can occur if the average intensity of light exceeds 1.0 × 10−2 W/cm2. By what factor has the intensity of this laser beam exceeded the safe value?
Solution:
Chapter 25 Electromagnetic Waves Q.62P
Find the rms electric and magnetic fields at a point 2.50 m from a lightbulb that radiates 75.0 W of light uniformly in all directions.
Solution:
Chapter 25 Electromagnetic Waves Q.63P
A 0.50-mW laser produces abeam of light with a diameter of 1.5 mm. (a) What is the average intensity of this beam? (b) At what distance does a 150-W lightbulb have the same average intensity as that found for the laser beam in part (a)? (Assume that 5.0% of the bulb’s power is converted to light.)
Solution:
Chapter 25 Electromagnetic Waves Q.64P
A laser emits a cylindrical beam of light 2.4 mm in diameter. If the average power of the laser is 2.8 mW, what is the rms value of the electric field in the laser beam?
Solution:
Chapter 25 Electromagnetic Waves Q.65P
(a) If the laser in Problem 64 shines its light on a perfectly absorbing surface, how much energy does the surface receive in 12 s? (b) What is the radiation pressure exerted by the beam?
Solution:
Chapter 25 Electromagnetic Waves Q.66P
Laser Surgery Each pulse produced by an argon-fluoride excimer laser used in PRK and LASIK ophthalmic surgery lasts only 10.0 ns but delivers an energy of 2.50 mJ. (a) What is the power produced during each pulse? (b) If the beam has a diameter of 0.850 mm, what is the average intensity of the beam during each pulse? (c) If the laser emits 55 pulses per second, what is the average power it generates?
Solution:
Chapter 25 Electromagnetic Waves Q.67P
A pulsed laser produces brief bursts of light. One such laser emits pulses that carry 0.350 J of energy but last only 225 fs. (a) What is the average power during one of these pulses? (b) Assuming the energy is emitted in a cylindrical beam of light 2.00 mm in diameter, calculate the average intensity of this laser beam. (c) What is the rms electric field in this wave?
Solution:
Chapter 25 Electromagnetic Waves Q.68P
Consider the two polarization experiments shown in Figure 25–28. (a) If the incident light is unpolar-ized, is the transmitted intensity in case A greater than, less than, or the same as the transmittedintensity in case B? (b) Choose the best explanation from among the following:
I. The transmitted intensify is the same in either case; the first polarizer lets through one-half the incident intensity, and the second polarizer is at an angle θ relative to the first.
II. Case A has a smaller transmitted intensity than case B because the first polarizer is at an angle θ relative to the incident beam.
III. Case B has a smaller transmitted intensity than case A because the direction of polarization is rotated by an angle θ in the clockwise direction in case B.
Solution:
Chapter 25 Electromagnetic Waves Q.69P
Consider the two polarization experiments shown in Figure 25–28. (a) If the incident light is polarized in the horizontal direction, is the transmitted intensity in case A greater than, less than, or the same as the transmitted intensity in case B? (b) Choose the best explanation from among the following:
I. The two cases havethe same transmitted intensity because the angle between the polarizers is θ in each case.
II. The transmitted intensity is greater in case B because all of the initial beam gets through the first polarizer.
III. The transmitted in tensi ty in case B is smaller than in case A; in fact, the transmitted intensity in case B is zero because the first polarizer is oriented vertically.
Solution:
Chapter 25 Electromagnetic Waves Q.70P
Suppose linearly polarized light is incident on the polarization experiments shown in Figure25–28. In what direction, relative to the vertical, must the incident light be polarized if the transmitted intensity is to be the same in both experiments? Explain.
Solution:
Chapter 25 Electromagnetic Waves Q.71P
An incident beam of light with an intensityI0 passes through a polarizing filter whose transmissionaxis is at an angle θ to the vertical. As the angle is changed from θ = 0to θ = 90°, the intensity as a function of angle is given by one of the curves in Figure 25–29. Give the color of the curve corresponding to an incident beam that is (a) unpolarized, (b) vertically polarized, and (c) horizontally polarized.
Solution:
a) The color of the curve corresponding to an incident beam that is unpolarized is Green. Unpolarized light will be reduced in intensity by a factor of 2 regardless of the angle of the transmission axis.
b) The color of the curve corresponding to an incident beam that is vertically polarized is Red. At θ = 0 we have complete transmission, meaning that the light is vertically polarized.
c) The color of the curve corresponding to an incident beam that is horizontally polarized is Blue. At θ = 0 there is no transmission, which means the incident light is horizontally polarized.
Chapter 25 Electromagnetic Waves Q.72P
Vertically polarized light with an intensity of 0.55 W/m2 passes through a polarizer whose transmission axis is at an angle of 65.0° with the vertical. Whatis the intensity of the transmitted light?
Solution:
Chapter 25 Electromagnetic Waves Q.73P
A person riding in a boat observes that the sunlight reflected by the water is polarized parallel to the surface of the water. The person is wearing polarized sunglasses with the polarization axis vertical. If the wearer leans at an angle of 21.5° to the vertical, what fraction of the reflected light intensity will pass through the sunglasses?
Solution:
Chapter 25 Electromagnetic Waves Q.74P
Unpolarized light passes through two polarizers whose transmission axes are at an angle of 30.0° with respect to each other. What fraction of the incident intensity is transmitted through the polarizers?
Solution:
Chapter 25 Electromagnetic Waves Q.75P
In Problem, what should be the angle between the transmission axes of the polarizers if it is desired that one-tenth of the incident intensity be transmitted?
Solution:
Chapter 25 Electromagnetic Waves Q.76P
Unpolarized iight with intensity I0 falls on a polarizing filter whose transmission axis is vertical. The axis of a second polarizing filter makes an angle of θ with the vertical. Plot a graph that shows the intensity of light transmitted by the second filter (expressed as a fraction of I0) as a function of θ. Your graph should cover the range θ = 0° to θ = 360°.
Solution:
Chapter 25 Electromagnetic Waves Q.77P
A beam of vertically polarized light encounters two polarizing filters, as shown in Figure 25–30. (a) Rank the three cases, A, B, and C, in order of increasing transmitted intensity. Indicate ties where appropriate. (b) Calculate the transmitted intensity for each of the cases in Figure 25–30, assuming that the incident intensity is 37.0 W/m2. Verify that your numerical results agree with the rankings in part (a).
Solution:
Chapter 25 Electromagnetic Waves Q.78P
This time assuming that the polarizers to the left in Figure 25–30 are at an angle of 22.5° to the vertical rather than 45°. The incident intensity is again 37.0 W/m2.
Solution:
Chapter 25 Electromagnetic Waves Q.79P
BIO Optical Activity Optically active molecules have the property of rotating the direction of polarization of linearly polarized light. Many biologically important molecules have this property, some causing a counterclockwise rotation (negative rotation angle), others causing a clockwise rotation (positive rotation angle). For example, a 5.00 gram per 100 mL solution of l-leucine causes a rotation of −0.550°; the same concentration of d-glutamic acid causes a rotation of 0.620”. (a) If placed between crossed polarizers, which of these solutions transmits the greater intensity? Explain. (b) Find the transmitted intensity for each of these solutions when placed between crossed polarizers. The incident beam is unpolarized and has an intensity of 12.5 W/m2.
Solution:
Chapter 25 Electromagnetic Waves Q.80P
A helium-neon laser emits a beam of unpolarized light that passes through three Polaroid filters, as shown in Figure 25–31. The intensity of the laser beam is I0. (a) What is the intensity of the beam at point A? (b) What is the intensity of the beam at point B? (c) What is the intensity of the beam at point C? (d) If filter 2 is removed, what is the intensity of the beam at point C?
Solution:
Chapter 25 Electromagnetic Waves Q.81P
Referring to Figure 25–31, suppose that filter 3 is at a general angle θ with the vertical, rather than the angle 90°. (a) Find an expression for the transmitted intensity as a function of θ. (b) Plot your result from part (a), and determine the maximum transmitted intensity. (c) At what angle θ does maximum transmission occur?
Solution:
Chapter 25 Electromagnetic Waves Q.82GP
Suppose the magnitude of the electric field in an electro-magnetic wave is doubled. (a) By what factor does the magnitude of the magnetic field change? (b) By what factor does the maximum intensity of the wave change?
Solution:
Chapter 25 Electromagnetic Waves Q.83GP
If “sailors” of the future use radiation pressure to propel their ships, should the surfaces of their sails be absorbing or re fleeting? Explain.
Solution:
We know that there will be a greater transfer of momentum when the beam is reflected than when it is merely absorbed. Thus if the sailors of the future use the radiation pressure to propel their ships then surface of their sails should be reflecting.
Chapter 25 Electromagnetic Waves Q.84GP
Sunlight at the surface of Earth has an average intensity of about 1.00 × 103 W/m2. Find the rms values of the electric and magnetic fields in the sunlight.
Solution:
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Chapter 25 Electromagnetic Waves Q.85GP
A typical medical X-ray has a frequency of 1.50 × 1019 Hz. What is the wavelength of such an X-ray?
Solution:
Chapter 25 Electromagnetic Waves Q.86GP
How many hydrogen atoms, 0.10 nm indiameter, must be placed end to end to fit into one wavelength of 410-nm violet light?
Solution:
Chapter 25 Electromagnetic Waves Q.87GP
Radiofrequency Ablation In radiofrequency (RF) ablation, a small needle is inserted into a cancerous tumor. When radiofrequency oscillating currents are sent into the needle, ions in the neighboring tissue respond by vibrating rapidly, causing local heating to temperatures as high as 100°C. This kills the cancerous cells and, because of the small size of the needle, relatively few of the surrounding healthy cells. A typical RF ablation treatment uses a frequency of 750 kHz. What is the wavelength that such radio waves would have in a vacuum?
Solution:
Chapter 25 Electromagnetic Waves Q.88GP
Figure 25–32 shows four polarization experiments in which unpolarized incident light passes throughtwo polarizing filters with different orientations. Rank the four cases in order of increasing amount of transmitted light. Indicate ties where appropriate.
Solution:
Chapter 25 Electromagnetic Waves Q.89GP
(a) What minimum intensity must a laser beam have if it is to levitate a tiny black (100% absorbing) sphere of radius r = 0.5 μ m and mass = 1.6 × 10−15 kg? Comment on the feasibility of such levitation. (b) If the radius of the sphere is doubled but its mass remains the same, will the minimum intensity be greater than, less than, or equal to the value found in part (a)? Explain. (c) Find the minimum intensity for the situation described in part (b).
Solution:
Chapter 25 Electromagnetic Waves Q.90GP
The Apollo 11 Reflector One of the experiments placed on the Moon’s surface by Apollo 11 astronauts was a reflector that is used to measure the Earth–Moon distance with high accuracy. A laser beam on Earth is bounced off the reflector, and its round-trip travel time is recorded. If the travel time can be measured to within an accuracy of 0.030 ns, what is the uncertainty in the Earth–Moon distance?
Solution:
Chapter 25 Electromagnetic Waves Q.91GP
The Hβ line of the hydrogen atom’s spectrum has a normal wavelength λβ = 486 nm. This same line is observed in the spectrum of a distant quasar, but lengthened by 20.0 nm. What is the speed of the quasar relative to Earth, assuming it is moving along our line of sight?
Solution:
Chapter 25 Electromagnetic Waves Q.92GP
Suppose the distance to the fixed mirror in Figure 25–25 is decreased to 20.5 km. (a) Should the angular speed of the rotating mirror be increased or decreased to ensure that the experiment works as described in Problem 17? (b) Find the required angular speed, assuming the speed of light is 3.00 × 108 m/s.
Solution:
Chapter 25 Electromagnetic Waves Q.93GP
Suppose the speed of the galaxy in Problem is increased by a factor of 10; that is, V = 3.600 × 106 m/s. The speed of the arms, v, and the frequency of the light remain the same. (a) Does the arm near the top of Figure 25–26 show a red shift (toward lower frequency) or a blue shift (toward higher frequency)? Does the lower arm show a red or a blue shift? Explain. What frequency is detected by astronomers observing (b) the upper arm and (c) the lower arm?
Solution:
Chapter 25 Electromagnetic Waves Q.94GP
Consider the physical situation illustrated in Figure 25–27. (a) Is Erms in the incident laser beam greater than, less than, or the same as Erms where the beam hits the retina? Explain. (b) If the intensity of the beam at the retina is equal to the damage threshold, 1.0 × 10−2 W/cm2, what is the value of Erms at that location? (c) If the diameter of the spot on the retina is reduced by a factor of 2, by what factor does the intensity increase? By what factor does Erms increase?
Solution:
Chapter 25 Electromagnetic Waves Q.95GP
Polaroid Vision in a Spider Experiments show that the ground spider Drassodes cupreus uses one of its several pairs of eyes as a polarization detector. In fact, the two eyes in this pair have polarization directions that are at right angles to one another. Suppose linearly polarized light with an intensity of 825 W/m2 shines from the sky onto the spider, and that the intensity transmitted by one of the polarizing eyes is 232 W/m2. (a) For this eye, what is the angle between the polarization direction of the eye and the polarization direction of the incident light? (b) What is the intensity transmitted by the other polarizing eye?
Solution:
Chapter 25 Electromagnetic Waves Q.96GP
A state highway patrol car radar unit uses a frequency of 9.00 × 109 Hz. What frequency difference will the unit detect from a car approaching a parked patrol car with a speed of 35.0 m/s?
Solution:
Chapter 25 Electromagnetic Waves Q.97GP
What is the ratio of the sunlight intensity reaching Mercury compared with the sunlight intensity reaching Earth? (On average, Mercury’s distance from the Sun is 0.39 that of Earth’s.)
Solution:
Chapter 25 Electromagnetic Waves Q.98GP
What area is needed for a solar collector to absorb 45.0 kW of power from the Sun’sradiation if the collector is 75.0% efficient? (At the surface of Earth, sunlight has an average intensity of 1.00 × 103 W/m2.)
Solution:
Chapter 25 Electromagnetic Waves Q.99GP
Near-Infrared Brain Scans Light in the near-infrared (close to visible red) can penetrate surprisingly far through human tissue, a fact that is being used to “illuminate” the interior of the brain in a noninvasive technique known as near-infrared spectroscopy (NIRS). In this procedure, illustrated in Figure 25–33, an optical fiber carrying a beam of infrared laser light with a power of 1.5 mW and a cross-sectional diameter of 1.2 mm is placed against the skull. Some of the light enters the brain, where it scatters from hemoglobin in the blood. The scattered light is picked up by a detector and analyzed by a computer. (a) According to the Beer-Lambert law, the intensity of light, I, decreases with penetration distance, d, as I = I0e−μd where I0 is the initial intensify of the beam and μ = 4.7 cm−1 for a typical case. Find the intensity ofthe laser beam after it penetrates through 3.0 cm of tissue. (b) Find the electric field of the initial light beam.
Solution:
Chapter 25 Electromagnetic Waves Q.100GP
Three polarizers are arranged as shown in Figure 25–31. If the incident beam of light is unpolarized and has an intensity of 1.60 W/m2, find the transmitted intensify (a) when θ2 = 25.0° and θ3 = 50.0°, and (b) when θ2 = 50.0° and θ3 = 25.0°.
Solution:
Chapter 25 Electromagnetic Waves Q.101GP
This time assuming an incident beam that is vertically polarized. The intensity of the incident beam is 1.60 W/m2.
Solution:
Chapter 25 Electromagnetic Waves Q.102GP
A lightbulb emits light uniformly in all directions. If the rms electric field of this light is 16.0 N/C at a distance of 1.35 m from the bulb, what is the average total power radiated by the bulb?
Solution:
Chapter 25 Electromagnetic Waves Q.103GP
A beam of light is a mixture of unpolarized light with intensity Iu and linearly polarized light with intensity Ip. The polarization direction for the polarized light is vertical. When this mixed beam of light is passed through a polarizer that is vertical, the transmitted intensity is 1.6.8 W/m2; when the polarizer is at an angle of 55.0° with the vertical, the transmitted intensity is 8.68 W/m2. (a) Is Iu greater than, less than, or equal to Ip? Explain. (b) Calculate Iu and Ip.
Solution:
Chapter 25 Electromagnetic Waves Q.104GP
As mentioned in Problem 95, one pair of eyes in a particular species of ground spider has polarization directions that are at right angles to one another. Suppose that linearly polarized light is incident on such a spider. (a) Prove that the transmitted intensity of one eye plus the transmitted intensity from the other eye is equal to the incident intensity. (b) If the transmitted intensities for the two eyes are 163 W/m2 and 662 W/m2, through what angle must the spider rotate to make the transmitted intensities equal to one another?
Solution:
Chapter 25 Electromagnetic Waves Q.105GP
A typical home may require a total of 2.00 × 103 kWh of energy per month. Suppose you would like to obtain this energy from sunlight, which has an average daylight intensity of 1.00 × 103 W/m2. Assuming that sunlight is available 8.0 h per day, 25 d per month (accounting for cloudy days), and that you have a way to store energy from your collector when the Sun isn’t shining, determine the smallest collector size that will provide the needed energy, given a conversion efficiency of 25%.
Solution:
Chapter 25 Electromagnetic Waves Q.106GP
At the top of Earth’s atmosphere, sunlight has an average intensify of 1360 W/m2. If the average distance from Earth to the Sun is 1.50 × 1011 m, at what rate does the Sun radiate energy?
Solution:
Chapter 25 Electromagnetic Waves Q.107GP
A typical laser used in introductory physics laboratories produces a continuous beam of light about 1.0 mm in diameter. The average power of such a laser is 0.75 mW. What are (a) the average intensity, (b) the peak intensity, and (c) the average energy density of this beam? (d) If the beam is reflected from a mirror, what is the maximum force the laser beam can exert on it? (e) Describe the orientation of the laser beam relative to the mirror for the case of maximum force.
Solution:
Chapter 25 Electromagnetic Waves Q.108GP
Four polarizers are set up so that the transmission axis of each successive polarizer is rotated clockwise by an angle θ relative to the previous polarizer. Find the angle θ for whichunpolarized light is transmitted through these four polarizers with its intensity reduced by a factor of 25.
Solution:
Chapter 25 Electromagnetic Waves Q.109GP
Optical Activity of Sugar The sugar concentration in a solution (e.g., in a urine specimen) can be measured conveniently by vising the optical activity of sugar and other asymmetric molecules. In general, anoptically active molecule, like sugar, will rotate the plane of polarization through an angle that is proportional to the thickness of the sample and to the concentration of the molecule. To measure the concentration of a given solution, a sample of known thickness is placed between two polarizing filters that are at right angles to each other, as shown in Figure 25–34. The intensity of light transmitted through the two filters can be compared with a calibration chart to determine the concentration. (a) What percentage of the incident (unpolarized) light will pass through the first filter? (b) If no sample is present, what percentage of the initial light will pass through the second filter? (c) When a particular sample is placed between the twofilters, the intensity of light emerging from the second filter is 40.0% of the incident intensity. Through what angle did the sample rotate the plane ofpolarization? (d) A second sample has half the sugar concentration of the first sample. Find the intensity of light emerging from the second filter in this
Solution:
Chapter 25 Electromagnetic Waves Q.110PP
An essential part of modern dentistry is visible-light curing (VLC), a procedure that hardens the restorative materials used in fillings, veneers, and other applications. These “curing lights” work by activating molecules known as photoinitiators within the restorative materials. The photoinitiators, in turn, start a process of polymerization that causes monomers to link together to form a tough, solid polymer network. Thus, with VLC a dentist can apply and shape soft restorative materials as desired, shine a bright light on the result as shown in Figure 25–35, and in 20 seconds have a completely hardened—or cured—final product.
The most common photoinitiator is camphoroquinone (CPQ). To cure CPQ in the least time, one should illuminate it with light having a wavelength of approximately 465 nm. Many VLC units use a halogen light, but there are some draw-backs to this approach. First, the filament in a halogen light is heated to a temperature of about 3000 K, which can cause heat degradation of components in the curing unit itself. Second less than 1 % of the energy given off by a halogen bulb is visible light, so a halogen bulb must have a high power rating to produce the desired light intensity.
More recently, VLC units have begun to use LEDs as their light source. These lights stay cool, emit their energy output as visible light at the desired wavelength, and provide light with an intensity as high as 1000 mW/cm2, which is about 10 times the intensity of sunlight on the surface of the Earth.
Solution:
Chapter 25 Electromagnetic Waves Q.111PP
An essential part of modern dentistry is visible-light curing (VLC), a procedure that hardens the restorative materials used in fillings, veneers, and other applications. These “curing lights” work by activating molecules known as photoinitiators within the restorative materials. The photoinitiators, in turn, start a process of polymerization that causes monomers to link together to form a tough, solid polymer network. Thus, with VLC a dentist can apply and shape soft restorative materials as desired, shine a bright light on the result as shown in Figure 25–35, and in 20 seconds have a completely hardened—or cured—final product.
The most common photoinitiator is camphoroquinone (CPQ). To cure CPQ in the least time, one should illuminate it with light having a wavelength of approximately 465 nm. Many VLC units use a halogen light, but there are some draw-backs to this approach. First, the filament in a halogen light is heated to a temperature of about 3000 K, which can cause heat degradation of components in the curing unit itself. Second less than 1 % of the energy given off by a halogen bulb is visible light, so a halogen bulb must have a high power rating to produce the desired light intensity.
More recently, VLC units have begun to use LEDs as their light source. These lights stay cool, emit their energy output as visible light at the desired wavelength, and provide light with an intensity as high as 1000 mW/cm2, which is about 10 times the intensity of sunlight on the surface of the Earth.
Solution:
Chapter 25 Electromagnetic Waves Q.112PP
An essential part of modern dentistry is visible-light curing (VLC), a procedure that hardens the restorative materials used in fillings, veneers, and other applications. These “curing lights” work by activating molecules known as photoinitiators within the restorative materials. The photoinitiators, in turn, start a process of polymerization that causes monomers to link together to form a tough, solid polymer network. Thus, with VLC a dentist can apply and shape soft restorative materials as desired, shine a bright light on the result as shown in Figure 25–35, and in 20 seconds have a completely hardened—or cured—final product.
The most common photoinitiator is camphoroquinone (CPQ). To cure CPQ in the least time, one should illuminate it with light having a wavelength of approximately 465 nm. Many VLC units use a halogen light, but there are some draw-backs to this approach. First, the filament in a halogen light is heated to a temperature of about 3000 K, which can cause heat degradation of components in the curing unit itself. Second less than 1 % of the energy given off by a halogen bulb is visible light, so a halogen bulb must have a high power rating to produce the desired light intensity.
More recently, VLC units have begun to use LEDs as their light source. These lights stay cool, emit their energy output as visible light at the desired wavelength, and provide light with an intensity as high as 1000 mW/cm2, which is about 10 times the intensity of sunlight on the surface of the Earth.
Solution:
Chapter 25 Electromagnetic Waves Q.113PP
An essential part of modern dentistry is visible-light curing (VLC), a procedure that hardens the restorative materials used in fillings, veneers, and other applications. These “curing lights” work by activating molecules known as photoinitiators within the restorative materials. The photoinitiators, in turn, start a process of polymerization that causes monomers to link together to form a tough, solid polymer network. Thus, with VLC a dentist can apply and shape soft restorative materials as desired, shine a bright light on the result as shown in Figure 25–35, and in 20 seconds have a completely hardened—or cured—final product.
The most common photoinitiator is camphoroquinone (CPQ). To cure CPQ in the least time, one should illuminate it with light having a wavelength of approximately 465 nm. Many VLC units use a halogen light, but there are some draw-backs to this approach. First, the filament in a halogen light is heated to a temperature of about 3000 K, which can cause heat degradation of components in the curing unit itself. Second less than 1 % of the energy given off by a halogen bulb is visible light, so a halogen bulb must have a high power rating to produce the desired light intensity.
More recently, VLC units have begun to use LEDs as their light source. These lights stay cool, emit their energy output as visible light at the desired wavelength, and provide light with an intensity as high as 1000 mW/cm2, which is about 10 times the intensity of sunlight on the surface of the Earth.
Solution:
Chapter 25 Electromagnetic Waves Q.114IP
Suppose the incident beam of light is linearly polarized in the same direction θ as the transmission axis of the analyzer. The transmission axis of the polarizer remains vertical. (a) What value must θ have if the transmitted intensity is to be 0.200 I0? (b) If θ is increased from the value found in part (a), docs the transmittedintensity increase, decrease, or stay the same? Explain.
Solution:
Chapter 25 Electromagnetic Waves Q.115IP
Suppose the incident beam of light is linearly polarized in the vertical direction. In addition, the transmission axis of the analyzer is at an angle of 80.0° to the vertical. What angle should the transmission axis of the polarizer make with the vertical if the transmitted intensity is to be a maximum?
Solution:
