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Homework 6Due: 11:59pm on Sunday, April 24, 2011

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A Driven Series L-C Circuit

Learning Goal: To understand why a series L-C circuit acts like a short circuit at resonance.An AC source drives a sinusoidal current of amplitude and frequency into an inductor having inductance

and a capacitor having capacitance that are connected in series. The current as a function of time is

given by .

Part A

Recall that the voltages and across the inductor and capacitor are not in phase with the

respective currents and . In particular, which of the following statements is true for a sinusoidal

current driver?

ANSWER: and both lag their respective currents.

and both lead their respective currents.

lags and leads .

leads and lags

Correct

The phase angle between voltage and current for inductors and capacitors is 90 degrees, or

radians. Among other things, this means that no power is dissipated in either the inductor or thecapacitor, since the time average of current times voltage, , is zero.

MasteringPhysics: Assignment Print View http://session.masteringphysics.com/myct/assignmentPrintView?assignme...

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Part B

What is , the voltage delivered by the current source?

Hint B.1 Current in a series circuit

Note that the current through the current source, the capacitor, and the inductor are all equal at all times.

Hint B.2 Find the voltage across the capacitor

What is , the voltage across the capacitor as a function of time?

Hint B.2.1 Find the charge on the capacitor

What is the charge on the capacitor?

Express your answer in terms of the capacitance and the voltage across it.

ANSWER: =

Answer Requested

Hint B.2.2 What is the current in terms of the charge on the capacitor?

With the conventions in the circuit diagram, what is the current on the capacitor in terms of its

voltage?

Express your answer in terms of the capacitance and the voltage across it and/or its

derivative .

ANSWER: =

Answer Requested

Integrate both sides of the equation (once you substitute in the

appropriate expression for ). The constant of integration is zero because there is no average

(DC) current or voltage in a pure AC circuit.

Express your answer in terms of , , , and .

ANSWER: =

Answer Requested

Hint B.3 What is the voltage across the inductor?


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What is , the voltage across the inductor as a function of time?

Hint B.3.1 Voltage and current for an inductor

Hint not displayed

Express in terms of , , and .

ANSWER: = Answer not displayed

Hint B.4 Total voltage

You should have defined the voltages across the capacitor and inductor in such a way that the totalvoltage is equal to , that is, so that total voltage is obtained by adding the voltages

at time , with careful attention paid to the signs.

Express in terms of some or all of the variables , and , or and , ,

and, of course, time .

ANSWER: =

Correct

Part C

With and the amplitudes of the voltages across the inductor and capacitor, which of the following

statements is true?

ANSWER: At very high frequencies and at very low frequencies .

At very high frequencies and at very low frequencies .

for all frequencies.


and are about the same at all frequencies.

Correct

Part D

The behavior of the L-C circuit provides one example of the phenomenon of resonance. The resonant

frequency is . At this frequency, what is the amplitude of the voltage supplied by the current


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source?

Express your answer using any or all of the constants given in the problem introduction.

ANSWER: = 0Correct

Part E

Which of the following statements best explains this fact that at the resonant frequency, there is zerovoltage across the capacitor and inductor?

ANSWER: The voltage is zero at all times because .

The voltages ; ; and are zero at all times.

The voltage is zero only when the current is zero.

The voltage is zero only at times when the current is stationary (at a max

or min).

Correct

At the resonant frequency of the circuit, the current source can easily push the current through theseries L-C circuit, because the circuit has no voltage drop across it at all! Of course, there is always avoltage across the inductor, and there is always a voltage across the capacitor, since they do have acurrent passing through them at all times; however, at resonance, these voltages are exactly out ofphase, so that the net effect is a current passing through the capacitor and the inductor without anyvoltage drop at all. The L-C series circuit acts as a short circuit for AC currents exactly at the resonantfrequency. For this reason, a series L-C circuit is used as a trap to conduct signals at the resonantfrequency to ground.

Voltage and Current in AC Circuits

Learning Goal: To understand the relationship between AC voltage and current in resistors, inductors, andcapacitors, especially the phase shift between the voltage and the current.

In this problem, we consider the behavior of resistors, inductors, and capacitors driven individually by asinusoidally alternating voltage source, for which the voltage is given as a function of time by

. The main challenge is to apply your knowledge of the basic properties of resistors,

inductors, and capacitors to these "single-element" AC circuits to find the current through each. The key

is to understand the phase difference, also known as the phase angle, between the voltage and the current. Itis important to take into account the sign of the current, which will be called positive when it flows clockwisefrom the b terminal (which has positive voltage relative to the a terminal) to the a terminal (see figure). Thesign is critical in the analysis of circuits containing combinations of resistors, capacitors, and inductors.


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Part A

First, let us consider a resistor with resistance connected to an AC source (diagram 1). If the AC source

provides a voltage , what is the current through the resistor as a function of time?

Hint A.1 Ohm's law

Hint not displayed

Express your answer in terms of , , , and .

ANSWER: =

Correct

Note that the voltage and the current are in phase; that is, in the expressions for and , the

arguments of the cosine functions are the same at any moment of time. This will not be the case forthe capacitor and inductor.

Part B

Now consider an inductor with inductance in an AC circuit (diagram 2). Assuming that the current in the

inductor varies as , find the voltage that must be driving the inductor.

Hint B.1 Kirchhoff's loop rule

Hint not displayed

Hint B.2 The derivative of

Hint not displayed

Hint B.3 The phase relationship between sine and cosine

Hint not displayed

Express your answer in terms of , , , and . Use the cosine function, not the sine function, in

your answer.


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ANSWER: =

Correct

Graphs of and are shown below. As you can see, for an inductor, the voltage leads (i.e.,

reaches its maximum before) the current by ; in other words, the current lags the voltage by . This

can be conceptually understood by thinking of inductance as giving the current inertia: The voltage"tries" to push current through the inductor, but some sort of inertia resists the change in current. Thisis another manifestation of Lenz's law. The difference is called the phase angle.

Part C

Again consider an inductor with inductance connected to an AC source. If the AC source provides a

voltage , what is the current through the inductor as a function of time?

Hint C.1 Using Part B

Hint not displayed

Express your answer in terms of , , , and . Use the cosine function, not the sine function, in

your answer.

ANSWER: =

Correct

For the amplitudes (magnitudes) of voltage and current, one can write (for the resistor) and

(for the inductor). If one compares these expressions, it should not come as a surprise that

the quantity , measured in ohms, is called inductive reactance; it is denoted by (sometimes

). It is called reactance rather than resistance to emphasize that there is no dissipation of energy.


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Using this notation, we can write (for a resistor) and (for an inductor). Also,

notice that the current is in phase with voltage when a resistor is connected to an AC source; in thecase of an inductor, the current lags the voltage by . What will happen if we replace the inductor with

a capacitor? We will soon see.

Part D

Consider the potentials of points a and b on the inductor in diagram 2. If the voltage at point b is greaterthan that at point a, which of the following statements is true?

Hint D.1 How to approach the problem

Hint not displayed

ANSWER: The current must be positive (clockwise).

The current must be directed counterclockwise.

The derivative of the current must be negative.

The derivative of the current must be positive.

Correct

It may help to think of the current as having inertia and the voltage as exerting a force that overcomesthis inertia. This viewpoint also explains the lag of the current relative to the voltage.

Part E

Assume that at time , the current in the inductor is at a maximum; at that time, the current flows frompoint b to point a. At time , which of the following statements is true?

Hint E.1 How to approach the problem

Hint not displayed

ANSWER: The voltage across the inductor must be zero and increasing.

The voltage across the inductor must be zero and decreasing.

The voltage across the inductor must be positive and momentarily constant.

The voltage across the inductor must be negative and momentarily constant.

Correct

Part F


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Now consider a capacitor with capacitance connected to an AC source (diagram 3). If the AC source

provides a voltage , what is the current through the capacitor as a function of

time?

Hint F.1 The relationship between charge and voltage for a capacitor

Hint not displayed

Hint F.2 The relationship between charge and current

Hint not displayed

Hint F.3 Mathematical details

Hint not displayed

Express your answer in terms of , , and . Use the cosine function with a phase shift, not the

sine function, in your answer.

ANSWER: =

Correct

For the amplitude values of voltage and current, one can write . If one compares this

expression with a similar one for the resistor, it should come as no surprise that the quantity ,

measured in ohms, is called capacitive reactance; it is denoted by (sometimes ). It is called

reactance rather than resistance to emphasize that there is no dissipation of energy. Using thisnotation, we can write , and voltage lags current by radians (or 90 degrees). The

notation is analogous to for a resistor, where voltage and current are in phase, and

for an inductor, where voltage leads current by radians (or 90 degrees). We see, then,

that in a capacitor, the voltage lags the current by , while in the case of an inductor, the current lags

the voltage by the same quantity . In a capacitor, where voltage lags the current, you may think of

the current as driving the change in the voltage.

Part G

Consider the capacitor in diagram 3. Which of the following statements is true at the moment the alternatingvoltage across the capacitor is zero?

Hint G.1 How to approach the problem

Try drawing graphs of the (displacement) current through the capacitor and voltage across the capacitoras functions of time.

Hint G.2 Graphs of and


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ANSWER: The current must be directed clockwise.


The magnitude of the current must be a maximum.

The current must be zero.

Correct

Part H

Consider the capacitor in diagram 3. Which of the following statements is true at the moment the charge ofthe capacitor is at a maximum?

Hint H.1 How to approach this problem

Hint not displayed

Hint H.2 Graphs of and

Hint not displayed



The magnitude of the current must be a maximum.

The current must be zero.

Correct

Part I


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Consider the capacitor in diagram 3. Which of the following statements is true if the voltage at point b isgreater than that at point a?

Hint I.1 How to approach the problem

Hint not displayed

Hint I.2 Graphs of and

Hint not displayed



The current may be directed either clockwise or counterclockwise.

Correct

Part J

Consider a circuit in which a capacitor and an inductor are connected in parallel to an AC source. Which ofthe following statements about the magnitude of the current through the voltage source is true?

Hint J.1 Driven AC parallel circuits

Hint not displayed

ANSWER: It is always larger than the sum of the magnitudes of the currents in thecapacitor and inductor.

It is always less than the sum of the magnitudes of the currents in thecapacitor and inductor.

At very high frequencies it will become small.

At very low frequencies it will become small.

Correct

This surprising result occurs because the currents in inductor and capacitor are exactly out of phasewith each other (i.e., one lags and the other leads the voltage), and hence they cancel to some extent.At a particular frequency, called the resonant frequency, the currents have exactly the same amplitude,and they cancel exactly; that is, no current flows from the voltage source to the circuit. (Lots of currentflows around the loop made by the inductor and capacitor, however.) If an L-C parallel circuit like thisone connects the wire between amplifier stages in a radio, it will allow frequencies near the resonancefrequency to pass easily, but will tend to short those at other freqeuncies to ground. This is the basicmechanism for selecting a radio station.

A High-Pass Filter


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A series L-R-C circuit consisting of a voltage source, a capacitor of capacitance , an inductor of inductance

, and a resistor of resistance is driven with an AC voltage of amplitude and frequency . Define

to be the amplitude of the voltage across the resistor and the inductor.

Part A

Find the ratio .

Hint A.1 Find

Hint not displayed

Hint A.2 Find

Hint not displayed

Express your answer in terms of either , , and , or , , and .

ANSWER:

=

Correct

For the following questions it will be useful to write the voltage ratio in the following form:

.

Part B

Which of the following statements is true in the limit of large ( )?


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Hint B.1 Implications of large

Hint not displayed

ANSWER: is proportional to .

is proportional to .


is close to 1.

Correct

Part C

Which of the following statements is true in the limit of small ( )?

Hint C.1 Implications of small

Hint not displayed




is close to 1.

Correct

When is large, , and when is small, . Therefore, this circuit has the property

that only the amplitude of the low-frequency inputs will be attenuated (reduced in value) at the output,while the amplitude of the high-frequency inputs will pass through relatively unchanged. This is whysuch a circuit is called a high-pass filter.


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A Low-Pass FilterThe figure shows a low-pass filter; the output voltage is taken across the capacitor in an L-R-C series circuit.

Part A

Derive an expression for , the ratio of the output and source voltage amplitudes, as a function of the

angular frequency of the source.

Hint A.1 Voltage over a capacitor

Hint not displayed

Hint A.2 Current through the circuit

Hint not displayed

Hint A.3 What is an expression for impedance?

Hint not displayed

Express your answer in terms of the capacitance , resistance , inductance , and angular

frequency of the circuit.

ANSWER:

=

Correct

You can see from the shape of the graph (see the figure) of why the circuit in this problem is

called a low-pass filter: Low frequencies pass through (i.e., have output voltage) at a much highermultple of their source voltage than high frequencies. No units are shown, since this graph is simplymeant to give a general idea of how the graph is shaped. The exact details depend on the inductance,resistance, and capacitance of the components of the circuit.


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Alternating Current, LC circuitA capacitor with capacitance is connected in parallel to two inductors: inductor 1 with inductance , and

inductor 2 with inductance , as shown in the figure. The capacitor is charged up to a voltage , at which

point it has a charge . There is no current in the inductors. Then the switch is closed.

Part A

Since the two inductors are in parallel, the voltage across them is the same at any time. Hence,, where and are the reactances of inductors 1 and 2, and and are the

currents through them. Use this equality to express in terms of .

Hint A.1 The reactance of an inductor

Hint not displayed

Express your answer in terms of .

ANSWER: =

Correct

Part B


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What is the effective inductance of the inductors 1 and 2 in the circuit?

Hint B.1 Formulas for effective inductance

Hint not displayed


ANSWER: =

Correct

Part C

Find the maximum current through inductor 1.

Hint C.1 Use conservation of energy to find the total maximum current

Hint not displayed

Express your answer in terms of , , and .

ANSWER:

=

Correct

Average Power in an L-R-C CircuitA circuit consists of a resistor (resistance ), inductor (inductance ), and capacitor (capacitance )

connected in series with an AC source supplying sinusoidal voltage . Assume that all circuit

elements are ideal, so that the only resistance in the circuit is due to the resistor. Also assume that is theresonant frequency of the circuit.

Part A

What is the average power supplied by the voltage source?

Hint A.1 Find the instantaneous power

What is the instantaneous power dissipated in the circuit? Note: You will need to use the fact that

the circuit is being driven at its resonant frequency to eliminate from your answer.

Hint A.1.1 A useful formula for power

The power dissipated in the circuit is equal to the voltage supplied by the AC source times the currentin the circuit: .


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Hint A.1.2 The current in a driven L-R-C circuit

The L-R-C circuit described in the problem introduction obeys the differential equation

,

where is the charge on one of the capacitor plates. After solving this equation for as a function of

time, you can take the derivative to find the current in the circuit: . The result is

, where , and .

Hint A.1.3 Determine the resonant frequency

The problem introduction states that the circuit is being driven at its resonant frequency . What is thevalue of in terms of other given quantities?

Express your answer in terms of any or all of the following quantities: , , .

ANSWER:

=

Answer Requested

Note that the current in the circuit is in phase with the voltage supplied by the voltage source onlyif the voltage source is in resonance with the circuit (i.e., oscillates at the resonant frequency ofthe circuit). In addition, the amplitude of the current in the circuit is at a maximum when the voltagesource is in resonance with the circuit. The combination of these two factors means that thepower dissipated in the circuit is maximum when the voltage source and the circuit are inresonance.

Express your answer in terms of any or all of the following quantities: , , , , .

ANSWER:

=

Answer Requested

Now finding the average power is a matter of calculating the average value of this time-dependentfunction over a single period.

Hint A.2 Average value of a periodic function

To find the average value of a periodic function , integrate the function over one period and

then divide by the period: . You will find that the average value of the function


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or is . This is a useful result to remember!

Express your answer in terms of any or all of the following quantities: , , , and .

ANSWER:

=

Correct

Constructing a Low-Pass FilterA series L-R-C circuit is driven with AC voltage of amplitude and frequency . Define to be the

amplitude of the voltage across the capacitor. The resistance of the resistor is , the capacitance of the

capacitor is , and the inductance of the inductor is .

Part A

What is the ratio ?

Hint A.1 Find

Hint not displayed

Hint A.2 Find

Hint not displayed

Express your answer in terms of either , , , and or , , and .


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ANSWER:

=

Correct

For the following questions it will be useful to write the voltage ratio in the following form:

.

Part B

Which of the following statements is true in the large limit (that is, for )?

Hint B.1 Implications of large

Hint not displayed




is close to 1.

Correct

Part C

Which of the following statements is true in the small limit (that is, for )?

Hint C.1 Implications of small

Hint not displayed



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is close to 1.

Correct

When omega is large, ; and when omega is small, . Therefore, the circuit of this

problem has the property that only the high-frequency inputs will be attenuated (reduced in value) atthe output, while low-frequency inputs will pass through relatively unchanged. That is why such a circuitis called a low-pass filter.

Reactance and CurrentConsider the two circuits shown in the figure. Thecurrent in circuit 1, containing an inductor ofself-inductance , has an angular frequency , while

the current in circuit 2, containing a capacitor ofcapacitance , has an angular frequency . If we

increase and decrease , both bulbs growdimmer.

Part A

If we keep and constant, we can achieve the exact same effect of decreasing the brightness of eachbulb by performing which of the following sets of actions?

Hint A.1 How to approach the problem

Hint not displayed

Hint A.2 Inductive reactance

Hint not displayed

Hint A.3 Capacitive reactance


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Hint not displayed

Hint A.4 Determine how can be changed

Hint not displayed

Hint A.5 Determine how can be changed

Hint not displayed

ANSWER: increasing and decreasing

increasing both and

decreasing and increasing

decreasing both and

Correct

As you found out, the reactance of these circuits can be changed not only by varying the frequency ofthe current, but also by changing the characteristics of the elements in them, i.e., by changing theinductance of the inductor and the capacitance of the capacitor.

Part B

Now combine the capacitor, the inductor, and the bulbs in a single circuit, as shown in the figure. Whathappens to the brightness of each bulb if youincrease the frequency of the current in the newcircuit while keeping and constant?

Hint B.1 How to approach the problem

Hint not displayed

Hint B.2 Find which element experiences a decrease in current at higher frequencies

Hint not displayed


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ANSWER: Both bulbs become brighter.

The brightness of each bulb remains constant.

Bulb 1 becomes brighter than bulb 2.

Bulb 2 becomes brighter than bulb 1.

Both bulbs grow dimmer.

Correct

Since inductive reactance is proportional to frequency, for a given voltage, high-frequency currents willhave a much smaller amplitude through the inductor than through the capacitor. That is, the inductortends to block high-frequency currents. The opposite situation occurs if the frequency is decreased.The capacitor will block low-frequency currents and bulb 2 will grow dimmer.

Resonating RLC Series Circuit Ranking Task

Six series RLC circuits are described below, where is the resistance, the inductance, and the

capacitance of the circuits.

Part A

Rank these circuits on the basis of their resonance frequencies.

Hint A.1 Resonance and the phase between voltage and current

Hint not displayed

Hint A.2 Resonance and impedance

Hint not displayed

Hint A.3 Equating inductive and capacitive reactance at resonance

Hint not displayed

Rank from largest to smallest. To rank items as equivalent, overlap them.

ANSWER:

View Correct


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Part B

Each circuit is driven at its resonance frequency by a 100 AC power supply. Rank these circuits on the

basis of their maximum current.

Hint B.1 Current at resonance

Hint not displayed

Hint B.2 Maximizing maximum current

Hint not displayed


ANSWER:

View Correct

± A Voltage-Driven Parallel L-C CircuitAn AC source that provides a voltage drives an inductor having inductance and a

capacitor having capacitance , all connected in parallel.

Part A

Recall that the currents and through the inductor and capacitor are not in phase with their

respective voltages and . In particular, for a sinusoidal voltage driver, which of the following

statements is true?


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ANSWER: and both lag their respective voltages.

and both lead their respective voltages.

leads and lags .

lags and leads .

Correct

The phase angle between voltage and current for inductors and capacitors is radians. Among other

things, this means that no power is dissipated in either, since the time-averaged product of current andvoltage (= power) is zero: .

Part B

What is the amplitude of the current through the voltage source? In other words, if the time-dependentcurrent is , what is the value of ?

Hint B.1 Voltage in a parallel circuit

Hint not displayed

Hint B.2 Current through the capacitor

Hint not displayed

Hint B.3 The current through the inductor

Hint not displayed

Hint B.4 Total current

Hint not displayed

Express the amplitude of the current in terms of , , , and .

ANSWER: =

Correct

Part C

With and representing the respective magnitudes of the currents through the inductor and

capacitor, which of the following statements is true?


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ANSWER: At very high frequency and at very low frequency .

At very high frequency and at very low frequency .



and are about the same at all frequencies.

Correct

Part D

The L-C circuit is one example of a system that can exhibit resonance behavior. The resonant frequency is

. At this frequency, what is the amplitude of the current supplied by the voltage source?

Express your answer in terms of and any other need terms from the problem introduction.

ANSWER: = 0Correct

Part E

Which of the following statements best describes the implications of the result that when

?

ANSWER: The currents ; ; are zero at all times.

The current is zero at all times although and are not.

The current is zero only when the voltage is zero.

The current is zero only when the amplitude of the voltage is not

changing.

Correct

At the resonant frequency of the circut, the voltage source can supply voltage easily because theparallel L-C circuit draws no current at all! (For this reason, a parallel L-C circuit can be used as a filterto pass signals at the resonance frequency.) Of course, the same current flows in the inductor and thecapacitor as the current that flows when they are driven separately by the voltage source. However,the currents are exactly out of phase, so that the net effect is a current circulating around the loopmade by the capacitor and the inductor without passing into or out of the voltage source.


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Resonance in an R-L-C CircuitIn an L-R-C series circuit, the resistance is 340 ohms, the inductance is 0.400 henrys, and the capacitance is1.40×10−2 microfarads.

Part A

What is the resonance angular frequency of the circuit?

Hint A.1 Definition of the resonance angular frequency

Hint not displayed

Hint A.2 Relationship between current and voltage amplitudes

Hint not displayed

Hint A.3 What is an expression for impedance?

Hint not displayed

Hint A.4 Finding the formula for the resonant frequency

Hint not displayed

Express your answer in radians per second to three significant figures.

ANSWER: = 1.34×104

Correct

Part B

The capacitor can withstand a peak voltage of 560 volts. If the voltage source operates at the resonancefrequency, what maximum voltage amplitude can the source have if the maximum capacitor voltage is

not exceeded?

Hint B.1 Voltage across a capacitor

Hint not displayed

Hint B.2 Current at the resonance frequency

Hint not displayed

Express your answer in volts to three significant figures.

ANSWER: = 35.6Correct

Triangle Electromagnetic Wave


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Learning Goal: To show how a propagating triangle electromagnetic wave can satisfy Maxwell's equationsif the wave travels at speed c.

Light, radiant heat (infrared radiation), X rays, and radio waves are all examples of traveling electromagneticwaves. Electromagnetic waves consist of mutually compatible combinations of electric and magnetic fields("mutually compatible" in the sense that changes in the electric field generate the magnetic field, and viceversa).The simplest form for a traveling electromagnetic wave is a plane wave. One particularly simple form for aplane wave is known as a "triangle wave," in which the electric and magnetic fields are linear in position andtime (rather than sinusoidal). In this problem we will investigate a triangle wave traveling in the x directionwhose electric field is in the y direction. This wave is linearly polarized along the y axis; in other words, theelectric field is always directed along the y axis. Its electric and magnetic fields are given by the followingexpressions:

and ,

where , , and are constants. The constant , which has dimensions of length, is introduced so that

the constants and have dimensions of electric and magnetic field respectively. This wave is pictured in

the figure at time . Note that we have only drawn

the field vectors along the x axis. In fact, this idealizedwave fills all space, but the field vectors only vary inthe x direction.

We expect this wave to satisfy Maxwell's equations.For it to do so, we will find that the following must betrue:The amplitude of the electric field must be directlyproportional to the amplitude of the magnetic field.The wave must travel at a particular velocity (namely,the speed of light).

Part A

What is the propagation velocity of the electromagnetic wave whose electric and magnetic fields are

given by the expressions in the introduction?

Hint A.1 Phase velocity

All points along the wave will propagate with the same velocity. You may find it easiest to concentrate onthe point where . At , this point occurs at . Where is this point when

is some later time ?

Express the location , where the fieldamplitude is zero at time , in terms of andany necessary quantities from the problemintroduction.


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ANSWER: =

Answer Requested

Express in terms of and the unit vectors , , and . The answer will not involve ; we have

not yet shown that this wave travels at the speed of light.

ANSWER: =

Correct

In the next few parts, we will use Faraday's law of induction to find a relationship between and .

Faraday's law relates the line integral of the electric field around a closed loop to the rate of change inmagnetic flux through this loop:

.

Part B

To use Faraday's law for this problem, you will need to constuct a suitable loop, around which you willintegrate the electric field. In which plane should the loop lie to get a nonzero electric field line integral and anonzero magnetic flux?

ANSWER: the xy plane

the yz plane

the zx plane

Correct

Part C

Consider the loop shown in the figure. It is a square loop with sides of length , with one corner at the

origin and the opposite corner at the coordinates , . Recall that .

What is the value of the line integral of the electricfield around loop at arbitrary time ?


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Hint C.1 Integrating along segments 1 and 2

Hint not displayed

Hint C.2 Integrating along segments 3 and 4

Hint not displayed

Hint C.3 Integrating around the entire loop

Hint not displayed

Express the line integral in terms of , , , , and/or .

ANSWER: =

Correct

Part D

Recall that . Find the value of the magnetic flux through the surface in the xy

plane that is bounded by the loop , at arbitrary time .

Hint D.1 Simplifying the integrand

The quantity appearing in the integrand is . The surface is in the xy plane and is bounded

by the curve . Because the curve is oriented in a counterclockwise direction, the unit vector that is

normal (i.e., perpendicular) to the surface is . Therefore, . When you take the dot

product of and , you find that .

Hint D.2 Evaluating the integral

Using the result from the previous hint,, we find that the integral reduces to

.

What is the value of the integral ?

Express your answer in terms of , , and .

ANSWER: =

Answer Requested


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Express the magnetic flux in terms of , , , , and/or .

ANSWER:

=

Correct

Part E

Now use Faraday's law to establish a relationship between and .

Hint E.1 Using Faraday's law

Hint not displayed

Express in terms of and other quantities given in the introduction.

ANSWER: =

Correct

If the electric and magnetic fields given in the introduction are to be self-consistent, they must obey all ofMaxwell's equations, including the Ampère-Maxwell law. In these last few parts (again, most of which arehidden) we will use the Ampère-Maxwell law to show that self-consistency requires the electromagneticwave described in the introduction to propagate at the speed of light.The Ampère-Maxwell law relates the line integral of the magnetic field around a closed loop to the rate ofchange in electric flux through this loop:

.

In this problem, the current is zero. (For to be nonzero, we would need charged particles moving

around. In this problem, there are no charged particles present. We assume that the electromagnetic waveis propagating through a vacuum.)

Part F

To use the Ampère-Maxwell law you will once again need to construct a suitable loop, but this time you willintegrate the magnetic field around the loop. In which plane should the loop lie to get a nonzero magneticfield line integral and hence nonzero electric flux?

ANSWER: the xy plane

the yz plane

the zx plane

Correct

Part G


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Use the Ampère-Maxwell law to find a new relationship between and .

Hint G.1 How to approach the problem

Hint not displayed

Hint G.2 Find an expression for the left-hand side of the equation

Hint not displayed

Hint G.3 Find an expression for the right-hand side of the equation

Hint not displayed

Hint G.4 Use the Ampère-Maxwell law

Hint not displayed

Express in terms of , , , and other quantities given in the introduction.

ANSWER: =

Correct

Part H

Finally we are ready to show that the electric and magnetic fields given in the introduction describe anelectromagnetic wave propagating at the speed of light. If the electric and magnetic fields are to beself-consistent, they must obey all of Maxwell's equations. Using one of Maxwell's equations, Faraday's law,we found a certain relationship between and . You derived this in Part E. Using another of Maxwell's

equations, the Ampère-Maxwell law, we found what appears to be a different relationship between and

. You derived this in Part I. If the results of Parts E and I are to agree, what does this imply that the

speed of propagation must be?

Express in terms of only and .

ANSWER: =

Correct

You have just worked through the details of one of the great triumphs of physics: Maxwell's equationspredict a form of traveling wave consisting of a matched pair of electric and magnetic fields moving ata very high velocity . We can measure and independently in the laboratory, and

these experimentally determined values lead to a speed of , the speed of light .

After thousands of years of speculation about the nature of light, Maxwell had developed a plausibleand quantitatively testable theory about it.Faraday had a hunch that light and magnetism were related, as demonstrated by the Faraday effect.(Glass, put in a large magnetic field, will rotate the plane of polarization of light that passes through it.)Now Maxwell had predicted an electromagnetic wave with the following properties:


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It was transverse, with two possible polarizations (which agreed with an already known characteristicof light).It had an extraordinarily high velocity (relative to waves in air or on strings) that agreed with theexperimentally determined value for the speed of light.

Any doubt that light waves were in fact electromagnetic waves vanished as various optical phenomena(such as the behavior of electromagnetic waves at glass surfaces) were predicted and found to agreewith the behavior of light. This theory showed that lower frequency waves could be created anddetected by their interactions with currents in wires (later called antennas) and paved the way to thecreation and detection of radio waves.

Electromagnetic Wave Vector DrawingAn antenna emits an electromagnetic wave. The electric field lines at a certain instant in time are shown.

Part A

Correct the orientation of the electric field and electromagnetic wave velocity vectors at points A and B.

ANSWER:

View Correct

If you wanted to place a receiving antenna at either point A or B, how would you orient it to maximizethe signal? The "signal" consists of a current generated in the antenna by the incident wave. The wavewill cause electrons in the receiving antenna to move in a direction parallel to the electric field. So ifyou orient the receiving antenna with the straight conducting rod parallel to the electric field, you willget maximum electron motion and therefore maximum current.

Part B

At point A, what is the direction of the magnetic field?


Hint not displayed

ANSWER:

Correct

Electromagnetic Waves Ranking Task


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Part A

Rank these electromagnetic waves on the basis of their speed (in vacuum).

Hint A.1 Relating speed, frequency, and wavelength

Hint not displayed

Rank from fastest to slowest. To rank items as equivalent, overlap them.

ANSWER:

View Correct

Part B

Rank these electromagnetic waves on the basis of their wavelength.

Hint B.1 Electromagnetic spectrum

Hint not displayed

Hint B.2 Radio waves

Hint not displayed

Hint B.3 Visible light

Hint not displayed

Rank from longest to shortest. To rank items as equivalent, overlap them.

ANSWER:

View Correct

Part C


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Rank these electromagnetic waves on the basis of their frequency.


ANSWER:

View Correct

Energy in Electromagnetic WavesElectromagnetic waves transport energy. This problem shows you which parts of the energy are stored in theelectric and magnetic fields, respectively, and also makes a useful connection between the energy density ofa plane electromagnetic wave and the Poynting vector.In this problem, we explore the properties of a plane electromagnetic wave traveling at the speed of light along the x axis through vacuum. Its electric and magnetic field vectors are as follows:

.

Throughout, use these variables ( , , , , , , and ) in your answers. You will also need the

permittivity of free space and the permeability of free space .

Note: To indicate the square of a trigonometric function in your answer, use the notation sin(x)^2 NOTsin^2(x).

Part A

What is the instantaneous energy density in the electric field of the wave?

Hint A.1 Energy density in an electric field

Hint not displayed

Give your answer in terms of some or all of the variables in .

ANSWER:

=

Correct

Part B

What is the instantaneous energy density in the magnetic field of the wave?


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Hint B.1 Energy density in a magnetic field

Hint not displayed

Give your answer in terms of some or all of the variables in .

ANSWER:

=

Correct

Part C

What is the average energy density in the electric field of the wave?

Hint C.1 Average value of

Hint not displayed

Give your answer in terms of and .

ANSWER: =

Correct

Part D

What is the average energy density in the magnetic field of the wave?

Hint D.1 Average value of

Hint not displayed

Give your answer in terms of and .

ANSWER: =

Correct

Part E

From the previous results, derive an expression for , the average energy density in the whole wave.

Hint E.1 Relationship among , , and


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Hint not displayed

Hint E.2 Relationship between and for electromagnetic waves in vacuum

Hint not displayed

Hint E.3 Relationship among , and for electromagnetic waves in vacuum

Hint not displayed

Express the average energy density in terms of and only.

ANSWER: =

Correct

Part F

The Poynting vector gives the energy flux per unit area of electromagnetic waves. It is defined by the

relation

.

Calculate the time-averaged Poynting vector of the wave considered in this problem.

Hint F.1 Relationship between and for electromagnetic waves in vacuum

Hint not displayed

Hint F.2 Relationship among , and for electromagnetic waves in vacuum

Hint not displayed

Give your answer in terms of , and and unit vectors , , and/or . Do not use or .

ANSWER: =

Correct

If you compare this expression for the time-averaged Poynting flux to the one obtained for the overallenergy density, you find the simple relation

.

Thus, the energy density of the electromagnetic field times the speed at which it moves gives theenergy flux, which is a logical result.


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Magnetic Field and Poynting Flux in a Charging CapacitorWhen a circular capacitor with radius and plate separation is charged up, the electric field , and hence

the electric flux , between the plates changes.

According to Ampère's law as extended by Maxwell,this change in flux induces a magnetic field that can befound from

,

where is the permittivity of free space and is thepermeability of free space. We can solve this equationto obtain the field inside a capacitor:

,

where is the radial distance from the axis of thecapacitor.

Part A

You might know already that it is possible to think of the energy stored in a charged capacitor as beingstored in the electric field between the plates. We will explore this idea by considering the flow of energyinto the space between the plates during the charging process. The capacitor is charged by a constantcurrent , which flows for a time . At the beginning of this charging process ( ), there is no charge on

the plates.The Poynting vector gives the flow of electromagnetic energy per unit area per unit time and is defined in

terms of the electric field vector and the magnetic field vector by the relation

.

Find an expression for the magnitude of the Poynting vector on the surface that connects the edges

of the two circular plates.

Hint A.1 Find the electric field

Hint not displayed

Express the magnitude of the Poynting vector in terms of , , , , , and other variables and

parameters of the problem. Ignore all fringing effects.

ANSWER: =

Correct

Part B

Calculate the the total amount of energy that flows into the space between the capacitor plates from

to , by first integrating the Poynting vector over the surface that connects the edges of the two

circular plates, and then integrating over time.


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Hint not displayed

Hint B.2 Integration over the surface

Hint not displayed

Hint B.3 Surface area of a cylinder

Hint not displayed

Hint B.4 Time integral

Hint not displayed

Hint B.5 A helpful integral

Hint not displayed

Express the total amount of energy in terms of , , , , , and other variables and parameters

of the problem. Ignore all fringing effects.

ANSWER: =

Correct

Recalling that the capacitance of a parallel plate capacitor given in terms of the surface area of theplates and the distance between the plates is

,

and also recalling that the charge on the plates at time is given by

,

we can see that we have expressed the energy stored by the capacitor in the familiar way,

,

even though we derived it in a different way using the Poynting vector.

Poynting Flux and Power Dissipation in a ResistorWhen a steady current flows through a resistor, the resistor heats up. We say that "electrical energy isdissipated" by the resistor, that is, converted into heat. But if energy is dissipated, where did it come from?Did it come from the voltage source through the wires?This problem will show you an alternative way to think about the flow of energy and will introduce a picture inwhich the energy flows in many unexpected places--but not through the wires!We will calculate the Poynting flux, the flow of electromagnetic energy, across the surface of the resistor. ThePoynting flux, or Poynting vector , has units of energy per unit area per unit time and is related to the

electric field vector and the magnetic field vector by the equation


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,

where is the permeability of free space.

Consider a cylindrical resistor of radius , length ,

and resistance with a steady current flowing along

the axis of the cylinder.

Part A

Which of the following is the most accurate qualitative description of the the magnetic field vector inside

the cylindrical resistor?

ANSWER: The magnetic field vector points radially away from the axis of the cylinder.

The magnetic field vector is everywhere tangential to circles centered on theaxis of the cylinder.

The magnetic field vector points inward toward the axis of the cylinder.

The magnetic field vector points along the axis of the cylinder in the directionof the current.

Correct

Part B

Find the magnitude of the magnetic field inside the cylindrical resistor, where is the distance from

the axis of the cylinder, in terms of , , , , and other given variables. You will also need and .

Ignore fringing effects at the ends of the cylinder.

Hint B.1 Ampère's law

To calculate the magnetic field, you need Ampère's law, which relates the integral of the magnetic fieldvector around a loop to the flow of current through it:

.

Hint B.2 How to set up the integral


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What kind of loop do you need? You could take any loop, and it might be interesting for you to takedifferent loops inside the resistor. However, the better way to proceed is to exploit the symmetry of thesituation, choosing a loop in a plane perpendicular to the axis of the cylinder, making it circular, andcentering it on the axis of the resistor.The magnetic field always reflects the symmetry of the problem.

Hint B.3 Amount of current through a loop

The amount of current that passes through a loop of radius centered on the axis is simply

.

ANSWER: =

Correct

Part C

What can you say about the electric field vector inside the resistor?

ANSWER: The electric field vector points along the axis of the resistor in the direction ofthe current.

The electric field vector is zero inside the resistor and on its surface.

The electric field vector is confined to the surface of the resistor and points inthe direction.

The electric field vector points radially outward--away from the axis of thecylinder.

The electric field vector is everywhere tangential to circles centered on theaxis of the resistor that lie in the plane perpendicular to the current direction.

Correct

Part D

What is the magnitude of the electric field vector ?

Hint D.1 Use Ohm's law

Use Ohm's law to express the potential drop across the resistor in terms of the current through it.

Give you answer in term of resistance and current

ANSWER: = Answer not displayed


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Hint D.2 Relationship between and

Inside the resistor, the electric field is uniform, just like the electric field inside a parallel plate capacitor;therefore the field is defined by the relationship

.

Give the magnitude of the electric field vector in terms of , , and other parameters of the

problem.

ANSWER: =

Correct

Part E

In what direction does the Poynting vector point?

Hint E.1 Cross products in cylindrical coordinates

Hint not displayed

ANSWER:

The Poynting vector is zero inside the resistor including its surface.

Correct

Part F

Calculate , the magnitude of the Poynting vector at the surface of the resistor (not at the circular ends of

the cylinder). To answer this you need to take .

Hint F.1 Definition of the Poynting vector

Hint not displayed

Give your answer in terms of , , and other parameters of the problem.

ANSWER: =


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Correct

Multiplying this value of the Poynting flux by the surface area of the resistor (which in this case isequivalent to integrating the Poynting vector over the surface of the resistor), we recover the familiarexpression for the power dissipated in a resistor through which a current flows:

.

Satellite Television TransmissionA satellite in geostationary orbit is used to transmit data via electromagnetic radiation. The satellite is at aheight of 35,000 km above the surface of the earth, and we assume it has an isotropic power output of 1 kW(although, in practice, satellite antennas transmit signals that are less powerful but more directional).

Part A

Reception devices pick up the variation in the electric field vector of the electromagnetic wave sent out bythe satellite. Given the satellite specifications listed in the problem introduction, what is the amplitude of

the electric field vector of the satellite broadcast as measured at the surface of the earth? Use for the permittivity of space and for the speed of light.

Hint A.1 How to approach this problem

Hint not displayed

Hint A.2 Find the Poynting Vector

Hint not displayed

Hint A.3 Find the energy flux through a sphere

Hint not displayed

Express the amplitude of the electric field vector in microvolts per meter to three significantfigures.

ANSWER: = 7.00

Correct

Part B

Imagine that the satellite described in the problem introduction is used to transmit television signals. Youhave a satellite TV reciever consisting of a circular dish of radius which focuses the electromagnetic

energy incident from the satellite onto a receiver which has a surface area of 5 .

How large does the radius of the dish have to be to achieve an electric field vector amplitude of 0.1

at the receiver?

For simplicity, assume that your house is located directly beneath the satellite (i.e. the situation youcalculated in the first part), that the dish reflects all of the incident signal onto the receiver, and that there


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are no losses associated with the reception process. The dish has a curvature, but the radius refers to

the projection of the dish into the plane perpendicular to the direction of the incoming signal.

Hint B.1 How to approach this problem

The power incident on the dish is equal to the magnitude of the Poynting vector at the surface of the dishtimes the dish area: . Similarly for the reciever. You also know that all of the power incident

on the dish is reflected onto the receiver: . Express and in terms of the radii (of the

receiver and the dish ) and the amplitude of the electric field incident on each. Then solve for in

terms of and the field amplitudes.

Hint B.2 The relationship between and

We know that . This implies that and, similarly, that ,

where and are the electric field amplitudes at the dish and receiver, respectively.

Using these two equations and the fact that , find in terms of .

Give your answer in terms of , , and .

ANSWER: =

Answer Requested

Give your answer in centimeters, to two significant figures.

ANSWER: = 18Answer Requested

Solar SailA solar sail allows a spacecraft to use radiation pressure for propulsion, similar to the way wind propels asailboat. The sails of such spacecraft are made out of enormous reflecting panels. The area of the panels ismaximized to catch the largest number of incident photons, thus maximizing the momentum transfer from theincident radiation.For such spacecraft to work, the force from the radiation pressure exerted by the photons must be greaterthan the gravitational attraction to the star emitting the photons. The critical parameter is the area density(mass per unit area) of the sail.

Part A

Consider a perfectly reflecting mirror oriented so that solar radiation of intensity is incident upon, and

perpendicular to, the reflective surface of the mirror. If the mirror has surface area , what is , the

magnitude of the average force due to the radiation pressure of the sunlight on the mirror?



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Radiation pressure arises from the photon momentum transfer as the photons strike the mirror. Thus, ifyou find an expression for the total momentum transferred to the mirror by the photons that strike it, youcan determine the average force exerted on the mirror. Notice that when writing an expression for themomentum transfer you'll need to take into account the fact that the mirror reflects the photons, ratherthan absorbs them.

Hint A.2 Find the total momentum transfer

What is the total momentum transferred to the mirror by the photons in a time interval ?

Hint A.2.1 Energy of the photons and their momentum

The momentum of a photon can be expressed in terms of the photon energy as

,

where is the speed of light in vacuum. This ratio also holds for the total momentum and energy of thephotons striking the mirror.

Hint A.2.2 Radiation intensity and energy

The total energy of the photons striking the mirror during a time interval is given by

,

where is the intensity of the radiation and is the surface area of the mirror.

Hint A.2.3 Reflection vs. absorption

When an object absorbs a photon of energy , it receives momentum equal to . When an object

reflects a photon of energy , the object must not only stop the photon (as is the case when the

photon is absorbed) but also send it back in the opposite direction. Thus, the total momentum transferfor photon reflection is twice as much as in the case of photon absorption.

Express your answer in terms of the time interval , the intensity , the mirror's surface area

, and the speed of light .

ANSWER: =

Answer Requested

Hint A.3 Force and change in momentum

Let be the total momentum transferred to the mirror by the photons that strike the mirror during a

time interval . Then the magnitude of the average force exerted on the mirror is

Express your answer in terms of the intensity , the mirror's surface area , and the speed of light

.


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ANSWER: =

Correct

To solve the second part of this problem you will need to know the following:the mass of the sun, ,

the intensity of sunlight as a function of the distance from the sun,

,

andthe gravitational constant .

Part B

Suppose that the mirror described in Part A is initially at rest a distance away from the sun. What is the

critical value of area density for the mirror at which the radiation pressure exactly cancels out thegravitational attraction from the sun?

Hint B.1 Find the force due to gravity

Suppose the mirror has mass . Find a general expression for , the magnitude of the gravitational

force due to the sun that acts on the mirror.

Express your answer symbolically in terms of the gravitational constant , the mass of the sun,

, the mass of the mirror, , and the mirror's distance from the sun, .

ANSWER: =

Answer Requested

Hint B.2 Solving for area density

By equating the force due to the sun's radiation ( found in Part A) and the force due to the sun's

gravitational pull, you should be able to solve for the area density of the mirror. Note that the expressionfor the intensity, given in the problem, has a factor of , just like the expression for the gravitational

force, so the critical value of the area density turns out to be independent of .

Express your answer numerically, to two significant figures, in units of kilograms per metersquared.

ANSWER: mass/area = 1.60×10−3

Correct

In selecting the material for a solar sail, area density, strength, and reflectivity are the principalconcerns. Given a representative thickness of the sail of 1 , one of the few currently existing


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materials with a sufficiently low density and high strength can be made from carbon fibers. These havea density of 1.60 , roughly one-fifth that of iron.

The Electromagnetic SpectrumElectromagnetic radiation is more common than you think. Radio and TV stations emit radio waves when theybroadcast their programs; microwaves cook your food in a microwave oven; dentists use X rays to checkyour teeth. Even though they have different names and different applications, these types of radiation arereally all the same thing: electromagnetic (EM) waves, that is, energy that travels in the form of oscillatingelectric and magnetic fields.

Consider the following:radio waves emitted by a weather radar system to detect raindrops and ice crystals in the atmosphere tostudy weather patterns;microwaves used in communication satellite transmissions;infrared waves that are perceived as heat when you turn on a burner on an electric stove;the multicolor light in a rainbow;the ultraviolet solar radiation that reaches the surface of the earth and causes unprotected skin to burn; andX rays used in medicine for diagnostic imaging.

Part A

Which of the following statements correctly describe the various forms of EM radiation listed above?

Hint A.1 The electromagnetic spectrum

Hint not displayed

Hint A.2 Frequency and wavelength of an EM wave

Hint not displayed

Check all that apply.

ANSWER: They have different wavelengths.

They have different frequencies.

They propagate at different speeds through a vacuum depending on theirfrequency.

They propagate at different speeds through nonvacuum media depending onboth their frequency and the material in which they travel.

They require different media to propagate.

Correct

The frequency and wavelength of EM waves can vary over a wide range of values. Scientists refer to the fullrange of frequencies that EM radiation can have as the electromagnetic spectrum.Electromagnetic waves are used extensively in modern technology. Many devices are built to emit and/orreceive EM waves at a very specific frequency, or within a narrow band of frequencies. Here are someexamples followed by their frequencies of operation:garage door openers: 40.0 ,


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standard cordless phones: 40.0 to 50.0 ,

baby monitors: 49.0 ,

FM radio stations: 88.0 to 108 ,

cell phones: 800 to 900 ,

Global Positioning System: 1227 to 1575 ,

microwave ovens: 2450 ,

wireless Internet technology: 2.4 to 2.6 .

Part B

Which of the following statements correctly describe the various applications listed above?

Hint B.1 Frequency and wavelength of an EM wave

Hint not displayed

Hint B.2 Hertz, megahertz, and gigahertz

Hint not displayed

Hint B.3 Meters and kilometers

Hint not displayed


ANSWER: All these technologies use radio waves, including low-frequency microwaves.

All these technologies use radio waves, including high-frequency microwaves.

All these technologies use a combination of infrared waves andhigh-frequency microwaves.

Microwave ovens emit in the same frequency band as some wireless Internetdevices.

The radiation emitted by wireless Internet devices has the shortest wavelengthof all the technologies listed above.

All these technologies emit waves with a wavelength in the range 0.10 to 10.0.

All the technologies emit waves with a wavelength in the range 0.01 to 10.0 .

Correct

The frequency band used in wireless technology is strictly regulated by government agencies to avoidundesired interference effects. In the United States, the Federal Communications Commission (FCC) isresponsible for assigning specific radio frequency bands to different wireless communication systems.

Despite their extensive applications in communication systems, radio waves are not the only form of EMwaves present in our atmosphere. Another form of EM radiation plays an even more important role in our life(and the life of our planet): sunlight.The sun emits over a wide range of frequencies; however, the fraction of its radiation that reaches the


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earth's surface is mostly in the visible spectrum. (Note that about 35% of the radiation coming from the sunis absorbed directly by the atmosphere before even reaching the earth's surface.) The earth, then, absorbsthis radiation and reemits it as infrared waves.

Part C

Based on this information, which of the following statements is correct?

Hint C.1 Relation between frequency and wavelength

Hint not displayed


ANSWER: The earth absorbs visible light and emits radiation with a shorter wavelength.

The earth absorbs visible light and emits radiation with a longer wavelength.

The earth absorbs visible light and emits radiation with a lower frequency.

The earth absorbs visible light and emits radiation with a higher frequency.

Correct

Even though our atmosphere absorbs a very small amount of visible light, it strongly reflects andabsorbs infrared waves. Therefore the radiation emitted by the earth does not leave the atmosphere.Instead, it is reflected back into it, contributing to a warming effect known as the greenhouse effect.

Part D

A large fraction of the ultraviolet (UV) radiation coming from the sun is absorbed by the atmosphere. Themain UV absorber in our atmosphere is ozone, . In particular, ozone absorbs radiation with frequencies

around 9.38×1014 . What is the wavelength of the radiation absorbed by ozone?

Hint D.1 Frequency and wavelength of an EM wave

Hint not displayed

Hint D.2 Meters and nanometers

Hint not displayed

Express your answer in nanometers.

ANSWER: = 320Correct

± Charges at Rest on a Standing WaveAn electromagnetic standing wave in air of frequency 750 is set up between two conducting planes 80.0

apart.


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Part A

At how many positions between the planes could a point charge be placed at rest so that it would remain atrest?


A point charge will remain at rest only if it is placed where the electric field is always zero. This is true atany node on a standing wave. Thus you need to find the number of nodes between the plates.

Hint A.2 Formula for a standing wave

The magnitude of the electric field in a standing electromagnetic wave can be expressed in the followingform:

,

where is the wavenumber, is the wavelength, and where is the frequency (750

in our case).

Hint A.3 Locate the nodes

Take to be the location of the first conductor, and take to be the location of the

second conductor. At any point that satisfies the equation , the electric field is zero there for

all points in time. The point at is one value that satisfies this equation. This is important because

the electric field inside a conductor is always zero.Using the fact that the electric field inside a conductor is always zero, at what other value of do youexpect to find a node no matter what the frequency or wavelength of the standing wave is?

Express your answer in centimeters.

ANSWER: = 80

Answer Requested

Note that we now have a boundary condition in which any standing wave that exists between the twoconductors must have a node at the conductor walls. This limits the number of standing waves tothose that have a frequency and wavelength that satisfy these boundary conditions.

Hint A.4 Find the distance between nodes

In terms of the wavelength , what is , the distance between two adjacent nodes on a standing wave?

Express your answer as a multiple of

ANSWER: =

Answer Requested

Hint A.5 Find


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Using with , determine .

Express your answer in centimeters.

ANSWER: = 40Answer Requested

ANSWER: 3All attempts used; correct answer displayed

Poynting FluxAn electromagnetic wave is traveling through vacuum. Its electric field vector is given by

,

where is the unit vector in the y direction.

Part A

If is the amplitude of the magnetic field vector, find the complete expression for the magnetic field vector

of the wave.

Hint A.1 Relative orientation of and for a wave in vacuum

Hint not displayed

Hint A.2 Orientation of and relative to the direction of propagation

Hint not displayed

Hint A.3 Determine the direction of propagation of the wave

Hint not displayed

Hint A.4 Phase relationship between and

Hint not displayed

ANSWER:


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Correct

Part B

What is the Poynting vector , that is, the power per unit area associated with the electromagnetic

wave described in the problem introduction?

Hint B.1 Definition of the Poynting vector

Hint not displayed

Give your answer in terms of some or all of the variables , , , , , , and . Specify the

direction of the Poynting vector using the unit vectors , , and as appropriate.

ANSWER: =

Correct

Polarization of Light and Malus's Law

Learning Goal: To understand polarization of light and how to use Malus's law to calculate the intensity of abeam of light after passing through one or more polarizing filters.

The two transverse waves shown in the figure bothtravel in the +z direction. The waves differ in that thetop wave oscillates horizontally and the bottom waveoscillates vertically. The direction of oscillation of awave is called the polarization of the wave. The upperwave is described as polarized in the +x directionwhereas the lower wave is polarized in the +ydirection. In general, waves can be polarized along anydirection.Recall that electromagnetic waves, such as visiblelight, microwaves, and X rays, consist of oscillatingelectric and magnetic fields. The polarization of anelectromagnetic wave refers to the oscillation directionof the electric field, not the magnetic field. In thisproblem all figures depicting light waves illustrate onlythe electric field.A linear polarizing filter, often just called a polarizer, isa device that only transmits light polarized along a specific transmission axis direction. The amount of light thatpasses through a filter is quantified in terms of its intensity. If the polarization angle of the incident lightmatches the transmission axis of the polarizer, 100 of the light will pass through, so the transmitted intensity


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will equal the incident intensity. More generally, the intensity of light emerging from a polarizer is described byMalus's law:

,

where is the intensity of the polarized light beam just before entering the polarizer, is the intensity of the

transmitted light beam immediately after passing through the polarizer, and is the angular difference

between the polarization angle of the incident beam and the transmission axis of the polarizer. After passingthrough the polarizer, the transmitted light is polarized in the direction of the transmission axis of the polarizingfilter.

In the questions that follow, assume that all angles are measured counterclockwise from the +x axis in thedirection of the +y axis.

Part A

A beam of polarized light with intensity and polarization angle strikes a polarizer with transmission axis

. What angle should be used in Malus's law to calculate the transmitted intensity ?

This process is illustrated in the figure , where thepolarization of the light wave is visually illustrated bya magenta double arrow oriented in the direction ofpolarization, the transmission axis of the polarizer isrepresented by a blue double arrow, and thedirection of motion of the wave is illustrated by apurple arrow.

ANSWER:

Correct

Part B

What is the polarization angle of the light emerging from the polarizer?


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ANSWER:

Correct

Part C

If = 20.0 , = 25.0 , and = 40.0 , what is the transmitted intensity ?

Express your answer numerically in watts per square meter .


If the polarization axis of the incident light and the transmission axis of the filter are aligned (so) or if they point in exactly opposite directions (so ), then in

Malus's law, indicating that 100 of the incident intensity will be transmitted.

Most natural light sources emit "unpolarized" light, perhaps better described as "randomly polarized" light.These light sources emit numerous brief bursts of light whose polarization directions are unrelated, so onaverage the resulting beam has all polarization angles equally represented. When unpolarized light withintensity passes through a polarizer, its intensity is cut in half, regardless of the orientation of the

transmission axis of the polarizer:

.

Part D

One way to produce a beam of polarized light with intensity and polarization angle would be to pass

unpolarized light with intensity through a polarizer whose transmission axis is oriented such that .

How large must be if the transmitted light is to have intensity ?

Express your answer as a decimal number timesthe symbol . For example, if , enter

0.25 * I.


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ANSWER: =

Correct

Part E

A beam of unpolarized light with intensity falls first upon a polarizer with transmission axis then

upon a second polarizer with transmission axis , where (in other words the

two axes are perpendicular to one another). What is the intensity of the light beam emerging from the

second polarizer?

Hint E.1 How to approach the problem

Hint not displayed

Hint E.2 Determine the intensity of light between the two polarizers

Hint not displayed

Hint E.3 Determine the angle to be used in Malus's law

Hint not displayed

Express your answer as a decimal number times the symbol . For example, if , enter

0.25 * I_0.

ANSWER: =

Correct

Adjacent polarizers with perpendicular transmission axes are said to be "crossed." No light can getthrough both polarizers, regardless of the light's initial polarization state.


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Part F

Notice that a polarizer modifies the light intensity according to Malus's law and also reorients thepolarization angle of the beam to match its own transmission axis. Hence it is possible for light to passthrough a pair of crossed polarizers if a third polarizer is inserted between them with an intermediatetransmission axis direction. What is the new intensity of the light emerging from the final polarizer in Part

E if a third polarizer (Polarizer A in the figure ),whose transmission axis is offset 45 from

each of the others, is inserted between the originaltwo polarizers?

Hint F.1 How to approach the problem

Hint not displayed

Hint F.2 Relate to

Hint not displayed

Hint F.3 Relate to

Hint not displayed

Hint F.4 Relate to

Hint not displayed

Express your answer as a decimal number times the symbol . For example, if , enter

0.25 * I_0.

ANSWER:

=

All attempts used; correct answer displayed

Polarizing filters for visible light are made of Polaroid, which contains long molecular chains that havebeen aligned by stretching the material during production. Polaroid is commonly used in sunglassesbecause it reduces the intensity of unpolarized sunlight light by 50 . Glare is often at least partially

polarized, so Polaroid sunglasses, when properly oriented, can selectively reduce glare by even morethan 50 .

A Sparkling Diamond IIA beam of white light is incident on the surface of a diamond at an angle . Since the index of refraction

depends on the light's wavelength, the different colorsthat comprise white light will spread out as they passthrough the diamond. For example, the indices ofrefraction in diamond are for red light and

for blue light. Thus, blue light and red

light are refracted at different angles inside thediamond, as shown in the picture. The surrounding airhas . Note that the angles in the figure are

not to scale.

Part A

Now consider , the angle at which the blue refracted ray hits the bottom surface of the diamond. If is

larger than the critical angle , the light will not be refracted out into the air, but instead it will be totally

internally reflected back into the diamond. Find .

Hint A.1 Find the refracted angle when

Hint not displayed

Hint A.2 Apply Snell's law

Hint not displayed

Express your answer in degrees to four significant figures.


Part B

A diamond is cut such that the angle between its top surface and its bottom surface is . For , find

the largest possible value of the incident angle such that the blue light is totally internally reflected off the

bottom surface.

Hint B.1 Determine

Hint not displayed

Hint B.2 Apply Snell's law


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Hint not displayed

Hint B.3 When is maximized?

Hint not displayed



The angle at which a diamond is cut plays an important role in how brightly it sparkles. The properchoice for will ensure that a large fraction of the light gets totally internally reflected back towardyour eyes, maximizing the diamond's "fire." Generally is chosen to be between 39 and 42 . For

more information on the physics of diamonds, check out http://www.folds.net/diamond/.

Birefringence in CalciteCalcite ( ) is a crystal with abnormally large birefringence. The index of refraction for light with electric

field parallel to the optical axis (called extraordinary waves or e-waves) is 1.4864. The index of refraction forlight with electric field perpendicular to the optical axis (called ordinary waves or o-waves) is 1.6584.

Part A

Find the critical angle for e-waves in calcite.

Hint A.1 Snell's law

Hint not displayed

Hint A.2 Definition of critical angle

Hint not displayed



Part B

Find the critical angle for o-waves in calcite.




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Part C

Assume that the optical axis of the crystal is oriented horizontally (perpendicular to the direction of theincident ray). If unpolarized light shines into the calcite crystal shown and strikes the far side with ,

which of the following pictures correctly shows whatwill happen?

Hint C.1 Thinking about angles

Hint not displayed

Please make sure you know which circle corresponds to which figure.

ANSWER:

Correct

Because the difference in index of refraction is based on polarization, the two rays are both polarized.

Part D


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Is the transmitted ray (the one that passes into the air) polarized parallel to or perpendicular to the opticalaxis?

ANSWER: parallel

perpedicular

Correct

Part E

Now assume that the optical axis of the crystal is vertical (parallel to the direction of the incident ray). (Thesame figure still applies; just assume now that theoptical axis is vertical.) Which of the following figuresshows what would happen to the incident ray fromPart C?

Hint E.1 A property of light waves

Hint not displayed

ANSWER:


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Correct

Diffuse ReflectionThe law of reflection is quite useful for mirrors and other flat, shiny surfaces. (This sort of reflection is calledspecular reflection). However, you've likely been told that when you look at something, you are seeing lightreflected from the object that you are looking at. This is reflection of a different sort: diffuse reflection. In thisproblem, you will see how diffuse reflection actually arises from the same law of reflection that you areaccustomed to for reflections from mirrors.

Part A

Consider a spotlight shining onto a horizontal mirror .If the light from the spotlight strikes the mirror at anangle to the normal, what angle to the normal

would you expect for the reflected rays?


ANSWER: =

Correct

This simple rule of reflection no longer seems to hold for diffuse reflection. Consider the same spotlight butnow reflecting from the surface of a table . Unlike thelight reflected from the mirror, the light reflected fromthe table seems to go in all directions. If it didn't, thenyou'd only be able to see tables when you were at aspecific angle to the lights above you! To understandwhy the light reflects in all directions, you must firstlook at a slightly simpler problem.Consider a flat surface, inclined downward from thehorizontal by an angle . The red line represents thesurface and the red dotted line indicates the normalto this surface (the normal line). The two blue dashedlines represent horizontal and vertical. The angle


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between the incoming ray and the vertical is .

Throughout this problem, assume that is larger

than but smaller than . (If you wish, you can

determine the correct sign rules to generalize yourresults later.)

Part B

Find the angle between the reflected ray and the vertical.


Hint not displayed

Hint B.2 Find the angle between the normal line and vertical

Hint not displayed

Hint B.3 Find the angle between the incoming ray and the normal line

Hint not displayed

Hint B.4 Find the angle between the normal line and the reflected ray

Hint not displayed

Express the angle between the reflected ray and the vertical in terms of and .

ANSWER: =

Correct

Part C

Suppose that the spotlight shines so that different parts of the beam reflect off of different two surfaces,one inclined at an angle (from the horizontal) and one inclined at an angle . What would the angular

separation be between the rays reflected from the two surfaces? Assume that the light comes at an

angle to the vertical.

Hint C.1 Finding the angular separation

You already know the angle between the reflected ray and the vertical for the surface tilted at an angle . To find the angle between the reflected ray and the vertical for the surface tilted at an angle , just


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substitute for in the formula from the previous part. The difference between these two angles will be

the angular separation between the two reflected angles.

Express your answer in terms of some or all of the angles , , and .

ANSWER: =


Most surfaces are not smooth like the surface of a mirror. Instead, on a microscopic scale, they arecovered in ridges and pits. Thus, when light from a spotlight (or any other source) strikes a seeminglyflat surface, the light is reflected in all directions. You can see that if and (the

negative angle indicates that the second surface is inclined above the horizontal), then the angularseparation between the reflected rays would be 180 . In most surfaces, the roughness can be

represented as a large number of small, flat surfaces inclined at different angles to the horizontal. Inthis way, light reflects in all directions, frequently sending rays that began very close to one another invery different directions. This microscopic roughness is why most surfaces do not form images asmirrors do.

Huygens' PrincipleHuygens' principle (first described by the Dutch scientist Christiaan Huygens in 1678) uses geometry todetermine the shape of a wavefront at a time , given some initial wavefront at an earlier time. This can be

constructed by imagining that the initial wavefront is a source of wavelets that propagate from each point atthe speed of light. From this principle, one can determine the angles of reflection and refraction by consideringthe speed of light at each point on the wavefront. In this problem, we will explore some of these concepts.

Part A

First, let's look at a plane wave that is incident on a flat piece of material with an index of refraction . Partof the light is transmitted through the material and part of it is reflected. For the moment, let's just look atthe part that is transmitted. At time , the wavefront is a distance away from the surface of the

material . At time , the wavefront is at theposition of the material interface . Use the Huygens'principle to determine how far into the material ( )

the wavefront has propagated by time .


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Hint A.1 Speed of light in a material

Recall that the speed of light in a material is less than the speed of light in a vacuum. The index ofrefraction is defined by the relation . Use this relation to find the speed of light in a material,

which you will need to answer this part.

Express your answer in terms of the variables , , and the speed of light in a vacuum .

ANSWER: =

Correct

Note that there would be no change in the direction of the wavefront at any point because thewaverfront encountered the material interface at the same time at all points. Thus all of the individualwavefronts all propagated at the same speed, thereby maintaining the flat wavefront.

Now, instead of having a flat wavefront propagating normal to the material interface we have a flat wavefrontpropagating toward the material at an angle of relative to the axis perpendicular to the material

interface. In this part, we will look at the relative positions of a few points--A, B, and C--on the wavefront toillustrate Huygens' principle. Point C touches thevacuum/material interface at time whereas

point B is a distance and point A is a distance

away from the interface.

Part B

What is the time it will take for point B of the wavefront to encounter the vacuum/material interface?


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Hint B.1 Looking at the direction

Point B propagates along a flat wavefront in the direction that's relative to the axis perpendicular to

the material surface. As a result, one can use standard geometry and the fact that point B is originallyonly a distance away from the surface.

Express your answer numerically to two decimal places in units of (the time it takes light to travela distance in a vacuum).

ANSWER: = 1.74

Correct

It should be noted that the accuracy of the method increases the smaller the distance is before makinga new wavefront and redoing the wavelets. If you tried to just draw a large wavelet from point B then itwould hit the surface after only traveling a distance . This large wavelet would make it difficult to

visualize how to add up all of the other wavelets to make a coherent wavefront. As a result, one shouldjust draw a line perpendicular to the wavefront at point B until it hits the surface, at which point thewavelets will change owing to the new index of refraction.

Part C

How far did point C go into the material interface in the time that it took for point B to get to theinterface? For this part we are looking for the distance traversed by the point C in the material ( ).

Give your answer in terms of the time , , and .

ANSWER: =

Correct

Part D

What is the new angle at which point C of the wavefront is propagating (relative to a line perpendicular to

the vacuum/material interface)? Try to use the fact that you have a spherical wavefront propagating from at the point where C met the vacuum/material interface until time when the wavefront at point B

reached the interface.

Hint D.1 Geometry of the problem

Hint not displayed

Express your answer in terms of inverse trig functions and .

ANSWER: =



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Refracted Waves in Unknown Materials

Part A

When monochromatic light passes through the interface between two unknown materials at an angle

where , no changes in the direction of propagation of light are observed. What can be said

about the two materials?


Hint not displayed


ANSWER: The two materials have matching indexes of refraction.

The second material through which light propagates has a lower index ofrefraction.

The second material through which light propagates has a higher index ofrefraction.

The two materials are identical.

Correct

By simply observing no change in the direction of propagation of light at the interface between twomaterials, you cannot be certain that the two materials are identical. In fact, different materials withmatching indexes of refraction would produce a similar effect. For example, at the interface betweenglass Pyrex ( ) and glycerin ( ), no relevant change in the direction of propagation

of light can be observed.

Part B

The same monochromatic light passes through the interface between two other unknown materials. Thistime the transmitted wave is observed to be farther from the normal to the interface than the incident wave.What can be said about these two materials and the light traveling through them?


Hint not displayed

Hint B.2 Law of refraction

Hint not displayed

Hint B.3 Find how the sines of the angles relate

Hint not displayed

Hint B.4 Find in which material light travels faster

Hint not displayed


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ANSWER: The second material through which the light propagates has a higher index ofrefraction.

The second material through which the light propagates has a lower index ofrefraction.

As the light passes into the second material, its speed increases.

As the light passes into the second material, its speed decreases.

Correct

When a ray of light passes into a material having a lower index of refraction, and hence a higher wavespeed, the refracted ray bends away from the normal to the interface and will appear closer to theinterface than the incident ray.

Scattering and polarized lightThe process of scattering of light by a molecule (Rayleigh scattering) is an important physical phenomenon.Instead of thinking of scattering as light simply bouncing off the molecule, one should think of it as anabsorption followed by reradiation of light.The probability for light to be scattered is proportional to the inverse of the wavelength to the fourth power,

. This means that the shorter wavelengths (toward blue) get scattered more strongly than the longer

wavelengths (toward red).Rayleigh scattering can explain why the daytime sky looks blue, the sunset looks red, and clouds are white. Inthe afternoon you observe mostly scattered light (blue); in the evening you see mostly transmitted light (red).The clouds have higher concentration of water and ice droplets. This means that light gets rescattered manytimes and all wavelengths get a chance to scatter out of the clouds, adding up to white light.Another effect that can be explained by light scattering is polarization. When you look at the sky with Polaroidsunglasses it appears darker or brighter from different angles. This is because the scattered light is partiallypolarized. The white light scattered from the clouds is unpolarized, because the light scatters randomly,multiple times. The direction of its polarization becomes random and thus the light is unpolarized. This effectcan be useful for making dramatic photographs of the sky.Consider a photographer who wants to take a picture of an interesting cloud formation. To increase the ratioof the clouds' intensity to that of the blue sky the photographer uses a polarizing filter.

Part A

How would the photographer use the polarizing filter to find out the direction of polarization of the lightcoming from the blue sky? Her only reference is the polarization axis of the filter.

ANSWER: Rotate the filter until the light's intensity is minimum; light's polarization isalong filter's axis.

Rotate the filter until the light's intensity is maximum; light's polarization isalong filter's axis.

Correct

Part B


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Find the angle between the filter's polarizing axis and the direction of polarization of light necessary to

increase the ratio of the clouds' intensity to that of the blue sky so that it is three times the normal value.

Hint B.1 Find the intensity of light from the sky through the polarizing filter

Hint not displayed

Hint B.2 Find the intensity of light from the the clouds through the polarizing filter

Hint not displayed

Hint B.3 Ratio

Hint not displayed



Total Internal Reflection Conceptual QuestionConsider scenarios A to F in which a ray of light traveling in material 1 is incident onto the interface withmaterial 2.

Material 1 ( ) Material 2 ( )A air (1.00) water (1.33)B water (1.33) air (1.00)C diamond (2.42) air (1.00)D air (1.00) quartz (1.46)E benzene (1.50) water (1.33)F diamond (2.42) water (1.33)

Part A

For which of these scenarios is total internal reflection possible?


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Hint A.1 Total internal reflection

Hint not displayed

Hint A.2 Angle of refraction

Hint not displayed

Hint A.3 Determine the role of the indices of refraction

Hint not displayed

List all correct answers in alphabetical order. For example, if scenarios A and E are correct, enterAE.

ANSWER: BCEFCorrect

Part B

For the scenarios in which total internal reflection is possible, rank the scenarios on the basis of the criticalangle, the angle above which total internal reflection occurs. At this angle, the refracted ray is at 90

from the normal.

Hint B.1 Apply Snell's law

At the critical angle, the refracted ray is at 90 from the normal. Therefore, Snell’s law can be

used to determine the critical angle. Snell's law states that,

where is the angle of the incident ray from the normal, and is the angle of the refracted ray from the

normal. Solve this equation for the sine of the critical angle .

Express your answer in terms of and .

ANSWER: =

Correct

Since increases as increases, the relative ranking of the critical angle will be the same

as the relative ranking of , which is equal to .


ANSWER:


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View Correct

Wavelength, Frequency, and the Speed of Light in Different MediaA beam of light from a monochromatic laser shines into a piece of glass. The glass has thickness and index

of refraction . The wavelength of the laser light in vacuum is and its frequency is . In this

problem, neither the constant nor its numerical value should appear in any of your answers.

Part A

How long does it take for a short pulse of light to travel from one end of the glass to the other?


Hint not displayed

Hint A.2 The speed of light

Hint not displayed

Hint A.3 Find the speed of light in the glass

Hint not displayed

Express your answer in terms of the frequency, . Use the numeric value given for in the

introduction.

ANSWER: =

Correct

When light travels from air into another medium, the wavelength changes, but the frequency remainsthe same. As a result, the speed of the light wave changes (since it is the product of the frequency andwavelength).

± Light Refracted through a PrismLight is incident along the normal to face AB of a glass prism of refractive index 1.65, as shown in the figure.


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Part A

Find , the largest value of the angle such that no light is refracted out of the prism at face AC if theprism is immersed in air.


Hint not displayed

Hint A.2 Drawing the incident ray and the reflected ray

Hint not displayed

Hint A.3 Snell's law

Hint not displayed

Hint A.4 Total internal reflection

Hint not displayed

Express your answer in degrees. Ignore any reflections from the surface BC.

ANSWER: = 52.7

Correct

Part B

Find , the largest value the angle can have without any light being refracted out of the prism at faceAC if the prism is immersed in water (with index of refraction 1.33).


Hint not displayed

Express your answer in degrees. Ignore any reflections from the surface BC.

ANSWER: = 36.3

Correct


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Score Summary:Your score on this assignment is 85.7%.You received 85.66 out of a possible total of 100 points.


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