Announcements
Review
Polarization
Energy in Electromagnetic Waves
Radiation Pressure
Electromagnetic Waves
Sections 22.4 - 22.7
Electromagnetic Waves
Final Questions
Announcements
Review
Polarization
Energy in Electromagnetic Waves
Reading Assignment
Read sections 23.1 - 23.4
Homework Assignment 7
Homework for Chapter 22 (due at the beginning of class on Friday, October 15)
Q: 4, 7, 10
P: 2, 16, 20, 40
PP: 32.1, 32.3
Electromagnetic Waves
Radiation Pressure
Final Questions
Announcements
Review
Polarization
Energy in Electromagnetic Waves
Radiation Pressure
Maxwell’s equations
X
E⊥ ∆A =
Qenc
closed
surface
0
(Gauss’s law)
Electric fields lines emerge from positive charge and terminate at negative charge
X
B⊥ ∆A = 0 (Gauss’s law for magnetism)
closed
surface
Magnetic field lines form closed loops
X
Ek ∆` = ε = −
∆ΦB
∆t
(Faraday’s law)
An electric field can be created by a changing magnetic flux
The electric field lines due to this electric field form closed loops
X
Bk ∆` = µ0 Ienc + µ0 0
∆ΦE
∆t
(Ampére-Maxwell law)
A magnetic field can be created by a current or a changing electric flux
Electromagnetic Waves
Final Questions
Announcements
Review
Polarization
Energy in Electromagnetic Waves
Radiation Pressure
Final Questions
Electromagnetic waves
James Clerk Maxwell’s crowning achievement was to show that a beam of light is a traveling wave of electric and
magnetic fields (an electromagnetic wave) and thus that optics is a branch of electromagnetism
Properties of electromagnetic waves
The fundamental mechanism behind electromagnetic radiation is the acceleration of a charged particle
→
−
→
−
The electric and magnetic fields E and B are perpendicular to the direction of travel of the wave (the
wave is a transverse wave)
The electric field is perpendicular to the magnetic field
The fields vary sinusoidally, just like the transverse waves you studied in PHY 120; moreover, the fields vary
with the same frequency and are in phase with each other
Remarkably, Maxwell predicted that, in vacuum, both electromagnetic waves and light traveled at the same
speed, namely
1
8
8
= 2.997 × 10 m/s ≈ 3.00 × 10 m/s
c = √
µ0 0
Electromagnetic Waves
Announcements
Review
Electromagnetic Waves
Polarization
Energy in Electromagnetic Waves
Radiation Pressure
Final Questions
Announcements
Review
Electromagnetic Waves
Polarization
Energy in Electromagnetic Waves
Radiation Pressure
Final Questions
Announcements
Review
Polarization
Energy in Electromagnetic Waves
Radiation Pressure
Final Questions
Question
VHF (very high frequency) television antennas in England are oriented vertically, but those in North America are
horizontal. Why?
Electromagnetic Waves
Announcements
Review
Polarization
Energy in Electromagnetic Waves
Radiation Pressure
Final Questions
Question
VHF (very high frequency) television antennas in England are oriented vertically, but those in North America are
horizontal. Why?
Polarization
Polarization is a property of waves that describes the orientation of their oscillations
Electromagnetic Waves
Announcements
Review
Polarization
Energy in Electromagnetic Waves
Radiation Pressure
Final Questions
Question
VHF (very high frequency) television antennas in England are oriented vertically, but those in North America are
horizontal. Why?
Polarization
Polarization is a property of waves that describes the orientation of their oscillations
Polarization and electromagnetic waves
The polarization of electromagnetic waves is described by the oscillating electric field (not the oscillating
magnetic field)
If the electric field is oscillating vertically, the wave is said to be vertically polarized (these waves are
produced by vertical transmitting antenna)
If the electric field is oscillating horizontally, the wave is said to be horizontally polarized (these waves are
produced by horizontal transmitting antenna)
Questions
If a vertically oriented antenna intercepted vertically polarized electromagnetic waves, what would happen?
If a horizontally oriented antenna intercepted vertically polarized electromagnetic waves, what would
happen?
Electromagnetic Waves
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Review
Polarization
Energy in Electromagnetic Waves
Radiation Pressure
Final Questions
Energy in electromagnetic waves
Electromagnetic waves transport energy
→
−
The rate of flow of energy in an electromagnetic wave is described by the Poynting vector S whose
magnitude is given by
1
EB sin θ
S =
µ0
→
−
→
−
where θ is the angle between E and B
To find the direction of the Poynting vector, use the following right hand rule:
→
−
Point your hand in the direction of the electric field E
→
−
Curl your fingers in the direction of the magnetic field B
Your thumb will naturally point in the direction of the Poynting vector (this is also gives the
direction of travel of the wave)
The magnitude of the Poynting vector represents the rate at which energy flows through a unit surface
area perpendicular to the direction of wave propagation (power per area)
The SI units of the Poynting vector are J/s·m2 = W/m2
→
−
→
−
→
−
Because E and B are perpendicular to one another in an electromagnetic wave, the magnitude of S is
S =
in which S, E , and B are instantaneous values
Electromagnetic Waves
1
µ0
EB =
1
cµ0
E
2
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Final Questions
Energy transport
For an electromagnetic wave, the energy density associated with the electric field is
uE =
1
2
2
0 E =
1
2
2 2
0 B c =
B2
2µ0
= uB
Therefore, the instantaneous energy density associated with the electric field of an electromagnetic wave
equals the instantaneous energy density associated with the magnetic field
The average total energy density u is
u = uE + uB =
1
2
2
0 Emax =
2
Bmax
2µ0
Though we have an equation for the energy transport rate as a function of time, it is more useful to find
the average energy transported over time
The time-averaged value of S (written S) is also called the intensity I
I =S =
1
cµ0
E2 =
1
cµ0
2
Emax
cos2 (kx − ωt) =
1
2cµ0
2
Emax =
2
cBmax
2µ0
Therefore, the intensity of an electromagnetic wave is equal to the average energy density multiplied by the
speed of light
I = S = cu
Electromagnetic Waves
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Energy in Electromagnetic Waves
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Final Questions
Variation of intensity with distance
How intensity varies with distance from a real source of electromagnetic radiation is often complex
However, in some situations we can assume that the source is a point source that emits light isotropically
(equal intensity in all directions)
Assuming that the energy of the waves is conserved as they spread out from the source, we conclude that
at a distance r from the source, the intensity I is
I =
Ps
4πr 2
where Ps is the power of the source
Therefore, the intensity of the electromagnetic radiation emitted by an isotropic point source decreases
with the square of the distance r from the source
Electromagnetic Waves
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Review
Polarization
Energy in Electromagnetic Waves
Radiation Pressure
Final Questions
Question #1
A 3.0-mW laser pointer creates a spot on the screen that is 2.0 mm in diameter. What is the intensity of the laser?
(The 3.0-mW value is a time-averaged value.)
Electromagnetic Waves
Announcements
Review
Polarization
Energy in Electromagnetic Waves
Radiation Pressure
Final Questions
Question #1
A 3.0-mW laser pointer creates a spot on the screen that is 2.0 mm in diameter. What is the intensity of the laser?
(The 3.0-mW value is a time-averaged value.)
Answer
While a laser pointer may be approximated as a point source, it does not radiate light isotropically
Assuming that the laser pointer transmits all of its energy in the spot on the screen, then
I =
Electromagnetic Waves
P
A
=
P
πr 2
=
3.0 × 10−3 W
π(1.0 × 10−3 m)2
= 955 W/m
2
Announcements
Review
Polarization
Energy in Electromagnetic Waves
Radiation Pressure
Final Questions
Question #1
A 3.0-mW laser pointer creates a spot on the screen that is 2.0 mm in diameter. What is the intensity of the laser?
(The 3.0-mW value is a time-averaged value.)
Answer
While a laser pointer may be approximated as a point source, it does not radiate light isotropically
Assuming that the laser pointer transmits all of its energy in the spot on the screen, then
I =
P
A
=
P
πr 2
=
3.0 × 10−3 W
π(1.0 × 10−3 m)2
= 955 W/m
2
Question #2
The Sun has a luminosity (power) of 3.8 × 1026 W and a mean distance of 1.5 × 108 km from the Earth. What is
the mean intensity of the Sun at the Earth?
Electromagnetic Waves
Announcements
Review
Polarization
Energy in Electromagnetic Waves
Radiation Pressure
Final Questions
Question #1
A 3.0-mW laser pointer creates a spot on the screen that is 2.0 mm in diameter. What is the intensity of the laser?
(The 3.0-mW value is a time-averaged value.)
Answer
While a laser pointer may be approximated as a point source, it does not radiate light isotropically
Assuming that the laser pointer transmits all of its energy in the spot on the screen, then
I =
P
A
=
P
πr 2
=
3.0 × 10−3 W
π(1.0 × 10−3 m)2
= 955 W/m
2
Question #2
The Sun has a luminosity (power) of 3.8 × 1026 W and a mean distance of 1.5 × 108 km from the Earth. What is
the mean intensity of the Sun at the Earth?
Answer
The Sun may be approximated as an isotropic point source
Therefore, its intensity at the Earth is
I =
Electromagnetic Waves
P
4πr 2
=
3.8 × 1026 W
4π(1.5 × 1011 m)2
3
2
= 1.3 × 10 W/m
Announcements
Review
Polarization
Energy in Electromagnetic Waves
Radiation Pressure
Final Questions
Radiation pressure
Electromagnetic waves have linear momentum as well as energy
This means that we can exert a pressure (a radiation pressure Pr = F /A) on an object by shining light on
it!
If radiation is entirely absorbed (taken in) by an object, that object gained an energy ∆U and a linear
momentum ∆p over the time interval ∆t
The change in linear momentum is
∆p =









∆U
c
2∆U
c
(total absorption)
(total reflection back along path)
If a flat surface of area A, perpendicular to the path of radiation, intercepts the radiation, then the
magnitude of the force exerted on that area is
F =









IA
c
2IA
c
(total absorption)
(total reflection back along path)
In general, if the radiation is partly absorbed and partly reflected, the radiation pressure is somewhere
between these two extremes
Electromagnetic Waves
Announcements
Review
Polarization
Energy in Electromagnetic Waves
Radiation Pressure
Final Questions
Question #1
A 3.0-mW laser pointer creates a spot on the screen that is 2.0 mm in diameter. (The 3.0-mW value is a
time-averaged value.) What is the radiation pressure on the screen if it reflects all of the light that strikes it?
Electromagnetic Waves
Announcements
Review
Polarization
Energy in Electromagnetic Waves
Radiation Pressure
Final Questions
Question #1
A 3.0-mW laser pointer creates a spot on the screen that is 2.0 mm in diameter. (The 3.0-mW value is a
time-averaged value.) What is the radiation pressure on the screen if it reflects all of the light that strikes it?
Answer
Recall that in an earlier problem, we found that the intensity of this laser was 955 W/m2
Using that result, the radiation pressure on the screen (for total reflection) is
P =
Electromagnetic Waves
2I
c
=
2(955 W/m2 )
3.0 × 108 m/s2
−6
= 6.4 × 10
N/m
2
Announcements
Review
Polarization
Energy in Electromagnetic Waves
Radiation Pressure
Final Questions
Question #1
A 3.0-mW laser pointer creates a spot on the screen that is 2.0 mm in diameter. (The 3.0-mW value is a
time-averaged value.) What is the radiation pressure on the screen if it reflects all of the light that strikes it?
Answer
Recall that in an earlier problem, we found that the intensity of this laser was 955 W/m2
Using that result, the radiation pressure on the screen (for total reflection) is
P =
2I
c
=
2(955 W/m2 )
3.0 × 108 m/s2
−6
= 6.4 × 10
N/m
2
Question #2
Light of uniform intensity shines perpendicularly on a totally absorbing surface. If the area of the surface is
decreased, what happens to the radiation pressure on the surface?
Electromagnetic Waves
Announcements
Review
Polarization
Energy in Electromagnetic Waves
Radiation Pressure
Final Questions
Question #1
A 3.0-mW laser pointer creates a spot on the screen that is 2.0 mm in diameter. (The 3.0-mW value is a
time-averaged value.) What is the radiation pressure on the screen if it reflects all of the light that strikes it?
Answer
Recall that in an earlier problem, we found that the intensity of this laser was 955 W/m2
Using that result, the radiation pressure on the screen (for total reflection) is
P =
2I
c
=
2(955 W/m2 )
3.0 × 108 m/s2
−6
= 6.4 × 10
N/m
2
Question #2
Light of uniform intensity shines perpendicularly on a totally absorbing surface. If the area of the surface is
decreased, what happens to the radiation pressure on the surface?
Answer
It stays the same
Electromagnetic Waves
Announcements
Review
Polarization
Energy in Electromagnetic Waves
Radiation Pressure
Final Questions
Question #1
A 3.0-mW laser pointer creates a spot on the screen that is 2.0 mm in diameter. (The 3.0-mW value is a
time-averaged value.) What is the radiation pressure on the screen if it reflects all of the light that strikes it?
Answer
Recall that in an earlier problem, we found that the intensity of this laser was 955 W/m2
Using that result, the radiation pressure on the screen (for total reflection) is
P =
2I
c
=
2(955 W/m2 )
3.0 × 108 m/s2
−6
= 6.4 × 10
N/m
2
Question #2
Light of uniform intensity shines perpendicularly on a totally absorbing surface. If the area of the surface is
decreased, what happens to the radiation pressure on the surface?
Answer
It stays the same
Question #3
Light of uniform intensity shines perpendicularly on a totally absorbing surface. If the area of the surface is
decreased, what happens to the radiation force on the surface?
Electromagnetic Waves
Announcements
Review
Polarization
Energy in Electromagnetic Waves
Radiation Pressure
Final Questions
Question #1
A 3.0-mW laser pointer creates a spot on the screen that is 2.0 mm in diameter. (The 3.0-mW value is a
time-averaged value.) What is the radiation pressure on the screen if it reflects all of the light that strikes it?
Answer
Recall that in an earlier problem, we found that the intensity of this laser was 955 W/m2
Using that result, the radiation pressure on the screen (for total reflection) is
P =
2I
c
=
2(955 W/m2 )
3.0 × 108 m/s2
−6
= 6.4 × 10
N/m
2
Question #2
Light of uniform intensity shines perpendicularly on a totally absorbing surface. If the area of the surface is
decreased, what happens to the radiation pressure on the surface?
Answer
It stays the same
Question #3
Light of uniform intensity shines perpendicularly on a totally absorbing surface. If the area of the surface is
decreased, what happens to the radiation force on the surface?
Answer
It decreases
Electromagnetic Waves
Announcements
Review
Polarization
Energy in Electromagnetic Waves
Reading Assignment
Read sections 23.1 - 23.4
Homework Assignment 7
Homework for Chapter 22 (due at the beginning of class on Friday, October 15)
Q: 4, 7, 10
P: 2, 16, 20, 40
PP: 32.1, 32.3
Electromagnetic Waves
Radiation Pressure
Final Questions