Download Waves & Oscillations Physics 42200 Spring 2014 Semester Lecture 30 – Electromagnetic Waves

Physics 42200
Waves & Oscillations
Lecture 30 – Electromagnetic Waves
Spring 2014 Semester
Matthew Jones
Electromagnetism
• Geometric optics overlooks the wave nature of
light.
– Light inconsistent with longitudinal waves in an
ethereal medium
– Still an excellent approximation when feature sizes are
large compared with the wavelength of light
• But geometric optics could not explain
– Polarization
– Diffraction
– Interference
• A unified picture was provided by Maxwell c. 1864
Forces on Charges
• Coulomb’s law of electrostatic force:
Charles-Augustin de
Coulomb
(1736 - 1806)
• The magnitude of the attractive/repulsive force is
= where
=
and therefore
4
1
= 8.99 × 10 ∙
̂
∙
= 8.85 × 10 ∙
∙
(This constant is called the “permittivity of free space”)
Electric Field
• An electric charge changes the properties of the
space around it.
– It is the source of an “electric field”.
– It could be defined as the “force per unit charge”:
=
̂
=0
!"#$
– Quantum field theory provides a deeper description…
• Gauss’s Law:
1
%'
&∙
*
() =
+,-+./
Electric “flux” through surface S
Magnetic Field
• Moving charges (ie, electric current) produce a
magnetic field:
2
3(ℓ × ̂
%
1=
4
A moving charge in a magnetic field
experiences a force:
= 05 × 1
Lorentz force law:
= 0 + 05 × 1
Gauss’s Law for Magnetism
• Electric charges produce electric fields:
'∙
%&
*
() =
+,-+./
• But there are no “magnetic charges”:
2
' ∙ 1() =
%&
*
0
Magnetic “flux” through surface S
Ampere’s Law
• An electric current produces a magnetic field that
curls around it:
7
7
(Close, but not quite the whole story)
Faraday’s Law of Magnetic Induction
• Magnetic flux: 89 = :* 1 ∙ &; ()
• Faraday observed that a changing
magnetic flux through a wire loop
induced a current
– It transferred energy to the charge
carriers in the wire
(89
ℇ = −
(>
(
?
∙ (ℓ = − % &; ∙ 1()
(> *
7
3
The Problem with Ampere’s Law
Current through @ is 3
Current through @ is 0!
The current through @ is zero,
but the electric flux is not zero.
The electric flux changes as
charge flows onto the capacitor.
Maxwell’s Displacement Current
• We can think of the changing electric flux
through @ as if it were a current:
(8/
(
3. =
=
%
∙ &; ()
(>
(> *
? A ∙ Bℓ = DE F + FB
C
B
' BN
= DE G + DE HE % J ∙ K
BI LM
4
Maxwell’s Equations (1864)
OPKQPBR
' ∙ J BN = ? K
HE
L
' ∙ A BN = E
? K
L
BST
? J ∙ Bℓ = −
BI
C
BSR
? A ∙ Bℓ = DE F + DE UE
BI
C
1
2
3
4
Maxwell’s Equations in Free Space
In “free space” where there are no electric charges or
sources of current, Maxwell’s equations are quite
symmetric:
' ∙ J BN = E
? K
L
' ∙ A BN = E
? K
L
BST
? J ∙ Bℓ = −
BI
C
BSR
? A ∙ Bℓ = DE UE
BI
C
Maxwell’s Equations in Free Space
BST
? J ∙ Bℓ = −
BI
C
BSR
? A ∙ Bℓ = DE UE
BI
C
A changing magnetic flux induces an electric field.
A changing electric flux induces a magnetic field.
Will this process continue indefinitely?
Light is an Electromagnetic Wave
• Faraday’s Law:
?
7
∮7
∙ (ℓ =
∙ (ℓ =
≈
`
.WX
−
.Y
_ Δ^ −
=−
Z[\
` (_
)Δ^
ZY
∆^∆_
x
Zef
∆_∆^
Zg
Ex
z1
y
By
Δ^
Ex
Δ_
z2
z
Light is an Electromagnetic Wave
• Ampere’s law:
∮7 1 ∙ (ℓ = 2
.Wh
.Y
=2
? 1 ∙ (ℓ = 1j _ Δi − 1j (_ )Δi
7
≈−
Z[\
Zg
x
∆_∆i
Zef
∆i∆_
ZY
Ex
∆_
z1 z2
y
By
By
∆i
z
Putting these together…
k1j
k `
=−
k_
k>
k1j
k `
−
=2
k_
k>
Differentiate the first with respect to _:
k 1j
k `
=−
k_
k_k>
Differentiate the second with respect to >:
k 1j
k `
−
=2
k_k>
k>
lM J m
lM Jm
= DE HE
M
ln
lIM
Velocity of Electromagnetic Waves
k i
1 k i
=
k^
5 k>
lM J m
lM J m
= DE HE
M
ln
lIM
Speed of wave propagation is
1
5=
2
1
=
4 × 10 o /) 8.854 × 10
= M. qqr × sEr t/u
(speed of light)
/ ∙
Light is an Electromagnetic Wave
Speed of light was measured by Fizeau in 1849:
5 = 315,300km/s
Maxwell wrote:
This velocity is so nearly that of light, that it seems we have
strong reason to conclude that light itself (including radiant
heat, and other radiations if any) is an electromagnetic
disturbance in the form of waves propagated through the
electromagnetic field according to electromagnetic laws.
Electromagnetic Waves
lM Jm
lM Jm
= DE H E
M
ln
lIM
• A solution is ` _, > = sin _ − }>
where } = ~ = 2 ~/€
• What is the magnetic field?
k1j
k `
=−
=−
cos _ − }>
k>
k_
1j ^, > =
}
sin ^ − }>
Electromagnetic Waves
^x
_z
iy
• , 1 and 5 are mutually perpendicular.
• In general, the direction is
ƒ̂ = „ × 1„
Energy in Electromagnetic Waves
Energy density of electric and magnetic fields:
1
1
†/ =
†9 =
1
2
22
For an electromagnetic wave, 1 = ⁄~ = 2
1
1
†9 =
1 =
= †/
22
2
The total energy density is
† = †9 + †/ =
Intensity of Electromagnetic Waves
• Intensity is defined as the average power
transmitted per unit area.
Intensity = Energy density × wave velocity
3=
~
=
2 ~
2 ~ = 377Ω ≡ Š
(Impedance of free space)
Polarization
• Light is an oscillating electromagnetic field
• The electric field has a direction
^, > = cos ∙ − }>
• No need to specify the magnetic field direction:
@ = ׋
‹ = ( „ × )/Š
where Š =
2/Œ
• ‹ refers to the magnetic field due to the light, not
including any induced magnetic fields in the presence of
matter.
• Coherent light has the same phase over macroscopic
distances and time
• Polarized light has the electric field aligned over
macroscopic distances and time
Polarization
Sources of un-polarized light
• Hot atoms transfer kinetic energy to electrons
randomly
• Electrons randomly de-excite, emitting incoherent
light – uncorrelated in phase and polarization
Polarization
Sources of polarized light
• Lasers produce light by stimulated emission
– A photon causes an excited atom to emit another photon
– The photon is emitted in phase and with the same
polarization
• The resulting beam is highly coherent and polarized
Polarization by Absorption
• A polarizer absorbs the component with oriented
along a particular axis.
• The light that emerges is linearly polarized along the
perpendicular axis.
• If the light is initially un-polarized, half the light is
absorbed.
3
1
3
2
Example with Microwaves
^
microwave
source
LP
i
• The electric field in the ^-direction
induces currents in the metal plate
and loses energy:
– Horizontally polarized microwaves are
absorbed
• No current can flow in the idirection because of the slots
– Vertically polarized microwaves are
transmitted
slotted
metal
plate
Polarization by Reflection
• Reflected light is preferentially polarized
• The other component must be transmitted
• Transmission and reflection coefficients must
depend on the polarization
Boundary Conditions
• In the presence of matter, the components electric and
magnetic fields perpendicular to a surface change
abruptly
• The component parallel to a surface is the same on
both sides.
• Summary:
– Perpendicular to surface
Œ
2 ‹
– Parallel to surface
•
•
=Œ •
=2 ‹ •
= ∥
‹∥=‹ ∥
∥
Reflection From a Surface
First case:
is parallel to the surface…
= Y
++
‹ has components parallel
and perpendicular to the
surface
‹∥+ + ‹∥ = ‹∥Y
But ‹∥ = ‹ ∙ •̂ …
Reflection From a Surface
'
J is perpendicular to K
&× „
•̂ =
‹ can be written
„×
‹=
Š
So we can write
1
„ × + ∙ & × „+
‹+ ∙ •̂ =
Š +
=
+
Š
„ ∙ &; = −
+
Š
cos •+
„+
‹+
„
&;
•+ •
•̂
‹
is out of the page
Likewise,
‹ ∙ •̂ =
Š
cos •
perpendicular to
• Boundary condition for ‹:
‹∥+ + ‹∥ = ‹∥Y
• Previous results:
‹+∥ =
e‘
−
’
cos •+
−
+
‹
∥
cos •+ +
Š
=
e“
cos •
’
cos •+
=
=
−
+
= Y
• Two equations in two unknowns…
+
e“
cos •+
’
Y
cos •Y
Š
perpendicular to
cos •+ + cos •+ − Y cos •Y
=
Š
Š
cos •Y
+ cos •Y +
Y cos •Y
=
Š
Š
• Solve for / + :
J” •M –—u ˜P − •s –—u ˜I
=
JI •M ™šQ ˜P + •s ™šQ ˜I
• Solve for Y / + :
JI
M•M ™šQ ˜P
=
JI •M ™šQ ˜P + •s ™šQ ˜I
−
+
Reflection From A Surface
› parallel to surface
‹+ + ‹ = ‹Y
∥+ + ∥ = ∥Y
cos • = Y cos •Y
+ cos •+ −
+
+
=
Y
Š
Š
Š
• Two equations in two unknowns…
„+
+
„
&;
•+ •
•̂
‹ is out of the page
perpendicular to
cos •+ − cos •+
Y cos •Y
=
Š
Š
cos •Y
+ cos •Y +
Y cos •Y
=
Š
Š
• Solve for / + :
J” •s –—u ˜P − •M –—u ˜I
=
JI •s ™šQ ˜P + •M ™šQ ˜I
• Solve for Y / + :
JI
M•s ™šQ ˜P
=
JI •s ™šQ ˜P + •M ™šQ ˜I
+
Fresnel’s Equations
• In most dielectric media, 2 = 2 and therefore
Š
=
Š
2 Œ
=
2 Œ
Œ
&
sin •+
=
=
Œ
&
sin •Y
• After some trigonometry…
sin •+ − •Y
=−
sin •+ + •Y
+ •
ež
e‘ •
=
, /, Ÿ ¡ ¢‘
Ÿ ¡ ¢‘ £¢ž
ež
e‘ ∥
+ ∥
=
tan •+ − •Y
=
tan •+ + •Y
¤¥Ÿ ¢‘ Ÿ ¡ ¢ž
Ÿ ¡ ¢‘ £¢ž ¤¥Ÿ ¢‘ ¢ž
For perpendicular and parallel to plane of incidence.
(not the same as perpendicular or parallel to the surface)
Normal Incidence
• At normal incidence, cos • = 1 there really is no
component parallel to the plane of incidence.
• In this case
& −&
=
& +&
+ •
2&
Y
=
& +&
+ •
• This what we arrived at previously using only the
difference in the speed of light in the two
materials.