lectures 20,21 (ch. 32) electromagnetic waves
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Lectures 20,21 (Ch. 32) Electromagnetic waves. Maxwell’s equations Wave equation General properties of the waves Sinusoidal waves Travelling and standing waves Energy characteristics: the Pointing vector, intensity, power, energy - PowerPoint PPT PresentationTRANSCRIPT
Lectures 20,21 (Ch. 32)Electromagnetic waves
1. Maxwell’s equations2. Wave equation3. General properties of the waves4. Sinusoidal waves5. Travelling and standing waves6. Energy characteristics: the Pointing
vector, intensity, power, energy7. Generation, transmission and
receiving of electromagnetic waves
Maxwell’s equations
James Clerk Maxwell (1831 –1879)
)(dt
dildB Eencl
dt
dldE B
enclq
AdE
0 AdB
Two Gauss’s laws + Faraday’s law +Amper’s law
Maxwell introduced displacement current, wrote these four equations together, predicted the electromagnetic waves propagating in vacuum with velocity of light and shown that light itself is e.m. wave.
1865 Maxwell’s theory of electro-magnetism1887 Hertz’s experiment1890 Marconi radio (wireless communication)
Mechanical waves
Transverse waves: oscillation is in the direction perpendicular to the propagation direction (waves on the rope, on the surface of water)Longitudinal waves : oscillation in the direction of the propagation (sound, spring)E.M. waves are transverse waves
In mechanical waves there is collective oscillations of particles.E and B oscillate in e.m. waves. Matter is not required. E.M waves may propagate in vacuum.
Wave equation and major characteristics of the wave
kxAtxy
tAtxy
cos)0,(
cos),0(
)cos(),(
0),(1),(
222
kxtAtxyt
xty
vx
xty
2
,2
kT
vkkT
v ,
TTkdt
dxvkdxdt
dconst
kxt
2
2,
0
0 AdE
0 AdB
dt
dldE B
dt
dldB E
Maxwell’s equations in the absence of charges and currents take particular symmetric form
Look for solution in the form:
To satisfy Gauss’s laws it is necessary to have:
!, vBvE
If there is a component of E or B parallel to v Gauss’s laws are not satisfied . It may be verified choosing the front of the Gaussian surface ahead of the wave front.
Faraday’s law:
dt
vdtBaEa
Amper’s law:
1,
122
vvEvE
EBdt
vdtEaBa
s
mcvvacuumIn 8
00
1031
nKKv
cm
00
vBE
),1,( alwaysnotbutKnKtypically m
n
cv )1( alwaysnotbutcvntypically
Derivation of the wave equationLook for plane waves: Ey(x,t) and Bz(x,t)
Faraday’s law:
tx
B
x
E
t
B
x
E
xat
BxExxEa
zyzy
zyy
2
2
2
,
)]()([
Amper’s law:
2
22
,
)]()([
t
E
tx
B
t
E
x
B
xat
ExBxxBa
yzyz
yzz
01
01
2
2
22
2
2
2
22
2
t
B
vx
B
t
E
vx
E
zz
yy
1
vvBE
E and B in e.m. wave
)cos(
)cos(
0
0
kxtBB
kxtEE
z
y
)cos(
)cos(
0
0
kxtBkB
kxtEjE
or
)cos(
)cos(
0
0
kxtBB
kxtEE
z
y
)cos(
)cos(
0
0
kxtBkB
kxtEjE
or
This is y-polarized wave. The direction of E oscillations determines polarization of the wave. Do not confuse polarization of the wave with polarization of dielectric (i.e.separation of charges in E).
The frequency range (spectrum) of e/m. waves
Radio waves, microwaves, IR radiation, light, UV radiation, x-rays and gamma-rays are e/m waves of different frequencies. All of them propagate in vacuum with v=c=3x108m/s
)(11
][
2
1
HertzHs
f
Tf
Frequency of e.m.wave does not depend on the medium where it
propagates. It is determined by the frequency of charge oscillations. Both the speed of propagation and the wavelength do depend on the medium: v=c/n,
vacuuminlengthwavef
c
nf
c
f
vvT
0
,
n0
Example. A carbon-dioxide laser emits a sinusoidal e.m. wave that travels in vacuum in the negative x direction. The wavelength is 10.6 μm and the wave is z-polarized. Maximum magnitude of E is 1.5MW/m. Write vector equations for E and B as functions of time and position. Plot the wave in a figure.
sradmradsmck
mradm
radk
Tsm
mV
c
EB
kxtBjB
kxtEkE
/1078.1/1093.5)/103(
/1093.5106.10
17.322
105/103
/105.1
)cos(
)cos(
1458
56
38
60
0
0
0
NB1: Since B=E/c→B (in T) <<E (in V/m)NB2: in general, arbitrary initial phase may be added :
)cos(
)cos(
0
0
kxtBjB
kxtEkE
To find initial phase one needs to know either initial conditions E(x,t=0) or boundary condition E(t,x=0).
• Example. Nd:YAG laser emits IR radiation in vacuum at the wavelength 1.062μm.• The pulse duration is 30ps(picos). How many oscillations of E does the pulse contain?
10000103
103
3
30
)(3103/103
10062.1
15
12
158
6
s
s
fs
ps
TN
femtosfsssm
m
cT
pulse
The shortest pulses (~100 as (attos),1as=10-18s) obtained today consist of less then 1 period of E oscillations.They allow to visualize the motion of e in atoms and molecules.
Ends of string are fixed→nodes on the ends
Max possible wavelength is determined by the length of string
nfL
vnfn
n
L
L
vvfLL
nn min
maxminmax
max
2...2,1,
2
22
2
Reflection from a perfect conductor. Standing wavesTotal E is the superposition of the incoming and reflected waves. On the surface of the conductor E total parallel to the surface should be zero. Perfect conductor is a perfect reflector with E in ref. wave oscillating in opposite phase.
tkxBtxBtxBtxB
kxtBtxB
kxtBtxB
tkxEtxEtxEtxE
kxtEtxE
kxtEtxE
refzinzz
refz
inz
refyinyy
refy
iny
coscos2),(),(),(
)cos(),(
)cos(),(
sinsin2),(),(),(
sinsincoscos)cos(
)cos(),(
)cos(),(
0
0
0
0
0
E(x)=0 at arbitrary moment of time in the positions where sinkx=0, that is kx=πn, n=0,1,2,3,..
)(,...2,2
3,,
2,0
,..2,1,0,2
Eofplanesnodalx
nn
x
Example.In a microwave oven a wavelength 12.2cm (strongly absorbed by a water) is used. What is the minimum size of the oven? What are the other options? Why in the other options rotation is required?
nfL
vnfn
n
L
L
vvfLL
nn min
maxminmax
max
2...2,1,
2
22
2
If two conductors are placed parallel to each other the nodes of E should be on the ends just as on the string with fixed ends
...3.18
2.12
1.62min
cmL
middletheinnodeonecmL
cmL
The Energy Characteristics of e.m. waves
The energy density:
1,,
2,
2
22
vvBEB
uE
u magel
The Poynting vector is the energy transferred per unite time per unite cross-section, i.e. power per unite area=the energy flow rate in the direction of propagation
,
,1
EBEBvS
uvdtAudVdUdt
dU
AA
PS
BE
S
EBB
EBE
u
2
222
22
Intensity is the power per unite area averaged over the period of oscillationsFor travelling waves:
22][][
m
W
sm
JIS
RMSRMSrms
BEI
EkxtEEE
BEkxt
BEkxt
BEI
,2
)(cos
2))(2cos1(
2)(cos
020
2
0000200
200BE
I
TA
PdtUAdSP ,
Standing waves do not transfer the energy:
02sincossin2
cossincossin4 0000 tkxkx
BEttkxkx
BEI
Example. The distance from the sun to the earth is 1.5x1011m.1) What is the power of radiation of the sun if it’s intensity measured by the earth orbiting satellite is 1.4 kW/m. 2) If the area of the panels of the satellite is 4m and is perpendicular to the radiation of the sun, what is the power received by satellite?
Wmm
WIRAdSPsun
2622222
32 10410)5.1(14.34104.14
NB: the life on the earth is due to this power of radiation received from the sun!
kWmm
WIAAdSPpanels 6.54104.1 2
23
Example A radio station on the surface of the earth radiates a sinusoidal wave with an average total power 50kW. Assuming
that transmitter radiates equally in all directions, find the amplitudes of E and B detected by a satellite at a distance 100km.
27
210
4
21096.7
1028.6
105
2 m
W
mR
PI
Tc
EB
mVcIE
c
EBEI
1100
200
0
20
0
00
102.8
/105.22
22
E.m. waves are produced by oscillating charge or current
2)sin(~,
sin~~
rI
rBE
v
Richard Feynman ( 1918 – 1988)
Optimal position of antenna (maximizing the induced current in antenna) corresponds to the wire parallel to E
Optimal position of antenna (maximizing the induced current in antenna) corresponds to the loop perpendicular to B.
Optimal size of antenna~λ/2
Radiation Pressure
S
t
p
t
pF
p
S
S t
p
t
pF
2
Reflecting plane
Absorbing plane PaatmPam
NP
c
P
c
AIF
c
I
dt
Ud
Acdt
pd
AA
FP
c
Up
rad
rad
52
101,1][
11
c
IPrad
c
IPrad
2
Example. Find the force due to a radiation pressure on the solar panels. I=1.4kW/m2,A=1m2.
NAPF
Pasm
mW
c
IP
rad
rad
6
68
23
105
105/103
/104.1
However over long time it influences the satellite orbit!Comet tails, some stars formation
EMW carry both energy and momentum
r
Laser coolingNobel Prize,1997
Steven Chu,Claude Cohen –Tannoudji,Bill Phillips
s
mmvKT
s
kmvKT
kTmv
1~10~,1~300~
:~2
7
2
kv
laseratom
Photons: sJhh
kpE
2410626.6,2
,
Photon
Photon
Photon Photon
atom
atom
atom
atom
1E
2E
1E
2E
Polarization
Dichroism (dependence of absorption on polarization) is used for construction of the polarization filters for em wavesA grid of wires is a polarization flter for radio waves
When E in a radio wave is parallel to the wiresthe currents are induced in the wires and wave is absorbed. Long molecules play a role of wires for light and used for building of polarization filters (polaroids)
Linear polarized, namely, y-plz e.m.wave
Axis of the filter. If em wave is polarized along this axis it goes through without asborption. Linear plz em wave with orthoginal to this axis in not transmitted (fully absorbed by the filter).
Unpolarized em wave (random polarization)
Ein
Eout
cosinout EE
Malus’s law (1809)
2cosinout II
In general case when linear plz wave goes through the filter only its projection on the axis of the filter goes through.
2cos2 in
inout
III
Sun, lamp and other thermal sources produce unpolarized light
NB: After the filter em wave is always linear polarized
along the axis of the filter.
How to check polaroid glasses?
Crossed polaroids do not transmit light
Circular polarization
x
y
z
Ey
Ez])sin()[cos(
)sin(
)cos(
0
0
0
kkxtjkxtEE
kxtEE
kxtEE
t
z
y
Ey
Ez
Left circular polarization
If elliptic polarization
oxoy EE
Birefrigent materials: refractive index depends on polarization: )(),( 21 nn
x
x
2)()( 2121
xnn
cxkk