PHYS 219
Spring semester 2014
Lecture 19: EM Waves (continued)
Ron Reifenberger
Birck Nanotechnology Center
Purdue University
Lecture 19
1
Recap - Electromagnetic Wave
E=E(x,t)
B=B(x,t)
Eo
Bo

Relating f and λ:
Units: [c] in m/s;
Relating E and B:
c = f 
[λ] in m; [f] in s-1 = Hz
E
c
B
(at any position and at any time)
2
1
Electromagnetic Spectrum
3
Sunburn implies energy is transported by an EM wave!
The Poynting vector S specifies the instantaneous power
per unit area transported by an EM wave at a point in
space at an instant of time [E(t) and B(t) are
instantaneous values].
 1   E(t) B(t)
S = E ×B =
μo
μo
c=
E
B
=
1
εoμo
 E2  t 
S=
= cεoE2  t 
cμo

E2
S
= cεo o
AVERAGE
2
2
= cεoErms
Units: [W/m2]
4
2
The time-averaged value of S is called the
intensity I of the wave

I S
1
2
= cεo Eo2 = cεo Erms
AVERAGE
2
Eo
1
since c =
=
Bo
εo μo

I S
AVERAGE
 Bo2 
1
1
2 2
= cεo  c Bo  = cεo 

2
2
 εo μo 
1 c 2 c 2
=
Bo = Brms
2 μo
μo
The intensity I specifies the power (in
Watts per m2) carried by an EM wave in
free space averaged over time.
5
Be able to distinguish between
closely related concepts
 Intensity (in W/m2) of an EM wave: I
 Power (in W or J/s) carried (or transmitted) by EM wave through
an area A: P=IA
 Energy (in J) carried by an EM wave in a time t: U=Pt=(IA)t
 Energy density (in J/m3) of an EM wave: utot=I/c
These quantities are used to define the momentum
and the radiation pressure exerted by an EM wave.
6
3
The time averaged energy density of an EM wave
KEY IDEA: Energy is stored in
both E and B fields
Total absorber
Hole in mask,
area A
time aver. power thru hole : P = IA W 
time aver. energy thru hole : ΔU = PΔt J 
 J 
Define time aver. energy density utot in 3  :
 m 
ct
ΔU
PΔt
IAΔt
utot =
=
=
AcΔt  AcΔt  AcΔt 
I 11
 1 c 2
=  cεoEo2  = 
Bo 
c c2
 c  2μo 
1
1 2
= εoEo2 =
Bo
2
2μo
utot =
utot
2
utot = εoErms
=
1 2
Brms
μo
utot
Energy passing
thru hole in
time t
Time aver.
radiation,
intensity I
Units: J/m3
Important to distinguish between what we
can measure and what we can calculate.
7
An EM wave exerts a net force on absorber
absorber
Must conclude that a net force is
exerted on charged particles at
the surface of the absorber. A net
force means a change in momentum
over time.
Recall:
Δv Δ  mv  Δp
=
=
Δt
Δt
Δt
where Δp is change in momentum.
In general, F = ma = m
8
4
Consequences of net force on absorber
1. For EM radaition*, U = pc, so
ΔU = cΔp (valid for total adsorption)
If EM wave has time aver. power P, intensity I, we have
ΔU PΔt IAΔt
=
=
c
c
c
time aver. net momentum delivered by EM wave when totally adsorbed
Δp =
2. If EM wave with time aver. power P, intensity I, we also have a
time aver. radiation force
PΔt
Δp
P IA
Fradiation =
 c = =
Δt
Δt c c
F
I
Pradiation = radiation =
A
c
time aver. radiation pressure exerted by EM wave when totally adsorbed
* We use U for energy because E is already used for electric field. As an example,
for particles U=½ mv2 =½ pv
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SUMMARY
1
εoμo
Speed of EM Waves (m/s)
c=
Ratio of Peak Fields
Eo = cBo
2π
λ
Definition of Wavenumber (m-1)
k
Frequency-Wavelength
relationship
λf = c
 1  
S = E ×B
μo
Definition of Poynting Vector
(W/m2)
Energy density (J/m3)
 1  E 2 1 Bo2 
Eo2 1 Bo2
1
1
utotal =  εoE2 + B2  =  εo o +
=
 = εo
2
μo  2  2 μo 2 
2 μo 2
Intensity of EM Wave (W/m2)
(time average of Poynting vector for
sinusoidal, linearly polarized , plane
EM wave)
Momentum Transfer (kg m/s)
(complete absorption)
Radiation Pressure ( N/m2 or Pa)
(complete absorption)
I= S =
p=
EoBo
E2
cB2 ε c
= o = o = o Eo2
2μo 2μo c 2μo
2
U
; U = u × volume
c
I
P=
c
10
5
Propagation of Radiation
 In general, radiation can propagate outward in all
directions
 The ideal case of a very small source producing spherical
wave fronts is called a point source
 The intensity of a spherical wave decreases with
distance: I  1/r2
 The intensity must decrease as the constant amount of
energy spreads out over greater areas
 The intensity relationship (I  1/r2) applies to many
situations, including the strength of a radio signal from a
distant station
11
Intensity vs. Distance for Point Source?
Ideal case:
Uniform
radiation from a
point source
emitting EM
radiation at an
average power P
1. The radiation moves outward
at the speed of light
2. Represents
a sphere of
radius r=ct
P
3. Surface
area of a
sphere is
4r2
I=
P
4πr2
12
6
Example: What is the intensity, radiation pressure, energy density and
amplitude of E at 1 m from a 100 W light bulb? Assume all 100 W goes
into radiation.
Fradiation I
8W / m2
= =
A
c 3 ×108 m / s
Ws
N
= 2.67 ×10-8 3 = 2.67 ×10-8 2
m
m
Spherical surface
100 W bulb
1. I(r = 1 m) =
2. Pradiation =
r=1m
P
100 W
=
= 8 W m2
4πr2 12.56 m2
2
3. utot =
I
8W / m2
J
=
= 2.67 ×10-8 3
8
c 3 ×10 m / s
m
1
4. I = cεoEo2  8W / m2
2
2  8W / m2 
Eo2 =
cεo
=
16 W / m2
3 ×10 m / s 8.85 ×10-12 C2 / Nm2 
8
2
 1m 
= 6026 N 2
= 8 W m2 × 
= 8 ×10-4 W cm2
C

 100 cm 
Eo = 77.6 N
C
= 0.8 mW cm2
13
Polarization of EM Radiation
• For any EM wave, the electric field is perpendicular to the
direction of propagation
• There are many directions of the electric field in an EM
wave
• Knowing the direction of the electric field in an EM wave is
important to determine how the wave interacts with matter
• Most light is unpolarized
• Polarized light can be created using a polarizer
• A polarizer often consists of a thin, plastic film that allows
an EM wave to pass through it only if the electric field of
the wave is parallel to a particular direction called the axis
of the polarizer
• A polarizer strongly absorbs EM radiation with electric
fields that are not along the axis
14
7
The Effect of a Polarizer
15
The Law of Malus’
(Etienne-Louis Malus discovers polarized light around ~1810;
Edwin Land invents polaroid filter in 1929)
Recall.......... I = S =
cBo2 εo c 2
=
Eo
2μo
2
εo c 2
Ein
2
εc 2
εc
Iout = o Eout
= o Ein2 cos2 θ
2
2
= Iin cos2 θ
Iin =
Iout = Iin cos2 θ
Note: Ein, Eout represents amplitudes or maximum values
16
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