1. Physics 102: Lecture 15, Slide 1
Electromagnetic Waves
and Polarization
Physics 102: Lecture 15

2. Physics 102: Lecture 15, Slide 2
Today: Electromagnetic Waves
• Energy
• Intensity
• Polarization

3. Physics 102: Lecture 15, Slide 3
x
z
y
E
B
loop in xy
plane
loop in xz
plane
loop
in
yz
plane
1 2 3
Preflight 15.1, 15.2
“In order to find the loop that dectects the electromagnetic
wave, we should find the loop that has the greatest flux
through the loop.”

4. Physics 102: Lecture 15, Slide 4

5. Physics 102: Lecture 15, Slide 5
Propagation of EM Waves
• Changing B field creates E field
• Changing E field creates B field
E = c B
x
z
y
If you decrease E, you also
This is
important !

6. Physics 102: Lecture 15, Slide 6
Preflight 15.4
Suppose that the electric field of an
electromagnetic wave decreases in
magnitude. The magnetic field:
1 increases
2 decreases
3 remains the same
E=cB

7. Physics 102: Lecture 15, Slide 7
Energy in EM wave
Light waves carry energy but how?
Electric Fields
• Recall Capacitor Energy:
U = ½ C V2
• Energy Density (U/Volume):
uE = ½ ε0E2
• Average Energy Density:
uE = ½ (½ ε0E0
2
)
= ½ ε0E2
rms
Magnetic Fields
• Recall Inductor Energy:
U = ½ L I2
• Energy Density (U/Volume):
uB = ½ B2
/µ0
• Average Energy Density:
uB = ½ (½ B0
2
/µ0)
= ½ B2
rms/µ0

8. Physics 102: Lecture 15, Slide 8

9. Physics 102: Lecture 15, Slide 9
Energy Density
Calculate the average electric and magnetic energy density of
sunlight hitting the earth with Erms = 720 N/C

10. Physics 102: Lecture 15, Slide 10
Energy Density
Calculate the average electric and magnetic energy density of
sunlight hitting the earth with Erms = 720 N/C
2
0
2
1
rmsE Eu ε=
22
12
2
1 C N
8.85 10 720
2 Nm C
−  
= × ÷ ÷
  
0
2
2
1
µ
rms
B
B
u = 2
0
2
2
1
c
Erms
µ
=
00
1
µε
=cUse
ErmsB uEu == 2
0
2
1
ε 3
6
106.42
m
J
uuuu EBEtotal
−
×==+=
6
3
J
2.3 10
m
−
= ×

11. Physics 102: Lecture 15, Slide 11
Energy in EM wave
Light waves carry energy but how?
Electric Fields
• Recall Capacitor Energy:
U = ½ C V2
• Energy Density (U/Volume):
uE = ½ ε0E2
• Average Energy Density:
uE = ½ (½ ε0E0
2
)
= ½ ε0E2
rms
Magnetic Fields
• Recall Inductor Energy:
U = ½ L I2
• Energy Density (U/Volume):
uB = ½ B2
/µ0
• Average Energy Density:
uB = ½ (½ B0
2
/µ0)
= ½ B2
rms/µ0
In EM waves, E field energy = B field energy! ( uE = uB )
utot = uE + uB = 2uE = ε0E2
rms

12. Physics 102: Lecture 15, Slide 12

13. Physics 102: Lecture 15, Slide 13
Intensity (I or S) = Power/Area
• Energy (U) hitting flat surface in time t
= Energy U in red cylinder:
U = u x Volume
= u (AL) = uAct
• Power (P): A
L=ct
P = U/t
= uAc
• Intensity (I or S):
S = P/A [W/m2
]
= uc = cε0E2
rms
23
U = Energy
u = Energy Density
(Energy/Volume)
A = Cross section Area of light
L = Length of box

14. Physics 102: Lecture 15, Slide 14
Polarization
• Transverse waves have a polarization
– (Direction of oscillation of E field for light)
• Types of Polarization
– Linear (Direction of E is constant)
– Circular (Direction of E rotates with time)
– Unpolarized (Direction of E changes randomly)
x
z
y

15. Physics 102: Lecture 15, Slide 15
Linear Polarizers
• Linear Polarizers absorb all electric fields
perpendicular to their transmission axis.

16. Physics 102: Lecture 15, Slide 16

17. Physics 102: Lecture 15, Slide 17
Unpolarized Light on
Linear Polarizer
• Most light comes from electrons accelerating in random
directions and is unpolarized.
• Averaging over all directions: Stransmitted= ½ Sincident
Always true for unpolarized light!

18. Physics 102: Lecture 15, Slide 18
Linearly Polarized Light on
Linear Polarizer (Law of Malus)
Etranmitted = Eincident cos(θ)
Stransmitted = Sincident cos2
(θ)
T
A
θ
θ is the angle
between the
incoming light’s
polarization, and the
transmission axis
θ
Transmission
axisIncident E
ETransmitte
d
E
absorbed
=Eincidentcos(θ)

19. Physics 102: Lecture 15, Slide 19
ACT/Preflight 15.6
Unpolarized light (like the light from the
sun) passes through a polarizing
sunglass (a linear polarizer). The
intensity of the light when it emerges is
1. zero
2. 1/2 what it was before
3. 1/4 what it was before
4. 1/3 what it was before
5. need more information

20. Physics 102: Lecture 15, Slide 20

21. Physics 102: Lecture 15, Slide 21
ACT/Preflight 15.7
Now, horizontally polarized light passes
through the same glasses (which are
vertically polarized). The intensity of the
light when it emerges is
• zero
• 1/2 what it was before
• 1/4 what it was before
• 1/3 what it was before
• need more information

22. Physics 102: Lecture 15, Slide 22
Law of Malus – 2 Polarizers
Cool Link
unpolarized
light
E1
45°
I =I0
TA
TA
90°
TA
E0
I3
B1
unpolarized
light
E1
45°
I =I0
TA
TA
90°
TA
E0
I3
B1
1) Intensity of unpolarized light incident on
linear polarizer is reduced by ½ . S1 = ½ S0
S =
S0
S1
S2
2) Light transmitted through first polarizer is
vertically polarized. Angle between it and
second polarizer is θ=90º. S2 = S1 cos2
(90º) =
0

23. Physics 102: Lecture 15, Slide 23
How do polaroid sunglasses work?
incident light unpolarized
reflected light partially polarized
he sunglasses reduce the glare from reflected light

24. Physics 102: Lecture 15, Slide 24

25. Physics 102: Lecture 15, Slide 25
unpolarized
light
E1
45°
I =I0
TA
TA
90°
TA
E0
I3
B1
unpolarized
light
E1
45°
I =I0
TA
TA
90°
TA
E0
I3
B1
Law of Malus – 3 Polarizers
2) Light transmitted through first polarizer is vertically
polarized. Angle between it and second polarizer is θ=45º.
I2 = I1 cos2
(45º) = ½ I0 cos2
(45º)
3) Light transmitted through second polarizer is polarized
45º from vertical. Angle between it and third polarizer is
θ=45º. I3 = I2 cos2
(45º)
I2= I1cos2
(45)
= ½ I0 cos4
(45º) = I0/8
I1= ½ I0

26. Physics 102: Lecture 15, Slide 26
90°
TA
TA
S1
S2
S0
60°°
TATA
S1
S2
S0
60°°
ACT: Law of Malus
A B
1) S2
A
> S2
B
2) S2
A
= S2
B
3) S2
A
< S2
B
S1= S0cos2
(60)
S2= S1cos2
(30)= S0 cos2
(60) cos2
(30)
S1= S0cos2
(60)
S2= S1cos2
(60)= S0
cos4
(60)
Cool Link
E0
E0

27. Physics 102: Lecture 15, Slide 27