1. Electromagnetic Waves and Polarization
Physics 102: Lecture 15
Electromagnetic Waves and Polarization
Hour Exam 2 is Monday!
Material covered
lectures through 11
discussions through 6
homework through 6
Review Sunday, 3 PM, 141
Fall 2007 exam, 1-17 1

2. Today: Electromagnetic Waves
Energy
Intensity
Polarization

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.” x z y E B
loop in yz plane
loop in xy plane
loop in xz plane
Note wavelength must be much large than loop size
Radio waves: 3 meters
Demo 158
Only the loop in the xy plane will have a magnetic flux through it as the wave passes. The flux will oscillate with time and induce an emf. (Faraday’s Law!!!) 7

4. Propagation of EM Wavesx z y
Changing B field creates E field
Changing E field creates B field
E = c B
Note E=cB is only true for EM wave, not in general
This is important !
If you decrease E, you also decrease B! 10

5. 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
19%
72% 9%
E=cB

6. 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 e0E2
Average Energy Density:
uE = ½ (½ e0E02)
= ½ e0E2rms
Magnetic Fields
Recall Inductor Energy:
U = ½ L I2
Energy Density (U/Volume):
uB = ½ B2/m0
Average Energy Density:
uB = ½ (½ B02/m0)
= ½ B2rms/m0 13

7. U = ½ CV2
V = Ed
C=eA/d
U=1/2eAdE2 = 1/2eVE2
u = U/V = 1/2eE2

8. Energy Density Example
Calculate the average electric and magnetic energy density of sunlight hitting the earth with Erms = 720 N/C
Example
Again note energy density is same only for EM wave, not in general
Use
Note: This is true only for EM waves. 18

9. 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 e0E2
Average Energy Density:
uE = ½ (½ e0E02)
= ½ e0E2rms
Magnetic Fields
Recall Inductor Energy:
U = ½ L I2
Energy Density (U/Volume):
uB = ½ B2/m0
Average Energy Density:
uB = ½ (½ B02/m0)
= ½ B2rms/m0
In EM waves, E field energy = B field energy! ( uE = uB )
utot = uE + uB = 2uE = e0E02 13

10. Intensity (I or S) = Power/Area
Energy (U) in box:
U = u x Volume
= u (AL)
Power (P):
P = U/t
= U (c/L)
= u A c
Intensity (I or S):
S = P/A [W/m2]
= uc = ce0E2rms A
L=ct L
U = Energy
u = Energy Density (Energy/Volume)
A = Cross section Area of light
L = Length of box 23 23

11. 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 25

12. Linear Polarizers
Linear Polarizers absorb all electric fields perpendicular to their transmission axis.
Molecular View 30

13. Unpolarized Light on Linear Polarizer
demo 324
Most light comes from electrons accelerating in random directions and is unpolarized.
Averaging over all directions: Stransmitted= ½ Sincident
Always true for unpolarized light! 33

14. Linearly Polarized Light on Linear Polarizer (Law of Malus)TA
Etranmitted = Eincident cos(q)
Stransmitted = Sincident cos2(q) q
q is the angle between the incoming light’s polarization, and the transmission axis
Transmission axis
Incident E
Eabsorbed q
ETransmitted
=Eincidentcos(q) 36

15. 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
zero
     1/2 what it was before
     1/4 what it was before
     1/3 what it was before
     need more information 37

16. 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 38

17. Law of Malus – 2 Polarizers
Example
S = S0 S1 S2
1) Intensity of unpolarized light incident on linear polarizer is reduced by ½ . S1 = ½ S0
Demo
2) Light transmitted through first polarizer is vertically polarized. Angle between it and second polarizer is q=90º. S2 = S1 cos2(90º) = 0
Cool Link 39

18. Law of Malus – 3 Polarizers
Example
I1= ½ I0
I2= I1cos2(45)
2) Light transmitted through first polarizer is vertically polarized. Angle between it and second polarizer is q=45º. I2 = I1 cos2 (45º) = ½ I0 cos2 (45º)
Demo
3) Light transmitted through second polarizer is polarized 45º from vertical. Angle between it and third polarizer is q=45º. I3 = I2 cos2 (45º)
= ½ I0 cos4 (45º) = I0/8 43

19. ACT: Law of Malus A B Cool Link S2 S2 E0 E0 S0 S0 S1 S190 TA S1 S2 S0 60 TA S1 S2 S0 60 E0 E0 A B
Cool Link
S1= S0cos2(60)
S1= S0cos2(60)
S2= S1cos2(30)
= S0 cos2(60) cos2(30)
S2= S1cos2(60)
= S0 cos4(60)
1) S2A > S2B 2) S2A = S2B 3) S2A < S2B 48

20. See You Monday!