Physics 102: Lecture 15, Slide 1

Electromagnetic Wavesand Polarization

• Today’s lecture will cover Textbook Sections 22.7-8

Physics 102: Lecture 15

Physics 102: Lecture 15, Slide 2

xz

yE

B

loop in xy plane

loop in xz plane

loop in yz

plane

1 2 3

Which orientation will have the largest induced emf? Hint: Loops use B not E.

Preflight 15.1, 15.2

xz

yE

B

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!)

loop in xy plane

loop in xz plane

loop in yz

plane

1 2 3

Which orientation will have the largest induced emf?

Preflight 15.1, 15.2

Physics 102: Lecture 15, Slide 4

Propagation of EM Waves

• Changing B field creates E field• Changing E field creates B field

E = c B

xz

y

If you decrease E, you also decrease B!

This is important !

Physics 102: Lecture 15, Slide 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

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

Physics 102: Lecture 15, Slide 7

Energy in EM waveLight waves carry energy but how?

Electric Fields • Recall Capacitor Energy:

U = ½ C V2

• Energy Density (U/Volume): uE = ½ 0E2

• Average Energy Density: uE = ½ (½ 0E0

2)

= ½ 0E2rms

Magnetic Fields • Recall Inductor Energy:

U = ½ L I2

• Energy Density (U/Volume):uB = ½ B2/0

• Average Energy Density: uB = ½ (½ B0

2/0)

= ½ B2rms/0

Physics 102: Lecture 15, Slide 8

Energy Density

Calculate the average electric and magnetic energy density of sunlight hitting the earth with Erms = 720 N/C

202

1rmsE Eu

0

2

21

rms

BBu

00

1

c

Note: This is true only for EM waves.

Energy Density

Calculate the average electric and magnetic energy density of sunlight hitting the earth with Erms = 720 N/C

202

1rmsE Eu

2212

2

1 C N8.85 10 720 2 Nm C

0

2

21

rms

BBu 2

0

2

21

cErms

00

1

cUse

ErmsB uEu 202

1 36106.42

mJuuuu EBEtotal

Note: This is true only for EM waves.

63

J2.3 10m

Physics 102: Lecture 15, Slide 10

Energy in EM waveElectric and magnetic fields carry equal amounts of energy.

Electric Fields • Average Energy Density:

uE = ½ (½ 0E02)

= ½ 0E2rms

Magnetic Fields

• Average Energy Density: uB = ½ (½ B0

2/0)

= ½ B2rms/0

= ½ E2rms/(c20)

= ½ 0E2rms

In EM waves, E field energy = B field energy! ( uE = uB )

Physics 102: Lecture 15, Slide 11

Intensity (I or S) = Power/Area• Energy (U) in box:

U = u x Volume = u (AL)

• Power (P):

L

A

L=ct

P = U/t = U (c/L) = u A c

• Intensity (I or S): S = P/A = uc = c0E2

rms23

U = Energy u = Energy Density (Energy/Volume)A = Cross section Area of lightL = Length of box 23

Physics 102: Lecture 15, Slide 12

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)

xz

y

Physics 102: Lecture 15, Slide 13

Linear Polarizers• Linear Polarizers absorb all electric fields

perpendicular to their transmission axis.

Molecular View (link)

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!

Physics 102: Lecture 15, Slide 15

Linearly Polarized Light on Linear Polarizer (Law of Malus)

Etranmitted =Stransmitted =

TA

is the angle between the incoming light’s polarization, and the transmission axis

Transmission axisIncident E

ETransmitted

Eabsorbed

=Eincidentcos()

Physics 102: Lecture 15, Slide 16

Linearly Polarized Light on Linear Polarizer (Law of Malus)

Etranmitted = Eincident cos()Stransmitted = Sincident cos2()

TA

is the angle between the incoming light’s polarization, and the transmission axis

Transmission axisIncident E

ETransmitted

Eabsorbed

=Eincidentcos()

Physics 102: Lecture 15, Slide 17

Preflight 15.6, 15.7Unpolarized 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. ½ what it was before 3. ¼ what it was before 4. ⅓ what it was before 5. need more information

Now, horizontally polarized light passes through the same glasses (which are vertically polarized). The intensity of the light when it emerges is

1. zero2. ½ what it was before3. ¼ what it was before4. ⅓ what it was before5. Need more information

Physics 102: Lecture 15, Slide 18

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

Physics 102: Lecture 15, Slide 19

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

Physics 102: Lecture 15, Slide 20

Law of Malus – 2 Polarizers

S1 =

S2 =

unpolarized light

E1

I = I0

TATA

TA

E0

I3

B1

unpolarized light

E1

I = I0

TATA

TA

E0

I3

B1S = S0

S1

S2

Physics 102: Lecture 15, Slide 21

Law of Malus – 2 Polarizers

Cool Link

unpolarized light

E1

I = I0

TATA

TA

E0

I3

B1

unpolarized light

E1

I = I0

TATA

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 =90º. S2 = S1 cos2(90º) = 0

Physics 102: Lecture 15, Slide 22

unpolarized light

E1

I = I0

TATA

TA

E0

I3

B1

unpolarized light

E1

I = I0

TATA

TA

E0

I3

B1

Law of Malus – 3 Polarizers

I2 =

I3 =

I1 =

unpolarized light

E1

I = I0

TATA

TA

E0

I3

B1

unpolarized light

E1

I = I0

TATA

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 =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 =45º. I3 = I2 cos2 (45º)

I2= I1cos2(45)

= ½ I0 cos4 (45º)

I1= ½ I0

Physics 102: Lecture 15, Slide 24

TA

TA

S1

S2

S0

TATA

S1

S2

S0

ACT: Law of Malus

A B

1) S2A > S2

B 2) S2A = S2

B

3) S2A < S2

B

E0E0

Physics 102: Lecture 15, Slide 25

TA

TA

S1

S2

S0

TATA

S1

S2

S0

ACT: Law of Malus

A B

1) S2A > S2

B 2) S2A = S2

B

3) S2A < S2

B

S1= S0cos2(60)

S2= S1cos2(30)= S0 cos2(60) cos2(30)

S1= S0cos2(60)

S2= S1cos2(60)= S0 cos4(60)

Cool Link

E0E0

Physics 102: Lecture 15, Slide 26

See next time!• Read Sections 23.1-2, 7-8