physics 212 lecture 23, slide 1 physics 212 lecture 23: polarization

Physics 212 Lecture 23, Slide Physics 212 Lecture 23, Slide 11

Physics 212Physics 212Lecture 23: PolarizationLecture 23: Polarization


What makes EM waves

0 0

1

dBdA Edl

dtd

EdA Bdldt

Which of the following do not create EM waves?

A) Constant current

B) Accelerating electron.

C) Electron orbiting a nucleus.

D) Charged pith ball rotating on a string around a circle.

E) Electron moving with constant velocity.


So far we have considered plane waves that look like this:So far we have considered plane waves that look like this:

From now on just draw E and remember that B is still there:From now on just draw E and remember that B is still there:


Linear PolarizationLinear Polarization


The intensity of a wave The intensity of a wave does not depend on it’sdoes not depend on it’spolarization:polarization:

2 20 x yc E E

2 2 2 20 0 sin ( ) (sin cos )cE kz t

20I c E

1 2 1

SoSo Just like beforeJust like before20 0

1

2I cE

2 20 x yc E E

I could have also rotated the coordinate system such that x points along hat e!


The molecular structure of a polarizer causes the component of The molecular structure of a polarizer causes the component of

the the EE field perpendicular to the Transmission Axis to be absorbed. field perpendicular to the Transmission Axis to be absorbed.

PolarizersPolarizers

AnimationAnimation


Two Polarizers

since the second polarizer is orthogonal to the first, no light will come through. cos(90) = 0


Two Polarizers

you can use a polarizer at an angle so that it takes a portion of the polarized light and basically lets the light through AT THAT ANGLE. Then, when the light reaches the second polarizer, it is no longer orthogonal to it.


What if we put many polarizers between?What if we put many polarizers between?

Applet polarizer


There is no reason that There is no reason that has to be the same for has to be the same for EExx and and EEyy::

Making Making xx different from different from yy causes circular or elliptical polarization: causes circular or elliptical polarization:

902

x y

45 4

0 cos( )2

x

EE kz t

0 sin( )2

y

EE kz t

ExampleExample:: At t=0At t=0


Circular polarization• In Physics and Astronomy, we define left/right circular

polarization if the light rotates cw/ccw as seen from the receiver (looking at the incoming light). In engineering we define it as seen from the emitter!

As seen by emitter


Q: How do we change the relative phase between Ex and Ey?

A: Birefringence:Anisotropic material. Light incident from differentdirections propagates at different speeds.

By picking the right thickness we can changethe relative phase byexactly 90o.

This changes linear to circular polarization and is called a quarter wave plate

Right hand Right hand rulerule


This modification does notThis modification does notchange the intensity either:change the intensity either:

2 20 x yc E E

22 20

0 sin ( ) cos ( )2

E

c kz t kz t

20I c E

1

SoSo Just like beforeJust like before20 0

1

2I cE

Right Handed Circular PolarizationRight Handed Circular Polarization(at t = 0)(at t = 0)


Q:Q: What happens when circularly What happens when circularlypolarized light is put through apolarized light is put through apolarizer along the polarizer along the yy (or (or xx) axis ?) axis ?

2 20 x yc E E

220

0 cos ( )2

E

c kz t

20I c E

1 2

HalfHalf of before of before

20 0

1 1

2 2 cE

A) I = 0B) I = ½ I0C) I = I0

XX

Circular Light on Linear Polarizer


In case A you lose your horizontal component of the In case A you lose your horizontal component of the wave, however in B you retain both, just at a shift of wave, however in B you retain both, just at a shift of pi/2. pi/2.


QW PlateQW Plate

2 20 x yI c E E

Both Ex and EyBoth Ex and Eyare still there, soare still there, so

intensity is the sameintensity is the same


PolarizerPolarizer

2 20 x yI c E E

Ex is missing, soEx is missing, sointensity is lowerintensity is lower


What is the polarization of the light in Case 2 after it passed through the quarter wave plate?A)Linearly polarizedB)Right circularly polarized (as seen by receiver)C)Left circularly polarized (as seen by receiver)


Right circularly polarized (CCW) as seen by emitterLeft circularly polarized (CW) as seen by receiver.


Executive Summary:Executive Summary:

Circularly or Un-polarized LightPolarized Light

Polarizers & QW Plates:Polarizers & QW Plates:

BirefringenceBirefringence

0 cos( )2

x

EE kx

0 sin( )2

y

EE kx

RCP


CalculationCalculation

• Conceptual Analysis • Linear Polarizers: absorbs E field component perpendicular to TA• Quarter Wave Plates: Shifts phase of E field components in fast-slow directions

• Strategic Analysis• Determine state of polarization and intensity reduction after each object• Multiply individual intensity reductions to get final reduction.

x

y

z

45o fast

slow

II11

60o

II22

II33

Light is incident on two linear polarizers and a quarter wave plate (QWP) as shown.

What is the intensity I3 in terms of I1?


CalculationCalculationLight is incident on two linear polarizers and a quarter wave plate (QWP) as shown.

x

y

z

45o fast

slow

II11

60o

II22

II33

• What is the polarization of the light after the QWP as seen by emitter?

(A) LCP (B) RCP (C) (D) (E) unpolarizedx

y

x

y

Light will be circularly polarized after QWP

Light incident on QWP is linearly polarized at 45o to fast axis

E1

LCP or RCP? Easiest way: Right Hand Rule:

/4

Ex Ey

Curl fingers of RH back to frontThumb points in dir of propagationif right hand polarized.

RCP

RCP



x

y

z

45o fast

slow

II11

60o

II22

II33

No absorption: Just a phase change !

• What is the intensity I2 of the light after the QWP?

(A) I2 = I1 (B) I2 = ½ I1 (C) I2 = ¼ I1

/4

Ex Ey

BEFORE:

1 sin( )2

x

EE kz t

1 sin( )2

y

EE kz t

AFTER:

1 cos( )2

x

EE kz t

1 sin( )2

y

EE kz t

RCP

2 20 x yI c E E

Same before & after !

E1



x

y

z

45o fast

slow

II11

60o

II33

Absorption: only passes components of E parallel to TA ( = 60o)

/4

E1

Ex Ey

• What is the polarization of the light after the 60o polarizer? (A) LCP (B) RCP (C) (D) (E) unpolarizedx

y

x

y60o 60o

60o

Ex

3Ey

13 cos( )sin sin( )cos

2

EE kz t kz t

13 sin( )

2

EE kz t

E3

3 sin cosx yE E E E3

RCP

II2 2 = I= I11



x

y

z

45o fast

slow

II11

60o

NOTE: This does not depend on !!

/4

Ex Ey

60o

3

Ex

Ey

13

2

EE

E3

• What is the intensity I3 of the light after the 60o polarizer?

E3

II33 = ½ I = ½ I11

3 1

1

2I I

RCP

II2 2 = I= I11

E1

2I E

(A) I3 = I1 (B) I3 = ½ I1 (C) I3 = ¼ I1

physics 212 lecture 23, slide 1 physics 212 lecture 23: polarization

Documents

receiver slide

emitter slide

lower slide

linear polarization

linear polarizer slide

polarizers animation

applet polarizer slide

birefringence rcp slide