Lect 24

Physics 212

Lecture L ctur 24: Polarization P l riz ti n

Physics 212 Lecture 24, Slide 1

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Stephane Grappelli Pearl Django Mark O’Connor’s Hot Swing Trio Bob Wills Hot Club of Cowtown

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Your Comments

“This was quite confusing. confusing There are a lot of things that happen when light is polarized. It would be nice to break those down into smaller steps one at a time.”” “The linearly polarized stuff made sense but the circular things were terrifying. Could we go over those and the checkpoints more in depth? Also, Also and I can't can t believe I'm asking for this, could we do a calculation with the circular birefringent or a combo of both polarizers?”” “I I'm m not sure what the phase angle angle'ss for. Aren't the x and y components always perpendicular to each other?”” “I don't g get quarter q wave slabs. Are they y like, magic or something?”” “how can a firebinging material make two different velocities. Isn't it the same uniform material, so how are the components velocities different. Also what is a quarter wave plate”” 05

Lots of questions about circular polarization – we will dissect it!

When light passes through any medium it is absorbed b b d and d rere-emitted: itt d f for one component of light, it happens more often Physics 212 Lecture 24, Slide 3

Quantum Communication “I understand d t d th there are instances i t where h polarization l i ti is i used d in i the th reall world, ld b butt what are some special cases where circular polarization is used? I HAVE to get at least one question on the powerpoint in my lifetime. COME ON!”

• Communicate and know for sure if message has been looked at by someone else (quantum mechanics): P. Kwiat, et al.


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

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


Linear Polarization “I was a bit confused by the introduction of the "e-hat" vector ( iin it (as its purpose/usefulness)” / f l )”


Polarizers

The molecular structure of a polarizer causes the component of

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


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

the E field perpendicular to the Transmission Axis to be absorbed. Eo

Suppose we have a beam traveling in the + zz-direction. At t = 0 and z = 0, 0 the electric field is aligned along the y positive x x--axis and has a magnitude equal to Eo What is the component of Eo along a direction in the x x--y plane l that th t makes k an angle l of f  with ith respectt to t the th xxaxis? A) Eosin

B) Eocos

C) 0

D) Eo/sin

z Eo 

E) Eo/cos

What is the only angle for which Eo has no (zero) projection?

y


Linear Polarizers

“This stuff is so wicked awesome! Go over the Law of Malice though.”


An unpolarized EM wave is incident on two orthogonal polarizers.

Checkpoint 1a

Two Polarizers

What percentage of the intensity gets through both polarizers? A. 50% B. 25% C. 0%

“the horizontal and vertical components would go through” “after the 1st one only half of it remains and then after the 2nd is half of it” “Because the polarizers are at 90 degrees to each other intensity is zero because cos(pie/2) =0.” Physics 212 Lecture 24, Slide

10


Checkpoint 1a

Two Polarizers

What percentage of the intensity gets through both polarizers? A. 50% B. 25% C. 0%

The second polarizer is orthogonal to the first no light will come through.

cos(90o) = 0 Physics 212 Lecture 24, Slide 11


Checkpoint 1b

Two Polarizers

Is it possible to increase this percentage by inserting another Polarizer between the original two? A. Yes B. No

“any diagonal will do” “It wouldn't help any because it is all in the y direction after the first polarizer” Physics 212 Lecture 24, Slide

12


Checkpoint 1b

Two Polarizers

Is it possible to increase this percentage by inserting another Polarizer between the original two? A. Yes B. No

Any non-horizontal polarizer after the first polarizer will produce polarized light AT THAT ANGLE Part of that light will make it through the horizontal polarizer


There is no reason that  has to be the same for Ex and Ey:

Making g x different from y causes circular or elliptical p polarization: p At t=0

Example:: Example

x   y  90  

  45   4

 2



E0 Ex  cos(kz  t ) 2 E0 Ey  sin( i (kz k  t ) 2

RCP Physics 212 Lecture 24, Slide 14

Q: How do we change the relative phase between Ex and Ey? A Birefringence A: Bi f i

By picking p ck ng the right r ght thickness we can change the relative phase by exactly 90o.

Right hand rule

This changes linear to circular polarization and is called a

quarter wave plate


“Does Birefringent Material absorb intensity?”

NOTE: No Intensity is lost passing through the h QWP ! BEFORE QWP:

 iˆ  ˆj  E  Eo sin( kz  t )   2  

I  c o E 2  c o E x2  E y2

AFTER QWP: E E  o iˆ cos( kz  t )  ˆj sin( kz  t ) 2



 Eo2 Eo2   sin 2 (kz  t )  c o Eo2 1  c o    2 2  2 



I  c o E 2  c o E x2  E y2 Eo2  c o cos 2 (kz  t )  sin 2 (kz  t ) 2 Eo2 Eo2 1  c o  c o 2 2

THE SAME !!


Right or Left ???

Right circularly polarized Do right hand rule Fingers along slow direction Cross into fast direction If thumb th b points i t iin di direction ti of f propagation: ti RCP Physics 212 Lecture 24, Slide 17

Circular Light on Linear Polarizer Q: What happens when circularly polarized light is put through a polarizer along the x (or y) axis ?

A) I = 0 B) I = ½ I0 C) I = I0

I   0c E 2   0 c E x2  X E y2 E02   0c cos 2 (kz  t ) 2

12

1 1    0 cE02 2 2 Half of before Physics 212 Lecture 24, Slide 18

Identical linearly polarized light at 45o from the y-axis and propagating along the z axis is incident on two different objects. In Case A the light intercepts a linear polarizer with polarization along the y-axis In Case B, the light intercepts a quarter wave plate with vast axis along g the yy-axis.

A

B

Checkpoint 2a

Compare the intensities of the light waves after transmission. A. IA < IB B. IA = IB C. IA > IB

“In case 1, the polarizer absorbs some of the intensity while for case 2 it just changes its phase.” “The The polarizer and quarter wave plate have the same effect. effect ” “it looses more intensity when it passes through a birefringent material.” Physics 212 Lecture 24, Slide 19


A

B

Checkpoint 2a

Compare the intensities of the light waves after transmission. A. IA < IB B. IA = IB C. IA > IB

Case A: Ex is absorbed

Case B: (Ex,Ey) phase changed

I A  I 0 cos2 (45o ) I A  12 I 0

I B  I0 Physics 212 Lecture 24, Slide 20

Intensity:

I   0 c  Ex2  E y2 

QW Plate

Both Ex and Ey are still there, so intensity is the same


I   0 c  Ex2  E y2 

Polarizer

Ex is missing missing, so intensity is lower Physics 212 Lecture 24, Slide 22


A A

Checkpoint heckpo nt 2b b B B Checkpoint 2b

What is the polarization of the light wave in Case B after it passes through the quarter wave plate?. A. linearly polarized B. left circularly polarized C. right g circularly yp polarized D. undefined “the fast side will make the light polarize linearly” “the vertical component will be ahead of the horizontal component. Since the horizontal component is in the positive x direction it is left CP” “Right hand rule from slow to fast means that it travels in the positive z which in this case means right circularly polarized.” “It is at 45 degrees from the y-axis thus it is undefined.”



Checkpoint heckpo nt 2b b B B Checkpoint 2b

A A

What is the polarization of the light wave in Case B after it passes through the quarter wave plate?. A. linearly polarized B. left circularly polarized C. right g circularly yp polarized D. undefined

RCP 1/4 

Z



A A

Checkpoint heckpo nt 2c c B B Checkpoint 2c

If the thickness of the quarter-wave plate in Case B is doubled, what is the polarization of the wave after passing through the wave plate? A. linearly polarized B. circularly polarized C. undefined If it it'ss a half wave plate plate, the waves will be 180 degrees out of phase rather then 90 “If degrees and be linearly polarized” “I think it would still be the same it would just increase the difference between the component p phases.” p “The x component would lag the y component by half a period, so for half the period the wave would be linearly polarized in the y direction and then for the other half would be Physics 212 Lecture 24, Slide 25 linearly polarized in the x direction.”


Checkpoint heckpo nt 2c c B B Checkpoint 2c

A A

If the thickness of the quarter-wave plate in Case B is doubled, what is the polarization of the wave after passing through the wave plate? A. linearly polarized B. circularly polarized C. undefined

½ Z

Z Physics 212 Lecture 24, Slide 26

Executive Summary: Polarizers & QW Plates: Polarized Light

Birefringence

Circularly or Un-polarized Light

RCP Ex 

E0 cos(kx) 2

Ey 

E0 sin( kx) 2


Demos:

What else can we put in there to change the polarization? Physics 212 Lecture 24, Slide 28

Calculation Light is incident on two linear polarizers and a quarter wave plate (QWP) as shown. What is the intensity I3 in terms of I1? fast

45o

y x

60o

I1 I2

I3

z

• Conceptual Analysis • Linear Polarizers: absorbs E field component perpendicular to TA • Quarter Wave Plates: Shifts phase of E field components in fast-slow directions d • Strategic Analysis • Determine state of polarization and intensity reduction after each object reduction • Multiply individual intensity reductions to get final reduction. Physics 212 Lecture 24, Slide 29

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

fast

45o

y

RCP

E1

x

Ex E y

I1

60o

/4

I2

I3

z

• What is the polarization of the light after the QWP?

(A) LCP

(B) RCP

(C)

y

x

Light incident on QWP is linearly polarized at 45o to fast axis LCP or RCP? Easiest way: Right Hand Rule:

(D)

y

x

(E) unpolarized

Light will be circularly polarized after QWP

Curll fi C fingers of f RH b back k tto front f t (slow crossed into fast). Thumb RCP points in dir of propagation if right hand polarized. Physics 212 Lecture 24, Slide 30


fast

45o

y

RCP

E1

x

Ex E y

I1

60o

/4

I2

z

I3

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

(A) I2 = I1 BEFORE: Ex 

E1 sin(kz  t ) 2

E E y  1 sin(kz  t ) 2

(B) I2 = ½ I1

(C) I2 = ¼ I1

No absorption: Just a phase change !

I   0 c  E

2 x

 E  2 y

Same before & after !

AFTER: Ex 

E1 cos(kz  t ) 2

Ey 

E1 sin(kz  t ) 2



fast

45o

y

RCP

E1

x

Ex E y

I1

60o

E3

/4

I2 = I1

I3

z

• What is the polarization of the light after the 60o polarizer?

(A) LCP

(B) RCP

(C)

y

60o

x

(D)

y

60o

x

(E) unpolarized

Absorption: only passes components of E parallel to TA ( = 60o) E Ey E3  E x sin   E y cos  E3  1  sin(kz  t   )  3 E3 2 E 60o E3  1  cos(kz  t ) sin   sin(kz  t ) cos   2 Ex



fast

45o

y

RCP

E1

x

Ex E y

I1

60o

E3

/4

I2 = I1

II33 = ½ I1

z

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

(A) I3 = I1

Ey

E3

3

E E3  1 2

(B) I3 = ½ I1

I  E2

(C) I3 = ¼ I1 I3 

1 I1 2

NOTE: This does not depend on  !!

60o

Ex


Follow Up 1 Replace the 60o polarizer with another QWP as shown. fast

45o

y

RCP

E

x

slow

Ex E y

I1

fast Ey

/4

I2 = I1

I3

E3

Ex z

• What is the polarization of the light after the last QWP?

(A) LCP

(B) RCP

Easiest way: E Efast is /4 ahead of Eslow

(C)

y

x

(D)

y

x

(E) unpolarized

Brings Ex and Ey back in phase !!


Follow Up 2 Replace the 60o polarizer with another QWP as shown. fast

45o

y

RCP

E

x

slow

Ex E y

I1

/4

fast Ey

I2 = I1

I3 = I1

E3

Ex z

• What is the intensity I3 of the light after the last QWP?

(A) I1 BEFORE: Ex 

(B) ½ I1

No absorption: Just a phase change !

E1 cos(kz  t ) 2

E E y  1 sin(kz  t ) 2

(C) ¼ I1

Intensity = < E2 > I before

E12  2

I after

E12  2

AFTER: Ex 

E1 cos(kz  t ) 2

Ey 

E1 cos(kz  t ) 2


Follow Up 3 Consider light incident on two linear polarizers as shown. Suppose I2 = 1/8 I0 y x

I0

E1 60o

E2

I1 I1 = ½ I0

I2 = 1/8 I0

z

• What is the possible polarization of the INPUT light?

(A) LCP (B)

y

45o

x

(C) unpolarized

• After first polarizer: LP along y-axis with intensity I1 • After second polarizer: LP at 60o wrt y-axis • Intensity: y I2 = I1cos2(60o) = ¼ I1 • I2 = 1/8 I0 ﬂ I1 = ½ I0

(D) all of above

Question is: What kind of light loses ½ of its intensity after passing through vertical polarizer?

(E) none of above

Answer: Everything except LP at  other than 45o Physics 212 Lecture 24, Slide 36

Lect 24

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