Physics 212

Transcription

Physics 212

Physics 212
Lecture 24: Polarization
Physics 212 Lecture 24, Slide 1
Music
Who is the Artist?
A)
B)
C)
D)
E)
Stephane Grappelli
Pearl Django
Mark O’Connor’s Hot Swing Trio
Bob Wills
Hot Club of Cowtown
Hour Exam 3: 2 weeks from yesterday (Wed. Apr. 25) covers
L19-26 inclusive (LC circuits to lenses. Sign up for conflicts,
etc. NOW (no later than Mon. Apr. 23 at 10:00 p.m.)
Your Comments
“The prelecture was straightforward, but the checkpoints were challenging.”
“Please go through the handedness determination; I don’t understand it.”
“Why is it called a quarter-wave plate? Also,
please go over RCP vs LCP.”
Lots of questions about circular
“’Different speeds of light’? You all better
polarization – we will dissect it!
have a really good explanation for this or I
will just have to chalk quarter wave plates
up as ‘magic’.”
“I once came across a tv screen in a mall that I couldn’t see if I had my polarized
sunglasses on, because of the way the the glasses were oriented. Can you repeat a
kind of demonstration like that? that was cool.”
“Just something that came to my
mind, how do we know that the fast
v and the slow v in a bire-thingy
have to be at right angles with one
05
another?”
They don’t have to be in principle – some
materials have three separate axes
(like mica and topaz)
Quantum Communication
“WHY? Why would you ever want to do this?”
• 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
(as in its purpose/usefulness)”
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 + z-direction.
At t = 0 and z = 0, the electric field is aligned along the
y
positive x-axis and has a magnitude equal to Eo
What is the component of Eo along a direction in the x-y
plane that makes an angle of  with respect to the xaxis?
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%
“Orientation changes from vertical to horizontal”
“halved and halved again”
“There is no parallel component that passes through the second polarizer..”
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
Checkpoint 1b
Two Polarizers
Is it possible to increase this percentage by inserting another
Polarizer between the original two?
A. Yes
B. No
“If the middle one creates a non vertical polarization, the percent increases.”
“The percentage is already 0 and polarizers cannot increase intensity.”
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 x different from y causes circular or elliptical polarization:
At t=0
Example:
x   y  90 

  45   4

2

Ex 
Ey 
E0
2
E0
2
cos(kz  t )
sin(kz  t )
RCP
Q: How do we change
the relative phase
between Ex and Ey?
A: Birefringence
By picking the right
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 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
 c o
1  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 points in direction of propagation: RCP
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
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 the y-axis.
A
B
Checkpoint 2a
Compare the intensities of the light waves after transmission.
A. IA < IB
B. IA = IB
C. IA > IB
“intensity will decrease in case a and remain the same, just out of phase in
case B”
“They both reduce the intensity at the same rate because the transmission axis
angles are the same”
“The waves are in synch before the polarizer and out of it after, making the
intensity for Ib less.”
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
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, so
intensity is lower
A
A
Checkpoint 2b
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 circularly polarized D. undefined
“along the y axis with no x components”
“cant curl fingers from Ey to Ex using right hand rule so it must be Left Circularly Polarized”
“It is slow along the x-axis and fast along y-axis.”
“There is no component in the slow polarization, so none occurs.”
Checkpoint 2b
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?.
B. left circularly polarized
C. right circularly polarized D. undefined
RCP
1/4 
Z
A
A
Checkpoint 2c
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?
B. circularly polarized
C. undefined
“If the thickness is doubled then the relative phase is 180 degrees, which means it is
linearly polarized. I think.”
“Thickness does not affect how it is polarized.”
“Ex and Ey will be antiparallel if thickness is doubled, so the resulting light cannot be
defined.”
Checkpoint 2c
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?
B. circularly polarized
C. undefined
½
Z
Z
Executive Summary:
Polarizers & QW Plates:
Polarized Light
Birefringence
Circularly or Un-polarized Light
RCP
Ex 
E0
2
cos(kx)
Ey 
E0
2
sin(kx)
Demos:
What else can we put in there to change the polarization?
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
• Strategic Analysis
• Determine state of polarization and intensity reduction after each object
• Multiply individual intensity reductions to get final reduction.
Calculation
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
Curl fingers of RH back to front
(slow crossed into fast). Thumb
RCP
points in dir of propagation
if right hand polarized.
Calculation
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
Calculation
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
Calculation
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
Efast
(B) RCP
Easiest way:
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: 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

Transcription

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