chapter 3 polarization of light waves
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Chapter 3 Polarization of Light WavesLecture 1 Polarization3.1 The concept of polarizationIntroduction:1)Since F = qE, polarization determines force direction. The generation, propagation and control of lasers thus crucially depend on their state of polarization.2)The electric field E is used to define the polarization state of the light.3)In an anisotropic material the index of refraction depends on the polarization state of the light, which is used to manipulate light waves.
1
3.2 Polarization of monochromatic plane waves
For a monochromatic plane wave propagating in the z direction, the x and y components of the electric field oscillate independently.
)cos(),()cos(),(
)]](exp[)ˆˆRe[(
)]](exp[Re[),(
yyy
xxx
iy
ix
kztAtzEkztAtzE
kztieAeA
kztitzyx
ji
AE Ex
Ey
2
Ex
Ey
t
Ex
t
EEy
3
The trajectory of the end point of the electric field vector, at a fixed point as time goes, is
. ,
sincos2
) geliminatin()cos()cos(
2
22
xy
y
y
x
x
y
y
x
x
yyy
xxx
AE
AE
AE
AE
kztkztAEkztAE
Light is thus generally elliptically polarized.A complete description of the elliptical polarization needs1)Orientation angle 2)Ellipticity (shape) a/b, and3)Handedness (sense of revolution, can be combined to show the sign of ellipticity).
Ax
Ay E
y
x
x'y'
ab
They can be obtained by rotating the ellipse to its normal coordinates:
cos2
2tan
2
cos4 ,
1
22
2222222222
2'
2'
yx
yx
yxyxyx
yx
AAAA
AAAAAAba
bE
aE
4
Handedness (sense of revolution) of elliptical polarization:Looking at the approaching light, if the E vector revolves counterclockwise, the polarization is right-handed, with sin <0. If the E vector revolves clockwise, the polarization is left-handed, with sin >0. Many books use the opposite definition.
= - -3/4 -/2 -/4 0/4 /2 3/4
Linear polarization:
x
y
x
y
xy
AA
EE
,0
Circular polarization:
yx
xy
AA
2
= -/2 /2
5
3.4 Jones vector representationA plane wave can be uniquely described by a Jones vector in terms of its complex amplitudes on the x and y axes:
If we are only interested in the polarization state of the wave, we use the normalizedJones vector, with J+J=1:
y
x
iy
ix
eAeA
J
. ,angle) (auxiliary arctan ,sin
cos),( xy
x
yi A
Ae
J
Examples:1)Linearly polarized light:
2)Right- and left-handed circularly polarized light:
.10
ˆ ,01
ˆ especially ,sin
cos and
sincos
yx
. 1
21ˆ and
12
1ˆ
ii
LR
Some relations:
LRyLRx
yxLyxR
LRyx
ˆˆ2
ˆ ,ˆˆ2
1ˆ
ˆˆ2
1ˆ ,ˆˆ2
1ˆ
0ˆˆ ,0ˆˆ
i
ii
Lecture 2 Jones vector
6
Jones vectors are important when applied with Jones calculus, which enables us to track the polarization state and the intensity of a plane wave when traversing an arbitrary sequence of optical elements.
= - -3/4 -/2 -/4 0/4 /2 3/4
Examples of Jones vectors:
2
15
1)1(2
15
121
51
)1(21
51
21
51
)1(21
51
21
51
)1(21
51
21
51
iiiiii
J
Light intensity:
22** , yxiy
ixi
yi
xiy
ix AA
eAeA
eAeAIeAeA
y
xyx
y
x
JJJ
7
Solution 1: Suppose the ellipse is erect after rotating the x, y axes by an angle , that is
baccba
cbaB
cbaA
ByAx
cbayx
cbaycbax
yxyxcyxbyxacxybyax
ByAx yxyyxx
yxyyxx
2tan02cos2sin2sin
cossincossin
cossinsincos
1''
1)2cos2sin2sin(''
)cossincossin(')cossinsincos('
)cos'sin')(sin'cos'()cos'sin'()sin'cos'(
.1'' and ,cos'sin'sin'cos'
,cossin'
sincos'
22
22
22
222222
2222
22
(*Reading) What is the orientation angle of the ellipse How long are the principle axes?
? 122 cxybyax
8
2)()(
,
2tan
then,'' If
2)()(
2)()(1
)()(tan
tan1tan22tan
tan22tan
sin22sin2
1
2tan
2sin2
sin)(1cossinsincos
1
)2
or axes, principle thebe (Let
22
2222
2/12222
2
22
2
2
2
2222
2
cbabaBA
bac
ByAxcxybyax
cbabar
cbabar
ccbaba
bac
cacar
bac
cabarcba
r
r
m
mm
mm
mm
mm
mm
m
m
mmmmmmm
m
mm
2)()(
,
2tan
then,'' If
22
2222
cbabaBA
bac
ByAxcxybyax
m
9
2/1
22222222
2/1
22222222
2/1
222
22222222
2/1
2
22
2222
2/122
0'0'
22
22
2
22
2
cos4
cos4
2sin2
cos4
sin2/cos211112
)()(,
cos2
11
cos2
2tan
,sincos2 of case For the
yxyxyx
yxyxyxyx
yxyxyx
yxyxyxyx
yx
yx
yx
yxm
y
y
x
x
y
y
x
x
AAAAAA
AAAAAAAA
AAAAAA
AAAAAAcbaba
EE
AAAA
AA
AAba
c
AE
AE
AE
AE
10
Solution 2: Let us use a little linear algebra.
bacbac
cbaB
cbaA
cbabac
baccba
bcca
Rbc
caR
BA
yx
BA
yxByAx
yx
Ryx
yx
yx
bcca
yxcxybyax
2tan0cossin)()sin(cos2
cossincossin
cossinsincos
cossinsincoscossin)()sin(cos2
cossin)()sin(cos2
cossinsincos
cossinsincos
2/2/
cossinsincos
2/2/
00
1''
00
''1''
cossinsincos
''
12/
2/1
22
22
22
2222
2222
1
22
22
Solution 3: The problem is: given the condition for x and y, what is the maximum x2 + y2? This can be solved by the Lagrange multiplier method.
11
bac
cacb
cyaxcxby
xy
cxbyyyf
cyaxxxf
cxybyaxyxyxf
2tantan1tan2
tan2tan2
22tan
0)2(2
0)2(2
)1(),,(
2
2222
cossincossin
cossinsincoscossinsincos
1
.2tan02cos2sin2sin
0cossinsincos 010
.cossinsincos
11
22
22
222
222
22222
cbaB
cbaAcba
r
baccba
cbadd
rdd
ddr
cbarcxybyax
m
Solution 4: In polar coordinates, The curve is now .sin ,cos ryrx
122 cxybyax
12
indices). repeatedover (sum 1''''''
is,That
1'''
'''''''''
'''
as systems coordinate '-- and -- thein expressed is surface quadratica Surppose
333231
232221
131211
333231
232221
131211
rSrSrr TTjijijiji xSxxSx
zyx
SSSSSSSSS
zyxzyx
SSSSSSSSS
zyx
zy'x'zyx
Reading: Diagonalizing the tensor of a quadratic surfaceQuestion: What is the orientation angle of the ellipse
How long are the semiaxes?
Solution 5:Diagonalizing the tensor of a quadratic surface
? ''1 2222 ByAxcxybyax
y
x
x'y'
A-1/2
B-1/2
13
.100
,010
,001
have we,1 since Also
.' ,' ,'
'000'000'
''
, diagonizes tion tranformacoordinate othogonal theSuppose
133
123
113
132
122
112
131
121
111
1
133
123
113
31
33
123
113
132
122
112
21
32
122
112
131
121
111
11
31
121
111
3
2
1
133
132
131
123
122
121
113
112
111
133
132
131
123
122
121
113
112
111
111
RRR
RRR
RRR
RRR
SRRR
RRR
SRRR
RRR
SRRR
SS
S
RRRRRRRRR
RRRRRRRRR
RRR
RR
SSS
S
SRSRRSRSSR
a tensor. is surface quadratic thespecifies that .'
1'''')(')'()(')'()'(
then, ,' systems,
coordinate two therelates that tion tranformacoordinate othogonal theis Suppose
1
11111
1
SRSRSrSrrRSRrrRSRrrRSrRSrr
RRRrrR
TTTTTT
T
14
Therefore we conclude:If the orthogonal transformation R diagonalizes S into S',
then1) The diagonal elements of S' are the eigenvalues of S.2) The column vectors of R-1 (or the row vectors of R) are the eigenvectors of S.3) The new coordinate axes lie along the eigenvectors of S.
'000'000'
'
3
2
11
SS
SSRSR
.1 and 1
simply thenare semiaxes The .1'' in tscoefficien theare seigenvalue The 3)
. angle norientatio show the whichcurve, theof axes principle theare rseigenvecto The 2) . of seigenvalue and rseigenvecto thefind is,That . ng Diagonizi1)
:Strategy2/
2/1
2/2/
1
questionour Back to
22
22
BA
ByAx
bcca
yx
bcca
yxcxybyax
SS
S
15
bac
yyyyx
yxyx
cbabaB
rcbaba
B
cbabaA
rcbaba
A
cbabac
ycxcbaba
yx
bcca
cbababcca
m
m
2121
1121
21
11
2/122
2
22
2
2/122
1
22
1
222,1
22
22
2,1
)(222tan
2)()(1
2)()(
2)()(1
2)()(
)()( vectorsEigen
022
)()(0
2/2/
.2
)()( sEigenvalue0
2/2/
r