Solutions to the Wave Equation
(Part B)
Diffraction and Interferometry
OPTI 505 Fall 2002
© 2002 Tom D. Milster
Slide 3B-1
Polarization
Consider the case of an EM plane wave traveling in the +z direction.
E oscillates perpendicular to z.
General solution:
r
r
E (z, t) = E0 e j ( kz−ωt ) =  Ax xˆ + A ye jφyˆ  e j ( kz−ωt )
r
We will now trace the tip of the electric vector E for several special cases.
y
We will look at both:
r
E
-x
OPTI 505 Fall 2002
z
© 2002 Tom D. Milster
fixed z = z0 with t = variable
fixed t = t0 with z = variable
Slide 3B-2
1
Polarization – Linear
Ax = Ay = A0
φ=0
r
r
E (z, t) = E0 e j ( kz−ωt ) = A0 [ ˆx + yˆ ]e j (k z −ωt )
r
Re E (z , t ) = A0 [ ˆx + yˆ ] cos ( kz − ωt )
{
}
This is the physical wave.
r
Consider z = z0 , where Re E( z0 ,t ) = A0 [ ˆx + yˆ ] cos ( kz0 − ωt )
y
A0
r
E
Trace the tip of the electric vector
max extension = ± 2A0
in the (x,y) plane in time:
x
− A0
A0
This is true for all z0.
{
}
− A0
OPTI 505 Fall 2002
© 2002 Tom D. Milster
Slide 3B-3
Polarization – Left Circular
r
E (z, t) = A0  xˆ + e jπ / 2 ˆy  e j (k z −ωt )
Ax = Ay = A0
φ=π/2
Consider z = 0 , where
r
Re E (0, t ) = A0 [ cos( ωt )xˆ + cos(π / 2 − ωt ) yˆ ]
{
}
ωt = π
time
{
= A0 [cos( ωt) xˆ + sin( ωt ) yˆ ]
ωt = 3π/4
y
A0
− A0
Consider t = 0 , where
r
Re E ( z,0) = A 0 [ cos(kz )xˆ + cos(π / 2 + kz) yˆ ]
}
= A0 [ cos(kz )xˆ − sin(kz )yˆ ]
rotates CW with z
ωt = π/2
ωt = π/4
1
r
E
ωt = 0
r
E
A0
x
y
z
0
1
-1
− A0
0
rotates CCW with time
OPTI 505 Fall 2002
© 2002 Tom D. Milster
z
1
0
2 -1
x
Slide 3B-4
2
Polarization – Right Circular
r
E (z, t) = A0  xˆ + e − jπ / 2 yˆ  e j (k z −ωt )
Ax = Ay = A0
φ=−π/ 2
Consider z = 0 , where
Consider t = 0 , where
r
r
Re E (0, t ) = A0 [ cos(ωt ) xˆ + cos(−π / 2 − ωt ) ˆy ] Re E ( z,0) = A0 [ cos(kz) xˆ + cos( −π / 2 + kz) ˆy ]
{
}
{
= A0 [ cos(ωt )xˆ −sin( ωt ) yˆ ]
time
A0
y
}
= A0 [cos( kz) xˆ + sin(kz) yˆ ]
rotates CCW with z
rotates CW with time
1
ωt = π
ωt = 0
− A0
ωt = 3π/4
A0
r
E
− A0
x
y
ωt = π/4
ωt = π/2
OPTI 505 Fall 2002
0
-1
0
z
r
E
1
z
0
1
2 -1
© 2002 Tom D. Milster
x
Slide 3B-5
Polarization – Elliptical
r
E (z , t ) =  Ax xˆ + A ye jφ yˆ  e j ( kz−ωt ) ,where
Ax , Ay
φ0
Ex = Ax cos ( kz − ω t )
(1)
E y = Ay cos ( kz − ω t + φ )
(2)
Combination of (1) and (2) gives:
2
2
 Ex   Ey 
 Ex   Ey 
2
 A  +  A  − 2 A   A  cos φ = sin φ
 x  y
 x  y 
,
which is an equation of an ellipse.
In other words,
The tip of the electric vector specified by Ex and Ey trace
an ellipse in any plane z that is determined by Ax , Ay and
φ.
OPTI 505 Fall 2002
© 2002 Tom D. Milster
Slide 3B-6
3
Polarization – Elliptical
Ax = Ay = 1
Ax = Ay = 1
φ = π/ 4
φ = 5π / 4
1
y
y
0
-1
1
0
1
0
z
0
1
OPTI 505 Fall 2002
2 -1
1
-1
0
x
z
0
1
2 -1
© 2002 Tom D. Milster
x
Slide 3B-7
Polarization – Elliptical
(Arrows show trace in z direction)
LH
RH
This slide from Jim Wyant, 2000.
OPTI 505 Fall 2002
© 2002 Tom D. Milster
Slide 3B-8
4
Polarization – Ellipticity
Look into wave and observe the trace at z0 as a function of time
Ax
y
1
Ay
( 0 ≤ α ≤ π / 2)
Ax
b
tan ε = ±
( −π / 4 ≤ ε ≤ − π / 4 )
a
tan2θ = ( tan2α ) cos φ ( 0 ≤ θ ≤ π )
tan α =
Ay
0.5
b
θ
x
0
ε
a
sin2 ε = (sin2 α ) sin φ
-0.5
-1
-1
-0.5
OPTI 505 Fall 2002
0
0.5
ε is often called the ellipticity
1
© 2002 Tom D. Milster
Slide 3B-9
Polarization – Jones* Calculus
Look into wave and observe the trace at z0 as a function of time
The values of constants Ax , Ay and φ determine the state of
polarization of the plane wave. Various states include linear,
circular and elliptical.
We can write the wave conveniently with these constants in a
column vector:
v
 E x   Ax  j( kz−ωt)
E ( x, t ) =   = 
jφ  e
 E y   Ay e 
.
As the wave propagates, the state of polarization does not change
unless the wave encounters and optical element. The Jones vector
is the state of polarization written as as column vector.
 J x   Ax 
Jones vector = J =   = 
jφ 
 J y   Ay e 
*R. Clark Jones, JOSA A, 31, 488-493 (1941), provided on web site. Actually, there are a series of eight papers.
OPTI 505 Fall 2002
© 2002 Tom D. Milster
Slide 3B-10
5
Polarization – Jones Vector Examples
Common constants may be divided out of each Jones vector component.
 3e jπ / 2 
 1
J =  j π / 2  = 3e j π / 2  
9
e
 3


Example:
Various states of polarization:
 1
 0
 
 1
 1
 
linear x
linear +45o
 1 
− j
 
RHC
OPTI 505 Fall 2002
linear y
linear -45o
LHC
 0
 1
 
 1
 -1
 
 1
 j
 
© 2002 Tom D. Milster
Slide 3B-11
Polarization – Jones Matrices for Optical Elements
Optical elements can affect the state of polarization as a plane wave
passes through them.
optical element
state J1
state J2
If we assume that the element acts linearly, the change of state
through an element can be represented by a 2-by-2 matrix M.
m
J 2 = MJ 1 =  11
 m21
OPTI 505 Fall 2002
m12 
m12   J 1x   m11 J 1x + m12 J 1 y 
m
J = 11
  =

m22  1  m21 m22   J1 y   m21 J1 x + m22 J1 y 
© 2002 Tom D. Milster
Slide 3B-12
6
Polarization – Jones Matrices for Optical Elements
Consider optical elements with polarization-dependent refractive index.
E1y
Ny
E1x
Nx
E2y
N x = nx + jκ x
N y = ny + jκ y
E2x
z

e
M =


d
2π
2π
j nx d − κ x d
λ
λ
e
0


2π
2π

j n yd − κ y d

e λ e λ

0
Retardation plate: n x ≠ n y , κ x≈ κ y ≈ 0
Polarizer
OPTI 505 Fall 2002
plate: n x ≈ n y , κ x≠ κ y
© 2002 Tom D. Milster
Slide 3B-13
Polarization – Jones Matrix Examples
Uniaxial Crystals
E1y
E1x
ny
nx
E2y
nx = no
ny = ne (OA)
nz = no
E2x
z
d
Uniaxial crystals have two primary refractive indices, depending on how the
light beam is polarized. no (for ordinary) and ne (for extraordinary). A
common configuration is shown above. ne defines the optical axis OA. A
positive crystal, like quartz, has ne > no . A negative crystal, like calcite, has
no > ne .
OPTI 505 Fall 2002
© 2002 Tom D. Milster
Slide 3B-14
7
Polarization – Jones Matrix Examples
Quarter-Wave Plate
n0 = nx = 1.544
ne = n y = 1.553
A quartz quarter-wave plate is a uniaxial crystal with
κ e = κo ≈ 0
at λ = 500 nm
The thickness d is fabricated so that the phase difference between x and
y polarization components through the plate is π/2.
 j 2λπn od
0 
1
e
J2 = c
jπ/ 2  J 1 = M QWP J 1 = 
0
e


 0

where c is a constant. So,
 j 2 π no d
0  e λ
=
e jπ/ 2  
 0
1
c
0
OPTI 505 Fall 2002


2π
J1 ,
j n ed
e λ 
0

2π
1
0

 j λ no d 

=
e
2π
2π

j ( ne −n o )d
0 e λ

j ne d

λ


e

0
© 2002 Tom D. Milster
.
Slide 3B-15
Polarization – Jones Matrix Examples
Quarter-Wave Plate
Comparison of terms yields
π
2π
=
( n e − no ) d, or
2
λ
d =
OPTI 505 Fall 2002
λ
1
500 nm
3
=
= 14 × 10 nm = 14µ m
4 ne − no
4(1.553-1.544)
© 2002 Tom D. Milster
Slide 3B-16
8
Polarization – Rotated Elements
In order to describe the effect of a rotated optical element, we can project the
(x,y) electric fields onto the element with the rotation matrix, MR.
 cos θ sin θ 
MR =

 − sin θ cos θ 
OPTI 505 Fall 2002
Part of this slide from Jim Wyant, 2000.
© 2002 Tom D. Milster
Slide 3B-17
Polarization – Jones Matrix Examples
Rotated Quarter-Wave Plate
Consider rotating a quarter-wave plate by 45o. Follow these steps to find
the effect on transmitted light:
1.) Decompose the incident light into components. Application of the
rotation matrix produces a description of the light field in the coordinate
frame of the rotated plate.
J 1′ = M R J 1
2.) Apply the quarter-wave plate matrix.
J 2′ = M QWPJ1′ = M QWP M R J 1
3.) Rotate back into the (x,y) coordinate system.
R
J 2 = M R† J ′2 = M R† M QWP J ′1 = M R† M QWP M R J 1 = M QWP
J1
Notice that
OPTI 505 Fall 2002
R
M QWP
= M R† M QWP M R
(† = transpose)
© 2002 Tom D. Milster
Slide 3B-18
9
Polarization – Jones Matrix Examples
Rotated Quarter-Wave Plate (cont’d)
M
R
QWP
= M M QWP M R
†
R
 cos45o − sin45o   1 0   cos45o
=


o
cos45o   0 i   − sin45o
 sin45
 2
 2
2
2
−

 1 0  

2
2
2
2



=

 2
2  0 i  
2
2


−

 2
2 
 2
2 
1  1 − 1  1 0   1 1 
= 
2  1 1   0 i   − 1 1
1  1 −1  1 1 
= 
2  1 1   − i i 
1 1 + i 1 − i  1 + i
= 
=
2  1 − i 1 + i 
2
OPTI 505 Fall 2002
sin45 o 

cos45o 
A quarter-wave plate
rotated at 45o has a very
simple equivalent matrix.
 1 −i 
 −i 1 


© 2002 Tom D. Milster
Slide 3B-19
Polarization – Jones Matrix Examples
Rotated Quarter-Wave Plate (cont’d)
Look at the effect of a rotated QWP on x-polarized light.
R
J 2 = M QWP
J1
1 + i  1 −i   1 
2  − i 1   0 
1+ i  1 
=
2  − i 
=
The effect of a QWP (rotated at 45o) on x-polarized light is to change
the state of polarization to RHC.
OPTI 505 Fall 2002
© 2002 Tom D. Milster
Slide 3B-20
10
Polarization – Jones Matrix Examples
Reflection Off a Mirror
unfolded
representation
physical mirror
state J1
state J1
state J2
state J2
observer
Remember, we look into the wave to determine the state of polarization. A
π phase is introduced between x and y axes from the physical mirror due to
the right-handed coordinate system used in the wave description.
Therefore, the representation of a perfect mirror in the unfolded system is:
1 0 
MM = 

 0 −1
OPTI 505 Fall 2002
© 2002 Tom D. Milster
Slide 3B-21
Polarization – Jones Matrix Systems
Cascaded Components
component 1 component 2
component 3
state J1
state J2
M1
M2
J 2 = M 3 M 2M 1J 1 = M
M3
SYS
J1
Cascaded elements can be represented by algebraically
multiplying component matrices into a system matrix MSYS .
OPTI 505 Fall 2002
© 2002 Tom D. Milster
Slide 3B-22
11
Polarization – Jones Matrix Examples
Combination of a Quarter-Wave Plate and a Mirror
Unfolded System
Physical System
o
QWP at 45
QWP at 45o mirror QWP at -45o
mirror
state J1
state J1
state J2
state J2
R
MQWP
R
MQWP
MM
−R
R
J 2 = M QWP
M M M QWP
J1
1 0 
MM =

 0 −1 
−R
M QWP
OPTI 505 Fall 2002
MM
−R
MQWP
1 + i  1 −i 
2  − i 1 
1 + i 1 i 
=
2  i 1
R
M QWP
=
© 2002 Tom D. Milster
Slide 3B-23
Polarization – Jones Matrix Examples
Combination of a Quarter-Wave Plate and a Mirror (cont’d)
1 + i  1 i   1 0  1 + i  1 −i 




 J1
2  i 1  0 −1 2  −i 1 
i  1 i   1 0   1 −i 
= 


 J1
2  i 1  0 −1  −i 1 
J2 =
i 1
= 
2 i
i0
= 
2  2i
i   1 −i 
J
1  i −1 1
− 2i 
J
0  1
 0 1
=
 J1
 −1 0 
OPTI 505 Fall 2002
Notice the effect on x-polarized
light:
 1
If J1 =   ,
 0
 0 1  1
J2 = 
 
 − 1 0  0
0
= 
The state of polarization
 -1
is now y polarized.
© 2002 Tom D. Milster
Slide 3B-24
12
Polarization – Jones Matrix Examples
Linear polarizer x
Linear polarizer y
HWP
General retarder
OPTI 505 Fall 2002
 1 0


 0 0
 0 0
 0 1


1 0 
 0 − 1


0 
jφ / 2  1
e 
jφ 
0
e


© 2002 Tom D. Milster
Slide 3B-25
13