OPTICS BY THE NUMBERS
L’Ottica Attraverso i Numeri
Michael Scalora
U.S. Army Research, Development, and Engineering Center
Redstone Arsenal, Alabama, 35898-5000
&
Universita' di Roma "La Sapienza"
Dipartimento di Energetica
Rome, April-May 2004
SVEAT: The Slowly Varying Enevelope
Approximation In Time, and The Ability To
Inlcude Reflections To All Orders In The BPM
Algorithm
From the SVEAT to a Vector BPM:
Negative Refraction
2
2
2

Pnl
n

E
4

2
 E 2 2  2
2
c t
c t
E ( z, x, t )  E ( z, x, t )ei ( kz t )  c.c.
2
2
2
2
2



E

E
n
(
x
)

E
2
i

n
(
x
)

E

2
2
2
  E  2  2ik  2

  k  2 n ( x)  E
2
2
z
z
c t
c
t 
c

4
 2
c
 2

2
 2  2i    PNL
t
 t

 E
E n ( x)  E 2i n ( x) E
  E  2  2ik
 2

2
z
z
c t
c2
t
2
 2 2 2 

4 

2
  k  2 n ( x)  E  2  2  2i    PNL
c
c  t
t



2
2
2
2
2
Assuming steady state conditions…
n ( x)  n

E i 2
   E  i
 F
n0
2
2
0

(3)
0 2
4

0
E i
E E
in
n0 in
2
2
n
(
x
)

n

E i 2
4 0
0  0
   E  i
E i
PNL
 F
n0
in
n0 in
4 n0 0
F
in
  z / 0
Fresnel Number

k  n0
c
F 
F  small
Wave front does not distort:
Plane Wave propagation
Diffraction is very important
2
2
(3)
n
(
x
)

n


E i 2
4 0 2
0 0
   E  i
E i
E E
 F
n0
in
n0 in
This equation is of the form: E
 HE

Where:
2
2
n
(
x
)

n
i 2
4 (3) 0 2
0 0
H     i
i
E  D V
F
n0
in
n0 in



E ( , x)  e
 H ( ', x ) '
0
E (0, x)  e
Using the split-step
BPM algorithm
H (0, x )
E (0, x)
 eV (0, x ) / 2e D eV (0, x ) / 2 E (0, x)
2
2
n
(
z
,
x
,
y
)

E
2
 E
0
2
2
c
t
E ( z, x, t )  E ( z, x, t )ei ( kz t )  c.c.
2
2
2

E

E
n
(
z
,
x
,
y
)

E
2
  E  2  2ik


2
2
z
z
c
t
2
2


2i n ( z, x, y ) E
 2
2
  k  2 n ( z , x, y )  E  0
2
c
t 
c

2
2
2

E

E
n
(
z
,
x
,
y
)

E
2
  E  2  2ik


2
2
z
z
c
t
2
2


2i n ( z, x, y ) E
 2
2
  k  2 n ( z , x, y )  E  0
2
c
t 
c

Apply SVEAT, i.e., SVEA in time only:drop higher temporal
derivatives. This assumption means that pulse duration must remain
always much longer than the optical cycle at all times. In all kinds of
problems, if a pulse is as long as the optical cycle it means trouble for
any approximation. So this is a very good approximation almost
always.
2
2
2

E

E
n
(
z
,
x
,
y
)

E
2
  E  2  2ik


2
2
z
z
c
t
2
2


2i n ( z, x, y ) E
 2
2
  k  2 n ( z , x, y )  E  0
2
c
t 
c

2
2

E

E
2
i

n
( z , x, y ) E
2
  E  2  2ik

2
z
z
c
t
 2 2 2

  k  2 n ( z , x, y )  E  0
c


2
2

E

E
2
i

n
( z , x, y ) E
2
  E  2  2ik

2
z
z
c
t
 2 2 2

  k  2 n ( z , x, y )  E  0
c


This equation is first order in time. This suggests
writing equation in following form:
2
 2 2 2 
2in2 E

E
E
2
   k  2 n  E    E  2  2ik
2
c
t
c
z
z


2
 2 2 2 
2in2 E

E
E
2
   k  2 n  E    E  2  2ik
2
c
t
c
z
z


Now, adopting the usual kind of scaling:
  z / 0
x  x / 0
And choosing
y  y / 0
  ct / 0
k /c
2




E
i
i

E E
2
2
2
0
in
in
n
 i  n  1 E 
 E 

2

in
4 0
4 0 

2




E
i
i

E E
2
2
2
0
in
in
n
 i  n  1 E 
 E 

2

in
4 0
4 0 

This equation is of the form:
E
n
 ( D  V ) E  HE

2
Which we can ALMOST easily recognized and compare to:
E
 ( D  V ) E  HE

2




E
i
i

E E
2
2
2
0
in
in
n
 i  n  1 E 
 E 

2

in
4 0
4 0 

N.B.: the differential operator includes ALL longitudinal and
spatial derivatives, which means all boundary conditions are left
Intact. Integrating this equation must is therefore equivalent to
Including longitudinal and transverse reflections to all orders.
2




E
i
i

E E
2
2
2
0
in
in
n
 i  n  1 E 
 E 

2

in
4 0
4 0 

E
n
 ( D  V ) E  HE

2
E 1
 2 HE
 n
This is a nasty operator equation, which has this formal solution:
1
E (   , r)  e
n2
H 
E ( , r)
The exponential is the product
of two non-commuting operators
Why is it important to include the index in the denominator?
E 1
 2 HE
 n
E
 HE

Because that factor accounts for the correct group velocity.
Here is how I solve the problem:
1
E (   , r )  E ( , r )  2 HE ( , r )
n
Add zero:
1
1
1
E (   , r )  E ( , r )  2 HE ( , r )  2 E ( , r )  2 E ( , r )
n
n
n
Group terms as follows:
1 1

E (   , r )  E ( , r ) 1  2   2 E ( , r )  HE ( , r )
 n  n

And recognize…
1  1 H 

E (   , r )  E ( , r ) 1  2   2 e E ( , r )
 n  n
Solution is accurate up to first order in time

Algorithm:
E (1) (   , r)  e H E ( , r)
Which is solution of
E
 HE

Then algbreically manipulate solution to find
1  1 H 

E (   , r )  E ( , r ) 1  2   2 e E ( , r )
 n  n
Work very well in all cases except metals.
Special considerations must be made in that case.
1.25
n=1
n=1.42
1.00
|E|
2
0.75
0.50
0.25
0
-150
-50
50
position (microns)
150
1.2
n=1
n=1.42
|E|
2
0.8
0.4
0
-150
-50
50
position (microns)
Red: without the 1/n2 factor in the operator
150
Example:
Assume a PBG structure with cross sectional area as small as 1 mm2,
and a Thickness L~10 mirons. The volume is therefore
of order V~10-11 m3. I will further assume that the structure is
not solidly anchored to
the earth, i.e., it is free to move. The incident pulse can be tuned
anywhere in the pass band or band gap.
The rate of momentum transfer depends on tuning.
m
E,B
before
after
m
E,B
E,B
Pm
momentum density
The total momentum at
time t is given by:
In terms of the Poynting vector
1
g
EB
4c
P(t ) 

1
 4c E ( z, t )  B ( z, t )dz
1 
P (t )  2  S ( z , t )dz
c 
The momentum stored inside the object is the difference
between the initial total momentum and the
instantaneous momentum stored in the field, namely:
1 
Ppbg (t )  P(t  0)  2  S ( z , t )dz
c 
1 
P (t  0)  2  S ( z ,0)dz
c 
T~0
Tuned inside the gap...
1
Ppbg
gr cm/sec
2P0
P0
Pfield
-P0
time (/c=0.33x10-14 sec.)
...Mirror like interaction
2
Electromagnetic momentum as a function ot position
3
At different time shots
At band edge resonance: pulse is almost completely transmitted
With some reflections.
4
-6
3x10
field
-6
2x10
P(t) (gr cm/sec)
-6
1x10
pbg
0
0
-12
1x10
-12
2x10
-12
t (sec.)
3x10
5
P(z,t)
z
Electromagnetic momentum as a function ot position
6
At different time shots
0.10
0.08
acceleration
stage
deceleration
stage
0.06
Velocity
(cm/sec)0.04
<a> ~ 1011 m/sec2
0.02
0
0
-12
1x10
-12
2x10
Time (sec.)
-12
3x10
Input Pulse
Plane Wave: means no diffraction, even though beam width is
finite. I.e., each ray travels straight down.
9
8
7
0.4 0.6
EL\2
0.7
0.8
0.9
0.2
0.3
6
0.1
5
4
10
20
30
EL\1
Same as slide 1
40
50
/4
/4
Z (longitudinal)
x
The structure: Cross section of each column is nearly square.
In this case the square cross section does not
cause ray bending due to simple refraction
9
0.1
8
0.2
0.1
0.4
0.3
0.3
0.70.4
0.40.6
0.80.5
0.7
0.1 0.2
EL\2
7
0.5
0.1
0.2
0.1
0.6
0.5
0.4
0.6
0.5.7
0.4
0.8
0
0.8
0.4
0.6
0.5
0.3
0.7
0.5
0.4
0.4
0.5
0.3
0.4
0.2
0.1
0.2
0.1
0.6
0.2
6
0.7
0.2
0.1
5
0.2 0.3
0.1 0.2
0.1
0.1
4
10.00
18.75
27.50
EL\1
Same as slide 4.
36.25
45.00
The structure: Cross section of each column is nearly circular. The
discretization causes slight imperfections, which can be improved by
reducing the integration step. The diameters of each column is close to the
/4 condition, but not sure.
Another view: 12 columns long, or ~ 4 microns.
9
Transmitted portion
8
EL\2
7
0.8
0.7
6
0.7
0.6 0.5
0.4 0.3
0.2
5
0.2
0.1
4
11
13
15
17
19
EL\1
21
23
25
25.6
0.1
0.1
0.1
70
longitudinal position (microns)
80
0.60.9
0.4
0.
1
0.7
0.7 .4
0
0.3
0.3
0.6
0.1
60
0.7
0.3
0.
3
0
0.9.7
0.3
1.1
0.1
0.3 0.4 0.6
0.6
0.3 0.4
0
0.4.6
0.3
0.4 0.6
0.1
1.0
0.79
0.
1.1
0.6
0.4
0.1
0.1
50
01.9 0.7
.0
3
.
0
0.4
0.6
40
1.4
30
1
0.
1.00.9
0.7
0.4
20
1
0.
11..100.9
.3 .7
1
4
.
0
1 0 0.1 0.0.67.4
0.9 0.3
0.6
6.4
1.0
1.1
1.3
1.0
1.0 0.7
0.3.9
0 .0
1
0.7
0.9
12.8
0.0
10
0.1
0.4
0.3
0.
6
19.2
0.7
0.9
transvesre position (microns)
0.3
0.1
90
100
25.6
0.0
0.
2
0.2
0.2
00..12
1
.
0
00..1
0.1
0.0 0
19.2
1.1
0.
0
0.2
0.7
0.4 0.6
Input
0.3
4
0.
000....7
.677
00..6 6
6
0.5
0
0.3
0.3 .3
0.7
00. 8
00.6
. 5
0.
..77700.04...55
00
00.4
0.3
0.2
6.4
.21
0..2.1
0
0
0
1
.
0
1
0.
0.1
0.
0.1
1
.20.00.1
00.1
.1
00.0.00
0.0
0.1
EL\2
12.8
0.8
9
0.
0.0
00. .080.
1.90
80..9
8
00.0.0050
6.677
04.4.5..05.60.
0.2
1.0
.0
1
1.2
0.9
11.0.111.22.1
1
.8
000.7.7
00..3.44
3
0
0
000..0032...22 00...111
1
.
0
0
0.0
0.0
10
20
30
40
50
60
EL\1
70
80
90
100
Output
13
0.0
Output
1
0.
0.0
11
0.0
0.1
30
40
50
EL\1
60
0.30.4
0.3
0.1
0.0
0.1
Input Gaussian
20
Output
00..22
1
1 0.
0.
1
0.
0
0.
10
00..556
0.
0.3
0.4
00...654
0
3
0
0.7.6
0.1
5
0.20.1
1
0.
0. 0 0 0. 00. 0 0 00.60.7
1 .2.2 2 .33 .4.40.5.5 0.6
0.5
1.0
0.1
.778
.8
00.
.9
00.
0.9
7
0.
1
0.0
0.4 0..330.2
0
0.3 0.5
0.5
0.4
EL\2
0.1
9
1
0.1
5
0.
70
80
90
100
Transverse Coordinate (in microns)
110
0.4
90
50
10
30
50
70
90
0.4
0.4
22.1
6
8
264280
.7
0
..0
7
2
.82.4
912.2
23238
3399034
0.8
3.
6
0.04.18.2
2
1
..02.4
0.21382.4.42.4.648606.57.4098.4.1811.80111 4444748765543...56.931
6 8.52670..9.14.82125.91151
0762482
636321
7.3.16417.51
38
4
5.27.2942
972.43.18
03.0
26
6
57
2
.93
59
.3
7
..9
5
1
5127.
1
5
1.33721.2
.3..2
9.439
91
7.7
5
3
0.4
8
0.
130
0.
4
190
air
170
n=2 - i 0.02
gain
0.8
150
2. 37.0
350.04064
3.82 3
3
28048
46
6
2
3336
0
97
13298...04.8
4
3
4
440
70
6
3.2.0
12.
.
04
Reflections appear
to be suppressed
30
10
110
130
150
170
Longitudinal coordinate (in microns)
190
0.1
130
110
90
0.4 0.2
0.1
50
70
90
110
5.0
37
5
6
5555..12
890
566..1
1
11.5.67
30
0.5
.819 6.9
0.6 1
66...4
212..0
3
6
378
1
0
46
.2..2
7667.5
0. 00
22.0.3.30.4
1 .2.300
..0450.60.8191 2
.878787..7.1
0. 00
.20.12.131..14
20. .0.50071.1.112
.
3 .18692..034.21.825.2906.132.23.38.9.344 .90.6875
.734205.16.3
.01.264.34
2.43
.4
.9
.7
5845
7.453
73
.6.8
9
0.1
10
0.3 0.2
0.3
7
0.06. 0.2
70
0.
6
0.1 0.5
0.5 0.6
0.4 0.5
0.1
0.8
00.4.3.02.9
0
0.1
5 0
0. 0.8.7
3
00..20.9
0.1
0.04.3
0.2 0.1
0.6
0. 0.7
4
150
0
0..65
7
0.
Transverse Coordinate (in microns)
air
190
n=2 - i 0.01
gain
170
0.8
9
0.
1
1.
4
0.
1
0.
0.1
50
30
10
130
150
170
Longitudinal coordinate (in microns)
190
air
n=2
Transverse Coordinate (in microns)
190
170
150
130
110
90
70
50
30
10
10
30
50
70
90
110
130
150
170
Longitudinal coordinate (in microns)
190
air
n=2 + i 0.01
Transverse Coordinate (in microns)
190
loss
170
150
130
110
90
70
50
30
10
10
30
50
70
90
110
130
150
170
Longitudinal coordinate (in microns)
190
air
n=2 + i 0.02
loss
170
0
00.8.89.77
0
.
00..9
0..04.465.45.0
302..888
0
. .99
0000...0.2042.03..300
0
2
2
01
. 100.1.9
.02
000.00
.
.1
..550.8
40
0.1
0.
00.0.0 000000.00.3.4350.34.20.9
.0.02.0.21.02..899
00
2
00. . 1
00.010..110. 0..5.5.044
000.4.4..332
00.0
.0
00000..0.0.20.3
.23.02.
2
000.1.1
2.1
0
0
.
00..010
00.1
0.0
0.2
0.
0.2 000..1.104
0.10
.1
0.
2
00. 0.7
60
6.6.6
00.0.50..0
3 5.5
0.57
0000..7..8
0
150
110
90
2
0.
00.
0. .11
1
70
2 0.1 4
0.1
0.5 00.02.4.20.2
.
0
0
.3
0
0
0.6 0.3
0.140. 0.00.
0.
3 .2
0.4
0.0.3
770.2 0.5
0.
2
2
.
0.3
.
056.1
0.
000.
60.
0.
0.
1
1
.7 0.0
0.1 0.1 0.
0000..8.87 0.0
130
000..
0.5.66.07.7
00.0
0. 0.340.54.6
0.0
2
00.000. .7
0.00
0.1
0.
600.65.6.060.6
7
.0
0.1
0.1
.7505..6
01001
..06360.0.
0.6
0.2
0.0.
0.2
.7 5
0.
4050.1
0.1
.5
0.
0
0.2
00.
0.2
0.1
0
.3
.2
0.
0.6
000.5
0
0.1
.2.60.5
.0.223
0
.3
0.
3
0.
0.4.0330.4
0.4
010.2
0. 0.0.
0
0 1.10.1.0
0
.6.7.6
.77
000
000..88
Transverse Coordinate (in microns)
190
00.1.1
0.0
0.0
50
30
10
10
30
50
70
90
110
130
150
170
Longitudinal coordinate (in microns)
190
The symbols and the lines indicate
the location and direction of motion of
the baricenter of the wave packet.
190
170
150
EL\2
130
n=2+i0.01
110
90
n=2
n=2-i0.01
70
50
30
10
10
30
50
70
90
110
EL\1
130
150
170
190
62 nm of Ag
air
air
40
0.0000..770.7
5..56
00.89
.
000...9
8
0.67.00
..983
00
45..8
0.3
.04.4
.
0
0
4
0
.
2
0
0
.3
.
0
0.30.02.2 ..660.7
0.1
0.1
2
0
.
50.44.9
0
0.100.2
.10000.05
0.0
.9
.03.3.40
0..0.33
0.0 0000.2
0.6
0
2 0.5
.
.6.8
8.7 00.0 0.0
0.10
0
1
0
.
.
1
.
.
0
5
0
0
00.1 00
.00..404.0.33.3 .1
0.1
0..02.230.1
0
0
0.0 0.0
.2
0
0.2
0 0.4.4
1.01
0.33
0.0
0.0
0..0
.
00
.1.1102.1
0.0
00..668
0.
0.0
0.0
20
0.0
1
0. 0.1
0.4 .4
00.6 0
0.6
000...7
77
0. .6 .2
00
.55
0
00.0.4
4.33
3
0. .1
00
0
.2..2
0
01
1
0.0.
00.0
.1
0.
0
30
00. 00.70.7.7
0. 40.4.5
00.002.7
0. 00.
0.0.6.6.0.
67
000..440
30 5.5
48
0
0
0
000.0.0.5.45.5
0. 0 .4
.2 0.4.4 0.0 0.4
5
00 .2
7.7.07
0.20..4.55 00.
0.0.
0.
44 0
.0 000.2
.2
6.2
0.0
0
00.1
0
.
.
.1 2 0 0.12 0.1 0.
.0
0.0
0.1 0.1
30.220.1 0.300..33
10.
0.0.
00..55 0.2
0.7
0.1
0..33 0.4 ..33
0.6.060.3
0 0.3 0
7
0000..7....888
0.5 00
0.8.77
00.
Transverse Coordinate (in microns)
50
0.0
10
10
20
30
40
Longitudinal coordinate (in microns)
50
1 B
E  
c t
  (B /  ) 
 E
c t
A discontinuity in  gives a fundamental problem:
An infinite derivative for sudden chnages in 
However, the H field is continuous across
interfaces, just as E is continuous. Symmetrize
The equations of motion.
E  
 H
c t
 E
H 
c t
H y
H z


t
y
z
H y
E x

t
z
E x
H z

t
y
 Ex
c

c

c
E  i ( Ex ( y, z , t ) e
i ( k z z  k y y t )
H  j( H y ( y, z, t ) e
 c.c )
i ( k z z  k y y t )
 c.c )  k ( H z ( y, z, t ) e
i ( k z z  k y y t )
 c.c )
H y
H z
 i  Ex  ik y H z 
 ik z H y 
t
c
y
z
H y
Ex

 i  H y  ik z Ex 
t
c
z
Ex
H z

 i  H z  ik y Ex 
t
c
y
 Ex
c

c

c
E  i ( E ( y, z , t ) e
i ( k z z  k y y t )
H  j( H y ( y, z, t ) e
S

 c.c )
i ( k z z  k y y t )
 c.c )  k ( H z ( y, z, t ) e
i ( k z z  k y y t )
 c.c )
c
ExH
4
S  k ( E H y*  E* H y )  j( E H z*  E* H z )  k Sz ( y, z, t )  jS y ( y, z, t )
S  k ( E H y*  E* H y )  j( E H z*  E* H z )  k Sz ( y, z, t )  jS y ( y, z, t )
z 
Pz (t ) 
Py (t ) 

y 

z 
y 
z 
y 

z 

y 
S z ( y, z , t ) dy dz and
S y ( y, z , t ) dy dz
y
interface
E=Ex i
H
Hy
q
H=Hy j + Hz k
Hz
Ex
E points into the paper
H lays on the y-z plane, and so it
has components along z and y.
z
k
q
E  
 H
c t
 E
H 
c t
1
1
2
2
Input field is a gaussian in y and z, incident at 45 degrees
air
Transverse Coordinate (in microns)
190
n=2 (=4)
WAVE-FRONT
Incident angle=45 degrees
Refraction angle ~20.7 degrees
170
150
normal refraction
at a dielectric
interface
130
110
90
70
Red dashed line indicates
the major axis of the ellipse. It is
compressed in the direction of
propagation due to packet slowing
down in that direction to a velocity
of c/n
50
30
10
10
30
50
70
90
110
130
150
170
Longitudinal coordinate (in microns)
190
Sz
50
Sy
40
Pulse is incident on a Silver
half-space.
S
00.7.7
0.4
00.7.8
0
0..98
0.6
0..5.66 .4
00.05
.04
0
30.3
.
0
2
0.
0.2 0.1
.1
0.00
 = -8.98 + i 0.78
=1
0.3.3
0.2
0.001.2
0.1
0.30.4
0.34 0.2
0
0.5 0.7.7
0.3
0.43 0
.4
0.4
0.809.8
0.
0.076.5
0.
0.4
0.4.0
.3
00.4
0.3
5
02 2.2
0.1
.
0.
0
1
0.
0.0
For =~500nm
0.2
EL\2
30
0
0. .6 0.7
4 0.5
00.7.8
.6
05
. .4
0
0.4
0.20
03.3 .2
0.
0.1 .01
0
0.0
7
0.
0
0.5 .7
0.
3
0.6
0.7
10
0.6
0.7
04..45
0
.
0
0.40.3
0.02.2
0.1
0.1
0.0
10
This corresponds to an index
0.0
0.1
20
n = 0.13 + i 3
Sz
Sy
S
20
30
EL\1
40
50
Test 1
= -8.98 +i0.78
=1
For =~500nm
propagation from air into metal substrate
0.01
plus
Pz (t)
-0.01
plus
momenta
Py (t)
The maximum refraction angle into the metal
Is ~78 degrees at the time indicated, i.e., near peak.
Sz
-0.03
Sy
S
-0.05
0
10
20
30
time (/c)
40
50
propagation from air into metal substrate
75
This are the momenta as a function of time in the
incidence half-space. The reflection causes the zcomponent (red) to change sign. With both components
negative and nearly equal, it corresponds to a reflection
angle of 45 degree. There is some absorption, and so
each component is not conserved.
momenta
25
minus
Pz
minus
Py
Sz
(t)
(t)
S
-25
Sy
-75
0
10
20
30
time (/c)
40
50
Riassumendo:
Sriflesso
Sincidente
1
1
450
  8.98i 0.78
1
450
~79-800

Strasmesso
argento, con  incidente
a ~ 500nm
=1
1
1
1
0
0.6.7
0.8
0.9
7
0.
S
15
5.00
13.75
22.50
EL1\1
0.2
0.3
S
0.4
0.3
0.2
0.1
0.4 0
.6
0.8
0.7
S0y
0.10.2
25
0.2
0.1
0.3
00.4.60.7
0.3
0.6
0.9
0.
6
0.3
0.8
0.4.3 .9
0 .20
0
0.1
0.1
0.7
0.6
0.4
0.8
0.2
0.1
0.9
0.7
S0z
4
0.
0.8
EL1\2
35
0.6
0.7
0.8
0.4.3 0.9
0
0.2
0.1
S0y
S0z
The ellipse seems to be oriented
correctly, but propagation occurs in
the “wrong” direction.
31.25
40.00
60
momenta inside medium (plus side)
plus
Py (t)
plus
Pz (t)
40
S
Sy
20
Sz
0
-20
0
10
20
30
40
50
time (/c)
The initial momentum on the plus side is zero for both components. At the end of
the interaction, after the pulse has completely entered the medium, the momentum is equally split between
The x and z coordinates, and the momentum points in the direction of energy flow.
Riassumendo:
Sriflesso
Sincidente
1
1
450
  1
  1
450
450

Strasmesso
Quindi anche in questo caso, dove il momento y non e’
conservato (dovrebbe esserlo se l’assorbimento e’ zero)
trovo in ogni caso che il vettore di Poynting punta nella
direzione di propagazione.
Boundary
=1
1
2
2
0.6.8
0
0.9
0.6
0.5
0.8
35
5
0. .4 0.9
0 .3
0
0.1
15
5.00
13.75
22.50
EL\1
0.
5
0.4
0.4
0.3
0.1
0.3
0.3
0.4
0.4
00
.1.6
0.5
0.1
0.
4
0.3
0.1
0.3
25
0.6
0.1
0.5
0.1
0.6
5
0.
3
0.
0.5
5
0.
EL\2
0.8
0
0.4 .6
8
0.
.9
0.4 0
3
0.
0.1
31.25
40.00
CONCLUSIONS:
I AM WILLING TO
BELIEVE
ANYTHING!
(with a grain of salt, of course)