Transcript Lecture #9

PHYS 408
Applied Optics
(Lecture 9)
JA N - A PRIL 2 0 1 7 E DI T ION
JE F F YOUN G
AMPEL RM 113
Quick review of key points from last
lecture
There are only really two unknown complex amplitudes of the net forward and net backward
travelling plane waves inside a thin film.
Together with the overall reflected wave, and the overall transmitted wave amplitudes, there are
a total of four unknown complex amplitudes in a single thin-film/plane wave problem.
The continuity of the parallel electric and magnetic fields across the dielectric interfaces gives
two equations at each interface, so a total of four independent equations for the four unknowns
(linear algebra).
Wavevectors should ultimately be expressed in terms of frequency and refractive indices, not
wavelengths (advice that not all physicists or engineers follow!)
A reminder
Infinite series (homework, very intuitive)
U 0(  )
U 0(  )
More powerful approach
The solution
Note the number of indices: Which terms pertain to interfaces and which to
propagation?
Moving forward
Cast our result in the form of a matrix equation that yields the reflected and transmitted wave
amplitudes (the out-going wave amplitudes) when the incident wave amplitude from the left
hand medium is specified.
Moving forward
Cast our result in the form of a matrix equation that yields the reflected and transmitted wave
amplitudes (the out-going wave amplitudes) when the incident wave amplitude from the left
hand medium is specified.
0
0
0
Very general, the box could represent a single interface, or a 100 layer dielectric stack.
Moving forward
If you had an incident wave from the right hand side, the corresponding matrix representation of
the out-going field amplitudes is;
0
0
0
Moving forward
Combine these to generate a matrix equation that yields the outgoing waves for the general
case when you have two in-coming waves, one from each side.
out
in
S, or Scattering Matrix
The S and M matricies
Would this S matrix help you easily solve for the overall reflected and transmitted fields if you
had multiple dielectric layers up against each other?
No. Need a “transfer matrix” that can take you
from one side to the other, layer by layer,
interface by interface.
right
U 0(  )
U 0(  )
U 3(  )
left
The S and M matricies
This M matrix related to the S matrix via linear algegra:
Note the number of indices: Don’t get confused with the overall thin-film
expression earlier today. The former can be derived from the latter, as follows.
The S and M matricies
Let’s see how this works
U 0(  )
U1(  )
U 2(  )
n1
U 0(  )
n2
U1(  )
d1
Find the net, overall M and S
matrices assuming d2 is infinite
d2
Hint: start with the (intuitive)
individual S matrices for each
transition, and convert them to
M matrices
Step by step
U1  t01 r10  U 0 
   
  
r
t
U
 0   01 10  U1 
U 0(  )
U1(  )
n1
U 0(  )
?
U 2(  )
n2
t01 r10 
S01  

r01 t10 
1
M 01 
t10
M net  M 12 M 01 or M net  M 01M 12 or
t01t10  r01r10
 r
01

r10 
1 
?
U1(  )
d1
d2
U 2  t12
   
U1  r12
r21  U1 
 
t21  U 2 
t12
S12  
r12
r21 
t21 
1 t12t21  r12 r21 r21 
M 12  
1 
t21   r12
Step by step
U1  t01 r10  U 0 
   
  
r
t
U
 0   01 10  U1 
U 0(  )
U1(  )
U 2(  )
n1
U
()
0
U
n2
missing
()
1
d1
S11  ?
t01 r10 
S01  

r01 t10 
in1~d1
e
S11  
 0
1
M 01 
t10
t01t10  r01r10
 r
01

in1~d1
e
0 
M

11

in1~d1 
e
 0

r10 
1 
0 
in1~d1 
e

d2
U 2  t12
   
U1  r12
r21  U1 
 
t21  U 2 
t12
S12  
r12
r21 
t21 
1 t12t21  r12 r21 r21 
M 12  
1 
t21   r12
Step by step
U 0  U 2 
U 2 
    M 12 M 11M 01      
U 2 
U 0   0 
U 0(  )
U1(  )
U 2(  )
n1
U 0(  )
n2
U1(  )
d1
U 0 
U 0  U 2 
U 2 
    M 12 M 11M 01     M net      
U 2 
U 0 
U 0   0 
d2
1 t01t10  r01r10
M 01  
t10   r01
M 12 
r10 
1 
1 t12t21  r12 r21 r21 
1 
t21   r12
in1~d1
e
M 11  
 0
0 
~ 
e in1d1 
Next Step?
Bottom line
Consider Mnet as M, then Snet(A,B,C,D) yields t02,t20,r02,r20
U 0  U 2 
U 0
U 2
Or don’t be lazy, and just solve for  and  from M net      
U0
U0
U 0   0 
n2=1.3, d1=400 nm
Let’s play!
n1
n2
n1
n2
d1
d2
d1
d2
…
nlayers-1
n1
n3
n1
n2
n1
n2
d1
d3
d1
d2
d1
d2
…
nlayers-1
Uniform periodic multilayer stack
…
n1
n2
n1
n2
n1
n2
n1
n2
d1
d2
d1
d2
d1
d2
d1
d2
…
n1=1.3; n2=1.4
d1=580 nm; d2=20 nm
21 periods
Bragg reflection
0.14
0.07
X: 6522
Y: 0.1349
0.12
0.06
0.05
Reflectivity
Reflectivity
0.1
0.08
0.06
0.04
0.03
0.04
0.02
0.02
0.01
X: 304.7
Y: 7.109e-007
0
0
2000
4000
6000
Wavenumber 1/
8000
10000
0
0
200
400
600
Wavenumber 1/
800
1000
Physics of peak with max reflectivity