Crystallography 18

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X-Ray Reflectivity Measurement
(From Chapter 10 of Textbook 2)
http://www.northeastern.edu/nanomagnetism/downloads/Basic
%20Principles%20of%20Xray%20Reflectivity%20in%20Thin%20Films%20%20Felix%20JimenezVillacorta%20[Compatibility%20Mode].pdf
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X-ray is another light source to be used to perform
reflectivity measurements.
Refractive index of materials (: X-ray):
n  1    i
2

re  e
2


x
4
re: classical electron radius = 2.818 × 10-15 m-1
e: electron density of the materials
x: absorption coefficient
Definition in typical optics:
n1sin1 = n2sin2
In X-ray optics: n1cos1 = n2cos2
 > 1  n <1,
1
2
Critical angle for total reflection
n1cos1 = n2cos2, n1= 1; n2=1- ;
1 = c; 2 = 0 
cosc =1-  sinc = 1  (1   )2
 and c <<1  c  2
 ~ 10-5 – 10-6; and c ~ 0.1o – 0.5o
1
c
1-
1  (1   )2
X-ray reflectivity from thin films:
Single layer:
Path difference = BCD  2t sin 2
Snell’s law in X-ray optics: n1cos1 = n2cos2
cos1 = n2cos2=(1-)cos2.
cos 1
 cos  2 
1
(1   )2  cos2 1
 sin  2 
1
1-
2
(1   ) 2  cos2 1
cos1
cos2 1
2
2
2
 sin 2  1 

1

(
1

sin

)(
1





...)
1
2
(1   )
 (1  2  3 2  ...)  1  2
When 1 , 2, and  << 1
1  12
Ignore 
Constructive interference:  2t sin  2  2t 12  2  n
  2  12  2
 4t 2 (12  2 )  n 22
2 2
n

 12  2  2
4t
Si on Ta
 
2
1
2
2
n
 2
2
4t
y  ax  b
/180
2
b  2  re  e   e

2

a  2 t 
4t
2 a
Slope = a
b
use y 2  ax 2  b So that the horizontal axis is linear
Fresnel reflectivity: classical problem of reflection of an
EM wave at an interface – continuity of electric field and
magnetic field at the interface
Reflected k
3 3
beam
Incident k1  x
1
y
beam
Reflection and Refraction:
• Random polarized beam
travel in two homogeneous,
n1
n2
isotropic, nondispersive, and
nonmagnetic media (n1 and n2).
Snell’s law: n1 sin 1  n2 sin 2
and 1  3
 E2 x  t x 0   E1x 
E    0 t  E 
y   1y 
 2y 
and
 E3 x  rx
E    0
 3y  
Refracted
beam
k2
2
0   E1x 
ry   E1 y 
Continuity can be written for two different cases:
(a) TE (transverse electric) polarization: electric field
is  to the plane of incidence.
E1 E1x E3x
H1y 1 3
H2y
E3
H3y
E2x
E2
E2 x  tx E1x E3 x  rx E1x
 E1x  E3 x  E2 x
E1x  rx E1x  tx E1x
 tx  1  rx
2
H1 y cos 1  H 3 y cos 1  H 2 y cos  2 (horizontal field)
 E / H    0 / n (scalar)
n1E1x cos1 / 0  n1E3 x cos1 /0  n2 E2 x cos2 / 0
n1E1x cos1  n1rx E1x cos1  n2tx E1x cos2
1  rx  tx n2 cos2 / n1 cos1 & 1  rx  t x
2n1 cos 1
n1 cos 1  n2 cos  2
 tx 
; rx 
n1 cos 1  n2 cos  2
n1 cos 1  n2 cos  2
(b) TM (transverse magnetic) polarization: magnetic
field is  to the plane of incidence.
E1y
E3y
E1
H1x 1 3
H2x
2
E3
E2 y  t y E1 y
E3 y  ry E1 y
H3x
 H1x  H 3 x  H 2 x
E2y
E2
 n1E1 y  ry n1E1 y  t y n2 E1 y
& E / H    0 / n
 n1 (1  ry )  n2t y
E1 y cos 1  E3 y cos 1  E2 y cos  2
E1 y cos 1  ry E1 y cos 1  t y E1 y cos  2
1  ry  t y cos  2 / cos 1
1  ry  t y n2 / n1
2n1 cos 1
n2 cos 1  n1 cos  2
 ty 
; ry 
n2 cos 1  n1 cos  2
n2 cos 1  n1 cos  2
http://en.wikipedia.org/wiki/Image:Fresnel2.png
Rs: s-polarization; TE mode
Rp: p-polarization; TM mode
Another good reference (chapter 7)
http://www.ece.rutgers.edu/~orfanidi/ewa/
In X-ray arrangement n1 = 1, change cos  sin
n1 sin 1  n2 sin  2 sin 1  n2 sin  2
1
cos 
rx 

1
2
n1 sin 1  n2 sin  2 sin 1  n2 sin  2
n
cos1/n2
all angles are small; sin1 ~ 1.
cos 1

cos


Snell’s law obey  cos1 = n2 cos2.
2
2
1
2
2
n2
cos 1
2
2
 sin 2  1 

n
sin


n

cos
1
2
2
2
2
n2
2
 n22  (1    i )2  1  2  2i  2i   2   2
 n2 sin  2  1  2  2i  cos 1  sin 1  2  2i
2
2
12
2
2
sin 1  n2 sin  2 1  1   c  2i
rx 

sin 1  n2 sin  2 1  12   c2  2i
 c2
1      2i
R flat (1 )  r r 
1  12   c2  2i
2
1
*
x x
in term of q 
2
c
2
4 sin 

q1  q  q 
2
1
2
c
32 i
2
2
2

R flat ( q1 ) 
2
32 i
2
2
q1  q1  qc 
2
Effect of surface roughness is similar to Debye-Waller
factor
Rroughness(1 )  R flat (1 ) exp( 8 212 2 / 2 )
Rroughness(q1 )  R flat (q1 ) exp( 0.5q1q2 2 )
The result can be extended to multilayer. The treatment is
the same as usual optics except definition of geometry!
One can see that the roughness plays a major role at high
wave vector transfers and that the power law regime differs
from the Fresnel reflectivity at low wave vector transfers
X-ray reflection for multilayers
L. G. Parratt, “Surface studies of solids by total reflection
of x-rays”, Phys. Rev. 95 359 (1954).
y
z
Electric vector of the incident beam: E1 ( z1 )
Reflected beam: E1R ( z1 )
Refracted beam: E2 ( z2 )
E1 ( z1 )  E1 (0) expit  k1, y y1  k1,z z1 
E1R ( z1 )  E1R (0) expit  k1, y y1  k1, z z1 
E2 ( z2 )  E2 (0) expi t  k2, y y2  k2, z z2 
k1, k2 : wavevecto r in medium 1 and 2
Boundary conditions for the wave vector at the
interface between two media:
frequencies must be equal on either side of the
interface: 1 = 2 , n1 1 = n22  n2k1 = n1k2;
wave vector components parallel to the
interface are equal k1, ||  k2, ||
From first boundary condition
n1  1
k
k22, y  k22,z  k22  n22 k12  n22
2
n2  1   2  i 2
cos

1
2
k1, y
2
2
2
n2

k
(
1

2


2
i



)
1, y
2
2
1
2
cos 1
From second boundary condition k1, y  k2, y
k
2
2,z
 k k
2
1, y
 n k k  n k k k
 (n22  1)k12  k12 sin 2 1  k12 (12  2 2  2i2 )
2
2
2
2, y
2 2
2 1
2
1, y
2 2
2 1
2
1
2
1, z
k2, z  k1 (12  2 2  2i 2 )1 / 2  f 2k1
 E2 ( z2 )  E2 (0) exp[i(t  k2, y x2 )] exp[ik1 f 2 z2 ]
Shape of reflection curve: two media
The Fresnel coefficient for reflection
E1R n1 sin 1  n2 sin 2
F1, 2 

E1 n1 sin 1  n2 sin 2
Page 10
 n2 sin 2  1  2 2  2i 2  cos2 1  sin 2 1  2 2  2i 2
1  f 2 f1  f 2
F1, 2 

1  f 2 f1  f 2
f2
f1  (12  21  2i1 )1/ 2
A, B are real value
f 2  A  iB
1
A
{(12  2 2 )  [(12  2 2 )2  422 ]1/ 2 }1/ 2
2
1
B
{(12  2 2 )  [(12  2 2 )2  4 22 ]1/ 2 }1/ 2
2
From Snell’s law
c 2  2 2
Page 4
1
A
{(12  c22 )  [(12  c22 )2  422 ]1/ 2 }1/ 2
2
1
B
{(12  c22 )  [(12  c22 )2  422 ]1/ 2 }1/ 2
2
R 2
1
IR E

I
E1
(1  A)2  B 2 h  (1 / c 2 ) 2 (h  1)1 / 2


2
2
(1  A)  B
h  (1 / c 2 ) 2 (h  1)1 / 2
  
 1  
  1 
h         1   2 
c 2 
2 



 c 2  




2
2
2
1/ 2





N layers of homogeneous media
Thickness of nth layer: d n
medium 1: air or vacuum
an : the amplitude factor for half the perpendicular
depth
 ik n d n 
 if n k1d n 
 if n d n 
an  exp  
  exp  
  exp  

2 
2 
 



0
dn
n-1
1
n
a En
R
n
an E
En , E
n
an En
1
n
a E
R
n
R
n
The continuity of the tangential components of
the electric vectors for the n-1, n boundary
an1En1  a E
1
n 1
R
n 1
 a En  an E
1
n
R
n
(1)
The continuity of the tangential components of the
magnetic field for the n-1, n boundary
(an1H n1  an11H nR1 ) sin n1  (an1H n  an H nR ) sin n
(an1En1  an11EnR1 )nn1 sin n1  (an1En  an EnR )nn sin n
f n1k1
f n k1
 (an1En1  an11EnR1 ) f n1k1  (an1En  an EnR ) f n k1
Solve (1) and (2); (1)fn-1+(2), (1)fn-1-(2)
(2)
En 1 
E
R
n 1

1
2an 1 f n 1
1
[an1 En ( f n 1  f n )  an EnR ( f n 1  f n )]
[an1 En ( f n 1  f n )  an EnR ( f n 1  f n )]
2an11 f n 1
1
R
EnR1
[
a
E
(
f

f
)

a
E
2
n
n
n 1
n
n n ( f n 1  f n )]
 an1 1
En 1
[an En ( f n 1  f n )  an EnR ( f n1  f n )]
2
R
EnR1
2 [( f n 1  f n )  an ( En / En )( f n 1  f n )]
 an 1
En 1
[( f n 1  f n )  an2 ( EnR / En )( f n 1  f n )]
2
R
EnR1
[(
f

f
)
/(
f

f
)

a
(
E
2
n 1
n
n 1
n
n
n / En )]
 an1
En1
[1  an2 ( EnR / En )( f n1  f n ) /( f n 1  f n )]
Rn,n1  an2 ( EnR / En ) ; Fn1,n  ( f n1  f n ) /( f n1  f n )
Rn 1,n  a
2
n 1
EnR1
4 [ Fn 1, n  Rn , n 1 ]
 an 1
En 1
[1  Rn ,n 1 Fn 1,n ]
For N layers, starting at the bottom medium
RN , N 1  0 (N+1 layer: substrate)
R
R

E
Also, a1 = 1 (air or vacuum) 1, 2 1 E1
Finally, the reflectivity of the system is
R
1
2
IR
E
 R1, 2 
I0
E1
2
For rough interfaces:
Fn 1,n  [( f n 1  f n ) /( f n 1  f n )]
f n  (12  2 n  2i n )1/ 2  nn2  cos 2 1
 Fn1,n  [( f n1  f n ) /( f n1  f n )] exp( 8 2 f n1 f n n21 / 2 )
Can be calculated numerically!
Example of two layers with roughness
Au on Si substrate
Interface roughness
nn  1   n  i n
z
Probability density
nn1  1   n1  i n1
 z2 
Pn ( z ) 
exp   2 
2  n
 2 n 
Integration
 z  zn 
nn  nn 1 nn  nn 1

nn ( z ) 

erf 

2
2
 2 n 
1
Refractive index
Same roughness & refractive index profile

 / 1

 / 1
Félix Jiménez‐Villacorta
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