X-Ray Reflectivity Measurement (From Chapter 10 of Textbook 2) %20Principles%20of%20X-...
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X-Ray Reflectivity Measurement(From Chapter 10 of Textbook 2)
http://www.northeastern.edu/nanomagnetism/downloads/Basic%20Principles%20of%20X-ray%20Reflectivity%20in%20Thin%20Films%20-%20Felix%20Jimenez-Villacorta%20[Compatibility%20Mode].pdf
http://www.google.com/url?sa=t&rct=j&q=x-ray+reflectivity+amorphous&source=web&cd=1&cad=rja&ved=0CDEQFjAA&url=http%3A%2F%2Fwww.stanford.edu%2Fgroup%2Fglam%2Fxlab%2FMatSci162_172%2FLectureNotes%2F09_Reflectivity%2520%26%2520Amorphous.pdf&ei=L3zBUKfSEaLNmAX8vIC4AQ&usg=AFQjCNFfik-tSw8bSPGGyx1ckTK5WBTnSA
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X-ray is another light source to be used to performreflectivity measurements.
Refractive index of materials (: X-ray):
in 1 eer 2
2
x
4
re: classical electron radius = 2.818 × 10-15 m-1
e: electron density of the materials x: absorption coefficient
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Definition in typical optics: n1sin1 = n2sin2
In X-ray optics: n1cos1 = n2cos2
> 1 n <1,
1
2
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n1cos1 = n2cos2, n1= 1; n2=1- ;
1 = c; 2 = 0
cosc =1- sinc =
and c <<1
~ 10-5 – 10-6; and c ~ 0.1o – 0.5o
1c
1-
2)1(1 2)1(1
2c
Critical angle for total reflection
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X-ray reflectivity from thin films:
Path difference = BCD 2sin2 t
Single layer:
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Snell’s law in X-ray optics: n1cos1 = n2cos2
cos1 = n2cos2=(1-)cos2.
1
coscos 1
22
1-1
22 cos)1(
cos1
1
cos)1(sin 1
22
2
221
221
2
2 ...)1)(sin1(1)1(
cos1sin
When 1 , 2, and << 1
2212
21...)321( 2 2
11
ntt 22sin2 212 Constructive interference:
2221
2 )2(4 nt 24 2
222
1 t
n
Ignore
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24
22
22
1 nt
Si on Ta
baxy
eeerb 2
2
b
/180
at
ta
2
4 2
2
Slope = a
baxy 22use So that the horizontal axis is linear
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Reflection and Refraction: • Random polarized beam travel in two homogeneous, isotropic, nondispersive, and nonmagnetic media (n1 and n2). Snell’s law:
n1 n2
k1
k3 k2
1
3
2Incident
beam
Reflectedbeam
Refractedbeam
x
y
and
Fresnel reflectivity: classical problem of reflection of an EM wave at an interface – continuity of electric field and magnetic field at the interface
2211 sinsin nn 31
Continuity can be written for two different cases: (a) TE (transverse electric) polarization: electric field is to the plane of incidence.
y
x
y
x
y
x
y
x
y
x
y
x
E
E
r
r
E
E
E
E
t
t
E
E
1
1
3
3
1
1
2
2
0
0 and
0
0
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E1
E2
E3E1x
H1y
E3x
E2x
H3y
H2y
1 3
2
xxx EtE 12 xxx ErE 13
xxx EEE 231
xxxxx EtErE 111
221311 coscoscos yyy HHH
022201310111 /cos/cos/cos xxx EnEnEn nHE // 0
xx rt 1
(horizontal field)
(scalar)
212111111 coscoscos xxxxx EtnErnEn
1122 cos/cos1 nntr xx xx tr 1
2211
2211
2211
11
coscos
coscos ;
coscos
cos2
nn
nnr
nn
nt xx
&
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(b) TM (transverse magnetic) polarization: magnetic field is to the plane of incidence.
E1
E2
E3H1x
E1y
H3x
H2x
E3y
E2y
1 3
2
xxx HHH 231
yyy EtE 12 yyy ErE 13
221311 coscoscos yyy EEE
nHE //& 0
yyyyy EntEnrEn 121111
211111 coscoscos yyyyy EtErE
12 cos/cos1 yy tr 12 /1 nntr yy
2112
2112
2112
11
coscos
coscos ;
coscos
cos2
nn
nnr
nn
nt yy
yy tnrn 21 )1(
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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/
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221
221
2211
2211
sinsin
sinsin
sinsin
sinsin
n
n
nn
nnrx
In X-ray arrangement n1 = 1, change cos sin
all angles are small; sin1 ~ 1. Snell’s law obey cos1 = n2 cos2.
2
12
coscos
n
2
1
cos1/n2
22
12cos
1n
122
22222
12
2 cossincos
1sin nnn
2222
2 2221)1( iiin
iin 22sincos221sin 1
2
1
2
22
i
i
n
nr
c
cx
2
2
sinsin
sinsin22
11
22
11
221
221
21
2c
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2
22
11
22
11*
12
2)(
i
irrR
c
cxxflat
Effect of surface roughness is similar to Debye-Waller factor
)/8exp()()( 2221
211 flatroughness RR
in term of sin4
q2
2
222
11
2
222
11
132
32
)(
iqqq
iqqq
qR
c
c
flat
The result can be extended to multilayer. The treatment is the same as usual optics except definition of geometry!
)5.0exp()()( 22111 qqqRqR flatroughness
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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
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X-ray reflection for multilayers
L. G. Parratt, “Surface studies of solids by total reflection of x-rays”, Phys. Rev. 95 359 (1954).
Electric vector of the incident beam: )( 11 zE
z
Reflected beam: )( 11 zE R
Refracted beam: )( 22 zE
y
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1,11,1111 exp)0()( zkyktiEzE zy
1,11,1111 exp)0()( zkyktiEzE zyRR
2,22,2222 exp)0()( zkyktiEzE zy
2 and 1 medium inr wavevecto:, 21 kk
Boundary conditions for the wave vector at theinterface between two media:
frequencies must be equal on either side of the interface: 1 = 2 , n1 1 = n22 n2k1 = n1k2;
wave vector components parallel to the interface are equal || ,2|| ,1 kk
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1
2
2
,12
2
2
1
2
2
2
2
2
,2
2
,2 cos y
zy
knknkkk
)221(cos
2
122
2
,1
1
2
2
,12
2
ikk
n y
y
222
1
1
1
in
n
From first boundary condition
From second boundary condition yy kk ,2,1 2
,1
2
1
2
1
2
2
2
,1
2
1
2
2
2
,2
2
2
2
,2 zyyz kkknkknkkk )22(sin)1( 22
2
1
2
11
22
1
2
1
2
2 ikkkn
122/1
222
11,2 )22( kfikk z
]exp[)](exp[)0()( 2212,2222 zfikxktiEzE y
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Shape of reflection curve: two media
The Fresnel coefficient for reflection
2211
2211
1
12,1 sinsin
sinsin
nn
nn
E
EF
R
Page 10
221
2
1
2
2222 22sincos221sin iin
2f
21
21
21
212,1 ff
ff
f
fF
2/1
112
11 )22( if
iBAf 2
2/12/122
22
212
21 }]4)2[()2{(
2
1 A
2/12/122
22
212
21 }]4)2[()2({
2
1 B
A, B are real value
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From Snell’s law 22 2 c Page 4
2/121
2/121
221
221
2
1
1
)1(2)/(
)1(2)/(
)(
)(
hh
hh
BA
BA
E
E
I
I
c
cR
R
2/12/122
222
21
22
21 }]4)[(){(
2
1 ccA
2/12/122
222
21
22
21 }]4)[()({
2
1 ccB
2/12
2
2
2
2
1
2
2
1 1
cc
h
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N layers of homogeneous media
Thickness of nth layer: nd medium 1: air or vacuum
an : the amplitude factor for half the perpendicular depth
nnnnnn
n
dfidkifdika exp
2exp
2exp 1
0
nd
R
nn EE ,
nn Ea 1
n-1
n
nnEaR
nn Ea 1
R
nn Ea
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R
nnnn
R
nnnn EaEaEaEa
1
1
1
111
n
R
nnnnn
R
nnnn HaHaHaHa sin)(sin)( 1
11
1
111
The continuity of the tangential components of the magnetic field for the n-1, n boundary
nn
R
nnnnnn
R
nnnn nEaEanEaEa sin)(sin)( 1
111
1
111
11kfn 1kfn
1
1
111
1
111 )()( kfEaEakfEaEa n
R
nnnnn
R
nnnn
The continuity of the tangential components of the electric vectors for the n-1, n boundary
Solve (1) and (2); (1)fn-1+(2), (1)fn-1-(2)
(1)
(2)
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)]()([2
111
1
111 nn
Rnnnnnn
nnn ffEaffEa
faE
)]()([2
111
1
111
1 nnRnnnnnn
nn
Rn ffEaffEa
faE
)]()([
)]()([
111
111
21
1
1
nnRnnnnnn
nnRnnnnnn
nn
Rn
ffEaffEa
ffEaffEaa
E
E
)])(/()[(
)])(/()[(
12
1
12
121
1
1
nnnRnnnn
nnnRnnnn
nn
Rn
ffEEaff
ffEEaffa
E
E
)]/())(/(1[
)]/()/()[(
112
2112
11
1
nnnnnRnn
nRnnnnnn
nn
Rn
ffffEEa
EEaffffa
E
E
)/()( ; )/( 11,12
1, nnnnnnnRnnnn ffffFEEaR
]1[
][
,11,
1,,141
1
121,1
nnnn
nnnnn
n
Rn
nnn FR
RFa
E
EaR
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For N layers, starting at the bottom medium01, NNR (N+1 layer: substrate)
Also, a1 = 1 (air or vacuum) 112,1 EER R
2
1
12
2,10 E
ER
I
I RR
Finally, the reflectivity of the system is
For rough interfaces: )]/()[( 11,1 nnnnnn ffffF
)/8exp()]/()[( 2211
211,1 nnnnnnnnn ffffffF
1222/12
1 cos)22( nnnn nif
Can be calculated numerically!
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Example of two layers with roughness
Au on Si substrate
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Interface roughness
z
Probability density
2
2
2exp
2
1)(
nn
n
zzP
Refractive index
Integration
n
nnnnnn
zzerf
nnnnzn
222)( 11
111 1 nnn in
nnn in 1
Same roughness & refractive index profile
1/ 1/
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Félix Jiménez‐Villacorta