X-Ray Reflectivity Measurement (From Chapter 10 of Textbook 2)

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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. - PowerPoint PPT Presentation

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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

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

Definition in typical optics: n1sin1 = n2sin2

In X-ray optics: n1cos1 = n2cos2

> 1 n <1,

1

2

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

X-ray reflectivity from thin films:

Path difference = BCD 2sin2 t

Single layer:

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

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

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

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

&

(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(

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/

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

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

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).

Electric vector of the incident beam: )( 11 zE

z

Reflected beam: )( 11 zE R

Refracted beam: )( 22 zE

y

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

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

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

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

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

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)

)]()([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

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!

Example of two layers with roughness

Au on Si substrate

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/

Félix Jiménez‐Villacorta

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