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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.
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(From Chapter 10 of Textbook 2)
Refractive index of materials (: X-ray):
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,
n1cos1 = n2cos2, n1= 1; n2=1- ;
1 = c; 2 = 0
and c <<1
~ 10-5 – 10-6; and c ~ 0.1o – 0.5o
Path difference = BCD
cos1 = n2cos2=(1-)cos2.
When 1 , 2, and << 1
Slope = a
So that the horizontal axis is linear
EM wave at an interface – continuity of electric field and
magnetic field at the interface
Reflection and Refraction:
• Random polarized beam
travel in two homogeneous,
isotropic, nondispersive, and
nonmagnetic media (n1 and n2).
Continuity can be written for two different cases:
(a) TE (transverse electric) polarization: electric field
is to the plane of incidence.
field is to the plane of incidence.
Rs: s-polarization; TE mode
Rp: p-polarization; TM mode
Another good reference (chapter 7)
all angles are small; sin1 ~ 1.
Snell’s law obey cos1 = n2 cos2.
Effect of surface roughness is similar to Debye-Waller
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
L. G. Parratt, “Surface studies of solids by total reflection of x-rays”, Phys. Rev. 95 359 (1954).
Electric vector of the incident beam:
interface 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
From second boundary condition
The Fresnel coefficient for reflection
A, B are real value
Thickness of nth layer:
medium 1: air or vacuum
an : the amplitude factor for half the perpendicular depth
The continuity of the tangential components of the electric vectors for the n-1, n boundary
The continuity of the tangential components of the magnetic field for the n-1, n boundary
Solve (1) and (2); (1)fn-1+(2), (1)fn-1-(2)
(N+1 layer: substrate)
Also, a1 = 1 (air or vacuum)
Finally, the reflectivity of the system is
For rough interfaces:
Can be calculated numerically!
Au on Si substrate
Same roughness & refractive index profile