today’s outline - september 26, 2016csrri.iit.edu/~segre/phys570/16f/lecture_10.pdf ·...

Today’s Outline - September 26, 2016

• Reflectivity research topics

• Mirrors

• Refractive optics

APS Visits:10-ID: Friday, October 21, 201610-BM: Friday, October 28, 2016

Homework Assignment #03:Chapter 3: 1, 3, 4, 6, 8due Wednesday, October 05, 2016

C. Segre (IIT) PHYS 570 - Fall 2016 September 26, 2016 1 / 19

• Mirrors

Layering in liquid films

THEOS, tetrakis–(2-ethylhexoxy)–silane,a non-polar, roughlyspherical molecule, wasdeposited on Si(111) singlecrystals

Specular reflection mea-surements were made atMRCAT (Sector 10 atAPS) and at X18A (atNSLS).

Deviations from uniform density are used tofit experimental reflectivity

C.-J. Yu et al., “Observation of molecular layering in thin liquid films usingx-ray reflectivity”, Phys. Rev. Lett. 82, 2326–2329 (1999).

The peak below 10A appears in allbut the thickest film and dependson the interactions between filmand substrate.

There are always peaks between10-20A and 20-30A

A broad peak appears at free sur-face indicating that ordering re-quires a hard smooth surface.

Film growth kinetics

Gaussian roughness profilewith a “roughness” expo-nent 0 < h < 1.

As thefilm is grown by vapor de-position, the rms width σ,grows with a “growth ex-ponent” β and the correla-tion length in the plane ofthe surface, ξ evolves withthe “dynamic” scaling ex-ponent, zs = h/β.

g(r) ∝ r2h σ ∝ tβ

ξ ∝ t1/zs 〈h〉 ∝ t

h ≈ 0.33, β ≈ 0.25 for nodiffusion.

Ag/Si films: 10nm (A), 18nm (B),37nm (C), 73nm (D), 150nm (E)

C. Thompson et al., “X-ray-reflectivity study of the growth kinetics of vapor-deposited silver films”, Phys. Rev. B 49, 4902–4907 (1994).

Gaussian roughness profilewith a “roughness” expo-nent 0 < h < 1.

As thefilm is grown by vapor de-position, the rms width σ,grows with a “growth ex-ponent” β and the correla-tion length in the plane ofthe surface, ξ evolves withthe “dynamic” scaling ex-ponent, zs = h/β.

g(r) ∝ r2h

σ ∝ tβ

ξ ∝ t1/zs 〈h〉 ∝ t

Gaussian roughness profilewith a “roughness” expo-nent 0 < h < 1. As thefilm is grown by vapor de-position, the rms width σ,grows with a “growth ex-ponent” β

and the correla-tion length in the plane ofthe surface, ξ evolves withthe “dynamic” scaling ex-ponent, zs = h/β.

g(r) ∝ r2h

σ ∝ tβ

ξ ∝ t1/zs 〈h〉 ∝ t

Gaussian roughness profilewith a “roughness” expo-nent 0 < h < 1. As thefilm is grown by vapor de-position, the rms width σ,grows with a “growth ex-ponent” β

and the correla-tion length in the plane ofthe surface, ξ evolves withthe “dynamic” scaling ex-ponent, zs = h/β.

ξ ∝ t1/zs 〈h〉 ∝ t

Gaussian roughness profilewith a “roughness” expo-nent 0 < h < 1. As thefilm is grown by vapor de-position, the rms width σ,grows with a “growth ex-ponent” β and the correla-tion length in the plane ofthe surface, ξ evolves withthe “dynamic” scaling ex-ponent, zs = h/β.

ξ ∝ t1/zs 〈h〉 ∝ t

ξ ∝ t1/zs

〈h〉 ∝ t

ξ ∝ t1/zs 〈h〉 ∝ t

h can be obtained from thediffuse off-specular reflec-tion which should vary as

I (qz ) ∝ σ−2/hq−(3+1/h)z

This gives h = 0.63 but isthis correct?

Measure it directly usingSTM

g(r) = 2σ2[1− e(r/ξ)2h

]h = 0.78, ξ = 23nm,

σ = 3.2nm

Thus zs = h/β = 2.7 and diffraction data confirm ξ = 19.9〈h〉1/2.7 A

I (qz ) ∝ σ−2/hq−(3+1/h)z

g(r) = 2σ2[1− e(r/ξ)2h

]h = 0.78, ξ = 23nm,

σ = 3.2nm

I (qz ) ∝ σ−2/hq−(3+1/h)z

g(r) = 2σ2[1− e(r/ξ)2h

]h = 0.78, ξ = 23nm,

σ = 3.2nm

I (qz ) ∝ σ−2/hq−(3+1/h)z

g(r) = 2σ2[1− e(r/ξ)2h

]h = 0.78, ξ = 23nm,

σ = 3.2nm

I (qz ) ∝ σ−2/hq−(3+1/h)z

g(r) = 2σ2[1− e(r/ξ)2h

]h = 0.78, ξ = 23nm,

σ = 3.2nm

I (qz ) ∝ σ−2/hq−(3+1/h)z

g(r) = 2σ2[1− e(r/ξ)2h

]h = 0.78, ξ = 23nm,

σ = 3.2nm

I (qz ) ∝ σ−2/hq−(3+1/h)z

g(r) = 2σ2[1− e(r/ξ)2h

]h = 0.78, ξ = 23nm,

σ = 3.2nm

I (qz ) ∝ σ−2/hq−(3+1/h)z

g(r) = 2σ2[1− e(r/ξ)2h

h = 0.78, ξ = 23nm,

σ = 3.2nm

I (qz ) ∝ σ−2/hq−(3+1/h)z

g(r) = 2σ2[1− e(r/ξ)2h

]h = 0.78, ξ = 23nm,

σ = 3.2nm

Thus zs = h/β = 2.7

and diffraction data confirm ξ = 19.9〈h〉1/2.7 A

I (qz ) ∝ σ−2/hq−(3+1/h)z

g(r) = 2σ2[1− e(r/ξ)2h

]h = 0.78, ξ = 23nm,

σ = 3.2nm

Thus zs = h/β = 2.7

and diffraction data confirm ξ = 19.9〈h〉1/2.7 A

I (qz ) ∝ σ−2/hq−(3+1/h)z

g(r) = 2σ2[1− e(r/ξ)2h

]h = 0.78, ξ = 23nm,

σ = 3.2nm

Liquid metal surfaces

X-ray reflectivity using syn-chrotron radiation has madepossible the study of the sur-face of liquid metals

a liquid can be described ascharged ions in a sea of con-duction electrons

this leads to a well-definedsurface structure as can beseen in liquid gallium

contrast this with the scat-tering from liquid mercury

P. Pershan, “Review of the highlights of x-ray studies of liquid metal surfaces”, J. Appl. Phys. 116, 222201 (2014).

Liquid metal eutectics

High vapor pressure andthermal excitations limit thenumber of pure metals whichcan be studied but alloy eu-tectics provide many possi-bilities

tune x-rays around the Bi ab-sorption edge at 13.42 keVand measure a Bi43Sn57 eu-tectic

surface layer is rich in Bi(95%), second layer is defi-cient (25%), and third layeris rich in Bi (53%) once again

O. Shpyrko et al., “Atomic-scale surface demixing in a eutectic liquidBiSn alloy”, Phys. Rev. Lett. 95, 106103 (2005).

The MRCAT mirror

x-rays

Ultra low expansion glass polished to afew A roughness

One platinum stripe and one rhodiumstripe deposited along the length of themirror on top of a chromium buffer layer

A mounting system which permits angu-lar positioning to less than 1/100 of adegree as well as horizontal and verticalmotions

A bending mechanism to permit verticalfocusing of the beam to ∼ 60 µm

The MRCAT mirror

x-rays

The MRCAT mirror

x-rays

The MRCAT mirror

x-rays

The MRCAT mirror

x-rays

Mirror performance

When illuminated with 12 keVx-rays on the glass “stripe”, thereflectivity is measured as:

With the Rh stripe, the thinslab reflection is evident andthe critical angle is significantlyhigher.

The Pt stripe gives a higher crit-ical angle still but a lower reflec-tivity and it looks like an infiniteslab. Why?

0.1 0.2 0.3 0.4 0.5

α (degrees)

Reflectivity

12 keV

Mirror performance

0.1 0.2 0.3 0.4 0.5

α (degrees)

Reflectivity

12 keV

Mirror performance

0.1 0.2 0.3 0.4 0.5

α (degrees)

Reflectivity

12 keV

Mirror performance

The Pt stripe gives a higher crit-ical angle still but a lower reflec-tivity and it looks like an infiniteslab.

0.1 0.2 0.3 0.4 0.5

α (degrees)

Reflectivity

12 keV

Mirror performance

The Pt stripe gives a higher crit-ical angle still but a lower reflec-tivity and it looks like an infiniteslab.

0.1 0.2 0.3 0.4 0.5

α (degrees)

Reflectivity

12 keV

Mirror performance

The Pt stripe gives a higher crit-ical angle still but a lower reflec-tivity and it looks like an infiniteslab. Why? 0.1 0.2 0.3 0.4 0.5

α (degrees)

Reflectivity

12 keV

Mirror performance (cont.)

0.1 0.2 0.3 0.4 0.5

α (degrees)

Reflectivity

Pt stripe

12 keV

20 keV

30 keV

As we move up in energy thecritical angle for the Pt stripedrops.

The reflectivity at low anglesimproves as we are well awayfrom the Pt absorption edgesat 11,565 eV, 13,273 eV, and13,880 eV.

As energy rises, the Pt layer be-gins to show the reflectivity of athin slab.

0.1 0.2 0.3 0.4 0.5

α (degrees)

Reflectivity

Pt stripe

12 keV

20 keV

30 keV

0.1 0.2 0.3 0.4 0.5

α (degrees)

Reflectivity

Pt stripe

12 keV

20 keV

30 keV

0.1 0.2 0.3 0.4 0.5

α (degrees)

Reflectivity

Pt stripe

12 keV

20 keV

30 keV

Tangential focusing mirror

The shape of an ideal mirror is anellipse, where any ray coming fromone focus will be projected to thesecond focus.

Consider a 1:1 focus-ing mirror. For an ellipse the sumof the distances from any point onthe ellipse to the foci is a constant.

F1P + F2P = 2a

F1B = F2B = a

sin θ =b

The shape of an ideal mirror is anellipse, where any ray coming fromone focus will be projected to thesecond focus. Consider a 1:1 focus-ing mirror. For an ellipse the sumof the distances from any point onthe ellipse to the foci is a constant.

F1P + F2P = 2a

F1B = F2B = a

sin θ =b

F1P + F2P = 2a

F1B = F2B = a

sin θ =b

F1P + F2P = 2a

F1B = F2B = a

sin θ =b

F1P + F2P = 2a

F1B = F2B = a

sin θ =b

F1P + F2P = 2a

F1B = F2B = a

sin θ =b

F1P + F2P = 2a

F1B = F2B = a

sin θ =b

Saggital focusing mirror

Ellipses are hard figures to make,so usually, they are approximatedby circles. In the case of saggitalfocusing, an ellipsoid of revolutionwith diameter 2b, is used for focus-ing.

ρsaggital = b = 2f sin θ

The tangential focus is also usuallyapproximated by a circular cross-section with radius

ρtangential = a =2f

sin θ

Types of focusing mirrors

A simple mirror such as the one atMRCAT consists of a polished glassslab with two “legs”.

A force is appliedmechanically to push the legs apart andbend the mirror to a radius as small asR = 500m.

The bimorph mirror is designed to ob-tain a smaller form error than a simplebender through the use of multipleactuators tuned experimentally.

A cost effective way to focus in bothdirections is a toroidal mirror which hasa fixed bend in the transverse directionbut which can be bent longitudinally tochange the vertical focus.

A simple mirror such as the one atMRCAT consists of a polished glassslab with two “legs”. A force is appliedmechanically to push the legs apart andbend the mirror to a radius as small asR = 500m.

Dual focusing options

• Toroidal mirror — simple, moderate focus, good for initialfocusing element, easy to distort beam

• Saggittal focusing crystal & vertical focusing mirror —adjustable in both directions, good for initial focusingelement

• Kirkpatrick-Baez mirror pair — in combination with aninitial focusing element, good for final small focal spot andvariable energy

• K-B mirrors & zone plates — in combination with aninitial focusing element, gives smallest focal spot, but hardto vary energy

Refractive optics

Just as with visible, light, it is possible to make refractive optics for x-rays

visible light:

n ∼ 1.2− 1.5f ∼ 0.1m

x-rays:

n ≈ 1− δ, δ ∼ 10−5

f ∼ 100m!

x-ray lenses are complementary to those for visible lightgetting manageable focal distances requires making compound lenses

Refractive optics

visible light:

n ∼ 1.2− 1.5f ∼ 0.1m

x-rays:

n ≈ 1− δ, δ ∼ 10−5

f ∼ 100m!

Refractive optics

visible light:

n ∼ 1.2− 1.5f ∼ 0.1m

x-rays:

n ≈ 1− δ, δ ∼ 10−5

f ∼ 100m!

Refractive optics

visible light:

n ∼ 1.2− 1.5f ∼ 0.1m

x-rays:

n ≈ 1− δ, δ ∼ 10−5

f ∼ 100m!

x-ray lenses are complementary to those for visible light

getting manageable focal distances requires making compound lenses

Refractive optics

visible light:

n ∼ 1.2− 1.5f ∼ 0.1m

x-rays:

n ≈ 1− δ, δ ∼ 10−5

f ∼ 100m!

Focal length of a compound lens

f1 f2 f3

f→ 1

f− 1

f1− 1

o1→ 1

f→ i1 = f

f2− 1

o2→ 1

f→ i2 =

f2− 1

o2→ 1

f→ i2 =

Start with a 3-elementcompound lens, calculateeffective focal length

assuming each lens hasthe same focal length, f

f1 = f , o1 =∞

for the second lens, theimage i1 is a virtualobject, o2 = −i1

similarly for the third lens,o3 = −i2

so for N lenses feff = f /N

f1 f2 f3

f− 1

f1− 1

o1→ 1

f→ i1 = f

f2− 1

o2→ 1

f→ i2 =

f2− 1

o2→ 1

f→ i2 =

Start with a 3-elementcompound lens, calculateeffective focal lengthassuming each lens hasthe same focal length, f

f1 = f , o1 =∞

f1 f2 f3

f→ 1

f− 1

f1− 1

o1→ 1

f→ i1 = f

f2− 1

o2→ 1

f→ i2 =

f2− 1

o2→ 1

f→ i2 =

f1 = f , o1 =∞

f1 f2 f3

f→ 1

f− 1

f1− 1

f→ i1 = f

f2− 1

o2→ 1

f→ i2 =

f2− 1

o2→ 1

f→ i2 =

f1 = f , o1 =∞

f1 f2 f3

f→ 1

f− 1

f1− 1

o1→ 1

→ i1 = f

f2− 1

o2→ 1

f→ i2 =

f2− 1

o2→ 1

f→ i2 =

f1 = f , o1 =∞

f1 f2 f3

f→ 1

f− 1

f1− 1

o1→ 1

f→ i1 = f

f2− 1

o2→ 1

f→ i2 =

f2− 1

o2→ 1

f→ i2 =

f1 = f , o1 =∞

f1 f2 f3

f→ 1

f− 1

f1− 1

o1→ 1

f→ i1 = f

f2− 1

f→ i2 =

f2− 1

o2→ 1

f→ i2 =

f1 = f , o1 =∞

f1 f2 f3

f→ 1

f− 1

f1− 1

o1→ 1

f→ i1 = f

f2− 1

o2→ 1

→ i2 =f

f2− 1

o2→ 1

f→ i2 =

f1 = f , o1 =∞

f1 f2 f3

f→ 1

f− 1

f1− 1

o1→ 1

f→ i1 = f

f2− 1

o2→ 1

f→ i2 =

f2− 1

o2→ 1

f→ i2 =

f1 = f , o1 =∞

f1 f2 f3

f→ 1

f− 1

f1− 1

o1→ 1

f→ i1 = f

f2− 1

o2→ 1

f→ i2 =

f2− 1

f→ i2 =

f1 = f , o1 =∞

f1 f2 f3

f→ 1

f− 1

f1− 1

o1→ 1

f→ i1 = f

f2− 1

o2→ 1

f→ i2 =

f2− 1

o2→ 1

→ i2 =f

f1 = f , o1 =∞

f1 f2 f3

f→ 1

f− 1

f1− 1

o1→ 1

f→ i1 = f

f2− 1

o2→ 1

f→ i2 =

f2− 1

o2→ 1

f→ i2 =

f1 = f , o1 =∞

f1 f2 f3

f→ 1

f− 1

f1− 1

o1→ 1

f→ i1 = f

f2− 1

o2→ 1

f→ i2 =

f2− 1

o2→ 1

f→ i2 =

f1 = f , o1 =∞

Rephasing distance

A spherical surface is not the ideal lens as it introduces aberrations. Derivethe ideal shape for perfect focusing of x-rays.

λο λ (1+δ)ο

consider two waves, one traveling in-side the solid and the other in vacuum,λ = λ0/(1− δ) ≈ λ0(1 + δ)

if the two waves start in phase, they will be in phaseonce again after a distance

Λ = (N + 1)λ0 = Nλ0(1 + δ)

Nλ0 + λ0 = Nλ0 + Nδλ0 −→ λ0 = Nδλ0 −→ N =1

Λ = Nλ0 =λ0

λ0r0ρ≈ 10µm

Rephasing distance

λο λ (1+δ)ο

Λ = (N + 1)λ0 = Nλ0(1 + δ)

Nλ0 + λ0 = Nλ0 + Nδλ0 −→ λ0 = Nδλ0 −→ N =1

Λ = Nλ0 =λ0

λ0r0ρ≈ 10µm

Rephasing distance

λο λ (1+δ)ο

Λ = (N + 1)λ0 = Nλ0(1 + δ)

Nλ0 + λ0 = Nλ0 + Nδλ0 −→ λ0 = Nδλ0 −→ N =1

Λ = Nλ0 =λ0

λ0r0ρ≈ 10µm

Rephasing distance

λο λ (1+δ)ο

Λ = (N + 1)λ0 = Nλ0(1 + δ)

Nλ0 + λ0 = Nλ0 + Nδλ0 −→ λ0 = Nδλ0 −→ N =1

Λ = Nλ0 =λ0

λ0r0ρ≈ 10µm

Rephasing distance

λο λ (1+δ)ο

Λ = (N + 1)λ0 = Nλ0(1 + δ)

Nλ0 + λ0 = Nλ0 + Nδλ0

−→ λ0 = Nδλ0 −→ N =1

Λ = Nλ0 =λ0

λ0r0ρ≈ 10µm

Rephasing distance

λο λ (1+δ)ο

Λ = (N + 1)λ0 = Nλ0(1 + δ)

Nλ0 + λ0 = Nλ0 + Nδλ0 −→ λ0 = Nδλ0

−→ N =1

Λ = Nλ0 =λ0

λ0r0ρ≈ 10µm

Rephasing distance

λο λ (1+δ)ο

Λ = (N + 1)λ0 = Nλ0(1 + δ)

Nλ0 + λ0 = Nλ0 + Nδλ0 −→ λ0 = Nδλ0 −→ N =1

Λ = Nλ0 =λ0

λ0r0ρ≈ 10µm

Rephasing distance

λο λ (1+δ)ο

Λ = (N + 1)λ0 = Nλ0(1 + δ)

Nλ0 + λ0 = Nλ0 + Nδλ0 −→ λ0 = Nδλ0 −→ N =1

Λ = Nλ0 =λ0

λ0r0ρ≈ 10µm

Rephasing distance

λο λ (1+δ)ο

Λ = (N + 1)λ0 = Nλ0(1 + δ)

Nλ0 + λ0 = Nλ0 + Nδλ0 −→ λ0 = Nδλ0 −→ N =1

Λ = Nλ0 =λ0

λ0r0ρ

≈ 10µm

Rephasing distance

λο λ (1+δ)ο

Λ = (N + 1)λ0 = Nλ0(1 + δ)

Nλ0 + λ0 = Nλ0 + Nδλ0 −→ λ0 = Nδλ0 −→ N =1

Λ = Nλ0 =λ0

λ0r0ρ≈ 10µm

Ideal interface profile

λ ολ (1+δ)ο

The wave exits thematerial into vacuumthrough a surface ofprofile h(x), and istwisted by an angle α.Follow the path of twopoints on the wave-front, A and A′ as theypropagate to B andB ′.

from the AA′B ′ trian-gle

and from the BCB ′ tri-angle

using Λ = λ0/δ

λ0 (1 + δ) = h′(x)∆x −→ ∆x ≈ λ0

h′(x)

α(x) ≈ λ0δ

∆x= h′(x)δ = h′(x)

λ ολ (1+δ)ο

The wave exits thematerial into vacuumthrough a surface ofprofile h(x), and istwisted by an angle α.

Follow the path of twopoints on the wave-front, A and A′ as theypropagate to B andB ′.

using Λ = λ0/δ

λ0 (1 + δ) = h′(x)∆x −→ ∆x ≈ λ0

h′(x)

α(x) ≈ λ0δ

∆x= h′(x)δ = h′(x)

λ 0 λ (1+δ)0

using Λ = λ0/δ

λ0 (1 + δ) = h′(x)∆x −→ ∆x ≈ λ0

h′(x)

α(x) ≈ λ0δ

∆x= h′(x)δ = h′(x)

λ 0 λ (1+δ)0

using Λ = λ0/δ

λ0 (1 + δ) = h′(x)∆x −→ ∆x ≈ λ0

h′(x)

α(x) ≈ λ0δ

∆x= h′(x)δ = h′(x)

λ 0 λ (1+δ)0

using Λ = λ0/δ

λ0 (1 + δ) = h′(x)∆x

−→ ∆x ≈ λ0

h′(x)

α(x) ≈ λ0δ

∆x= h′(x)δ = h′(x)

λ 0 λ (1+δ)0

using Λ = λ0/δ

λ0 (1 + δ) = h′(x)∆x −→ ∆x ≈ λ0

h′(x)

α(x) ≈ λ0δ

∆x= h′(x)δ = h′(x)

λ 0 λ (1+δ)0

using Λ = λ0/δ

λ0 (1 + δ) = h′(x)∆x −→ ∆x ≈ λ0

h′(x)

α(x) ≈ λ0δ

∆x= h′(x)δ = h′(x)

λ 0 λ (1+δ)0

using Λ = λ0/δ

λ0 (1 + δ) = h′(x)∆x −→ ∆x ≈ λ0

h′(x)

α(x) ≈ λ0δ

= h′(x)δ = h′(x)λ0

λ 0 λ (1+δ)0

using Λ = λ0/δ

λ0 (1 + δ) = h′(x)∆x −→ ∆x ≈ λ0

h′(x)

α(x) ≈ λ0δ

∆x= h′(x)δ

= h′(x)λ0

λ 0 λ (1+δ)0

using Λ = λ0/δ

λ0 (1 + δ) = h′(x)∆x −→ ∆x ≈ λ0

h′(x)

α(x) ≈ λ0δ

∆x= h′(x)δ

= h′(x)λ0

λ 0 λ (1+δ)0

using Λ = λ0/δ

λ0 (1 + δ) = h′(x)∆x −→ ∆x ≈ λ0

h′(x)

α(x) ≈ λ0δ

∆x= h′(x)δ = h′(x)

If the desired focal length of this lens is f ,the wave must be redirected at an anglewhich depends on the distance from theoptical axis

α(x) =x

combining, we have

λ0h′(x)

f−→ h′(x)

this can be directly integrated

2f λ0=

2f λ0

α(x)f

a parabola is the ideal surfacefor focusing be refraction

α(x) =x

combining, we have

λ0h′(x)

f−→ h′(x)

2f λ0=

2f λ0

α(x)f

α(x) =x

combining, we have

λ0h′(x)

−→ h′(x)

2f λ0=

2f λ0

α(x)f

α(x) =x

combining, we have

λ0h′(x)

f−→ h′(x)

2f λ0=

2f λ0

α(x)f

α(x) =x

combining, we have

λ0h′(x)

f−→ h′(x)

2f λ0=

2f λ0

α(x)f

α(x) =x

combining, we have

λ0h′(x)

f−→ h′(x)

2f λ0

α(x)f

α(x) =x

combining, we have

λ0h′(x)

f−→ h′(x)

2f λ0=

2f λ0

α(x)f

α(x) =x

combining, we have

λ0h′(x)

f−→ h′(x)

2f λ0=

2f λ0

α(x)f

today’s outline - september 26, 2016csrri.iit.edu/~segre/phys570/16f/lecture_10.pdf ·...

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