today’s outline - september 25, 2013phys.iit.edu/~segre/phys570/13f/lecture_11.pdf · today’s...

Today’s Outline - September 25, 2013

• Ideal refractive lenses

• Zone plates

• Optics studies

• Kinematical diffraction

Homework Assignment #03:Chapter 3: 1, 3, 4, 6, 8due Monday, October 02, 2013

C. Segre (IIT) PHYS 570 - Fall 2013 September 25, 2013 1 / 20

• Zone plates

• Optics studies

• Zone plates

• Optics studies

• Zone plates

• Optics studies

• Zone plates

• Optics studies

• Zone plates

• Optics studies

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 ′.

λo (1 + δ) = h′(x)∆x −→ ∆x ≈ λoh′(x)

α(x) =λoδ

∆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 ′.

λo (1 + δ) = h′(x)∆x −→ ∆x ≈ λoh′(x)

α(x) =λoδ

∆x= h′(x)δ = h′(x)

λ ολ (1+δ)ο

λo (1 + δ) = h′(x)∆x −→ ∆x ≈ λoh′(x)

α(x) =λoδ

∆x= h′(x)δ = h′(x)

λ ολ (1+δ)ο

λo (1 + δ) = h′(x)∆x

−→ ∆x ≈ λoh′(x)

α(x) =λoδ

∆x= h′(x)δ = h′(x)

λ ολ (1+δ)ο

λo (1 + δ) = h′(x)∆x −→ ∆x ≈ λoh′(x)

α(x) =λoδ

∆x= h′(x)δ = h′(x)

λ ολ (1+δ)ο

λo (1 + δ) = h′(x)∆x −→ ∆x ≈ λoh′(x)

α(x) =λoδ

= h′(x)δ = h′(x)λoΛ

λ ολ (1+δ)ο

λo (1 + δ) = h′(x)∆x −→ ∆x ≈ λoh′(x)

α(x) =λoδ

∆x= h′(x)δ

= h′(x)λoΛ

λ ολ (1+δ)ο

λo (1 + δ) = h′(x)∆x −→ ∆x ≈ λoh′(x)

α(x) =λoδ

∆x= h′(x)δ = h′(x)

λ ολ (1+δ)ο

If the desired focallength of this lens isf , the wave must beredirected at an anglewhich, depends on thedistance from the op-tical axis

α(x) =x

combining, we have

λoh′(x)

f−→ h′(x)

f λothis can be easily integrated

2f λo=

2f λo

λ ολ (1+δ)ο

α(x) =x

combining, we have

λoh′(x)

f−→ h′(x)

2f λo=

2f λo

λ ολ (1+δ)ο

α(x) =x

combining, we have

λoh′(x)

−→ h′(x)

2f λo=

2f λo

λ ολ (1+δ)ο

α(x) =x

combining, we have

λoh′(x)

f−→ h′(x)

this can be easily integrated

2f λo=

2f λo

λ ολ (1+δ)ο

α(x) =x

combining, we have

λoh′(x)

f−→ h′(x)

2f λo=

2f λo

λ ολ (1+δ)ο

α(x) =x

combining, we have

λoh′(x)

f−→ h′(x)

2f λo

λ ολ (1+δ)ο

α(x) =x

combining, we have

λoh′(x)

f−→ h′(x)

2f λo=

2f λo

]2C. Segre (IIT) PHYS 570 - Fall 2013 September 25, 2013 3 / 20

Focal length of a compound lens

From the previous expression for the ideal parabolic surface, the focallength can be written in terms of the surface profile.

f =x2Λ

2λoh(x)=

h(x)or alternatively f =

h′(x)

If the surface is a circle instead of a parabola

h(x) =√R2 − x2 and for x � R h(x) ≈ R

(1− 1

x2 = R2 − h2(x) ≈ R2 so we have f ≈ R

for N circular lenses

fn ≈R

f =x2Λ

2λoh(x)

h′(x)

(1− 1

fn ≈R

f =x2Λ

2λoh(x)=

or alternatively f =1

h′(x)

(1− 1

fn ≈R

f =x2Λ

2λoh(x)=

h′(x)

(1− 1

fn ≈R

f =x2Λ

2λoh(x)=

h′(x)

(1− 1

fn ≈R

f =x2Λ

2λoh(x)=

h′(x)

h(x) =√R2 − x2

and for x � R h(x) ≈ R

(1− 1

fn ≈R

f =x2Λ

2λoh(x)=

h′(x)

(1− 1

fn ≈R

f =x2Λ

2λoh(x)=

h′(x)

(1− 1

x2 = R2 − h2(x) ≈ R2

so we have f ≈ R

fn ≈R

f =x2Λ

2λoh(x)=

h′(x)

(1− 1

fn ≈R

f =x2Λ

2λoh(x)=

h′(x)

(1− 1

fn ≈R

Focussing by a beryllium lens

unit lensf

For 50 holes of radius R = 1mm in beryllium (Be) at E = 10keV, we cancalculate the focal length, knowing δ = 3.41× 10−6

1× 10−3m

2(50)(3.41× 10−6)= 2.93m

depending on the wall thickness of the lenslets, the transmission can be upto 74%

H.R. Beguiristain, J.T. Cremer, M.A. Piestrup, C.K. Gary, and R.H. Pantell, Optics Letters, 27, 778 (2007).

unit lensf

=1× 10−3m

2(50)(3.41× 10−6)= 2.93m

unit lensf

1× 10−3m

2(50)(3.41× 10−6)

= 2.93m

unit lensf

1× 10−3m

2(50)(3.41× 10−6)= 2.93m

unit lensf

1× 10−3m

2(50)(3.41× 10−6)= 2.93m

unit lensf

1× 10−3m

2(50)(3.41× 10−6)= 2.93m

Alligator-type lenses

Perhaps one of the most original x-ray lenses has been made by using oldvinyl records in an “alligator” configuration.

Bjorn Cederstrom et al., “Focusing hard X-rays with old LPs”, Nature 404, 951 (2000).

This design has also beenused to make lenses out oflithium metal.

E.M. Dufresne et al., “Lithium metal for x-ray refractive optics”, Appl. Phys. Lett. 79, 4085 (2001).

Alligator-type lenses

Perhaps one of the most original x-ray lenses has been made by using oldvinyl records in an “alligator” configuration.

Bjorn Cederstrom et al., “Focusing hard X-rays with old LPs”, Nature 404, 951 (2000).

This design has also beenused to make lenses out oflithium metal.

E.M. Dufresne et al., “Lithium metal for x-ray refractive optics”, Appl. Phys. Lett. 79, 4085 (2001).

How to make a Fresnel lens

The ideal refracting lens has a parabolicshape (actually elliptical) but this is imprac-tical to make.

h(x) = Λ

2λo f

when h(x) = 100Λ ∼ 1000µm

x = 10√

2λo f ∼ 100µm

aspect ratio too large for a stable structureand absorption would be too large!

h(x) = Λ

2λo f

when h(x) = 100Λ ∼ 1000µm

x = 10√

2λo f ∼ 100µm

h(x) = Λ

2λo f

when h(x) = 100Λ ∼ 1000µm

x = 10√

2λo f ∼ 100µm

h(x) = Λ

2λo f

when h(x) = 100Λ ∼ 1000µm

x = 10√

2λo f ∼ 100µm

h(x) = Λ

2λo f

when h(x) = 100Λ ∼ 1000µm

x = 10√

2λo f ∼ 100µm

7ΛMark off the longitudinal zones (of thicknessΛ) where the waves inside and outside thematerial are in phase.

Each block of thickness Λ serves nopurpose for refraction but only attenuatesthe wave.

This material can be removed and theremaining material collapsed to producea Fresnel lens which has the same opticalproperties as the parabolic lens as long asf � NΛ where N is the number of zones.

7 Mark off the longitudinal zones (of thicknessΛ) where the waves inside and outside thematerial are in phase.

Fresnel lens dimensions

The outermost zones become very small andthus limit the overall aperture of the zoneplate. The dimensions of outermost zone, Ncan be calculated by first defining a scaledheight and lateral dimension

ν =h(x)

Λξ =

x√2λo f

Since ν = ξ2, the position of the Nth zoneis ξN =

√N and the scaled width of the Nth

(outermost) zone is

∆ξN = ξN − ξN−1 =√N −

√N − 1

ν =h(x)

ξ =x√

2λo f

(outermost) zone is

∆ξN = ξN − ξN−1 =√N −

√N − 1

ν =h(x)

Λξ =

x√2λo f

(outermost) zone is

∆ξN = ξN − ξN−1 =√N −

√N − 1

ν =h(x)

Λξ =

x√2λo f

(outermost) zone is

∆ξN = ξN − ξN−1 =√N −

√N − 1

ν =h(x)

Λξ =

x√2λo f

(outermost) zone is

∆ξN = ξN − ξN−1 =√N −

√N − 1

∆ξN = ξN − ξN−1 =√N −

√N − 1

√1− 1

≈√N

[1− 1

])∆ξN ≈

The diameter of the entire lens is thus

2ξN = 2√N =

∆ξN

∆ξN = ξN − ξN−1 =√N −

√N − 1

√1− 1

≈√N

[1− 1

])∆ξN ≈

2ξN = 2√N =

∆ξN

∆ξN = ξN − ξN−1 =√N −

√N − 1

√1− 1

≈√N

[1− 1

∆ξN ≈1

2ξN = 2√N =

∆ξN

∆ξN = ξN − ξN−1 =√N −

√N − 1

√1− 1

≈√N

[1− 1

])∆ξN ≈

2ξN = 2√N =

∆ξN

∆ξN = ξN − ξN−1 =√N −

√N − 1

√1− 1

≈√N

[1− 1

])∆ξN ≈

2ξN = 2√N =

∆ξN

Fresnel lens example

In terms of the unscaled variables

∆xN = ∆ξN√

2λo f

√λo f

dN = 2ξN =

√2λo f

∆ξN= 2√N√

2λo f =√

2Nλo f

If we take

λo = 1A = 1× 10−10m

f = 50cm = 0.5m

N = 100

∆xN = 5× 10−7m = 500nm dN = 2× 10−4m = 100µm

∆xN = ∆ξN√

2λo f =

√λo f

dN = 2ξN =

√2λo f

∆ξN= 2√N√

2λo f =√

2Nλo f

If we take

λo = 1A = 1× 10−10m

f = 50cm = 0.5m

N = 100

∆xN = 5× 10−7m = 500nm dN = 2× 10−4m = 100µm

∆xN = ∆ξN√

2λo f =

√λo f

dN = 2ξN

√2λo f

∆ξN= 2√N√

2λo f =√

2Nλo f

If we take

λo = 1A = 1× 10−10m

f = 50cm = 0.5m

N = 100

∆xN = 5× 10−7m = 500nm dN = 2× 10−4m = 100µm

∆xN = ∆ξN√

2λo f =

√λo f

dN = 2ξN =

√2λo f

∆ξN

= 2√N√

2λo f =√

2Nλo f

If we take

λo = 1A = 1× 10−10m

f = 50cm = 0.5m

N = 100

∆xN = 5× 10−7m = 500nm dN = 2× 10−4m = 100µm

∆xN = ∆ξN√

2λo f =

√λo f

dN = 2ξN =

√2λo f

∆ξN= 2√N√

2λo f =√

2Nλo f

If we take

λo = 1A = 1× 10−10m

f = 50cm = 0.5m

N = 100

∆xN = 5× 10−7m = 500nm dN = 2× 10−4m = 100µm

∆xN = ∆ξN√

2λo f =

√λo f

dN = 2ξN =

√2λo f

∆ξN= 2√N√

2λo f =√

2Nλo f

If we take

λo = 1A = 1× 10−10m

f = 50cm = 0.5m

N = 100

∆xN = 5× 10−7m = 500nm dN = 2× 10−4m = 100µm

∆xN = ∆ξN√

2λo f =

√λo f

dN = 2ξN =

√2λo f

∆ξN= 2√N√

2λo f =√

2Nλo f

If we take

λo = 1A = 1× 10−10m

f = 50cm = 0.5m

N = 100

∆xN = 5× 10−7m = 500nm

dN = 2× 10−4m = 100µm

∆xN = ∆ξN√

2λo f =

√λo f

dN = 2ξN =

√2λo f

∆ξN= 2√N√

2λo f =√

2Nλo f

If we take

λo = 1A = 1× 10−10m

f = 50cm = 0.5m

N = 100

∆xN = 5× 10−7m = 500nm dN = 2× 10−4m = 100µm

Making a Fresnel zone plate

The specific shape required for a zone plateis difficult to fabricate, consequently, itis convenient to approximate the nearlytriangular zones with a rectangular profile.

In practice, since the outermost zonesare very small, zone plates are generallyfabricated as alternating zones (rings for2D) of materials with a large Z-contrast,such as Au/Si or W/C.

This kind of zone plate is not as effi-cient as a true Fresnel lens would be in thex-ray regime. Nevertheless, efficiencies upto 35% have been achieved.

Variable focal length CRL

The compound refractive lenses (CRL) are useful for fixed focus but aredifficult to use if a variable focal distance and a long focal length isrequired.

Start with a 2 hole CRL. Rotate byan angle χ about vertical axis giv-ing an effective change in the num-ber of “lenses” by a factor 1/ cosχ.

at E = 5.5keV and χ = 0◦, heightis over 120µm

At χ = 30◦, it is under 50µm

Optimal focus is 20µm at χ = 40◦

B. Adams and C. Rose-Petruck, “Hybrid len/mirror x-ray focusing scheme and beam stabilization”, in prepration (2013).

Start with a 2 hole CRL.

Rotate byan angle χ about vertical axis giv-ing an effective change in the num-ber of “lenses” by a factor 1/ cosχ.

Start with a 2 hole CRL. Rotate byan angle χ about vertical axis

giv-ing an effective change in the num-ber of “lenses” by a factor 1/ cosχ.

Start with a 2 hole CRL. Rotate byan angle χ about vertical axis giv-ing an effective change in the num-ber of “lenses” by a factor 1/ cosχ.

The compound refractive lenses (CRL) are useful for fixed focus but aredifficult to use if a variable focal distance and a long focal length isrequired.Start with a 2 hole CRL. Rotate byan angle χ about vertical axis giv-ing an effective change in the num-ber of “lenses” by a factor 1/ cosχ.

Zone plate fabrication

Making high aspect ratio zone plates is challenging but a new process hasbeen developed to make plates with an aspect ratio as high as 25.

Start with Ultra nano crystallinediamond (UNCD) films on SiN.Coat with hydrogen silsesquioxane(HSQ). Pattern and develop theHSQ layer. Reactive ion etch theUNCD to the substrate. Plate withgold to make final zone plate.

The whole 150nm diameter zoneplate

Detail view of outer zones

M. Wojick et al., “High Aspect Ratio Zone Plate Fabrication Using a Bilayer Mold”, EIBPN 2011.

Start with Ultra nano crystallinediamond (UNCD) films on SiN.

Coat with hydrogen silsesquioxane(HSQ). Pattern and develop theHSQ layer. Reactive ion etch theUNCD to the substrate. Plate withgold to make final zone plate.

Start with Ultra nano crystallinediamond (UNCD) films on SiN.Coat with hydrogen silsesquioxane(HSQ).

Pattern and develop theHSQ layer. Reactive ion etch theUNCD to the substrate. Plate withgold to make final zone plate.

Start with Ultra nano crystallinediamond (UNCD) films on SiN.Coat with hydrogen silsesquioxane(HSQ). Pattern and develop theHSQ layer.

Reactive ion etch theUNCD to the substrate. Plate withgold to make final zone plate.

Start with Ultra nano crystallinediamond (UNCD) films on SiN.Coat with hydrogen silsesquioxane(HSQ). Pattern and develop theHSQ layer. Reactive ion etch theUNCD to the substrate.

Plate withgold to make final zone plate.

Scattering from two electrons

Consider systems where there is only weak scattering, with no multiplescattering effects. We begin with the scattering of x-rays from twoelectrons.

~Q = (~k − ~k ′)

|~Q| = 2k sin θ =4π

λsin θ

The scattering from the second electron willhave a phase shift of φ = ~Q ·~r .

A(~Q) = −ro(

1 + e i~Q·~r)

I (~Q) = A(~Q)A(~Q)∗ = r2o

(1 + e i

~Q·~r)(

1 + e−i~Q·~r)

= 2r2o

(1 + cos(~Q ·~r)

~Q = (~k − ~k ′)

|~Q| = 2k sin θ =4π

λsin θ

A(~Q) = −ro(

1 + e i~Q·~r)

I (~Q) = A(~Q)A(~Q)∗ = r2o

(1 + e i

~Q·~r)(

1 + e−i~Q·~r)

= 2r2o

(1 + cos(~Q ·~r)

~Q = (~k − ~k ′)

|~Q| = 2k sin θ =4π

λsin θ

A(~Q) = −ro(

1 + e i~Q·~r)

I (~Q) = A(~Q)A(~Q)∗ = r2o

(1 + e i

~Q·~r)(

1 + e−i~Q·~r)

= 2r2o

(1 + cos(~Q ·~r)

~Q = (~k − ~k ′)

|~Q| = 2k sin θ =4π

λsin θ

A(~Q) = −ro(

1 + e i~Q·~r)

I (~Q) = A(~Q)A(~Q)∗ = r2o

(1 + e i

~Q·~r)(

1 + e−i~Q·~r)

= 2r2o

(1 + cos(~Q ·~r)

~Q = (~k − ~k ′)

|~Q| = 2k sin θ =4π

λsin θ

A(~Q) = −ro(

1 + e i~Q·~r)

I (~Q) = A(~Q)A(~Q)∗ = r2o

(1 + e i

~Q·~r)(

1 + e−i~Q·~r)

= 2r2o

(1 + cos(~Q ·~r)

~Q = (~k − ~k ′)

|~Q| = 2k sin θ =4π

λsin θ

A(~Q) = −ro(

1 + e i~Q·~r)

I (~Q) = A(~Q)A(~Q)∗ = r2o

(1 + e i

~Q·~r)(

1 + e−i~Q·~r)

= 2r2o

(1 + cos(~Q ·~r)

~Q = (~k − ~k ′)

|~Q| = 2k sin θ =4π

λsin θ

A(~Q) = −ro(

1 + e i~Q·~r)

I (~Q) = A(~Q)A(~Q)∗ = r2o

(1 + e i

~Q·~r)(

1 + e−i~Q·~r)

= 2r2o

(1 + cos(~Q ·~r)

~Q = (~k − ~k ′)

|~Q| = 2k sin θ =4π

λsin θ

A(~Q) = −ro(

1 + e i~Q·~r)

I (~Q) = A(~Q)A(~Q)∗ = r2o

(1 + e i

~Q·~r)(

1 + e−i~Q·~r)

= 2r2o

(1 + cos(~Q ·~r)

~Q = (~k − ~k ′)

|~Q| = 2k sin θ =4π

λsin θ

A(~Q) = −ro(

1 + e i~Q·~r)

I (~Q) = A(~Q)A(~Q)∗ = r2o

(1 + e i

~Q·~r)(

1 + e−i~Q·~r)

= 2r2o

(1 + cos(~Q ·~r)

)C. Segre (IIT) PHYS 570 - Fall 2013 September 25, 2013 15 / 20

Scattering from many electrons

for many electrons

generalizing to a crystal

A(~Q) = −ro∑j

e i~Q·~rj

A(~Q) = −ro∑N

e i~Q· ~RN

e i~Q·~rj

Since experiments measure I ∝ A2, the phase information is lost. This is aproblem if we don’t know the specific orientation of the scattering systemrelative to the x-ray beam.

We will now look at the consequences of this orientation and generalize tomore than two electrons

for many electrons

A(~Q) = −ro∑j

e i~Q·~rj

A(~Q) = −ro∑N

e i~Q· ~RN

e i~Q·~rj

for many electrons

A(~Q) = −ro∑j

e i~Q·~rj

A(~Q) = −ro∑N

e i~Q· ~RN

e i~Q·~rj

for many electrons

A(~Q) = −ro∑j

e i~Q·~rj

A(~Q) = −ro∑N

e i~Q· ~RN

e i~Q·~rj

for many electrons

A(~Q) = −ro∑j

e i~Q·~rj

A(~Q) = −ro∑N

e i~Q· ~RN

e i~Q·~rj

for many electrons

A(~Q) = −ro∑j

e i~Q·~rj

A(~Q) = −ro∑N

e i~Q· ~RN

e i~Q·~rj

Two electrons — fixed orientation

The expression

I (~Q) = 2r2o

(1 + cos(~Q ·~r)

)assumes that the two electronshave a specific, fixed orienta-tion. In this case the intensityas a function of Q is.

Fixed orientation is not theusual case, particularly for solu-tion and small-angle scattering. 0 0.5 1 1.5 2

Q (units of 2π/r)

nsity (

units o

f r2 o

The expression

I (~Q) = 2r2o

(1 + cos(~Q ·~r)

Fixed orientation is not theusual case, particularly for solu-tion and small-angle scattering.

0 0.5 1 1.5 2

Q (units of 2π/r)

nsity (

units o

f r2 o

The expression

I (~Q) = 2r2o

(1 + cos(~Q ·~r)

Fixed orientation is not theusual case, particularly for solu-tion and small-angle scattering. 0 0.5 1 1.5 2

Q (units of 2π/r)

nsity (

units o

f r2 o

Orientation averaging

A(~Q) = f1 + f2ei ~Q·~r

I (~Q) = f 21 + f 22 + f1f2ei ~Q·~r + f1f2e

−i ~Q·~r

⟨I (~Q)

⟩= f 21 + f 22 + 2f1f2

⟨e i~Q·~r⟩

∫e iQr cos θ sin θdθdφ∫

sin θdθdφ

4π2π

∫ π

0e iQr cos θ sin θdθ

(− 1

)∫ −iQr

iQrexdx

sin(Qr)

Consider scattering from twoarbitrary electron distribu-tions, f1 and f2. A(~Q), isgiven by

and the intensity, I (~Q), is

if the distance between thescatterers,~r , remains constant(no vibrations) but is allowedto orient randomly in space⟨I (~Q)

⟩= f 21 +f 22 +2f1f2

sin(Qr)

A(~Q) = f1 + f2ei ~Q·~r

I (~Q) = f 21 + f 22 + f1f2ei ~Q·~r + f1f2e

−i ~Q·~r

⟨I (~Q)

⟩= f 21 + f 22 + 2f1f2

⟨e i~Q·~r⟩

sin θdθdφ

4π2π

∫ π

(− 1

)∫ −iQr

iQrexdx

sin(Qr)

⟩= f 21 +f 22 +2f1f2

sin(Qr)

A(~Q) = f1 + f2ei ~Q·~r

I (~Q) = f 21 + f 22 + f1f2ei ~Q·~r + f1f2e

−i ~Q·~r

⟨I (~Q)

⟩= f 21 + f 22 + 2f1f2

⟨e i~Q·~r⟩

sin θdθdφ

4π2π

∫ π

(− 1

)∫ −iQr

iQrexdx

sin(Qr)

⟩= f 21 +f 22 +2f1f2

sin(Qr)

A(~Q) = f1 + f2ei ~Q·~r

I (~Q) = f 21 + f 22 + f1f2ei ~Q·~r + f1f2e

−i ~Q·~r

⟨I (~Q)

⟩= f 21 + f 22 + 2f1f2

⟨e i~Q·~r⟩

sin θdθdφ

4π2π

∫ π

(− 1

)∫ −iQr

iQrexdx

sin(Qr)

⟩= f 21 +f 22 +2f1f2

sin(Qr)

A(~Q) = f1 + f2ei ~Q·~r

I (~Q) = f 21 + f 22 + f1f2ei ~Q·~r + f1f2e

−i ~Q·~r

⟨I (~Q)

⟩= f 21 + f 22 + 2f1f2

⟨e i~Q·~r⟩

sin θdθdφ

4π2π

∫ π

(− 1

)∫ −iQr

iQrexdx

sin(Qr)

if the distance between thescatterers,~r , remains constant(no vibrations) but is allowedto orient randomly in space

⟨I (~Q)

⟩= f 21 +f 22 +2f1f2

sin(Qr)

A(~Q) = f1 + f2ei ~Q·~r

I (~Q) = f 21 + f 22 + f1f2ei ~Q·~r + f1f2e

−i ~Q·~r

⟨I (~Q)

⟩= f 21 + f 22 + 2f1f2

⟨e i~Q·~r⟩

sin θdθdφ

4π2π

∫ π

(− 1

)∫ −iQr

iQrexdx

sin(Qr)

⟨I (~Q)

⟩= f 21 +f 22 +2f1f2

sin(Qr)

A(~Q) = f1 + f2ei ~Q·~r

I (~Q) = f 21 + f 22 + f1f2ei ~Q·~r + f1f2e

−i ~Q·~r

⟨I (~Q)

⟩= f 21 + f 22 + 2f1f2

⟨e i~Q·~r⟩

sin θdθdφ

4π2π

∫ π

(− 1

)∫ −iQr

iQrexdx

sin(Qr)

⟨I (~Q)

⟩= f 21 +f 22 +2f1f2

sin(Qr)

A(~Q) = f1 + f2ei ~Q·~r

I (~Q) = f 21 + f 22 + f1f2ei ~Q·~r + f1f2e

−i ~Q·~r

⟨I (~Q)

⟩= f 21 + f 22 + 2f1f2

⟨e i~Q·~r⟩

sin θdθdφ

4π2π

∫ π

(− 1

)∫ −iQr

iQrexdx

sin(Qr)

⟨I (~Q)

⟩= f 21 +f 22 +2f1f2

sin(Qr)

A(~Q) = f1 + f2ei ~Q·~r

I (~Q) = f 21 + f 22 + f1f2ei ~Q·~r + f1f2e

−i ~Q·~r

⟨I (~Q)

⟩= f 21 + f 22 + 2f1f2

⟨e i~Q·~r⟩

sin θdθdφ

4π2π

∫ π

(− 1

)∫ −iQr

iQrexdx

sin(Qr)

⟨I (~Q)

⟩= f 21 +f 22 +2f1f2

sin(Qr)

A(~Q) = f1 + f2ei ~Q·~r

I (~Q) = f 21 + f 22 + f1f2ei ~Q·~r + f1f2e

−i ~Q·~r

⟨I (~Q)

⟩= f 21 + f 22 + 2f1f2

⟨e i~Q·~r⟩

sin θdθdφ

4π2π

∫ π

(− 1

)∫ −iQr

iQrexdx

sin(Qr)

=sin(Qr)

⟨I (~Q)

⟩= f 21 +f 22 +2f1f2

sin(Qr)

A(~Q) = f1 + f2ei ~Q·~r

I (~Q) = f 21 + f 22 + f1f2ei ~Q·~r + f1f2e

−i ~Q·~r

⟨I (~Q)

⟩= f 21 + f 22 + 2f1f2

⟨e i~Q·~r⟩

sin θdθdφ

4π2π

∫ π

(− 1

)∫ −iQr

iQrexdx

sin(Qr)

⟨I (~Q)

⟩= f 21 +f 22 +2f1f2

sin(Qr)

A(~Q) = f1 + f2ei ~Q·~r

I (~Q) = f 21 + f 22 + f1f2ei ~Q·~r + f1f2e

−i ~Q·~r

⟨I (~Q)

⟩= f 21 + f 22 + 2f1f2

⟨e i~Q·~r⟩

sin θdθdφ

4π2π

∫ π

(− 1

)∫ −iQr

iQrexdx

sin(Qr)

⟩= f 21 +f 22 +2f1f2

sin(Qr)

Randomly oriented electrons

Recall that when we had a fixedorientation of the two electrons,we had and intensity variation ascos(Qr).

When we now replace the twoarbitrary scattering distributionswith electrons (f1, f2 → −ro),we change the intensity profilesignificantly.

0 0.5 1 1.5 2

Q (units of 2π/r)

nsity (

units o

f r2 o

0 0.5 1 1.5 2

Q (units of 2π/r)

nsity (

units o

f r2 o

0 0.5 1 1.5 2

Q (units of 2π/r)

nsity (

units o

f r2 o

Scattering from atoms

Single electrons are a good first example but a real system involvesscattering from atoms. We can use what we have already used to write anexpression for the scattering from an atom, ignoring the anomalous terms.

with limits in units of ro of

The second limit can be understoodclassically as loss of phase coher-ence when Q is very large and asmall difference in position resultsin an arbitrary change in phase.

ψ1sr =1√πa3

e−r/a a =ao

Z − zs

where zs is a screening correction

f 0(~Q) =

∫ρ(~r)e i

~Q·~rd~r

{Z for Q→ 0

0 for Q→∞

However, it is better to usequantum mechanics to calculatethe form factor, starting with ahydrogen-like atom as an example.The wave function of the 1s elec-trons is just

ψ1sr =1√πa3

e−r/a a =ao

Z − zs

f 0(~Q) =

∫ρ(~r)e i

~Q·~rd~r

{Z for Q→ 0

0 for Q→∞

ψ1sr =1√πa3

e−r/a a =ao

Z − zs

f 0(~Q) =

∫ρ(~r)e i

~Q·~rd~r

{Z for Q→ 0

0 for Q→∞

ψ1sr =1√πa3

e−r/a a =ao

Z − zs

f 0(~Q) =

∫ρ(~r)e i

~Q·~rd~r

{Z for Q→ 0

0 for Q→∞

ψ1sr =1√πa3

e−r/a a =ao

Z − zs

f 0(~Q) =

∫ρ(~r)e i

~Q·~rd~r

{Z for Q→ 0

0 for Q→∞

ψ1sr =1√πa3

e−r/a a =ao

Z − zs

f 0(~Q) =

∫ρ(~r)e i

~Q·~rd~r

{Z for Q→ 0

0 for Q→∞

ψ1sr =1√πa3

e−r/a a =ao

Z − zs

f 0(~Q) =

∫ρ(~r)e i

~Q·~rd~r

{Z for Q→ 0

0 for Q→∞

ψ1sr =1√πa3

e−r/a a =ao

Z − zs

f 0(~Q) =

∫ρ(~r)e i

~Q·~rd~r

{Z for Q→ 0

0 for Q→∞

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