Today’s Outline - February 13, 2018phys.iit.edu/~segre/phys570/18S/lecture_11.pdf · Today’s Outline - February 13, 2018 Ideal refractive surface Fresnel lenses and zone plates

$Page 1: Today’s Outline - February 13, 2018phys.iit.edu/~segre/phys570/18S/lecture_11.pdf · Today’s Outline - February 13, 2018 Ideal refractive surface Fresnel lenses and zone plates$
Today’s Outline - February 13, 2018

• Ideal refractive surface

• Fresnel lenses and zone plates

• Research papers on refraction

• Kinematical diffraction

Homework Assignment #02:Problems on Blackboarddue Tuesday, February 13, 2018

C. Segre (IIT) PHYS 570 - Spring 2018 February 13, 2018 1 / 23




































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

1

i+

1

o=

1

f→ 1

i=

1

f− 1

o

1

i1=

1

f1− 1

o1→ 1

i1=

1

f→ i1 = f

1

i2=

1

f2− 1

o2→ 1

i2=

1

f+

1

f→ i2 =

f

2

1

i2=

1

f2− 1

o2→ 1

i2=

1

f+

2

f→ i2 =

f

3

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

1

i+

1

o=

1

f

→ 1

i=

1

f− 1

o

1

i1=

1

f1− 1

o1→ 1

i1=

1

f→ i1 = f

1

i2=

1

f2− 1

o2→ 1

i2=

1

f+

1

f→ i2 =

f

2

1

i2=

1

f2− 1

o2→ 1

i2=

1

f+

2

f→ i2 =

f

3

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

f1 = f , o1 =∞






f1 f2 f3

1

i+

1

o=

1

f→ 1

i=

1

f− 1

o

1

i1=

1

f1− 1

o1→ 1

i1=

1

f→ i1 = f

1

i2=

1

f2− 1

o2→ 1

i2=

1

f+

1

f→ i2 =

f

2

1

i2=

1

f2− 1

o2→ 1

i2=

1

f+

2

f→ i2 =

f

3


f1 = f , o1 =∞






f1 f2 f3

1

i+

1

o=

1

f→ 1

i=

1

f− 1

o

1

i1=

1

f1− 1

o1

→ 1

i1=

1

f→ i1 = f

1

i2=

1

f2− 1

o2→ 1

i2=

1

f+

1

f→ i2 =

f

2

1

i2=

1

f2− 1

o2→ 1

i2=

1

f+

2

f→ i2 =

f

3


f1 = f , o1 =∞






f1 f2 f3

1

i+

1

o=

1

f→ 1

i=

1

f− 1

o

1

i1=

1

f1− 1

o1→ 1

i1=

1

f

→ i1 = f

1

i2=

1

f2− 1

o2→ 1

i2=

1

f+

1

f→ i2 =

f

2

1

i2=

1

f2− 1

o2→ 1

i2=

1

f+

2

f→ i2 =

f

3


f1 = f , o1 =∞






f1 f2 f3

1

i+

1

o=

1

f→ 1

i=

1

f− 1

o

1

i1=

1

f1− 1

o1→ 1

i1=

1

f→ i1 = f

1

i2=

1

f2− 1

o2→ 1

i2=

1

f+

1

f→ i2 =

f

2

1

i2=

1

f2− 1

o2→ 1

i2=

1

f+

2

f→ i2 =

f

3


f1 = f , o1 =∞






f1 f2 f3

1

i+

1

o=

1

f→ 1

i=

1

f− 1

o

1

i1=

1

f1− 1

o1→ 1

i1=

1

f→ i1 = f

1

i2=

1

f2− 1

o2

→ 1

i2=

1

f+

1

f→ i2 =

f

2

1

i2=

1

f2− 1

o2→ 1

i2=

1

f+

2

f→ i2 =

f

3


f1 = f , o1 =∞






f1 f2 f3

1

i+

1

o=

1

f→ 1

i=

1

f− 1

o

1

i1=

1

f1− 1

o1→ 1

i1=

1

f→ i1 = f

1

i2=

1

f2− 1

o2→ 1

i2=

1

f+

1

f

→ i2 =f

2

1

i2=

1

f2− 1

o2→ 1

i2=

1

f+

2

f→ i2 =

f

3


f1 = f , o1 =∞






f1 f2 f3

1

i+

1

o=

1

f→ 1

i=

1

f− 1

o

1

i1=

1

f1− 1

o1→ 1

i1=

1

f→ i1 = f

1

i2=

1

f2− 1

o2→ 1

i2=

1

f+

1

f→ i2 =

f

2

1

i2=

1

f2− 1

o2→ 1

i2=

1

f+

2

f→ i2 =

f

3


f1 = f , o1 =∞






f1 f2 f3

1

i+

1

o=

1

f→ 1

i=

1

f− 1

o

1

i1=

1

f1− 1

o1→ 1

i1=

1

f→ i1 = f

1

i2=

1

f2− 1

o2→ 1

i2=

1

f+

1

f→ i2 =

f

2

1

i2=

1

f2− 1

o2

→ 1

i2=

1

f+

2

f→ i2 =

f

3


f1 = f , o1 =∞






f1 f2 f3

1

i+

1

o=

1

f→ 1

i=

1

f− 1

o

1

i1=

1

f1− 1

o1→ 1

i1=

1

f→ i1 = f

1

i2=

1

f2− 1

o2→ 1

i2=

1

f+

1

f→ i2 =

f

2

1

i2=

1

f2− 1

o2→ 1

i2=

1

f+

2

f

→ i2 =f

3


f1 = f , o1 =∞






f1 f2 f3

1

i+

1

o=

1

f→ 1

i=

1

f− 1

o

1

i1=

1

f1− 1

o1→ 1

i1=

1

f→ i1 = f

1

i2=

1

f2− 1

o2→ 1

i2=

1

f+

1

f→ i2 =

f

2

1

i2=

1

f2− 1

o2→ 1

i2=

1

f+

2

f→ i2 =

f

3


f1 = f , o1 =∞






f1 f2 f3

1

i+

1

o=

1

f→ 1

i=

1

f− 1

o

1

i1=

1

f1− 1

o1→ 1

i1=

1

f→ i1 = f

1

i2=

1

f2− 1

o2→ 1

i2=

1

f+

1

f→ i2 =

f

2

1

i2=

1

f2− 1

o2→ 1

i2=

1

f+

2

f→ i2 =

f

3


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.

Λ

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

δ

Λ = Nλ0 =λ0δ

=2π

λ0r0ρ≈ 10µm


Rephasing distance


Λ

λο λ (1+δ)ο



Λ = (N + 1)λ0 = Nλ0(1 + δ)

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

δ

Λ = Nλ0 =λ0δ

=2π

λ0r0ρ≈ 10µm


Rephasing distance


Λ

λο λ (1+δ)ο



Λ = (N + 1)λ0 = Nλ0(1 + δ)

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

δ

Λ = Nλ0 =λ0δ

=2π

λ0r0ρ≈ 10µm


Rephasing distance


Λ

λο λ (1+δ)ο



Λ = (N + 1)λ0 = Nλ0(1 + δ)

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

δ

Λ = Nλ0 =λ0δ

=2π

λ0r0ρ≈ 10µm


Rephasing distance


Λ

λο λ (1+δ)ο



Λ = (N + 1)λ0 = Nλ0(1 + δ)

Nλ0 + λ0 = Nλ0 + Nδλ0

−→ λ0 = Nδλ0 −→ N =1

δ

Λ = Nλ0 =λ0δ

=2π

λ0r0ρ≈ 10µm


Rephasing distance


Λ

λο λ (1+δ)ο



Λ = (N + 1)λ0 = Nλ0(1 + δ)

Nλ0 + λ0 = Nλ0 + Nδλ0 −→ λ0 = Nδλ0

−→ N =1

δ

Λ = Nλ0 =λ0δ

=2π

λ0r0ρ≈ 10µm


Rephasing distance


Λ

λο λ (1+δ)ο



Λ = (N + 1)λ0 = Nλ0(1 + δ)

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

δ

Λ = Nλ0 =λ0δ

=2π

λ0r0ρ≈ 10µm


Rephasing distance


Λ

λο λ (1+δ)ο



Λ = (N + 1)λ0 = Nλ0(1 + δ)

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

δ

Λ = Nλ0 =λ0δ

=2π

λ0r0ρ≈ 10µm


Rephasing distance


Λ

λο λ (1+δ)ο



Λ = (N + 1)λ0 = Nλ0(1 + δ)

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

δ

Λ = Nλ0 =λ0δ

=2π

λ0r0ρ

≈ 10µm


Rephasing distance


Λ

λο λ (1+δ)ο



Λ = (N + 1)λ0 = Nλ0(1 + δ)

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

δ

Λ = Nλ0 =λ0δ

=2π

λ0r0ρ≈ 10µm


Ideal interface profile

α

+x0

h(x)

λ ολ (1+δ)ο

∆x

A’A

B

B’

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 ≈ λ0h′(x)

α(x) ≈ λ0δ

∆x= h′(x)δ = h′(x)

λ0Λ



α

+x0

h(x)

λ ολ (1+δ)ο

∆x

A’A

B

B’

α

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 ≈ λ0h′(x)

α(x) ≈ λ0δ

∆x= h′(x)δ = h′(x)

λ0Λ



α

+x0

h(x)

λ 0 λ (1+δ)0

∆x

A’A

B

B’

C




using Λ = λ0/δ

λ0 (1 + δ) = h′(x)∆x −→ ∆x ≈ λ0h′(x)

α(x) ≈ λ0δ

∆x= h′(x)δ = h′(x)

λ0Λ



α

+x0

h(x)

λ 0 λ (1+δ)0

∆x

A’A

B

B’

C




using Λ = λ0/δ

λ0 (1 + δ) = h′(x)∆x −→ ∆x ≈ λ0h′(x)

α(x) ≈ λ0δ

∆x= h′(x)δ = h′(x)

λ0Λ



α

+x0

h(x)

λ 0 λ (1+δ)0

∆x

A’A

B

B’

C




using Λ = λ0/δ

λ0 (1 + δ) = h′(x)∆x

−→ ∆x ≈ λ0h′(x)

α(x) ≈ λ0δ

∆x= h′(x)δ = h′(x)

λ0Λ



α

+x0

h(x)

λ 0 λ (1+δ)0

∆x

A’A

B

B’

C




using Λ = λ0/δ

λ0 (1 + δ) = h′(x)∆x −→ ∆x ≈ λ0h′(x)

α(x) ≈ λ0δ

∆x= h′(x)δ = h′(x)

λ0Λ



α

+x0

h(x)

λ 0 λ (1+δ)0

∆x

A’A

B

B’

C




using Λ = λ0/δ

λ0 (1 + δ) = h′(x)∆x −→ ∆x ≈ λ0h′(x)

α(x) ≈ λ0δ

∆x= h′(x)δ = h′(x)

λ0Λ



α

+x0

h(x)

λ 0 λ (1+δ)0

∆x

A’A

B

B’

C




using Λ = λ0/δ

λ0 (1 + δ) = h′(x)∆x −→ ∆x ≈ λ0h′(x)

α(x) ≈ λ0δ

∆x

= h′(x)δ = h′(x)λ0Λ



α

+x0

h(x)

λ 0 λ (1+δ)0

∆x

A’A

B

B’

C




using Λ = λ0/δ

λ0 (1 + δ) = h′(x)∆x −→ ∆x ≈ λ0h′(x)

α(x) ≈ λ0δ

∆x= h′(x)δ

= h′(x)λ0Λ



α

+x0

h(x)

λ 0 λ (1+δ)0

∆x

A’A

B

B’

C




using Λ = λ0/δ

λ0 (1 + δ) = h′(x)∆x −→ ∆x ≈ λ0h′(x)

α(x) ≈ λ0δ

∆x= h′(x)δ

= h′(x)λ0Λ



α

+x0

h(x)

λ 0 λ (1+δ)0

∆x

A’A

B

B’

C




using Λ = λ0/δ

λ0 (1 + δ) = h′(x)∆x −→ ∆x ≈ λ0h′(x)

α(x) ≈ λ0δ

∆x= h′(x)δ = h′(x)

λ0Λ



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

f

combining, we have

λ0h′(x)

Λ=

x

f−→ h′(x)

Λ=

x

f λ0

this can be directly integrated

h(x)

Λ=

x2

2f λ0=

[x√

2f λ0

]2

x0

α(x)f

a parabola is the ideal surfaceshape for focusing by refrac-tion




α(x) =x

f

combining, we have

λ0h′(x)

Λ=

x

f−→ h′(x)

Λ=

x

f λ0


h(x)

Λ=

x2

2f λ0=

[x√

2f λ0

]2

x0

α(x)f





α(x) =x

f

combining, we have

λ0h′(x)

Λ=

x

f

−→ h′(x)

Λ=

x

f λ0


h(x)

Λ=

x2

2f λ0=

[x√

2f λ0

]2

x0

α(x)f





α(x) =x

f

combining, we have

λ0h′(x)

Λ=

x

f−→ h′(x)

Λ=

x

f λ0


h(x)

Λ=

x2

2f λ0=

[x√

2f λ0

]2

x0

α(x)f





α(x) =x

f

combining, we have

λ0h′(x)

Λ=

x

f−→ h′(x)

Λ=

x

f λ0


h(x)

Λ=

x2

2f λ0=

[x√

2f λ0

]2

x0

α(x)f





α(x) =x

f

combining, we have

λ0h′(x)

Λ=

x

f−→ h′(x)

Λ=

x

f λ0


h(x)

Λ=

x2

2f λ0

=

[x√

2f λ0

]2

x0

α(x)f





α(x) =x

f

combining, we have

λ0h′(x)

Λ=

x

f−→ h′(x)

Λ=

x

f λ0


h(x)

Λ=

x2

2f λ0=

[x√

2f λ0

]2x0

α(x)f





α(x) =x

f

combining, we have

λ0h′(x)

Λ=

x

f−→ h′(x)

Λ=

x

f λ0


h(x)

Λ=

x2

2f λ0=

[x√

2f λ0

]2x0

α(x)f



Focal length of circular lens

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

f =x2Λ

2λ0h(x)=

1

2δ

x2

h(x)or alternatively f =

1

δ

x

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

2

x2

R2

)

x2 = R2 − h2(x) ≈ R2 so we have f ≈ R

2δ

for N circular lenses

fn ≈R

2Nδ




f =x2Λ

2λ0h(x)

=1

2δ

x2


1

δ

x

h′(x)



(1− 1

2

x2

R2

)


2δ


fn ≈R

2Nδ




f =x2Λ

2λ0h(x)=

1

2δ

x2

h(x)

or alternatively f =1

δ

x

h′(x)



(1− 1

2

x2

R2

)


2δ


fn ≈R

2Nδ




f =x2Λ

2λ0h(x)=

1

2δ

x2


1

δ

x

h′(x)



(1− 1

2

x2

R2

)


2δ


fn ≈R

2Nδ




f =x2Λ

2λ0h(x)=

1

2δ

x2


1

δ

x

h′(x)



(1− 1

2

x2

R2

)


2δ


fn ≈R

2Nδ




f =x2Λ

2λ0h(x)=

1

2δ

x2


1

δ

x

h′(x)


h(x) =√R2 − x2

and for x � R h(x) ≈ R

(1− 1

2

x2

R2

)


2δ


fn ≈R

2Nδ




f =x2Λ

2λ0h(x)=

1

2δ

x2


1

δ

x

h′(x)



(1− 1

2

x2

R2

)


2δ


fn ≈R

2Nδ




f =x2Λ

2λ0h(x)=

1

2δ

x2


1

δ

x

h′(x)



(1− 1

2

x2

R2

)

x2 = R2 − h2(x) ≈ R2

so we have f ≈ R

2δ


fn ≈R

2Nδ




f =x2Λ

2λ0h(x)=

1

2δ

x2


1

δ

x

h′(x)



(1− 1

2

x2

R2

)


2δ


fn ≈R

2Nδ




f =x2Λ

2λ0h(x)=

1

2δ

x2


1

δ

x

h′(x)



(1− 1

2

x2

R2

)


2δ


fn ≈R

2Nδ


Focussing by a beryllium lens

2R

d

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

fN =R

2Nδ=

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 et al., “X-ray focusing with compound lenses made from beryllium,” Optics Lett., 27, 778 (2007).



2R

d

unit lensf


fN =R

2Nδ

=1× 10−3m

2(50)(3.41× 10−6)= 2.93m





2R

d

unit lensf


fN =R

2Nδ=

1× 10−3m

2(50)(3.41× 10−6)

= 2.93m





2R

d

unit lensf


fN =R

2Nδ=

1× 10−3m

2(50)(3.41× 10−6)= 2.93m





2R

d

unit lensf


fN =R

2Nδ=

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).


Extruded Al lens

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.

Extruded aluminum lens withparabolic figure

Cut diagonally to exposevariable number of “lenses”to a horizontal beam

Horizontal translation allowschange in focal length but itis quantized, not continuous

A. Khounsary et al., “Fabrication, testing, and performance of a variable focus x-ray compound lens”, Proc. SPIE 4783, 49-54(2002).


Extruded Al lens







Extruded Al lens







Extruded Al lens







Extruded Al lens







Variable focal length CRL

A continuously variable focal length is very important for two specificreasons: tracking sample position, and keeping the focal length constantas energy is changed.

This can be achieved with a rotating lens system

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, “X-ray focusing scheme with continuously variable lens,” J. Synchrotron Radiation 22, 16-22(2015).



A continuously variable focal length is very important for two specificreasons: tracking sample position, and keeping the focal length constantas energy is changed. This can be achieved with a rotating lens system









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.






































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

(x√

2λo f

)2

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

(x√

2λo f

)2

when h(x) = 100Λ ∼ 1000µm

x = 10√

2λo f ∼ 100µm





h(x) = Λ

(x√

2λo f

)2

when h(x) = 100Λ ∼ 1000µm

x = 10√

2λo f ∼ 100µm





h(x) = Λ

(x√

2λo f

)2

when h(x) = 100Λ ∼ 1000µm

x = 10√

2λo f ∼ 100µm





h(x) = Λ

(x√

2λo f

)2

when h(x) = 100Λ ∼ 1000µm

x = 10√

2λo f ∼ 100µm




Λ

2Λ

3Λ

4Λ

5Λ

6Λ

7Λ

Mark 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.



1

2

3

4

5

6

7






1

2

3

4

5

6

7





Fresnel lens dimensions

1

2

3

4

5

6

7 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



1

2

3

4

5

6


ν =h(x)

Λ

ξ =x√

2λo f



(outermost) zone is

∆ξN = ξN − ξN−1 =√N −

√N − 1



1

2

3

4

5

6


ν =h(x)

Λξ =

x√2λo f



(outermost) zone is

∆ξN = ξN − ξN−1 =√N −

√N − 1



1

2

3

4

5

6


ν =h(x)

Λξ =

x√2λo f



(outermost) zone is

∆ξN = ξN − ξN−1 =√N −

√N − 1



1

2

3

4

5

6


ν =h(x)

Λξ =

x√2λo f



(outermost) zone is

∆ξN = ξN − ξN−1 =√N −

√N − 1



1

2

3

4

5

6

7

∆ξN = ξN − ξN−1 =√N −

√N − 1

=√N

(1−

√1− 1

N

)

≈√N

(1−

[1− 1

2N

])∆ξN ≈

1

2√N

The diameter of the entire lens is thus

2ξN = 2√N =

1

∆ξN



1

2

3

4

5

6

7

∆ξN = ξN − ξN−1 =√N −

√N − 1

=√N

(1−

√1− 1

N

)

≈√N

(1−

[1− 1

2N

])∆ξN ≈

1

2√N


2ξN = 2√N =

1

∆ξN



1

2

3

4

5

6

7

∆ξN = ξN − ξN−1 =√N −

√N − 1

=√N

(1−

√1− 1

N

)

≈√N

(1−

[1− 1

2N

])

∆ξN ≈1

2√N


2ξN = 2√N =

1

∆ξN



1

2

3

4

5

6

7

∆ξN = ξN − ξN−1 =√N −

√N − 1

=√N

(1−

√1− 1

N

)

≈√N

(1−

[1− 1

2N

])∆ξN ≈

1

2√N


2ξN = 2√N =

1

∆ξN



1

2

3

4

5

6

7

∆ξN = ξN − ξN−1 =√N −

√N − 1

=√N

(1−

√1− 1

N

)

≈√N

(1−

[1− 1

2N

])∆ξN ≈

1

2√N


2ξN = 2√N =

1

∆ξN


Fresnel lens example

In terms of the unscaled variables

∆xN = ∆ξN

√2λo f

=

√λo f

2N

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

2N

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

2N

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

2N

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

2N

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

2N

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

2N

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

2N

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

Λ

Λ/2

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.



Λ

Λ/2






Λ

Λ/2





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

SiN

UNCD

HSQ

M. Wojick et al., “X-ray zone plates with 25 aspect ratio using a 2-µm-thick ultrananocrystalline diamond mold,” Microsyst.Technol. 20, 2045-2050 (2014).




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.



SiN

UNCD

HSQ





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.



SiN

UNCD

HSQ





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.



SiN

UNCD

HSQ





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.



SiN

UNCD

HSQ








SiN

UNCD

HSQ








SiN

UNCD

HSQ








SiN

UNCD

HSQ



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.

r

2θ

k

k

2θ

Q

k

k

~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) = −r0(

1 + e i ~Q·~r)

I (~Q) = A(~Q)∗A(~Q)

= r20

(1 + e i ~Q·~r

)(1 + e−i ~Q·~r

)I (~Q) = r20

(1 + e i ~Q·~r + e−i ~Q·~r + 1

)= 2r20

(1 + cos(~Q ·~r)

)




r

2θ

k

k

2θ

Q

k

k

~Q = (~k − ~k ′)

|~Q| = 2k sin θ =4π

λsin θ


A(~Q) = −r0(

1 + e i ~Q·~r)

I (~Q) = A(~Q)∗A(~Q)

= r20

(1 + e i ~Q·~r

)(1 + e−i ~Q·~r

)I (~Q) = r20

(1 + e i ~Q·~r + e−i ~Q·~r + 1

)= 2r20

(1 + cos(~Q ·~r)

)




r

2θ

k

k

2θ

Q

k

k

~Q = (~k − ~k ′)

|~Q| = 2k sin θ =4π

λsin θ


A(~Q) = −r0(

1 + e i ~Q·~r)

I (~Q) = A(~Q)∗A(~Q)

= r20

(1 + e i ~Q·~r

)(1 + e−i ~Q·~r

)I (~Q) = r20

(1 + e i ~Q·~r + e−i ~Q·~r + 1

)= 2r20

(1 + cos(~Q ·~r)

)




r

2θ

k

k

2θ

Q

k

k

~Q = (~k − ~k ′)

|~Q| = 2k sin θ =4π

λsin θ


A(~Q) = −r0(

1 + e i ~Q·~r)

I (~Q) = A(~Q)∗A(~Q)

= r20

(1 + e i ~Q·~r

)(1 + e−i ~Q·~r

)I (~Q) = r20

(1 + e i ~Q·~r + e−i ~Q·~r + 1

)= 2r20

(1 + cos(~Q ·~r)

)




r

2θ

k

k

2θ

Q

k

k

~Q = (~k − ~k ′)

|~Q| = 2k sin θ =4π

λsin θ


A(~Q) = −r0(

1 + e i ~Q·~r)

I (~Q) = A(~Q)∗A(~Q)

= r20

(1 + e i ~Q·~r

)(1 + e−i ~Q·~r

)I (~Q) = r20

(1 + e i ~Q·~r + e−i ~Q·~r + 1

)= 2r20

(1 + cos(~Q ·~r)

)




r

2θ

k

k

2θ

Q

k

k

~Q = (~k − ~k ′)

|~Q| = 2k sin θ =4π

λsin θ


A(~Q) = −r0(

1 + e i ~Q·~r)

I (~Q) = A(~Q)∗A(~Q)

= r20

(1 + e i ~Q·~r

)(1 + e−i ~Q·~r

)I (~Q) = r20

(1 + e i ~Q·~r + e−i ~Q·~r + 1

)= 2r20

(1 + cos(~Q ·~r)

)




r

2θ

k

k

2θ

Q

k

k

~Q = (~k − ~k ′)

|~Q| = 2k sin θ =4π

λsin θ


A(~Q) = −r0(

1 + e i ~Q·~r)

I (~Q) = A(~Q)∗A(~Q)

= r20

(1 + e i ~Q·~r

)(1 + e−i ~Q·~r

)I (~Q) = r20

(1 + e i ~Q·~r + e−i ~Q·~r + 1

)= 2r20

(1 + cos(~Q ·~r)

)




r

2θ

k

k

2θ

Q

k

k

~Q = (~k − ~k ′)

|~Q| = 2k sin θ =4π

λsin θ


A(~Q) = −r0(

1 + e i ~Q·~r)

I (~Q) = A(~Q)∗A(~Q)

= r20

(1 + e i ~Q·~r

)(1 + e−i ~Q·~r

)I (~Q) = r20

(1 + e i ~Q·~r + e−i ~Q·~r + 1

)= 2r20

(1 + cos(~Q ·~r)

)




r

2θ

k

k

2θ

Q

k

k

~Q = (~k − ~k ′)

|~Q| = 2k sin θ =4π

λsin θ


A(~Q) = −r0(

1 + e i ~Q·~r)

I (~Q) = A(~Q)∗A(~Q)

= r20

(1 + e i ~Q·~r

)(1 + e−i ~Q·~r

)

I (~Q) = r20

(1 + e i ~Q·~r + e−i ~Q·~r + 1

)= 2r20

(1 + cos(~Q ·~r)

)




r

2θ

k

k

2θ

Q

k

k

~Q = (~k − ~k ′)

|~Q| = 2k sin θ =4π

λsin θ


A(~Q) = −r0(

1 + e i ~Q·~r)

I (~Q) = A(~Q)∗A(~Q)

= r20

(1 + e i ~Q·~r

)(1 + e−i ~Q·~r

)I (~Q) = r20

(1 + e i ~Q·~r + e−i ~Q·~r + 1

)

= 2r20

(1 + cos(~Q ·~r)

)




r

2θ

k

k

2θ

Q

k

k

~Q = (~k − ~k ′)

|~Q| = 2k sin θ =4π

λsin θ


A(~Q) = −r0(

1 + e i ~Q·~r)

I (~Q) = A(~Q)∗A(~Q)

= r20

(1 + e i ~Q·~r

)(1 + e−i ~Q·~r

)I (~Q) = r20

(1 + e i ~Q·~r + e−i ~Q·~r + 1

)= 2r20

(1 + cos(~Q ·~r)

)C. Segre (IIT) PHYS 570 - Spring 2018 February 13, 2018 19 / 23

Scattering from many electrons

for many electrons

generalizing to a crystal

A(~Q) = −r0∑

j

e i ~Q·~rj

A(~Q) = −r0∑

N

e i ~Q· ~RN∑

j

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) = −r0∑

j

e i ~Q·~rj

A(~Q) = −r0∑

N

e i ~Q· ~RN∑

j

e i ~Q·~rj





for many electrons


A(~Q) = −r0∑

j

e i ~Q·~rj

A(~Q) = −r0∑

N

e i ~Q· ~RN∑

j

e i ~Q·~rj





for many electrons


A(~Q) = −r0∑

j

e i ~Q·~rj

A(~Q) = −r0∑

N

e i ~Q· ~RN∑

j

e i ~Q·~rj





for many electrons


A(~Q) = −r0∑

j

e i ~Q·~rj

A(~Q) = −r0∑

N

e i ~Q· ~RN∑

j

e i ~Q·~rj





for many electrons


A(~Q) = −r0∑

j

e i ~Q·~rj

A(~Q) = −r0∑

N

e i ~Q· ~RN∑

j

e i ~Q·~rj




Two electrons — fixed orientation

The expression

I (~Q) = 2r20

(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)

0

1

2

3

4

Inte

nsity (

units o

f r2 o

)



The expression

I (~Q) = 2r20

(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)

0

1

2

3

4

Inte

nsity (

units o

f r2 o

)



The expression

I (~Q) = 2r20

(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)

0

1

2

3

4

Inte

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 i ~Q·~r

⟩=

∫e iQr cos θ sin θdθdφ∫

sin θdθdφ

=1

4π2π

∫ π

0e iQr cos θ sin θdθ

=2π

4π

(− 1

iQr

)∫ −iQr

iQrexdx

=1

22

sin(Qr)

Qr=

sin(Qr)

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 spaceand we take ~Q along the z-axis

substituting x = iQr cos θ anddx = −iQr sin θdθ⟨I (~Q)

⟩= f 21 +f 22 +2f1f2

sin(Qr)

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

⟩⟨e i ~Q·~r

⟩=


sin θdθdφ

=1

4π2π

∫ π


=2π

4π

(− 1

iQr

)∫ −iQr

iQrexdx

=1

22

sin(Qr)

Qr=

sin(Qr)

Qr





⟩= f 21 +f 22 +2f1f2

sin(Qr)

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

⟩⟨e i ~Q·~r

⟩=


sin θdθdφ

=1

4π2π

∫ π


=2π

4π

(− 1

iQr

)∫ −iQr

iQrexdx

=1

22

sin(Qr)

Qr=

sin(Qr)

Qr





⟩= f 21 +f 22 +2f1f2

sin(Qr)

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

⟩⟨e i ~Q·~r

⟩=


sin θdθdφ

=1

4π2π

∫ π


=2π

4π

(− 1

iQr

)∫ −iQr

iQrexdx

=1

22

sin(Qr)

Qr=

sin(Qr)

Qr





⟩= f 21 +f 22 +2f1f2

sin(Qr)

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

⟩⟨e i ~Q·~r

⟩=


sin θdθdφ

=1

4π2π

∫ π


=2π

4π

(− 1

iQr

)∫ −iQr

iQrexdx

=1

22

sin(Qr)

Qr=

sin(Qr)

Qr



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

and we take ~Q along the z-axis


⟩= f 21 +f 22 +2f1f2

sin(Qr)

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

⟩

⟨e i ~Q·~r

⟩=


sin θdθdφ

=1

4π2π

∫ π


=2π

4π

(− 1

iQr

)∫ −iQr

iQrexdx

=1

22

sin(Qr)

Qr=

sin(Qr)

Qr



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

and we take ~Q along the z-axis


⟩= f 21 +f 22 +2f1f2

sin(Qr)

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

⟩

⟨e i ~Q·~r

⟩=


sin θdθdφ

=1

4π2π

∫ π


=2π

4π

(− 1

iQr

)∫ −iQr

iQrexdx

=1

22

sin(Qr)

Qr=

sin(Qr)

Qr





⟩= f 21 +f 22 +2f1f2

sin(Qr)

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

⟩⟨e i ~Q·~r

⟩=


sin θdθdφ

=1

4π2π

∫ π


=2π

4π

(− 1

iQr

)∫ −iQr

iQrexdx

=1

22

sin(Qr)

Qr=

sin(Qr)

Qr





⟩= f 21 +f 22 +2f1f2

sin(Qr)

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

⟩⟨e i ~Q·~r

⟩=


sin θdθdφ

=1

4π2π

∫ π


=2π

4π

(− 1

iQr

)∫ −iQr

iQrexdx

=1

22

sin(Qr)

Qr=

sin(Qr)

Qr





⟩= f 21 +f 22 +2f1f2

sin(Qr)

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

⟩⟨e i ~Q·~r

⟩=


sin θdθdφ

=1

4π2π

∫ π


=2π

4π

(− 1

iQr

)∫ −iQr

iQrexdx

=1

22

sin(Qr)

Qr=

sin(Qr)

Qr




substituting x = iQr cos θ anddx = −iQr sin θdθ

⟨I (~Q)

⟩= f 21 +f 22 +2f1f2

sin(Qr)

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

⟩⟨e i ~Q·~r

⟩=


sin θdθdφ

=1

4π2π

∫ π


=2π

4π

(− 1

iQr

)∫ −iQr

iQrexdx

=1

22

sin(Qr)

Qr=

sin(Qr)

Qr





⟨I (~Q)

⟩= f 21 +f 22 +2f1f2

sin(Qr)

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

⟩⟨e i ~Q·~r

⟩=


sin θdθdφ

=1

4π2π

∫ π


=2π

4π

(− 1

iQr

)∫ −iQr

iQrexdx

=1

22

sin(Qr)

Qr

=sin(Qr)

Qr





⟨I (~Q)

⟩= f 21 +f 22 +2f1f2

sin(Qr)

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

⟩⟨e i ~Q·~r

⟩=


sin θdθdφ

=1

4π2π

∫ π


=2π

4π

(− 1

iQr

)∫ −iQr

iQrexdx

=1

22

sin(Qr)

Qr=

sin(Qr)

Qr





⟨I (~Q)

⟩= f 21 +f 22 +2f1f2

sin(Qr)

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

⟩⟨e i ~Q·~r

⟩=


sin θdθdφ

=1

4π2π

∫ π


=2π

4π

(− 1

iQr

)∫ −iQr

iQrexdx

=1

22

sin(Qr)

Qr=

sin(Qr)

Qr





⟩= f 21 +f 22 +2f1f2

sin(Qr)

Qr


Randomly oriented electrons

⟨I (~Q)

⟩= f 21 +f 22 +2f1f2

sin(Qr)

Qr

Recall that when we had a fixedorientation of the two electrons,we had and intensity variationI (~Q) = 2r20 (1 + cos(Qr)).

When we now replace the twoarbitrary scattering distributionswith electrons (f1, f2 → −r0),we change the intensity profilesignificantly.⟨I (~Q)

⟩= 2r20

(1 +

sin(Qr)

Qr

)0 0.5 1 1.5 2

Q (units of 2π/r)

0

1

2

3

4

Inte

nsity (

units o

f r2 o

)



⟨I (~Q)

⟩= f 21 +f 22 +2f1f2

sin(Qr)

Qr



⟩= 2r20

(1 +

sin(Qr)

Qr

)0 0.5 1 1.5 2

Q (units of 2π/r)

0

1

2

3

4

Inte

nsity (

units o

f r2 o

)



⟨I (~Q)

⟩= f 21 +f 22 +2f1f2

sin(Qr)

Qr



⟩= 2r20

(1 +

sin(Qr)

Qr

)

0 0.5 1 1.5 2

Q (units of 2π/r)

0

1

2

3

4

Inte

nsity (

units o

f r2 o

)



⟨I (~Q)

⟩= f 21 +f 22 +2f1f2

sin(Qr)

Qr


When we now replace the twoarbitrary scattering distributionswith electrons (f1, f2 → −r0),we change the intensity profilesignificantly.

⟨I (~Q)

⟩= 2r20

(1 +

sin(Qr)

Qr

)

0 0.5 1 1.5 2

Q (units of 2π/r)

0

1

2

3

4

Inte

nsity (

units o

f r2 o

)



⟨I (~Q)

⟩= f 21 +f 22 +2f1f2

sin(Qr)

Qr



⟩= 2r20

(1 +

sin(Qr)

Qr

)0 0.5 1 1.5 2

Q (units of 2π/r)

0

1

2

3

4

Inte

nsity (

units o

f r2 o

)



⟨I (~Q)

⟩= f 21 +f 22 +2f1f2

sin(Qr)

Qr



⟩= 2r20

(1 +

sin(Qr)

Qr

)0 0.5 1 1.5 2

Q (units of 2π/r)

0

1

2

3

4

Inte

nsity (

units o

f r2 o

)


Documents

Today’s Outline - February 13, 2018phys.iit.edu/~segre/phys570/18S/lecture_11.pdf · Today’s Outline - February 13, 2018 Ideal refractive surface Fresnel lenses and zone plates