Today’s Outline - April 23, 2015segre/phys570/15S/lecture_22.pdf · Today’s Outline - April 23,...

Today’s Outline - April 23, 2015

• Imaging

• Radiography and Tomography

• Microscopy

• Phase Contrast Imaging

• Grating Interferometry

• Coherent Diffraction Imaging

• Holography

Final Exam, Tuesday, May 5, 2015, Life Sciences 2404 sessions: 09:00-11:00; 11:00-13:00; 14:00-16:00; 16:00-18:00

Provide me with the paper you intend to present and a preferred sessionfor the examSend me your presentation in Powerpoint or PDF format before beforeyour session

C. Segre (IIT) PHYS 570 - Spring 2015 April 23, 2015 1 / 22

• Imaging

• Microscopy

• Holography

• Imaging

• Microscopy

• Holography

• Imaging

• Microscopy

• Holography

• Imaging

• Microscopy

• Holography

• Imaging

• Microscopy

• Holography

• Imaging

• Microscopy

• Holography

• Imaging

• Microscopy

• Holography

• Imaging

• Microscopy

• Holography

• Imaging

• Microscopy

• Holography

Provide me with the paper you intend to present and a preferred sessionfor the exam

Send me your presentation in Powerpoint or PDF format before beforeyour session

• Imaging

• Microscopy

• Holography

Problem 5.9

Consider a simple model of the surface roughness of a crystal in which allof the lattice sites of the z = 0 layer are fully occupied by atoms, but thenext layer out (z = −1) has a site occupancy of η, with η ≤ 1, the z = −2layer an occupancy of η2, etc. Show that midway between the Braggpoints, the so-called anti-Bragg points, the intensity of the crystaltruncation rods is given by

ICTR =(1− η)2

4(1 + η)2

What effect does a small, but finite, value of the roughness parameter ηhave on ICTR?

Problem 5.9

The general expression for the scattering factor of the crystal truncationrod in the a3 direction of the crystal, excluding absorption effects is

The rougness model addsterms with j < 0 of theform ηje−iQza3j so a sec-ond sum is added

evaluating these geometricsums in the usual way

at the anti-Bragg points,l = 0.5, and the exponen-tials all become -1

putting over a commondenominator and droppingthe A(~Q) gives

FCTR = A(~Q)∞∑j=0

e iQza3j

= A(~Q)

∞∑j=0

e iQza3j +∞∑j=0

ηje−iQza3j

= A(~Q)

1− e i2πl+

ηe−i2πl

1− ηe−i2πl

]= A(~Q)

2− η

1 + η

]= A(~Q)

(1 + η)− 2η

2(1 + η)= A(~Q)

(1− η)

2(1 + η)

Problem 5.9

e iQza3j

= A(~Q)

∞∑j=0

e iQza3j +∞∑j=0

ηje−iQza3j

= A(~Q)

1− e i2πl+

ηe−i2πl

1− ηe−i2πl

]= A(~Q)

2− η

1 + η

]= A(~Q)

(1 + η)− 2η

2(1 + η)= A(~Q)

(1− η)

2(1 + η)

Problem 5.9

e iQza3j

= A(~Q)

∞∑j=0

e iQza3j +∞∑j=0

ηje−iQza3j

= A(~Q)

1− e i2πl+

ηe−i2πl

1− ηe−i2πl

]= A(~Q)

2− η

1 + η

]= A(~Q)

(1 + η)− 2η

2(1 + η)= A(~Q)

(1− η)

2(1 + η)

Problem 5.9

e iQza3j

= A(~Q)

∞∑j=0

e iQza3j +∞∑j=0

ηje−iQza3j

= A(~Q)

1− e i2πl+

ηe−i2πl

1− ηe−i2πl

]= A(~Q)

2− η

1 + η

]= A(~Q)

(1 + η)− 2η

2(1 + η)= A(~Q)

(1− η)

2(1 + η)

Problem 5.9

e iQza3j

= A(~Q)

∞∑j=0

e iQza3j +∞∑j=0

ηje−iQza3j

= A(~Q)

1− e i2πl+

ηe−i2πl

1− ηe−i2πl

]= A(~Q)

2− η

1 + η

]= A(~Q)

(1 + η)− 2η

2(1 + η)= A(~Q)

(1− η)

2(1 + η)

Problem 5.9

e iQza3j

= A(~Q)

∞∑j=0

e iQza3j +∞∑j=0

ηje−iQza3j

= A(~Q)

1− e i2πl+

ηe−i2πl

1− ηe−i2πl

= A(~Q)

2− η

1 + η

]= A(~Q)

(1 + η)− 2η

2(1 + η)= A(~Q)

(1− η)

2(1 + η)

Problem 5.9

e iQza3j

= A(~Q)

∞∑j=0

e iQza3j +∞∑j=0

ηje−iQza3j

= A(~Q)

1− e i2πl+

ηe−i2πl

1− ηe−i2πl

= A(~Q)

2− η

1 + η

]= A(~Q)

(1 + η)− 2η

2(1 + η)= A(~Q)

(1− η)

2(1 + η)

Problem 5.9

e iQza3j

= A(~Q)

∞∑j=0

e iQza3j +∞∑j=0

ηje−iQza3j

= A(~Q)

1− e i2πl+

ηe−i2πl

1− ηe−i2πl

]= A(~Q)

2− η

1 + η

= A(~Q)(1 + η)− 2η

2(1 + η)= A(~Q)

(1− η)

2(1 + η)

Problem 5.9

e iQza3j

= A(~Q)

∞∑j=0

e iQza3j +∞∑j=0

ηje−iQza3j

= A(~Q)

1− e i2πl+

ηe−i2πl

1− ηe−i2πl

]= A(~Q)

2− η

1 + η

= A(~Q)(1 + η)− 2η

2(1 + η)= A(~Q)

(1− η)

2(1 + η)

Problem 5.9

e iQza3j

= A(~Q)

∞∑j=0

e iQza3j +∞∑j=0

ηje−iQza3j

= A(~Q)

1− e i2πl+

ηe−i2πl

1− ηe−i2πl

]= A(~Q)

2− η

1 + η

]= A(~Q)

(1 + η)− 2η

2(1 + η)

= A(~Q)(1− η)

2(1 + η)

Problem 5.9

e iQza3j

= A(~Q)

∞∑j=0

e iQza3j +∞∑j=0

ηje−iQza3j

= A(~Q)

1− e i2πl+

ηe−i2πl

1− ηe−i2πl

]= A(~Q)

2− η

1 + η

]= A(~Q)

(1 + η)− 2η

2(1 + η)= A(~Q)

(1− η)

2(1 + η)

Problem 5.9

The scattering factor thushas a dependence on the cov-erage factor, η, of

and the observed intensityhas a dependence of

if the coverage is very low,this can be approximated us-ing the first term of a bino-mial expansion

FCTR = A(~Q)(1− η)

2(1 + η)

ICTR = |A|2 (1− η)2

4(1 + η)2

(1− 2η

1 + 2η

)≤ 1

Thus any small value of the roughness parameter will dampen the intensityof the peaks

Problem 5.9

FCTR = A(~Q)(1− η)

2(1 + η)

ICTR = |A|2 (1− η)2

4(1 + η)2

(1− 2η

1 + 2η

)≤ 1

Problem 5.9

FCTR = A(~Q)(1− η)

2(1 + η)

ICTR = |A|2 (1− η)2

4(1 + η)2

(1− 2η

1 + 2η

)≤ 1

Problem 5.9

FCTR = A(~Q)(1− η)

2(1 + η)

ICTR = |A|2 (1− η)2

4(1 + η)2

(1− 2η

1 + 2η

)≤ 1

Problem 5.9

FCTR = A(~Q)(1− η)

2(1 + η)

ICTR = |A|2 (1− η)2

4(1 + η)2

(1− 2η

1 + 2η

)≤ 1

Problem 5.9

FCTR = A(~Q)(1− η)

2(1 + η)

ICTR = |A|2 (1− η)2

4(1 + η)2

(1− 2η

1 + 2η

Problem 5.9

FCTR = A(~Q)(1− η)

2(1 + η)

ICTR = |A|2 (1− η)2

4(1 + η)2

(1− 2η

1 + 2η

)≤ 1

Problem 5.9

FCTR = A(~Q)(1− η)

2(1 + η)

ICTR = |A|2 (1− η)2

4(1 + η)2

(1− 2η

1 + 2η

)≤ 1

Phase difference in scattering

All imaging can be broken into athree step process

1 x-ray interaction with sample

2 scattered x-ray propagation

3 interaction with detector

The spherical waves scattered offthe two points will travel differentdistances

In the far field, the phase differenceis ~Q ·~r with ~Q = ~k − ~k ′

Franuhofer, Fresnel, and contact regimes

In the far field (Fraunhofer) regime, the phase difference is ≈ ~k ′ ·~rThe error in difference computedwith the far field approximation is

∆ = R − R cosψ

≈ R(1− (1− ψ2/2))

= a2/(2R)

R � a2/λ Fraunhofer

R ≈ a2/λ Fresnel

R � a2/λ Contact

In the far field (Fraunhofer) regime, the phase difference is ≈ ~k ′ ·~r

The error in difference computedwith the far field approximation is

∆ = R − R cosψ

≈ R(1− (1− ψ2/2))

= a2/(2R)

R ≈ a2/λ Fresnel

R � a2/λ Contact

∆ = R − R cosψ

≈ R(1− (1− ψ2/2))

= a2/(2R)

R ≈ a2/λ Fresnel

R � a2/λ Contact

∆ = R − R cosψ

≈ R(1− (1− ψ2/2))

= a2/(2R)

R ≈ a2/λ Fresnel

R � a2/λ Contact

∆ = R − R cosψ

≈ R(1− (1− ψ2/2))

= a2/(2R)

R ≈ a2/λ Fresnel

R � a2/λ Contact

∆ = R − R cosψ

≈ R(1− (1− ψ2/2))

= a2/(2R)

R ≈ a2/λ Fresnel

R � a2/λ Contact

∆ = R − R cosψ

≈ R(1− (1− ψ2/2))

= a2/(2R)

R ≈ a2/λ Fresnel

R � a2/λ Contact

∆ = R − R cosψ

≈ R(1− (1− ψ2/2))

= a2/(2R)

R ≈ a2/λ Fresnel

R � a2/λ Contact

Contact to far-field imaging

Fourier slice theorem

I = I0e−

∫µ(x ,y)dy ′

∫µ(x , y)dy ′

p(x) =

∫f (x , y)dy

P(qx) =

∫p(x)e iqxxdx

F (qx , qy ) =

∫ ∫f (x , y)e iqxx+qyydxdy

F (qx , qy = 0) =

∫ [∫f (x , y)dy

]e iqxxdx =

∫p(x)e iqxxdx = P(qx)

I = I0e−

∫µ(x ,y)dy ′

∫µ(x , y)dy ′

p(x) =

∫f (x , y)dy

P(qx) =

∫p(x)e iqxxdx

F (qx , qy ) =

F (qx , qy = 0) =

∫ [∫f (x , y)dy

]e iqxxdx =

I = I0e−

∫µ(x ,y)dy ′

∫µ(x , y)dy ′

p(x) =

∫f (x , y)dy

P(qx) =

∫p(x)e iqxxdx

F (qx , qy ) =

F (qx , qy = 0) =

∫ [∫f (x , y)dy

]e iqxxdx =

I = I0e−

∫µ(x ,y)dy ′

∫µ(x , y)dy ′

p(x) =

∫f (x , y)dy

P(qx) =

∫p(x)e iqxxdx

F (qx , qy ) =

F (qx , qy = 0) =

∫ [∫f (x , y)dy

]e iqxxdx =

I = I0e−

∫µ(x ,y)dy ′

∫µ(x , y)dy ′

p(x) =

∫f (x , y)dy

P(qx) =

∫p(x)e iqxxdx

F (qx , qy ) =

F (qx , qy = 0) =

∫ [∫f (x , y)dy

]e iqxxdx =

I = I0e−

∫µ(x ,y)dy ′

∫µ(x , y)dy ′

p(x) =

∫f (x , y)dy

P(qx) =

∫p(x)e iqxxdx

F (qx , qy ) =

F (qx , qy = 0) =

∫ [∫f (x , y)dy

]e iqxxdx

I = I0e−

∫µ(x ,y)dy ′

∫µ(x , y)dy ′

p(x) =

∫f (x , y)dy

P(qx) =

∫p(x)e iqxxdx

F (qx , qy ) =

F (qx , qy = 0) =

∫ [∫f (x , y)dy

]e iqxxdx =

∫p(x)e iqxxdx

= P(qx)

I = I0e−

∫µ(x ,y)dy ′

∫µ(x , y)dy ′

p(x) =

∫f (x , y)dy

P(qx) =

∫p(x)e iqxxdx

F (qx , qy ) =

F (qx , qy = 0) =

∫ [∫f (x , y)dy

]e iqxxdx =

Fourier transform reconstruction

Sinograms

Medical tomography

Microscopy

Angular deviation from refraction

When x-rays cross an inter-face that is not normal totheir direction, there is re-fraction

The angle of refraction α canbe calculated

λn =λ

1− δ≈ λ(1 + δ)

α =λ(1 + δ)− λ

= δλ

∆x≈ δ tanω

λn =λ

1− δ≈ λ(1 + δ)

α =λ(1 + δ)− λ

= δλ

∆x≈ δ tanω

λn =λ

1− δ≈ λ(1 + δ)

α =λ(1 + δ)− λ

= δλ

∆x≈ δ tanω

λn =λ

1− δ≈ λ(1 + δ)

α =λ(1 + δ)− λ

= δλ

∆x≈ δ tanω

λn =λ

1− δ

≈ λ(1 + δ)

α =λ(1 + δ)− λ

= δλ

∆x≈ δ tanω

λn =λ

1− δ≈ λ(1 + δ)

α =λ(1 + δ)− λ

= δλ

∆x≈ δ tanω

λn =λ

1− δ≈ λ(1 + δ)

α =λ(1 + δ)− λ

= δλ

∆x≈ δ tanω

λn =λ

1− δ≈ λ(1 + δ)

α =λ(1 + δ)− λ

= δλ

≈ δ tanω

λn =λ

1− δ≈ λ(1 + δ)

α =λ(1 + δ)− λ

= δλ

∆x≈ δ tanω

Angular deviation from graded density

In a similar way, there isan angular deviation whenthe material density variesnormal to the propagationdirection

The angle of refraction αcan be calculated