Lecture31 Ch10 diffraction4.ppt
Transcript of Lecture31 Ch10 diffraction4.ppt
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Diffraction
Chapter 10
Phys 322Lecture 31
Diffraction applications: x-ray diffractionFresnel Diffraction
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X-ray optics
Beam of X-rays can be used to reveal the structure of a crystal.Why X-rays?
- they can penetrate deep into matter- the wavelength is comparable to interatomic distance
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X-ray diffraction
Diffraction = multi-source interference
lattice
X-ray
Electrons in atoms will oscillate causing secondary radiation.Secondary radiation from atoms will interfere.Picture is complex: we have 3-D grid of sources
We will consider only simple cases
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X-ray diffraction: constructive interference
Simple crystal: 3D cubic grid
first layer
Simple case: ‘reflection’ incident angle=reflected anglephase shift = 0
Note: angle here is in respect to the surface!
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X-ray diffraction: constructive interference
Reflection from the second layer will not necessarily be in phase
Path difference:
sin2dl
Each layer re-radiates. The total intensity of reflected beam depends on phase difference between waves ‘reflected’ from different layers
Condition for intense X-ray reflection:
where m is an integer md sin2
Note: angle is in respect to the surface!
Bragg’s law
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X-ray diffraction: simple experiment
crystalturn crystal
x-ray diffracted
md sin2
May need to observe several maxima to find m and deduce d
Note: angle is in respect to the surface!
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Example: X-ray diffraction of tungsten
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X-ray monochromator
Suppose you have a source of X-rays which has continuum spectrum of wavelengths.How can one make it monochromatic?
crystal
incident broadband X-ray
reflected single-wavelength X-ray
md sin2
Note: angle is in respect to the surface!
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X-ray crystallography
Braggs’s law
2 sind m
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X-ray diffraction (XRD)Single crystal
Powder diffraction
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X-ray of powdered crystals
Powder contains crystals in all possible orientations
polycrystalline LiF
Note: Incident angle doesn't have to be equal to scattering angle.Crystal may have more than one kind of atoms.Crystal may have many ‘lattices’ with different d
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X-ray of complex crystals
(Myoglobin) 1960, Perutz & Kendrew
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X-ray diffraction: sample problem
The spacing between neighboring layers in a particular crystal is 2 Å. A monochromatic X-ray beam of wavelength 0.96 Å strikes the crystal. At what angle might one expect to find a diffraction maximum?
md sin2
md
m 24.02
sin
m24.0arcsin
= 13.9o, 28.7o, 46.1o, 73.70
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Fraunhofer vs. Fresnel Diffraction
Far from
the slit
zClose to the slit
Incident wave
Slit
S P
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Fresnel diffraction
Reminders:Huygens-Fresnel principle: Every unobstructed point of a wave front serves as a source of spherical wavelets. The amplitude of the optical field at any point beyond is the superposition of all these wavelets, taking into account their amplitudes and phases.
Fraunhofer (far field) diffraction: Both the incoming and outgoing waves approach being planar. R > a2/.Fresnel (near field) diffraction: For any R1 and R2.
• Fraunhofer diffraction is a special case of Fresnel diffraction.• The integration for Fresnel diffraction is usually complicated. Fresnel zones will be introduced to estimate the diffraction pattern.
S
P
aR1
R2
Note: Huygens-Fresnel theory is an approximation of the more accurate Fresnel-Kirchhoff formulation.
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Fresnel diffraction(near-field diffraction)
Huygens: source of spherical wave
Problem: it must also go backwards, but that is not observed in experiment
Solution: Kirchhoff’s scalar diffraction theory
cos121
K
inclination factor
Intensity of the spherical wavelet depends on direction:
krtr
KE cosE
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Simple spherical wave: Fresnel half-period zones
SO
O’ r0
x
Z1
Zl
r
dS
P
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Fresnel half-period zones
rktrdSKdE A cosE
0EE QA source strength
unit area
sin2ddS
00
1 sin211
rktr
KdEE Allr
rl
l
l
EDisturbance due to zone l:
Alternating Note:1) Kl is almost constant within one zone.
3) . The contributions from adjacent zones tend to cancel each other.
constant)1( 1 l
ll KE
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Fresnel half-period zones
00
1 sin211
rktr
KdEE Allr
rl
l
l
E
m
m
ll EEEEEEE
...43211
If the number of zones m is even: 22
1 mEEE
If the number of zones m is odd: 22
1 mEEE
The optical disturbance contributed by the whole wavefront is2
1EE
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The vibration curve (phasor representation)Graphic method for qualitatively analyzing diffraction problems with circular symmetry.
SO
O’ r0x
Z1
P
/NOs
Zs1
E1
For the first zone:• Divide the zone into N subzones.• Each subzone has a phase shift of /N.• The phasor chain deviates from a circle due to the
inclination factor.• When N ∞, the phasor train composes a smooth
spiral called a vibration curve.
The first zone
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The vibration curve:• Each zone swings ½ turn and has a phase shift of .• The total disturbance at P is:
• The total disturbance is /2 out of phase with the primary wave (a drawback of Fresnel formulation).
• The contribution from O to any point A on the sphereis .
2||
2' 11 EZOOO ss
ss
ss AO
Os’
Zs1
Zs2
Zs3
As
Os
PS OO’ r0
x
Zm
Circular aperturesSpherical waves
P on axis:a) The aperture has m (integer, not very large) zones.
i) If m is even,
ii) If m is odd,
which is roughly twice as the unobstructed field.
b) If m is not an integer, . This can be seen from the vibration curve.
0 |)||(||)||(||)||| ( 14321 mm EEEEEEE
|| |)||(||)||(||| 11321 EEEEEEE mm
|| 0 1EE
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Circular aperture: Fresnel diffraction
m is even: Ecenter~0m is odd: Ecenter~E1
Diffraction from increasing size
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Circular aperture: plane waves (Fresnel)
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Fresnel zone plate
m
m
ll EEEEEEE
...43211
Unobstructed: E ~ E1/2
What would be irradiance in the center for a zone plate with 50 zones?
A device that modifies (in amplitude or phase) the light from the Fresnel zones.Example: Transparent only for odd (or even) zones.
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Fresnel diffraction on a slit
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Fresnel diffraction by a slit
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Fresnel diffraction on an edge
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Fresnel diffraction by a narrow obstacle
Interestingly, the hole fills in from the center first!
x0 x1
Stop
Beam after some distance
Input beam with hole
The Spot of Arago or Poisson
This irradiance can be quite high and can do some damage!
If a beam encounters a stop, it develops a hole, which fills in as it propagates and diffracts: