Lecture31 Ch10 diffraction4.ppt

Diffraction

Chapter 10

Phys 322Lecture 31

Diffraction applications: x-ray diffractionFresnel Diffraction

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

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

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!

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

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!

Example: X-ray diffraction of tungsten

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!

X-ray crystallography

Braggs’s law

2 sind m

X-ray diffraction (XRD)Single crystal

Powder diffraction

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

X-ray of complex crystals

(Myoglobin) 1960, Perutz & Kendrew

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

Fraunhofer vs. Fresnel Diffraction

Far from

the slit

zClose to the slit

Incident wave

Slit

S P

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.

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

Simple spherical wave: Fresnel half-period zones

SO

O’ r0

x

Z1

Zl

r

dS

P

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

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

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

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

Circular aperture: Fresnel diffraction

m is even: Ecenter~0m is odd: Ecenter~E1

Diffraction from increasing size

Circular aperture: plane waves (Fresnel)

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.

Fresnel diffraction on a slit

Fresnel diffraction by a slit

Fresnel diffraction on an edge

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: