Chapter 25

Waves and ParticlesMidterm 4 UTC 1.132  

Wave Phenomena

• Interference

• Diffraction

• Reflection

l – wavelength: distance between crests (meters)T – period: the time between crests passing fixed location (seconds)v – speed: the distance one crest moves in a second (m/s)f – frequency: the number of crests passing fixed location in one second (1/s or Hz) – angular frequency: 2f: (rad/s)

Tv

Tf

1 fv

Wave Description

E E0 cos t E0 cos2T

t

Wave: Variation in Time

xE

xEE

2

cos2

cos 00

Wave: Variation in Space

xEE

2

cos0

t

TEE

2cos0

xt

TEE

22

cos0

‘-’ sign: the point on wave moves to the right

Wave: Variation in Time and Space

xt

TEE

22

cos0

But E @ t=0 and x =0, may not equal E0

xtT

EE22

cos0

phase shift, =0…2

Two waves are ‘out of phase’

Wave: Phase Shift

tEt

TEE cos

2cos 00

(Shown for x=0)

In many cases we are interested only in E at certain location:can ignore dependence on x:

tEt

TEE cos

2cos 00

Using angular frequency makes equation more compact

Wave: Angular Frequency

t

y(x,t) Asin(kx t)

2T

k 2

tEE cos0

E0 is a parameter called amplitude (positive). Time dependenceis in cosine function

Often we detect ‘intensity’, or energy flux ~ E2. For example: Vision – we don’t see individual oscillations

Intensity I (W/m2):

20EI

Works also for other waves,such as sound or water waves.

Wave: Amplitude and Intensity

Superposition principle: The net electric field at any location isvector sum of the electric fields contributed by all sources.

Can particle model explain the pattern?

Laser: source of radiation which has the same frequency (monochromatic) and phase (coherent) across the beam.

Two slits are sources of two waves with the same phase and frequency.

Interference

Two emitters:

E1

E2

Fields in crossing point

tEE

tEE

cos

cos

02

01

Superposition: tEEEE cos2 021

Amplitude increases twice: constructive interference

Interference: Constructive

Two emitters:

E1

E2

tEEEE cos2 021

What about the intensity (energy flux)?

Energy flux increases 4 times while two emitters produce onlytwice more energy

There must be an area in space where intensity is smaller than thatproduced by one emitter

Interference: Energy

E1

E2

ttEEEE coscos021

tEE

tEE

cos

cos

02

01

tcos

0

Two waves are 1800 out of phase: destructive interference

Interference: Destructive

Superposition principle: The net electric field at any location isthe vector sum of the electric fields contributed by all sources.

Interference

tEE

tEE

cos

cos

02

01

tEEEE cos2 021

Amplitude increases twice

Constructive: Energy flux increases 4 times while two emitters produce only twice more energy

ttEEEE coscos021

tEE

tEE

cos

cos

02

01

Two waves are 1800 out of phase

Constructive: Destructive:

Intensity at each location depends on phase shift between twowaves, energy flux is redistributed.

Maxima with twice the amplitude occur when phase shift between two waves is 0, 2, 4, 6 …(Or path difference is 0, , 2 …)

Minima with zero amplitude occur when phase shift between two waves is , 3, 5 …(Or path difference is 0, /2, 3/2…)

Can we observe complete destructive interference if 1 2 ?

Interference

Predicting Pattern For Two SourcesPoint C on screen is very far from sourcesC

normal

Need to know phase difference

Very far: angle ACB is very small

Path AC and BC are equal

Path difference: )sin(dl

If l = 0, , 2, 3, 4 … - maximum

If l = /2, 3/2, 5/2 … - minimum

Predicting Pattern For Two SourcesC

normal

Path difference: )sin(dl

If l = 0, , 2, 3, 4 … - maximum

If l = /2, 3/2, 5/2 … - minimum

What if d < ?

complete constructive interferenceonly at =00, 1800

What if d < /2 ?

no complete destructive interference anywhere

Note: largest Dl for q= /2p

d = 4.5

Why is intensity maximum at =0 and 1800 ?

Why is intensity zero at =90 and -900 ?

What is the phase difference at Max3?

Intensity versus AnglePath difference: )sin(dl

If l = 0, , 2, 3, 4 … - maximum

If l = /2, 3/2, 5/2 … - minimum

Path difference: )sin(dl

If l = 0, , 2, 3, 4 … - maximum

If l = /2, 3/2, 5/2 … - minimum

d = /3.5

Two sources are /3.5 apart. What will be the intensity pattern?

Intensity versus Angle

Path difference:

If l = 0, , 2, 3, 4 … - maximum

If l = /2, 3/2, 5/2 … - minimum

)sin(dl

L=2 m, d=0.5 mm, x=2.4 mmWhat is the wavelength of this laser?

)sin( dd

)sin(

L

x)tan(

Small angle limit: sin() tan()

L

x

d

nm m 600106 7

L

xd

Two-Slit Interference

Using interference effect we can measure distances with submicronprecision

laser

Detector

Application: Interferometry

Coherent 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

Diffraction = multi-source interference

Multi-Source Interference: 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

Multi-Source Interference

Acceleratedelectrons

Copper

X-rays

Electrons knock out innerelectrons in Cu. When theseelectrons fall back X-rayis emitted.(Medical equipment)

Synchrotron radiation: Electrons circle around accelerator.Constant acceleration leads to radiation

Generating X-Rays

Simple crystal: 3D cubic grid

first layer

Simple case: ‘reflection’ incident angle = reflected anglephase shift = 0

X-Ray: 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 n is an integer nd sin2

X-Ray: Constructive Interference

crystal

turn crystal

x-ray diffracted

nd sin2

May need to observe several maxima to find n and deduce d

Simple X-Ray Experiment

X-ray of Tungsten

Suppose you have a source of X-rays which has a continuum spectrum of wavelengths.How can one make it monochromatic?

crystal

incident broadband X-ray

reflected single-wavelength X-ray

nd sin2

Using Crystal as Monochromator

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 Powdered Crystals

(Myoglobin) 1960, Perutz & Kendrew

X-Ray of Complex Crystals