coherent and incoherent scattering

7/29/2019 Coherent and Incoherent Scattering

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11. Light ScatteringCoherent vs. incoherent scattering

Radiation from an accelerated charge

Larmor formula

Why the sky is blue

Rayleigh formula

Reflected and refracted beams from water droplets

rainbows


2/22

Coherent vs. Incoherent light scattering

Coherentlight scattering: scattered wavelets have nonrandomrelative phases in the direction of interest.

Incoherentlight scattering: scattered wavelets have random

relative phases in the direction of interest.

Forward scatteringis coherenteven if the scatterers are randomlyarranged in the plane.

Path lengths are equal.

Off-axis scatteringis incoherentwhen the scatterers are randomlyarranged in the plane.

Path lengths are random.

Incidentwave

Example: Randomly spaced scatterers in a plane

Incidentwave


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Coherentvs. IncoherentScattering

1

exp( )

N

incoh mm

A jIncoherent scattering: Total complex amplitude,

2

2

1 1 1exp( ) exp( ) exp( )

N N N

incoh incoh m m nm m n

I A j j j

The irradiance:

So incoherent scattering is weaker than coherent scattering, but not zero.

1

1N

cohm

A N

Coherent scattering:

Total complex amplitude, . Irradiance, I A2. So: Icoh N2

m=n m n

1 1 1 1exp[ ( )] exp[ ( )]

N N N N

m n m nm n m nm n m n

j j N


4/22

Incoherent scattering: Reflection

from a Rough Surface

A rough surface scatters lightinto all directions with lots ofdifferent phases.

As a result, what we see is light

reflected from many differentdirections. Well see no glare,

and also no reflections.


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6/22


initial positionof a charge q,

at rest

{

tiny period of

acceleration,of duration t

{

coasting atconstant velocityv for a time t1

ct

r = ct1

In order to understand this scattering process, we will analyze it at amicroscopic level. With several simplifying assumptions:1. the scatterer is much smaller than the wavelength of the incident light

2. the frequency of the light is much less than any resonant frequency.


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ct

vt1

||EE

|| 1v t

1v t

By similar triangles: 1

||

v t

c t

E

E

But the velocity v can be related to theacceleration during the small interval t:

v = at

which implies: v a t

1|| || 2

a t a r

c cE E E

and therefore:

||EFinally, the field must be equal to the field of a static charge(this can be proved using Gauss Law):

||204 r

qE

2

0

a

4 rc

q

E


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20

a

4 rc

q

E

|| 204 r

qE

As r becomes large, the parallelcomponent goes to zero muchmore rapidly than the perpendicularcomponent. We can therefore

neglectE|| if we are far enoughaway from the moving charge.

Also: a asin

So, the radiated EM wave has a magnitude:

20

a sin,

4 rc

q tE r t

0 1 2 3 4 5 6 7 8 9 1010-6

10-5

10-4

10-3

10 -2

10-1

100

1/r

1/r2


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Spatial pattern of the radiation

60

240

30

210

0

180

330

150

300

120

270 90

a

S

2D slice 3D cutaway view

direction of the

acceleration

Magnitude of the Poynting vector: 2 2 2 22 2 3

0

a sin, sin

16 r c

q tS r t

No energy is radiated in the direction of the acceleration.


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This integral is equal to 4/3

Total radiated power - the Larmor formula

To find the total power radiated in all directions, integrate themagnitude of the Poynting vector over all angles:

22

0 0

2 23

3

0 0

sin ,

asin

8 c

P t r d d S r t

qd

2 2

30

a

6 c

qP t

Thus:

This is known as the Larmor formula.Total radiated power is independent of distance from the charge Power proportional to square of acceleration


11/22

Larmor formula - application to scattering

0

2 20

j te

e

eE m

x t e

Recall our derivation of the position of an electron, bound toan atom, in an applied oscillating electric field:

(we can neglect the damping

factor , for this analysis)

ae t^ h = dt2d2xe

=-eE0 me~02

~2

e-j~t

From the position we can compute the acceleration:

This is known asRayleighs Law:scattered power

proportional to4

xe t^ h .~0

2

eE0 me e-j~t

We assume that the frequency is much smaller than the resonantfrequency,


12/22

This is why the sky is blue.

Blue light ( = 400 nm) is scattered16times more efficiently than red

light ( =800 nm)

Total scattered power ~ 4th power of thefrequency of the incident light

Rayleighs Law:

sunlight

earth

scattered light thatwe see

For the same reason, sunsets are red.


13/22

The world of light scattering

is a very large oneParticle size/wavelength

Refractivein

dex

Mie Scattering

Ra

yleighScattering

Totally reflecting objects

Geometricaloptics

Rayleigh-Gans Scattering

Large

~1

~0

~0 ~1 Large

There are manyregimes of particlescattering,depending on the

particle size, thelight wavelength,and the refractiveindex.

As a result, thereare countlessobservable effectsof light scattering.


14/22

Another example of incoherent scattering:

the reason for rainbows

Light canenter adroplet atdifferent

distancesfrom its edge.

waterdroplet

One can compute the angle of the emerging light as a

function of the incident position.

Minimum deflection angle (~138);rainbow radius = 42

Input light paths

~180 deflection

Path leadingto minimum deflection


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Deflection angle vs. wavelength

Lots of light of all colors is deflected by more than 138,so the region below rainbow is bright and white.

Because nvaries with wavelength, theminimum deflection angle varies with color.

Lots of red deflected at this angle

Lots of violet deflected at this angle


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A rainbow, with supernumeraries

The sky is much brighter below the rainbow than above.

The multiple greenish-purple arcs inside the primary bow are calledsupernumeraries. They result from the fact that the raindrops are not all

the same size. In this picture, the size distribution is about 8% (std. dev.)


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Explanation of 2nd rainbow

Minimum deflection angle (~232.5)yielding a rainbow radius of 52.5.

Water droplet

Because the angular radius is larger, the 2nd bow is above the 1st one.

Because energy is lost at each reflection, the 2nd rainbow is weaker.

Because of the double bounce, the 2nd rainbow is inverted. And the

region above it (instead of below) is brighter.

A 2nd rainbow can result from light entering the dropletin its lower half and making 2 internal reflections.

Distance fromdroplet edge

D

eflectionangle


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The dark band between the two bows

is known as Alexanders dark band,after Alexander of Aphrodisias who firstdescribed it (200 A.D.)

A double rainbow

Note that the upper bow is inverted.

ray tracing


19/22

Multiple order bows

A simulation of thehigher order bows

3

4

5

6

Ray paths for thehigher order bows

3rd and 4th rainbows are weaker, more spread out, and toward the sun.

5th rainbow overlaps 2nd, and 6th is below the 1st.

There were no reliable reports of sightings of anything higher than a

second order bow, until 2011.


20/22

The first ever photo of a triple and a quad

from Photographic observation of a natural fourth-order rainbow,byM. Theusner,Applied Optics (2011)

(involving multiple superimposed exposures and significant image processing)


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Look here for lots ofinformation and pictures:

Other atmospheric optical effects

http://www.atoptics.co.uk


22/22

Six rainbows?

Explanation:http://www.atoptics.co.uk/rainbows/bowim6.htmhttp://www.atoptics.co.uk/rainbows/bowim6.htm

coherent and incoherent scattering

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