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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
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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
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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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Radiation from an accelerated charge
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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Radiation from an accelerated charge
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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Radiation from an accelerated charge
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
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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,
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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.
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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.
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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
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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.
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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
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Six rainbows?
Explanation:http://www.atoptics.co.uk/rainbows/bowim6.htmhttp://www.atoptics.co.uk/rainbows/bowim6.htm