11. Light Scattering Coherent vs. incoherent scattering Radiation from an accelerated charge – derivation of the Larmor formula Rayleigh scattering – why the sky is blue Reflection and refraction from water droplets – rainbows!
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11. Light Scattering
Coherent vs. incoherent scattering
Radiation from an accelerated charge
– derivation of the Larmor formula
Rayleigh scattering
– why the sky is blue
Reflection and refraction from water droplets– rainbows!
Coherent vs. Incoherent light scatteringCoherent light scattering: scattered wavelets have non-randomrelative phases in the direction of interest.
Incoherent light scattering: scattered wavelets have randomrelative phases in the direction of interest.
Forward scattering is coherent—even if the scatterers are randomly arranged in the plane.
Path lengths are equal.
Off-axis scattering is incoherentwhen the scatterers are randomly arranged in the plane.
Path lengths are random.
Example: Randomly spaced scatterers in a plane
Incident wave
Incident wave
Coherent vs. Incoherent Scattering
1exp( )
N
incoh mm
A jIncoherent scattering: Total complex amplitude,(for simplicity, we pay attention only
to the phase of the scattered wavelets)
22
1 1 1
exp( ) exp( ) exp( )
N N N
incoh incoh m m nm m n
I A j j j
The irradiance:
The intensity of incoherently scattered light is proportional to N, while coherent light is proportional to N2. Since N is often a very large number, incoherent scattering is much weaker than coherent scattering. But not zero.
1
1N
cohm
A N
Coherent scattering:Total complex amplitude, . Irradiance, I A2. So: Icoh N2
1 1 1 1exp[ ( )] exp[ ( )]
N N N N
m n m nm n m nm n m n
j j N
This term averages to zero. This one doesn’t!
Incoherent scattering: Reflection from a rough surface
A rough surface scatters light into all directions with lots of different phases.
As a result, what we see is light that came from many different
directions. We see no glare, and also no reflected images.
Most of what you see around you is light that has been incoherently scattered.
Coherent scattering: Reflection from a smooth surface
A smooth surface scatters light all into the same direction, thereby preserving the phase of the incident wave (and the amplitude too).
As a result, images are formed by the reflected light.
How smooth does the surface need to be? To be smooth, the roughness needs to be smaller than the wavelength of the light.
input wave front
preserved output wave front
One way to think about coherent scattering: it is a process which preserves images.
Wavelength-dependent incoherent scattering: Why the sky is blueAir molecules scatter light, and the scattering depends on frequency.
Shorter-wavelength light is scattered out of the beam, leaving longer-wavelength light behind.
Light from the sun
Air
Radiation from an accelerated charge
initial position of a charge q,
at rest
{tiny period of acceleration, of duration t
{
coasting at constant velocity v for a time t1
ct
r = ct1
In order to understand this scattering process, we will analyze it at a microscopic level. With several simplifying assumptions:1. the scatterer is much smaller than the wavelength of the incident light2. the frequency of the light is much less than any resonant frequency.
Remember this?
Radiation from an accelerated charge
ct
vt1
||EE
|| 1v t
1v t
By similar triangles: 1
||
v tc t
EE
But the velocity v can be related to the acceleration during the small interval t:
v = a twhich implies: v a t a sin t
1|| || 2
a t a r sinc c
E 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 rqE
20
a sin4 rcqE
Radiation from an accelerated charge
20
a sin4 rcqE || 2
04 rqE
So, we can describe the radiated EM wave as:
2
0
a sin,
4 rcq t
E r t
0 1 2 3 4 5 6 7 8 9 1010 -6
10 -5
10 -4
10 -3
10 -2
10 -1
10 0
1/r
1/r2
As r becomes large, the parallel component goes to zero muchmore rapidly than the perpendicular component. We can therefore neglect if we are far enough away from the moving charge.
Also, note that does not depend on time, but does.E
||E
||E
Spatial pattern of the radiationMagnitude of the Poynting vector: 2 2 2 2
22 2 3 2
0
a sin a, sin16 r c rq t
S r t
60
240
30
210
0
180
330
150
300
120
270 90
a
S
2D slice 3D cutaway view
direction of the acceleration
No energy is radiated in the direction of the acceleration.
This integral is equal to 4/3
Total radiated power - the Larmor formula
To find the total power radiated in all directions, integrate the magnitude of the Poynting vector over all angles:
2
2
0 0
2 23
30 0
sin ,
a sin8 c
P t r d d S r t
q d
2 2
30
a6 c
qP t
Thus:
This is known as the Larmor formula (1897) Total radiated power is independent of distance from the charge Total power proportional to square of acceleration
Sir Joseph Larmor1857-1942
Larmor formula: application to scattering
0
2 20
j tee
eE mx t e
Recall our derivation of the position of an electron, bound to an atom, in an applied oscillating electric field:
(we can neglect the damping factor , for this analysis)
This is known as Rayleigh scattering: scattered power is proportional to 4
(Rayleigh: 1871)
We assume that the light wave frequency is much smaller than the resonant frequency, << 0, so this is approximately:
02
0
j tee
eE mx t e
From the position we can compute the acceleration:
2 2
02 20
j tee e
d xa t eE m edt
Insert this into the Larmor formula to find:2 4
scat e incidentP a P
This is (mostly) why the sky is blue.Total scattered power ~ 4th power of the frequency of the incident lightRayleigh Scattering:
sunlight
earth
scattered light that we see
For the same reason, sunsets are red.
People here looking back at the sun see
the unscattered remaining light
Blue light ( = 400 nm) is scattered 16 times more efficiently than red light ( = 800 nm)
atmosphere
The world of light scattering is a very large one
Particle size/wavelength
Ref
ract
ive
inde
x
Mie Scattering
Ray
leig
h Sc
atte
ring
Totally reflecting objects
Geo
met
rical
opt
ics
Rayleigh-Gans Scattering
Larg
e
~1
~
0
~0 ~1 Large
There are many regimes of particle scattering, depending on the particle size, the light wavelength, and the refractive index.
As a result, there are countless observable effects of light scattering.
Another example of incoherent scattering: rainbows
Light can enter a droplet at different distances from its edge.
waterdroplet
One can compute the deflection angleof the emerging light as a function of the incident position.
Minimum deflection angle (~138°)
Input light paths
~180° deflection
Path leadingto minimum deflection
deflection angle (relative to the original direction)
Lots of light of all colors is deflected by more than 138°, so the region below rainbow is bright and white.
Because n varies with wavelength, the minimum deflection angle varies with color.
Lots of red deflected at this angle
Lots of violet deflected at this angle
Deflection angle vs. wavelength
The size of rainbowsRainbows are full circles, but often part of the circle is blocked by the ground.
If the scattered light’s source is lower than the viewer’s perspective, then you can see more than half an arc.
The minimum deviation angle of 138 is what determines the size of the circle seen by the viewer: 180 – 138 = 42 opening angle.
A rainbow, with supernumerariesThe sky is much brighter below the rainbow than above.
The multiple greenish-purple arcs inside the primary bow are called “supernumeraries”. 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.)
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 droplet in its lower half and making 2 internal reflections.
Distance from droplet edge
Def
lect
ion
angl
e
The dark band between the two bows is known as Alexander’s dark band, after Alexander of Aphrodisias who first described it (200 A.D.)
A double rainbowNote that the upper bow is inverted.
“ray tracing”
Multiple order bows
A simulation of the higher order bows
3
4
5
6
Ray paths for the higher 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 natural rainbow, until…
The first ever photo of a triple and a quad
from “Photographic observation of a natural fourth-order rainbow,” by M. Theusner, Applied Optics (2011)
(involving multiple superimposed exposures and significant image processing)
Look here for lots of information and pictures:
Other atmospheric optical effects
http://www.atoptics.co.uk
Six rainbows?
Explanation: http://www.atoptics.co.uk/rainbows/bowim6.htm
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