11. Light Scattering
Coherent vs. incoherent scattering
Radiation from an accelerated charge
Larmor formula
Why the sky is blue
Rayleigh scattering
Reflected and refracted beams from water droplets
Rainbows
Coherent vs. Incoherent light scattering
Coherent light scattering: scattered wavelets have nonrandom
relative phases in the direction of interest.
Incoherent light scattering: scattered wavelets have random
relative phases in the direction of interest.
Example:
Randomly spaced scatterers in a plane
Incident
wave
Forward scattering is coherent—
even if the scatterers are randomly
arranged in the plane.
Path lengths are equal.
Incident
wave
Off-axis scattering is incoherent
when the scatterers are randomly
arranged in the plane.
Path lengths are random.
Coherent vs. Incoherent Scattering
Incoherent scattering: Total complex amplitude, Aincoh 
(paying attention only to the phase
of the scattered wavelets)
The irradiance:
I incoh  Aincoh 
2
2
N
 exp( j
m 1
m
)

N
N
 exp( j
m 1
)
N
 exp( j ) exp( j
m 1
m
m
n 1
n
)
 N N

 N N

   exp[ j ( m   n )]     exp[ j ( m   n )] 
N
 m 1 n 1
 m  n  m 1 n 1
m n
m = n
m  n
Coherent scattering:
N
Total complex amplitude, Acoh  1  N. Irradiance, I  A2. So: Icoh  N2
m 1
So incoherent scattering is weaker than coherent scattering, but not zero.
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
reflected from many different
directions. We’ll see no glare,
and also no reflections.
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.
How smooth does the surface need to be?
To be smooth, the roughness needs to be
smaller than the wavelength of the light.
As a result, images are formed
by the reflected light.
Wavelength-dependent incoherent
scattering: Why the sky is blue
Air molecules scatter light, and the scattering depends on frequency.
Light from the sun
Air
Shorter-wavelength light is scattered out of the beam, leaving longerwavelength light behind, so the sun appears yellow. In space, the sun
is white, and the sky is black.
Radiation from an accelerated charge
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 light
2. the frequency of the light is much less than any resonant frequency.
ct
coasting at
constant velocity
v for a time t1
tiny period of
acceleration,
of duration t

{
{
initial position
of a charge q,
at rest
r = ct1
Radiation from an accelerated charge
By similar triangles:

vt1
v  t1
E
E||
ct
E v  t 1

E|| ct
But the velocity v can be related to the
acceleration during the small interval t:
v = a t
which implies: v   a  t
v|| t1
and therefore:
a t 
a r
E  E||   1   E||  2 
 c 
 c 
Finally, the field E|| must be equal to the field of a static charge
(this can be proved using Gauss’ Law):
E|| 
q
4 0 r 2
qa 
E 
4 0 rc 2
Radiation from an accelerated charge
E|| 
q
qa 
E 
4 0 rc 2
4 0 r 2
10 0
As r becomes large, the parallel
component goes to zero much
more rapidly than the perpendicular
component. We can therefore
neglect E|| if we are far enough
away from the moving charge.
10 -1
1/r
10 -2
10 -3
10 -4
1/r2
10 -5
10 -6 0
1
2
3
4
5
6
Also: a   a sin 
So, the radiated EM wave has a magnitude:
qa  t  sin 
E  r, t  
4 0 rc 2
7
8
9
10
Spatial pattern of the radiation

q 2 a 2  t  sin 2 
2
Magnitude of the Poynting vector: S  r , t  
sin


2
2 3
16  0 r c
direction of the
acceleration
0
330

a
30
60
300

S
270
240
90
120
210
150
180
2D slice
3D cutaway view
No energy is radiated in the direction of the acceleration.
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

P  t    r sin  d  d S  r , t 
2
0
0

q 2a 2
3

sin
 d
3 
8 0 c 0
Sir Joseph Larmor
1857-1942
This integral is equal to 4/3
q 2a 2
Thus: P  t  
6 0 c3
 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
Larmor formula - application to scattering
Recall our derivation of the position of an electron, bound to
an atom, in an applied oscillating electric field:
xe  t  

eE0 me
0 2   2

e  jt
(we can neglect the damping
factor , for this analysis)
We assume that the light wave frequency is much smaller than the
resonant frequency,  << 0, so this is approximately:
xe  t  
eE0 me
0 2
e  jt
From the position we can compute the acceleration:
d 2 xe
 2  jt
ae  t   2    eE0 me  2 e
0
dt
Insert this into the Larmor formula to find:
Pscat  ae 2   4  Pincident
This is known as
Rayleigh scattering:
scattered power
proportional to 4
(Rayleigh: 1871)
This is (mostly) why the sky is blue.
Rayleigh Scattering:
Total scattered power ~ 4th power of
the frequency of the incident light
Blue light ( = 400 nm) is scattered
16 times more efficiently than red
light (
= 800 nm)
sunlight
scattered light that
we see
earth
For the same reason,
sunsets are red.
People here see the
unscattered
remaining light
The world of light scattering
is a very large one
~0
Particle size/wavelength
~0
~1
Large
Mie Scattering
Totally reflecting objects
Geometrical optics
Rayleigh Scattering
~1
Large
Refractive index
Rayleigh-Gans Scattering
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
water
Input light paths
Light can
enter a
droplet at
different
distances
from its edge.
droplet
Path leading
to minimum deflection
One can compute
the deflection angle
of the emerging light
as a function of the
incident position.
~180° deflection
Minimum deflection
angle (~138°)
deflection angle
(relative to the
original direction)
Deflection angle vs. wavelength
Because n varies with wavelength, the
minimum deflection angle varies with color.
Lots of violet deflected at this angle
Lots of red deflected at this angle
Lots of light of all colors is deflected by more than 138°,
so the region below rainbow is bright and white.
The size of rainbows
If the light 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 supernumeraries
The 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
Water droplet
Minimum deflection angle (~232.5°)
yielding a rainbow radius of 52.5°.
Deflection angle
A 2nd rainbow can result from light entering the droplet
in its lower half and making 2 internal reflections.
Distance from
droplet edge
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 double rainbow
Note that the upper bow is inverted.
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.)
“ray tracing”
Multiple order bows
3
5
4
6
Ray paths for the
higher order bows
A simulation of 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
(involving multiple superimposed exposures and significant image processing)
from “Photographic observation of a natural fourth-order rainbow,” by
M. Theusner, Applied Optics (2011)
Other atmospheric optical effects
Look here for lots of
information and pictures:
http://www.atoptics.co.uk
Six rainbows?
Explanation: http://www.atoptics.co.uk/rainbows/bowim6.htm