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 dropletsRainbows

Coherent vs. Incoherent light scatteringCoherent light scattering: scattered wavelets have nonrandomrelative 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.

Incident wave

Example: Randomly spaced scatterers in a plane

Incident wave

Coherent vs. Incoherent Scattering

1exp( )

N

incoh mm

A jIncoherent scattering: Total complex amplitude,(paying 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:

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

11

N

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

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.

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.

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, so the sun appears yellow. In space, the sun is white, and the sky is black.

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.

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 t

which implies: v a t

1|| || 2

a t a rc 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 r

qE

2

0

a4 rc

qE

Radiation from an accelerated charge

20

a4 rc

qE

|| 204 r

qE

As r becomes large, the parallel component goes to zero muchmore rapidly than the perpendicular component. We can therefore neglect E|| if we are far enough away from the moving charge.

Also: a a sin

So, the radiated EM wave has a magnitude:

2

0

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

10 0

1/r

1/r2

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 22

2 2 30

a sin, sin

16 r cq t

S r t

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 proportional to 4

(Rayleigh: 1871)

We assume that the light wave frequency is much smaller than theresonant 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.

Blue light ( = 400 nm) is scattered 16 times more efficiently than red light ( = 800 nm)

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 see the unscattered

remaining light

The world of light scattering is a very large one

Particle size/wavelength

Ref

ract

ive

inde

x

Mie Scattering

Ray

leig

h S

catte

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 rainbowsIf 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 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.htmhttp://www.atoptics.co.uk/rainbows/bowim6.htm