Coherent vs. incoherent scattering Radiation from an accelerated charge Larmor formula Why the is Reflected and refracted beams from water droplets 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 Incident wave

Forward scattering is coherent— Off-axis scattering is incoherent even if the scatterers are randomly when the scatterers are randomly arranged in the plane. arranged in the plane.

Path lengths are equal. Path lengths are random. Coherent vs. Incoherent Scattering N Incoherent scattering: Total complex amplitude, Aincoh exp(j m ) (paying attention only to the phase m1 of the scattered wavelets) The irradiance: NNN2 2 IAincoh incohexp( j m )  exp( j m ) exp(  j n ) mmn111

NN NN exp[jjN (mn  )]  exp[ ( mn )] mn11mn mn  11 mn

m = n mn  Coherent scattering: N 2 2 Total complex amplitude, A coh   1  N . Irradiance, I  A . So: Icoh  N 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 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


Shorter-wavelength light is scattered out of the beam, leaving longer- wavelength light behind, so the sun appears yellow. In space, the sun is , and the sky is . 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.


coasting at constant velocity 

v for a time t1

tiny period of { acceleration, { r = ct of duration t 1 initial position of a charge q, at rest Radiation from an accelerated charge

E vt By similar triangles:  1 E|| ct  E E|| But the velocity v can be related to the acceleration during the small interval t: vt  1 ct v = a t vt1

vt|| 1 which implies: vat 

at ar and therefore: 1 EE || E || 2 cc

Finally, the field E || must be equal to the field of a static charge (this can be proved using Gauss’ Law): q qa  E||  E  4r 2  2 0 4rc 0 Radiation from an accelerated charge q qa  E||  E  4r 2  2 0 4rc 0 10 0

As r becomes large, the parallel 10 -1 component goes to zero much 10 -2 1/r more rapidly than the perpendicular -3 component. We can therefore 10 10 -4 2 neglect E|| if we are far enough 1/r away from the moving charge. 10 -5

-6 10 0 1 2 3 4 5 6 7 8 9 10

Also: aasin   So, the radiated EM wave has a magnitude: qtasin   Ert,  2 4rc 0 Spatial pattern of the radiation  qt22asin  2 Magnitude of the Poynting vector: 2 Srt,sin223  160 r c direction of the  acceleration a 0 330 30

300 60

 270 S 90

240 120

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  Pt  r2 sin d  d Srt ,   00 q22a  sin3 d 3   8c0 0  This integral is equal to 4/3 Sir Joseph Larmor 1857-1942 q22a Thus: Pt 3 6c 0 . 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 , bound to an atom, in an applied oscillating electric field: eE m x te 0 e  jt (we can neglect the damping e 2 2 factor , for this analysis) 0  We assume that the light wave frequency is much smaller than the

resonant frequency,  << 0, so this is approximately:

eE0 me  jt xe te 2 0 From the position we can compute the acceleration: 2 2 dxe   jt atee22 eEm0 e This is known as dt 0 Rayleigh scattering: Insert this into the Larmor formula to find: scattered power proportional to 4 24 (Rayleigh: 1871) Pascat e P incident This is (mostly) why the sky is blue.

th Rayleigh Scattering: Total scattered power ~ 4 power of the frequency of the incident light

Blue light ( = 400 nm) is scattered 16 times more efficiently than light ( = 800 nm) scattered light that we see


For the same reason, People here see the are red. unscattered remaining light The world of light scattering is a very large one size/wavelength ~0 ~1 Large There are many Rayleigh-Gans Scattering regimes of particle scattering, depending on the particle size, the ~0 light wavelength, and the .

Refractive index As a result, there Geometrical Rayleigh Scattering are countless observable effects Totally reflecting objects of light scattering. Large ~1 Large Another example of incoherent scattering: rainbows water droplet Input light paths

Light can Path leading enter a to minimum deflection droplet at different ~180° deflection distances from its edge.

One can compute the deflection angle of the emerging light Minimum deflection as a function of the angle (~138°) deflection angle incident position. (relative to the original direction) Deflection angle vs. wavelength

Because n varies with wavelength, the minimum deflection angle varies with .

Lots of violet deflected at this angle Lots of red deflected at this angle

Lots of light of all is deflected by more than 138°, so the region below 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- 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

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

Water droplet Deflection angle

Minimum deflection angle (~232.5°) yielding a rainbow radius of 52.5°. 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.) Multiple order bows

3 5

4 6

A simulation of the Ray paths for the higher order bows 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