Coherent vs. Incoherent light scattering

Transcription

1 11. Light Scattering Coherent vs. incoherent scattering Radiation from an accelerated charge Larmor formula Rayleigh scattering Why the sky is blue Reflected and refracted beams from water droplets Rainbows

2 Coherent vs. Incoherent light scattering Coherent light scattering: scattered wavelets have non-random 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 Incident wave Forward scattering is coherent even if the scatterers are randomly arranged in the plane. Path lengths are equal. Off-axis scattering is incoherent when the scatterers are randomly arranged in the plane. Path lengths are random.

3 Coherent vs. Incoherent Scattering N A Incoherent scattering: Total complex amplitude, (here we pay attention only to the phase of the scattered wavelets) The irradiance: 2 incoh N 2 N N m= 1 exp( jθ ) I A exp( jθ ) = exp( jθ ) exp( jθ ) incoh incoh m m n m= 1 m= 1 n= 1 N N N N = exp[ j( θ θ )] + exp[ j( θ θ )] = N m n m n m= 1 n= 1 m n m= 1 n= 1 m= n m This term averages to zero. This one doesn t! Coherent scattering: N Total complex amplitude, A 1 = N. Irradiance, I A 2. So: I coh N 2 coh m= 1 The intensity of incoherently scattered light is proportional to N, while coherent light is proportional to N 2. Since N is often a very large number, incoherent scattering is much weaker than coherent scattering. But not zero.

4 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.

5 Coherent scattering: Reflection from a smooth surface preserved output wave front input wave front How smooth does the surface need to be? To be smooth, the roughness needs to be smaller than the wavelength of the light. 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.

6 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.

7 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 t 1 tiny period of acceleration, of duration t { { θ initial position of a charge q, at rest r = ct 1

8 Radiation from an accelerated charge vt 1 θ v t 1 E E c t By similar triangles: E E v t = c t But the velocity v can be related to the acceleration during the small interval t: v = a t 1 v t1 which implies: v = a t and therefore: a t1 a r E = E = 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 2 4πε 0r a E = q 4πε rc 0 2

9 Radiation from an accelerated charge E = q 2 4πε 0r 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. a E = q 4πε rc /r 2 1/r Also: a = a sinθ So, the radiated EM wave has a magnitude: (, ) E r t ( ) a sin = q t θ 2 4πε rc 0

10 Spatial pattern of the radiation Magnitude of the Poynting vector: S ( r t) direction of the acceleration 0 a ( t) q 2 a 2 sin 2 θ, = sin 16π ε r c θ S D slice 150 3D cutaway view No energy is radiated in the direction of the acceleration.

11 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: Sir Joseph Larmor π 2π P t r d d S r t 2 ( ) = sin θ θ φ (, ) q a = 8πε c Thus: π ( ) P t sin 3 θdθ 2 2 a = q 6πε c 0 3 This integral is equal to 4/3 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

12 Larmor formula: application to scattering Recall our derivation of the position of an electron, bound to an atom, in an applied oscillating electric field: ( ) ee m 0 e j t xe t = e ω ( 2 2 ω ) 0 ω (we can neglect the damping factor Γ, for this analysis) We assume that the light wave frequency is much smaller than the resonant frequency, ω << ω 0, so this is approximately: ee 0 ( ) m e xe t e 2 ω0 From the position we can compute the acceleration: jωt 2 2 d xe ω ae ( t) = = ( ee m ) e dt 2 0 e 2 ω0 Insert this into the Larmor formula to find: P a ω P 2 4 scat e incident jωt This is known as Rayleigh scattering: scattered power is proportional to ω 4 (Rayleigh: 1871)

13 This is (mostly) why the sky is blue. Rayleigh Scattering: Total scattered power ~ 4 th power of the frequency of the incident light Blue light (λ = 400 nm) is scattered 16 times more efficiently than red light (λ = 800 nm) sunlight scattered light that we see earth For the same reason, sunsets are red. People here looking back at the sun see the unscattered remaining light

14 The world of light scattering is a very large one Refractive index Large ~1 ~0 Rayleigh Scattering Particle size/wavelength ~0 ~1 Large Rayleigh-Gans Scattering Mie Scattering Totally reflecting objects Geometrical optics 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.

15 Another example of incoherent scattering: rainbows Input light paths water droplet Light can enter a droplet at different distances from its edge. Path leading to minimum deflection ~180 deflection One can compute the deflection angle of the emerging light as a function of the incident position. Minimum deflection angle (~138 ) deflection angle (relative to the original direction)

16 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.

17 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: = 42 opening angle.

18 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.)

19 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 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.

20 A double rainbow ray tracing 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.)

21 Multiple order bows 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 1 st. There were no reliable reports of sightings of anything higher than a second order natural rainbow, until

22 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)

23 Other atmospheric optical effects Look here for lots of information and pictures:

24 Six rainbows? Explanation: