11. light scattering - brown university .11. light scattering coherent vs. incoherent scattering

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  • 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 coherenteven 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. Well 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:

    20

    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 2

    22 2 3

    0

    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,

  • 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 viewers 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 Alexanders 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

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