MET 4410 Remote Sensing: Radar and Satellite MeteorologyMET 5412 Remote Sensing in Meteorology

Lecture 9: Reflection and Refraction(Petty Ch4)

Whentousethelawsofreflectionandrefraction?

� Thelawsofreflection&refractioncanbeusedtoEMwavesincidentonaplanar(flat)boundarybetweentwohomogeneousmediums.Herewemustconsiderthemediumasawholesurface,nottheindividualparticleswithinthemedium.

� Definitionof“homogeneous”medium:themediumissmoothanduniformonscalescomparabletothewavelengthoftheradiation,i.e.,thesizeandspacingoftheindividualmoleculesandotherirregularitiesaremuchsmallerthanthewavelength.

� Homogeneousmediumexamples:water,air,glass,etc.arealleffectivelyhomogeneoustoradiationinvisible,IR,andmicrowaveband.

� Inhomogeneousmediumexamples:◦ Milkforvisiblelight◦ Mostobjects(water,glass,milk,etc.)areinhomogeneous forx-raysandgammarays.◦ Cloudswith10µmdiameterdropletsarehomogeneous inmicrowaveband,butinhomogeneous inIRandvisiblebands.

Thecomplexindexofrefractionofamedium� Definition:N=nr+i ni� Fromamacroscopicpointofview,allthecomplexitybetweenradiationandmediaishiddeninN.

� Foranysubstance,N isafunctionofwavelength,temperature,pressure,etc.

� TherealpartnrcontrolsthephasespeedoftheEMwavethroughthemedium.nrisdefinedastheratioofthespeedoflightinvacuumtothespeedofEMwavesthroughthemedium:𝑛" =

$$%

◦ Forallrealsubstance,nr>1.◦ nrwater≈1.33atvisibleband◦ nrair≈1.0003atvisibleband,atsealevel

ImaginarypartofN� Theimaginarypartni describestherateofabsorption.� ni ≠0meansabsorptionoccursasanEMwavepassesamedium

� ni isalsocalledabsorptionindex.Therelationshipbetweenabsorptioncoefficientβα𝝺 andabsorptionindexni is:

𝛽(𝝺 =4𝜋𝑛+𝝺

� nr and ni arenotindependent.Ifyouknowoneofthem,youcangettheotherthroughsomecomplicatedrelationships.

1

1.1

1.2

1.3

1.4

1.5

1.6

1.7

1.8

0.1 0.15 0.2 0.3 0.5 0.7 1 1.5 2 3 4 5 7 10 15 20 30 50

nr

Wavelength [µm]

(a) Index of Refraction of Water and Ice (Real Part)

Water (10o C)

Ice (-5o C)

1e-09

1e-08

1e-07

1e-06

1e-05

0.0001

0.001

0.01

0.1

1

0.1 0.15 0.2 0.3 0.5 0.7 1 1.5 2 3 4 5 7 10 15 20 30 50

ni

Wavelength [µm]

(b) Index of Refraction of Water and Ice (Imag. Part)

Water (10o C)

Ice (-5o C)

ReflectionandRefraction� Whatcausesthephenomenaofreflectionandrefraction?– Thesuddenchangeofphasespeedatboundariesbetweenmedia.

� Reflection:WhenanEMwaveencountersaplanarboundarybetweentwohomogeneousmediahavingdifferentindicesofrefraction,someoftheenergyoftheradiationisreflected;whiletheremainderpassesthroughtheboundaryintothesecondmedium.

� Refraction:Thedirectionofthetransmittedwaveinmedium2mayalteredfromtheoriginaldirectioninmedium1,aphenomenonknownasRefraction.

AngleofReflection� Thereflectedrayliesinthesameplaneasthedirectionofincidentrayandlocalnormal,butontheoppositesidefromtheincidentray.

� Thereflectedangleequalstheincidentangle:𝜃+ = 𝜃"� Thewavelengthofreflectedrayequalsthewavelengthofincidentray:𝝺+ = 𝝺"

incident angle= 0°

AngleofRefractionandSnell’sLaw� Inrefraction,thetransmittedraychangesdirectionaccordingtoSnell’sLaw:

-+./01%

= -+./213

,N1 andN2aretheindicesofrefractionofmedium1and2.𝜃4 istheangleofthetransmittedrayrelativetolocalnormal.

� If𝜃+ = 0, then𝜃4 = 0.� If𝜃+ > 0𝑎𝑛𝑑N1 <N2 (fromairtowater),then𝜃4 < 𝜃+ andthewavelength

oftransmittedrayislessthanthewavelengthofincidentray:𝝺4 < 𝝺+

incident angle 𝜃+ > 0and N1 < N2

CriticalAngleofRefraction� If𝜃+ > 0𝑎𝑛𝑑N1 >N2 (fromwatertoair),then𝜃4 > 𝜃+ andthewavelengthoftransmittedrayisgreaterthanthewavelengthofincidentray:𝝺4 > 𝝺+

� Underthesamecondition,Criticalangle𝜃A ≡ arcsin(131%).When𝜃+ = 𝜃A, 𝜃4 =90°.It’simpossibletohave𝜃+ >θAbecasueitwouldmake𝑠𝑖𝑛𝜃4 > 1.Sowavesincidentontheinterfaceatananglegreaterthanthecriticalanglecan’tpassthroughtheinterfaceatallbutratherexperiencetotalreflection.

� Inthevisibleband,N1 ≈1.33forwaterandN2 ≈1.0003forair,thenθA ≈49°.

ApplicationstoMeteorology� Forplanarboundariesbetweenhomogenousmediums,wecouldusetherulesforreflectionandrefraction.

� Thenhowaboutwhenweconsiderindividualparticles?◦ Therulesforreflectionandrefraction canbeappliednotonlytoplanarboundaries,buttoanysurfacewhoseradius(r)ofcurvatureismuchgreaterthanthewavelength(𝝺)oftheradiation.

� Therefore,foratmosphericparticles,whenr>>𝝺 (orsizeparameterx>50~2000,wherex=2πr/𝝺),wecanusereflection/refractionrules(i.e.raytracingorgeometricoptics)tosolvescattering/absorptionpropertiesintheradiativetransfercalculations.

� Ontheotherhand,whenr~𝝺 (orsizeparameter0.2<x<50~2000),wehavetouseMie/Rayleighscatteringtheorytosolvescatteringpropertiesintheradiativetransfercalculations.

ScatteringRegimes

MieRayleigh

UseGeometricOpticstoExplainRainbow� Condition:scattering(reflectionandrefraction)ofvisiblesunlight(0.3µm<𝝺<0.7µm )bylargecloudiceparticles(radius>50µm)orraindrops(100µmto3mm).

From:https://scijinks.gov/rainbow/

� Whendoyouseerainbow:◦ Arainbowrequireswaterdroplets tobefloating intheair.That’swhyweseethemrightafteritrains.TheSunmustbebehindyouandthecloudsclearedawayfromtheSunfortherainbowtoappear.

� Whyisarainbowabow—orarc?◦ Afullrainbowisactuallyacompletecircle,butfromtheground weseeonlypartofit.Fromanairplane,intherightconditions, onecanseeanentirecircularrainbow.

UseGeometricOpticstoExplainRainbow

PettyFig.4.7

� Thepathofasingleincidentray:◦ Externalreflection◦ Transmittedthrough refraction◦ Atthebacksideof thedrop,afractionofenergygoesdirecttransmission,whileanother fractionreflectedinternally◦ Single internalreflection:theportion oftheoriginal raythatwasinternallyreflected.Itisresponsible fortheprimary rainbow.Thefirstinternally reflectedraywillbetransmittedbyrefractionandreflectedagain.◦ Double internalreflection:repeattheaboveprocess,responsible forsecondaryrainbow.◦ Separationofcolorsinarainbow:becausenr forwaterincreasesslightly fromtheredendtothevioletend,causingdifferentrefractionanglesfordifferentwavelengths.

Whydoweseeprimaryrainbowfromsingleinternalreflection,notthedirecttransmission?

� Directtransmission:◦ separatedcoloredraysaretransmitted/refractedoutoftheraindropatthesamedirection

� Singleinternalreflection:◦ Theanglebetweentheincidentrayandthesinglereflectedrayisalways42deg.◦ Theseparatedcoloredrayswerecombinedinthemiddleofthepath,andseparatedagainatthebottomoftheraindrop,thenrefractedoutoftheraindrop,causinganevenlargerseparationofdifferentcolors.

From:https://scijinks.gov/rainbow/

SecondaryRainbow� Thesecondaryrainbowiscausedbyasecondreflectioninsidethedroplet,andthis“re-reflected”lightexitsthedropatadifferentangle(50°insteadof42° fortheredprimarybow).Thisiswhythesecondaryrainbowappearsabovetheprimaryrainbow.

� Thesecondaryrainbowwillhavetheorderofthecolorsreversed,too,withredonthebottomandvioletonthetop.

From:https://scijinks.gov/rainbow/

Transmitivity (transmittance)andReflectivity(reflectance)

� Recallthedefinitionofabsorptivity(absorptance):

𝑎𝝺=WX-Y"XZ["W[+W4+Y.W4\W]Z^Z._4`𝝺+.$+[Z.4"W[+W4+Y.W4\W]Z^Z._4`𝝺

� Similarly,reflectivity(reflectance)isdefinedas:

𝑟𝝺="Zb^Z$4Z["W[+W4+Y.W4\W]Z^Z._4`𝝺+.$+[Z.4"W[+W4+Y.W4\W]Z^Z._4`𝝺

� Transmitivity (transmittance)isdefinedas:

𝑡𝝺=4"W.-d+44Z["W[+W4+Y.W4\W]Z^Z._4`𝝺+.$+[Z.4"W[+W4+Y.W4\W]Z^Z._4`𝝺

Relationshipbetweenabsorptivity,transmitivity,andreflectivity

� Case1:sunlightincidentonEarth’ssurface,notransmission,so:𝑎𝝺 +𝑟𝝺 =1

� Case2:Whensunlightincidentonthetopofacloudlayer.Aphotonofsunlightwillexperienceoneofthefourpossiblefates:◦ Directtransmittedwithoutbeingscatteredorabsorbed:directtransmittance𝑡𝝺dir◦ Itmaybescatteredoneormoretimesandemergefromthebottomofthecloudlayer:diffusetransmittance𝑡𝝺diff◦ Itmaybescatteredoneormoretimesandemergefromthetopofthecloudlayer:reflectance𝑟𝝺◦ Whateverisneithertransmittednorreflectedmustbeabsorbed:absorptance 𝑎𝝺◦ Obviouslyforcase2:𝑡𝝺dir+𝑡𝝺diff+𝑎𝝺 +𝑟𝝺 =1,

where𝑡𝝺dir+𝑡𝝺diff=𝑡𝝺