lecture 9: reflection and refraction (petty...
TRANSCRIPT
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