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红外波段的发射率、透过率和反射率的关系EmissivityintheInfraredPhysicsofEmissivityInfrared(thermal)energy,whenincidentuponmatter,beitsolid,liquidorgas,willexhibitthepropertiesofabsorption,reflection,andtransmissiontovaryingdegrees.AbsorptionAbsorptionisthedegreetowhichinfraredenergyisabsorbedbyamaterial.Materialssuchasplastic,ceramic,andtextilesaregoodabsorbers.Thermalenergyabsorbedbyreal-worldobjectsisgenerallyretransferredtotheirsurroundingsbyconduction,convection,orradiation.TransmissionTransmissionisthedegreetowhichthermalenergypassesthroughamaterial.Therearefewmaterialsthattransmitenergyefficientlyintheinfraredregionbetween7and14µm.Germaniumisoneofthefewgoodtransmittersofinfraredenergyandthusitisusedfrequentlyaslensmaterialinthermalimagingsystems.ReflectionReflectionisthedegreetowhichinfraredenergyreflectsoffamaterial.Polishedmetalssuchasaluminum,goldandnickelareverygoodreflectors.Conservationofenergyimpliesthattheamountofincidentenergyisequaltothesumoftheabsorbed,reflected,andtransmittedenergy.IncidentEnergy=AbsorbedEnergy+TransmittedEnergy+ReflectedEnergy[1]EmittedEnergy=AbsorbedEnergyConsiderequation1foranobjectinavacuumataconstanttemperature.Becauseitisinavacuum,therearenoothersourcesofenergyinputtotheobjectoroutputfromtheobject.Theabsorbedenergybytheobjectincreasesitsthermalenergy-thetransmittedandreflectedenergydoesnot.Inorderforthetemperatureoftheobjecttoremainconstant,theobjectmustradiatethesameamountofenergyasitabsorbs.EmittedEnergy=AbsorbedEnergy[2]Therefore,objectsthataregoodabsorbersaregoodemittersandobjectsthatarepoorabsorbersarepooremitters.Applyingequation2,Equation1canberestatedasfollows:IncidentEnergy=EmittedEnergy+TransmittedEnergy+ReflectedEnergy[3]Settingtheincidentenergyequalto100%,theequation3becomes:100%=%EmittedEnergy+%TransmittedEnergy+%ReflectedEnergy[4]Becauseemissivityequalstheefficiencywithwhichamaterialradiatesenergy,equation4canberestatedasfollows:100%=Emissivity+%TransmittedEnergy+%ReflectedEnergy[5]Applyingsimilartermsto%TransmittedEnergyand%ReflectedEnergy,100%=Emissivity+Transmissivity（透过率）+Reflectivity[6]Accordingtoequation6,thereisabalancebetweenemissivity,transmissivity,andreflectivity.Increasingthevalueofoneoftheseparametersrequiresadecreaseinthesumoftheothertwoparameters.Iftheemissivityofanobjectincreases,thesumofitstransmissivityandreflectivitymustdecrease.Likewise,ifthereflectivityofanobjectincreases,thesumofitsemissivityandtrasmissivitymustdecrease.Mostsolidobjectsexhibitverylowtransmissionofinfraredenergy-themajorityofincidentenergyiseitherabsorbedorreflected.Bysettingtransmissivityequaltozero,equation6canberestatedasfollows:100%=Emissivity+Reflectivity[7]Forobjectsthatdonottransmitenergy,thereisasimplebalancebetweenemissivityandreflectivity.Ifemissivityincreases,reflectivitymustdecrease.Ifreflectivityincreases,emissivitymustdecrease.Forexample,aplasticmaterialwithemissivity=0.92hasreflectivity=0.08.Apolishedaluminumsurfacewithemissivity=0.12hasreflectivity=0.88.Theemissiveandreflectivebehaviorofmostmaterialsissimilarinthevisibleandinfraredregionsoftheelectromagneticspectrum.Polishedmetals,forexample,havelowemissivityandhighreflectivityinboththevisibleandinfrared.Itisimportanttounderstand,however,thatsomematerialsthataregoodabsorbers,transmitters,orreflectorsinthevisible,mayexhibitcompletelydifferentcharacteristicsintheinfrared.