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1、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. AbsorptionAbsorptionisthedegreetowhichinfraredenergyisabsorb
2、edbyamaterial. Materialssuchasplastic,ceramic,andtextilesaregoodabsorbers. Thermalenergyabsorbedbyreal-worldobjectsisgenerallyretransferredtotheirsurroundingsbyconduction,convection,orradiation. TransmissionTransmissionisthedegreetowhichthermalenergypassesthroughamaterial.
3、 Therearefewmaterialsthattransmitenergyefficientlyintheinfraredregionbetween7and14µm. Germaniumisoneofthefewgoodtransmittersofinfraredenergyandthusitisusedfrequentlyaslensmaterialinthermalimagingsystems. ReflectionReflectionisthedegreetowhichinfraredenergyreflectsoffamater
4、ial. Polishedmetalssuchasaluminum,goldandnickelareverygoodreflectors. Conservationofenergyimpliesthattheamountofincidentenergyisequaltothesumoftheabsorbed,reflected,andtransmittedenergy. IncidentEnergy = AbsorbedEnergy + TransmittedEnergy + ReflectedEnergy[1] EmittedE
5、nergy=AbsorbedEnergyConsiderequation1foranobjectinavacuumataconstanttemperature. Becauseitisinavacuum,therearenoothersourcesofenergyinputtotheobjectoroutputfromtheobject. Theabsorbedenergybytheobjectincreasesitsthermalenergy-thetransmittedandreflectedenergydoesnot. Inorder
6、forthetemperatureoftheobjecttoremainconstant,theobjectmustradiatethesameamountofenergyasitabsorbs. EmittedEnergy = AbsorbedEnergy[2] Therefore,objectsthataregoodabsorbersaregoodemittersandobjectsthatarepoorabsorbersarepooremitters. Applyingequation2,Equation1canberestateda
7、sfollows: IncidentEnergy = EmittedEnergy + TransmittedEnergy + ReflectedEnergy[3] Settingtheincidentenergyequalto100%,theequation3becomes: 100% = %EmittedEnergy + %TransmittedEnergy + %ReflectedEnergy[4] Becauseemissivityequalstheefficiencywithwhichamaterialradiat
8、esenergy,equation4canberestatedasfollows: 100% = Emissivity + %T
