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1、EnglishHomeworkforChapter11.Inancienttimestherectilinearpropagationoflightwasusedtomeasuretheheightofobjectsbycomparingthelengthoftheirshadowswiththelengthoftheshadowofanobjectofknownlength.Astaff2mlongwhenhelderectcastsashadow3.4mlong,whileabuilding’sshadowis170m
2、long.Howtallisthebuilding?Solution.Accordingtothelawofrectilinearpropagation,weget,x=100(m)Sothebuildingis100mtall.2.Lightfromawatermediumwithn=1.33isincidentuponawater-glassinterfaceatanangleof45o.Theglassindexis1.50.Whatangledoesthelightmakewiththenormalinthegla
3、ss?Solution.Accordingtothelawofrefraction,Weget,n′=1.50n=1.33water45oI′Sothelightmake38.8owiththenormalintheglass.3.Agoldfishswims10cmfromthesideofasphericalbowlofwaterofradius20cm.Wheredoesthefishappeartobe?Doesitappearlargerorsmaller?Solution.Accordingtotheequat
4、ion.andn’=1,n=1.33,r=-20AwecangetSothefishappearslarger.R2=-20cmR1=20cmA-10cm4.Anobjectislocated2cmtotheleftofconvexendofaglassrodwhichhasaradiusofcurvatureof1cm.Theindexofrefractionoftheglassisn=1.5.Findtheimagedistance.Solution.Refertothefigure.Accordingtotheequ
5、ationandn=1,n’=1.5,l1=-2cm,r1=1cm,wegetr1=1cmAA′-l1=2cml2′ EnglishHomeworkforChapter21.Anobject1cmhighis30cminfrontofathinlenswithafocallengthof10cm.Whereistheimage?Verifyyouranswerbygraphicalconstructionoftheimage.-l=30cmf′=10cmy=1cmSolution.AccordingtotheGauss’
6、sequation,andl=-30cmf’=10cm.wegetOthersareomitted.2.Alensisknowntohaveafocallengthof30cminair.Anobjectisplaced50cmtotheleftofthelens.Locatetheimageandcharacterizeit.-l=50cmf′=30cmSolution.AccordingtoGauss’sequation,andf′=30cml=-50cmwegetTheimageisareal,largerone.3
7、.Theobjectistransparentcube,4mmacross,placed60cminfrontof20cmfocallength.Calculatethetransverseandaxialmagnificationanddescribewhattheimagelookslike?Solution.FromGauss’sequation,wefindfortherearsurfaceofthecube(thefaceclosertothelens)that,Forthefrontsurface(thefac
8、efartherawayfromthelens),ThetransversemagnificationfortherearsurfaceisButtheaxialmagnificationisSince,thecubedoesn’tlooklikeacube.4.Abiconvexlensismadeo
