PHY 208 - wave-optics
Transcript of PHY 208 - wave-optics
Wave Optics
1. Lightasawave1. Huygens'Princinple2. Refraction3. LawofRefraction4. Thelightfromthesun5. ChromaticDispersion
2. Diffraction3. Thedoubleslitexperiment4. WavesandReflections5. Interferometer6. Thediffractionoflight
1. SingleSlit2. MultipleSlits3. CircularApertureDiffraction
1. Light as a wave
We'vespentsometimedevelopingapowerfulsetoftoolsforanalyzingthephysicsofwaves.Sofar,we'vetreatedcaseslikewavesonastring,orsoundwavesastheypropagatethroughair.But,therearemanymorephenomenainthenaturalworldthatcanbeexplainedandanalyzedusingwavephysics.Amajoronethatpersiststhroughoutthedevelopmentofscienceandhumanhistoryhasbeenlight.Fromancienttextstomoderncommunicationtechnology,thequesttounderstandlighthasbeenadrivingforceinscience.We'lltakealooknowathowlightcanbeunderstandasawave.
Intheearlyyearsofscientificthought,thestudyoflightwasveryproblematics.Severaloftheconvenienceswehavenowdidn'texists,andthismadelightaratherdifficultsubjecttostudy.Forexample,weknownowthespeedoflighttobe186,000milespersecond.(Or300,000,000m/s)So,imaginetryingtomeasuresomethingthatmovedthisfastusingonlypendulumclocksorsundials.Itwouldbehard.Also,whenwestudylightnow,weuselensesandothertoolstoshapeitandmakeitbehave.Suchthingswerenotavailableinthebeginningofnaturalscience.So,mostoftheworkonlightwasdoneusingreasoninganddeduction,ratherthenmeasurements.Ofcourse,someexperimentsweredone,buttheywereoftenhardtointerpretcorrectly.Amajorargumentthatpersistedinthescientificcommunitywasweatherlightwasaparticle,orawave.Attimes,itseemstoactlikeawavesfromapebbleonthesurfaceofapond,othertimes,likerubberballsmovinginstraightlines.We'llstudybothapproachesandseehowtheyofferdifferentadvantagesforunderstandingthenaturalworld.
Huygens' Princinple
ChristianHuygens[1629-1695]
1.Eachpointonawavefrontisthesourceofasphericalwaveletthatspreadsoutatthespeedofthewave.
2.Atalatertime,theshapeofthewaveisthecurvethatistangenttoallthewavelets.
Huygens'Sim
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plane wave
Oneofthefirstscientiststomakeagoodcasefor'lightasawave'wastheDutchphysicistChristianHuygens.Around1678,hewroteatreatiseaboutlightinwhichhedescribedawavebasedunderstandingaboutthepropagationoflight.Althoughthedetailsofthemechanismheproposedwerenotexactlycorrect,thebasicconceptualunderstandingprovidedbythemodelremainsuseful.Atthispointintime,peopleweren'tsureiflightmovedatinfinitespeed(instaneously)orifithadafinitevelocity.HuygensfirmlybelievedhadafinitespeedandwasconvincedofthisbyastronomicalobservationsprovidedbyOlausRoemer.Healsocouldexplaincertainaspectsbytreatinglightasawave.
Essentially,Huygen'sprinciplestatesthatawavefrontcanbecreatedbymanycloselyspaced,coherentpointsourcesthatareallspreadingoutinasphericalmanner.(Coherentmeanstheyhavethesameinitialphase)
1.Eachpointonawavefrontisthesourceofasphericalwaveletthatspreadsoutatthespeedofthewave.
2.Atalatertime,theshapeofthewaveisthecurvethatistangenttoallthewavelets.
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t = 0 t > 01.Eachpointonawavefrontisthesourceofasphericalwaveletthatspreadsoutatthespeedofthewave.
2.Atalatertime,theshapeofthewaveisthecurvethatistangenttoallthewavelets.
Index of Refraction
Theindexofrefraction(orrefractiveindex)ofamaterialisadimensionlessparameter, ,usedtodescribehowlightmovesinaparticularmedium.
n
n = =speedoflightinvacuum
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Theindexofrefractionofvacuumis andisexactly1.
Medium IndexVacuum 1(exactly)Air(0ºC,1atm) 1.00029Water 1.33Glass 1.52Saphire 1.77Diamond 2.42
[*Noteforlabs:Theindexofrefractioncanchangebasedonthewavelengthofthelight]Waves passing through media
Index of Refraction
Thevelocityofthewaveinamediumisgivenby: .
Fromthepreviousvisualization,wecanseethatthewavelengthwillalsochange.
Thuswecanwrite
Whataboutthefrequency?
Refraction
Refractioninvolvesthebendingoflightataninterfacebetweentwomediumwithdifferentindicesofrefraction.
We'llseethatthefollowingrelationholds:
n = =c
v
speedoflightinvacuumspeedoflightinmaterial
n = 1.000
v = c
n
=λnλ
n
=sinθ1sinθ2
n2
n1
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light
n1
n2vacuum (n = 1)
light
n1
n2vacuum (n = 1)
Refractioncanbeunderstoodasadirectconsequenceofthechangeinvelocityofawaveasitentersanewmediumatanangle.
Law of Refraction
Derive the law of refraction
Implicationsofrefraction
Twolightwavesentertwodifferentmaterials.
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out of phase!
Upon exiting the material:
Thewavesarenolongerinphase!
Yes,butbyhowmuch?
Quick Question 1
Asinglelightbeamissplitintotwoequalbeams,denotedAandB.BeamAtravelsthroughamediumwithahigherindexofrefractionthanthemediumthatbeamBtravelsthrough.Afterbothbeamsexittheirmediaandarebackintotheair,howdotheirwavelengthscompare?
1. BeamAhasalongerwavelength2. BeamBhasalongerwavelength3. Bothbeamshavethesamewavelength
Basedonthelengthofthematerials,theindicesofrefraction,andthewavelengthofthelightinvacuum,wecandeterminethephasedifference(actuallythedifferenceinnumberofwavelengths)betweenthetwolightwaves.
Wecangetthisrelationinthefollowingway.
Thewavelengthinthemediumwillbedifferentbywavelengthinvacuum:
Also,ifthemediumhasalength ,thentherewillbe wavelengthspresentinthemedium:
Similary,forthewavesinmedium2:
Subtracting - givesusthedesiredrelation.
Color
Thecoloroflightthatweperceiveisbasedonthefrequency(orwavelength)ofthelight.
− = ( − )N2 N1L
λn2 n1
=λn1
λ
n1
L N1
= =N1L
λn1
Ln1
λ
= =N2L
λn2
Ln2
λ
N2 N1
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10141018
10-10 10-610-7 10-2 1
1081010
radi
o/tv
mic
row
ave
x ra
ysfrequency of wave [Hz]
wavelength of wave [m]
780 Terahertz 400 Terahertz
~400 nm ~700 nm~500 nm
Thehumaneyehasthreetypesofcellsthataresensitivetodifferentfrequenciesoflight.Thetriggeringofthesecellsbythedifferentfrequenciesresultsincolorperceptions.Theyareresponsivetoonlyaverynarrowsetoffrequencieshowever.Thecompletespectrumoflightcontainfrequenciesandwavelengthsthatdonottriggerthecellsoureyes.We'llseethisspectrumagain,afterwegothroughelectricityandmagnetism,sincethosetopicswillcontainthematerialtoreallyunderstandwhatlightis.
The light from the sun
ThephysicalprocesseshappeningintheSuncreatelightofmanydifferent
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n > 1
Theimplicationsarethattheangleofrefractionisnowdependentonthewavelengthofthelightbecause carriesthisdependence: .
Fig.1
frequencies.Thischartshowstheintensitiesemittedfrequenciesasafunctionofwavelength.Itisanotherexampleofaspectrum.Wesawaspectrumgraphwhenwecoveredsoundwaves.Thereweplottedtheintensityofthesoundwavesasafunctionofthefrequency(pitch)ofthesound.Thisgraphcontainssimilarinformation.Fromit,wecanseethemostofthelightthatthesunproducesisaround500nmwavelength.Fortunately,thisisaroundthefrequencythatoureyesareverysensitivetoo.
Chromatic Dispersion
n = 1.55
n = 1.45
200 nm 400 nm 600 nm 800 nm 1000 nm 1200 nm 1400 nm wavelength
inde
x of
refra
ctio
n
Hereisaqualitativeplotoftheindexofrefractionforahypotheticalmaterialasafunctionofthewavelengthoftheincidentlight.
Whenlightentersamedium,theindexofrefractionitencounterswilldependslightlyonthewavelengthofthelight.ThiseffectiscalledChromaticDispersion.
Dispersion
2. Diffraction
Diffraction
Diffraction
n n → n(λ)
sin (λ) = sin (λ)θ1n1 θ2n2
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point 1
point 2
crest
trough
L1L2
L2L1
Thespreadingoutofthewavesafterpassingthroughtheopeningiscalleddiffraction.
Diffractionisageneralpropertiesofallwavephenomenon.Waterwavesareparticularlyeasytoseediffractioneffectsbecausethewavelengthsareaboutasbigastheobstacleswhichcausethediffraction.
CheckoutthePanamacanalentrance:LinktoMap
Thereweretwocompetingtheoriesaboutwhatexactlylightwas.Newton'sexperimentsledhimtopostulatethatlightwasaparticle(hecalledthem'corpuscles'.)
WhileHuygens'andothershaddoneworkwhichseemedtoshowthatlightactedasawave.
ThomasYoungwasanEnglishScientist.Around1801hedidsomeexperimentswithlightwiththegoalofdemonstratingitswave-likenature.
Wave interference
Twospeakers?
Thephysicsoflightborrowsmuchfromthetreatmentofsoundwaves.Wefiguredouthowtocalculatewhereloadandquitespotsshouldbeinthecaseoftwointerferingsoundwaves.Asimilarapproachwillgiveusinformationabouttheinterferenceoflight.
Thomas Young
3. The double slit experiment
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single slit (hole)
incident light
diffracted waves
Fig.2
double slit (hole)
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double slit (hole)
max
max
max
max
max
max
min
min
min
min
min
min
screen
Thebasicsetupisasfollows.Light(monochromaticsandcoherent)passesthroughtwosmallslits.Fromeachslightemergesadiffractedbeam.Thesebeamsundergowaveinterferenceandtheresultinglightthatappearsonascreenawayfromtheslitsshowstheinterferencepattern.
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r1
r2 Lslits
screen
P
m = 0
m = 1
m = 1
m = 2
m = 2
m = 3
m = 3
y0
y1 y’0
y’1
y’2y2
y3+y
r2
r1
path length r1
path length r2
L
slits
screen
P
y = 0
y = L tan
d
∆r = path length difference
Whenconstructive:
andfordestructive:
Quick Question 2
Abeamofmonochromaticlightwithawavelengthof660nmisdirectedatadoubleslit.Considerthefivelocationslabeledinthedrawing,(thecentralmaximumislabeled“B.”)Whichoneofthesefringesisproducedwhenthepathdifferenceis1320nm?
LightfromaLaserhasawavelengthof633nm.Thisispassedthroughtwosmallslitsspaced.4mmapart.Aviewingscreenis2.0mbehindtheslits.Whatarethedistancesbetweenthetwom=2brightfringesandbetweenthetwom=2darkfringes,onthescreen?
ΔL = d sinθ
d sinθ = mλform = 0,1,2
d sinθ = (m + 1/2)λform = 0,1,2
Example Problem #1:
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Quick Question 3
Alaboratoryexperimentproducesadouble-slitinterferencepatternonascreen.Ifthescreenismovedfartherawayfromtheslits,thefringesmarkedBandCwillbe:
1. Closertogether.2. Inthesamepositions.3. Fartherapart.4. Fuzzyandoutoffocus.
Adoubleslitinterferencepatternisobservedonascreen1.0mbehindtwoslitsspaced0.30mmapart.Tenbrightfringesspanadistanceof1.7cm.Whatisthewavelengthofthelight?
I
image data
Intensity plot
y
Ofcourse,asscientists,wewouldratherhaveourdatainamoreuseableform.Tothisend,wecanconvertour'screen'imagesofdarkandlightspotstoanintensityplot.Thisletsusputunits,like forintensityasafunctionofdistance,andusefunctionstodesribetheplot.
Thesameinformationispresent--it'sjusteasiertoworkwith.
Quick Question 4
Alaboratoryexperimentproducesadouble-slitinterferencepatternonascreenusingredlight.Ifgreenlightisused,witheverythingelsethesame,thebrightfringeswillbe:
1. Closertogether.2. Inthesamepositions.3. Fartherapart.4. Fuzzyandoutoffocus.
4. Waves and Reflections
Wesawwhathappenedwhenawaveonastringreflectedoffaboundary.
Boundary is not perfect
Example Problem #2:
W/m2
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incident wave
reflected wave
transmitted wave
Dependingonifthelightisreflectingoffahigherindexoralowerindex,wewillobtainthefollowingphaseshiftsforthereflectedwaves.
Reflection PhaseshiftReflectsoffalowerindex 0Reflectsoffahigherindex 0.5
incident wave
reflected wave
transmitted wave
incident wave
reflected wave
transmitted wave
n1 > n2
n1 < n2
Herewehaveawaveonaropetravelingtowardsapartiallyreflectingboundary.The'boundary'isreallyaconnectionbetweenalowdensityandhighdensitysectionofrope.
Wecanseethatmostofthewaveisreflected,andinverted,attheboundary,however,someoftheenergyofthewaveistransmittedacrosstheboundaryandcontinuesintothehigherdensityregion.Additionally,since ,thewavespeedisdifferentforthereflectedandthe
transmittedwaves.
Boundary is not perfect
Likewise,ifthewavetravelsfromahigherdensityregiontoalowerdensityregion,mostofthewave'senergywillbetransmitted,whilesomewillbereflected.
Again,thelowerdensitysectionofropewillhaveahigherwavespeedthanthehighdensityregion.Thistime,thereisnoamplitudeinversion.vid
Phase change due to reflection
Sincetheindexofrefractionofamaterialissomewhatanalogoustothedensityofastring,asfaraswavesgo,(theybothareparametersthataffectthespeedofthewaves),wewouldexpectasimilarphenomenawithlightasitreflectsfromboundaries.
v = τ
μ
−√
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n = 1 n = 1.5 n = 1.3
incident ray
reflected ray
1 2
incident wave
reflected wave (from first surface)
reflected wave (from glass surface)
transmitted wave
air thin film glass
t
nair= 1 nglass= 1.5nx= 1.33
incident ray
r1
n1
r2
n2 n3
12
3
Quick Question 5
Bluelaserlightistravelingthroughasandwichofmaterialsasshown.Ateachinterface,somelightwillbereflected.Comparingtheoriginalincidentraywiththefinalreflectedray,whatisthephaseshiftsduetoreflection?
1. 0rad2. rad
3. rad4. rad5. rad
Phase Shift Sources
Wenowhave3sourcesofphaseshiftsforalightwave:
1. Reflection:dependingonthe ofthetwomedia,thereflectedwavemayormaynotbephaseshifted
2. PathLength:Theanalysisofthedoubleslitshowedthatpathlengthdifferencecanalsoleadtoaphaseshift.
3. IndexofRefraction:Lighttravelingthroughmediawithdifferent canundergoaphaseshift.
Thin Film Interference
Thin Film Interference
①Theoriginalincidentrayisbothreflectedandtransmittedatthefirstboundary.Thisreflectionbecomes .Thetransmittedwavegoesuntilthenextboundary,②,whereitisalsotransmittedandreflected.Thereflectedrayproceedsbackthroughmaterial2,anditthentransmittedthroughtheboundaryat③.Thislasttransmittedrayis .
Byfiguringoutallthephaseshiftsalongtheway,wecantellifrays and willbeinphaseoroutofphase.Thatis,willtheyconstructivelyinterfereordestructivelyinterfere?
Constructive interference
Inthissituation,wehave .
π/2π
3π/22π
n
n
r1
r2
r1 r2
< >n1 n2 n3
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incident ray
r1
air
r2
air
12
3
n2
incident ray
r1
air
r2
air
12
3
n2
incident ray
r1
r2
12
3
n2n1 n3
t < .1 l
Sinceattheinterface①therayisreflectingoffalarge ,thereflectedray willhaveaphaseshiftof
.
willhaveapathlengthdifferencecomparedtoof2L.Thustogetconstructiveinterferencebetweenand ,weneed
Destructive Interference
Ifwewherelookingtohavedestructiveinterferencebetweenthetworeflectedrays,wewouldneed:
.
Thus,withthesamesubstitutions,wecanarriveat:
WhatisthethinnestfilmofMgF2(n=1.38)onglassthatproducesastrongreflectionforlightwithawavelengthof600nm.
Two15-cmlongflatglassplatesareseparatedbya10 mthickspaceratoneend,leavingathinwedgeofairbetweentheplates.Amonochromaticlight( nm)fromaboveilluminatestheplates.Alternatingbrightanddarkfringesareobserved.Whatisthespacingbetweentwobrightfringes.
Ifthefilmismuchsmallerthanthewavelengthofthelightpassingthrough,(i.e ),thenwecanignorethepathlengthcreatedbythefilmandonlyconcernouranalysiswiththereflectionsofthelightattheboundaries.
n r1λ/2
r2 r1
r1 r2
2L = ( ) = (m + ) form = 0,1,2,…oddnumber
2λn2
12
λ
n2
2L = integer × λ
2L = (integer) = m form = 0,1,2,…λn2
λ
n2
Example Problem #3:
Example Problem #4:
μ
λ = 589
t < .1λ
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incident
reflected
reflected
transmitted
transmitted
incident
soap filmSincethecriteriaforconstructiveanddestructiveinterferencedependsonwavelength,lightofdifferentcolorswillresponddifferentlytoathinfilm,likeasoapbubbleoroilslick.Somecolorswillexperienceconstructiveinterference,whileotherswillhavedestructiveinterferenceafterthetworeflections.
5. Interferometer
We’veseenthattheamountoflightvisibleisverysensitivetothepathdifferencebetweentworays.Thismeanswecanmakeadevicetomeasureverysmalldistances.
emitters detectorThesmallestlengthwecanmeasureusingthisdevicewillbegivenbythewavelengthofthelightused.
Interferometer
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L1
L2source splittermirror
detector
mirrorWeknowthepathlengthdifferenceis:
When isequaltoawholenumberofwavelengths,ie: ,thentheinterferencebetweenthetwopathswillbeperfectlyconstructive.Anychangeintheamplitudeofthefinal,detectedlight,impliesachangeineitherthewavelengthofthelightduringonearmoftravel,orachangeinthelengthofonearm.
L1
L2source
splittermirror
detector
mirror
Interferometer
Usesofinterferometry:
1. DisprovedtheLuminiferousAether2. Surfacetopographyofsamples.3. MeasuredNewton’sGravitationalConstant,G.4. Maybegravitationalwaves???
(lookupLIGO-http://www.ligo.caltech.edu/)
6. The diffraction of light
Ouruseoftheraymodelreliedontheassumptionthatthelightraysdidn'tinteractwithobjectsthathad
Δr = 2 − 2L2 L1
Δr
Δr = mλ
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ⒶThewaveletsfromeachpointontheinitialwavefrontoverlapandcreateaninterferencepattern
ⒷThesealltravelthesamedistanceandcreatethecentralmaximum
ⒸThesetraveldifferentdistancesandcreatethefringes.
featuresinthesamelengthscaleasthewavelengthofthelight.
Clearly,thislimitstherangeoftopicswecandiscuss,anditlimitsthedepthofphenomenawecanunderstand.
Let’sgobacktothis:Amonochromaticlightsourcepassingthroughasmallslit.
Thewavelengthofthislightisroughlythesamesizeastheslitthroughwhichitispassing.
Single Slit
Thesetupisthesameasthedoubleslit,exceptthereisonlyoneslit.
Again,weseeadiffractionpatternemergeonthescreen.Thisoneisabitdifferenthowever.
Huygens-like analysis of a single slit
A B C
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aa/2
1
2
3
4
5
6∆r
Thepathlengthdifferencebetweenwavelet1and2is:
Whichisanotherwayofsayingthatwavelet2travels moretogetthescreenattheangle .
If ,thensoarealltheother∆rvalues,andtheinterferenceatthescreenwillbedestructive,meaningtherewillbeadarkspotthere.Thelocationsofthedarkspotsarethen:
or,
Theotherminimumscanbefiguredoutinasimilarway.Wegetthefollowingfortheangularlocationsofthedarkregions.
Thewidthofthecentralmaximumiseasilyfiguredbylookingatthedistancebetweenthefirstminimums.
thus:
a
centralmaximum
m = 1
m = 1m = 2
m = 2
w
single slit screen y
LQuick Question 6
Lightofwavelength illuminatesasingleslitsothatthewidthofthecentraldiffractionmaximais .Theslitwidthisthendecreasedto .Whatisthewidthofthecentraldiffractionmaxima?
1.2.3.4.5.
Δ = sinθr12a
2
Δr θ
Δ = λ/2r12
sinθ =a
2λ
2
a sinθ = λ
a sinθ = mλ
y =mλL
a
w = + =1λL
a
1λL
a
2λL
a
λ
l a/2
l/22l
l/44l
l
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d
d towards screen
plane wavesapproaching from left
d
y1
y2
y1
0
y2
screen
5 slit grating
m = 2
m = 2
m = 1
m = 1
m = 0
Quick Question 7
Thistimeyoukeepthesameslitwidth,butuseanothermonochromaticlightofwavelength500nm.Howdoesthebroadness(width)ofthecentralbrightfringechangecomparedtothatproducedbythe600nmwavelength?
1. Increases2. Decreases3. Staysthesame4. Dependsontheexactvalueofslits
Multiple Slits
Whathappenswhenweincreasethenumberofslits?
Thissystemiscalledadiffractiongrating.Wecanthinkofitasamulti-slitdevice.
Wecanincreasethenumberofslitsperinch.Here's5slitsandasampleofwhatthediffractionpatternwouldlooklike.
Usingthesameanalysisasbefore,wecanfindthelocationsonthescreenwherethepathlengthdifferenceleadstoconstructiveinterference.
Again,notethatthisvalueisdependentonthewavelength.
Thequality(sharpnessandintensity)ofthediffractionpatternisdeterminedbythenumberofslitsinthegrating.Asthenumberofslits,n,increases,thewidthofthefringesdecreases,buttheirintensityincreases.
Δr = d sinθ = mλ
(form = 1,2,3)
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Differentwavelengthswillhavemaximumsatdifferentpositionsalongthescreen.
Awesome Tool: The spectrometer
Buildyourown:DIYspectrometerkithereCircular Aperture Diffraction
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D
circularaperture
screen
light intensity
incident light
L
Monochromaticlightwhichpassesthroughaverysmallcircularaperturewillcreateacirculardiffractionpatternlikethis.
Thelocationofthefirstminimumcanbefoundby:
The lens as an aperture
Alenscanbeconsideredatypeofaperture.Therewillbediffractionofthelightwavesasitpassesthrough.
Two distant stars
D
= 1.22θ1λ
D
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Rayleigh's Criterion
Twopointsourcescanonlyberesolvediftheirangularseparation, isgreaterthan:
Twolightbulbsare1.0mapart.Fromwhatdistancecantheselightbulbsberesolvedbyasmalltelescopewitha4.0cmdiameterobjectivelens.Assumethatthelensisdiffractionlimitedand =600nm.
ResolvingPower:Ingeneral,wecansaytheresolutionforanyopticalinstrumentislimitedbythewavelengthsofthelightused.
GuesswhatcolorthelightfromaBlu-rayplayerlaseris?Why?
α
= ≈θR sin−1 1.22λ
D
1.22λ
D
Example Problem #5:
λ
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