SolutionsPHYS1252Exam#1,Spring2017,V.200205-1807hbsProblemI:MultipleChoiceQuestions,20Ptotal=5P(Q.1)+5P(Q.2)+5P(Q.3)+5P(Q.4)ExamVersion: 1A 1B 1C 1D 1E 1FQ.1[5P] B B C D C BQ.2[5P] D D C E A BQ.3[5P] A B C B B DQ.4[5P] D D E E A E__:Corr.200205-1807DetailedsolutionsforverysimilarmultiplechoicequestionsarepostedonthePHYS1212/PHYS1252coursewebsitefor:--ConceptualPracticeProblemsforExam#1--PHYS1112Exam#1,ConceptualProblems,Spring2009,2010,2011,2014--PHYS1212andPHYS1252Exam#1,ProblemIQuestions,Spring2015--In-classquizzesQ1.01-Q1.04Alsosee:--LONCAPASolutionsforHWSets#1and#2.

SolutionsPHYS1252Exam#1,Spring2017,V.200205-1807hbsProblemII:RayDiagram, 10PtotalDetailedSolutionStepsforExamVersion#1A(seedrawingsbelowforallversions):1P:Drawopticalaxis,mirrorprincipalplane,H,andcorrectlyplacedimagearrowtotheleftofthemirror,withall3elementsproperlylabeled.Imagearrowcanbechosentopointeitherupwardordownward.1P:Correctlyidentifyandlabeloutgoingsideofmirror,totherightofmirror.Reasoning:i)imageisvirtualàbysignconvention,imageislocatednotonoutgoingsideii)imageistotheleftofmirroràtotheleftofmirrorisnottheoutgoingsideiii)àoutgoingsideistotherightofthemirror1P:Correctlyidentifyandlabelincomingside,totherightofthemirror:Reasoning:i)Thedeviceisamirrorwithreflectedrayssentbacktoincidentside,i.e.,leavingmirroronsametosidetheycomein.ii)àIncomingsideissameasoutgoingside:totherightofmirror.1P:CorrectlyplacefocalpointsFandF’:FandF’totheleftofthemirrorandtorightofimageontheopticalaxis.LabelFandF’.Reasoning:i)mirrorisdivergent,i.e.,hasnegativefocallengthf<0.ii)àbysignconvention,Fisnoton“in”andF’isnoton“out”side,i.e.,FandF’arebothtotheleftofmirror.iii)FandF’areatbothatadistance|f|frommirroràFandF’coincide,atsamedistancefromandonsamesideofthemirror.iv)absoluteimagedistancefrommirror,|d’|,isgreaterthanabsolutefocallength,|f|v)àFandF’arelocatedtotherightoftheimage,i.e.,betweenimageandmirror.2P:Correctlydrawoneofthethreeprincipalraypairs(seedrawingExamples1,2):

Either:F-F’-ray Or:P-P’-ray Or:C-C’-ray2P:Correctlydrawasecondprincipalraypair(seedrawingExamples1,2):

Either:P-P’-ray Or:C-C’-ray Or:F-F’-rayNote:Foramirror(unlikeamirror!)theC-rayandC’-rayaretwodistinctlines,eachintersectingtheopticalaxisatthesameangle,butonefromabove,theotherfrombelow.So,theC’-rayforamirrorwouldhavetobeconstructedfromtheC-ray,byreflectingtheC-rayattheopticalaxis.That’sunlikealens,wheretheC-rayissimplycontinuedstraightthroughthemirrortogettheC’-ray.1P:Correctlydrawobjectatintersectionoftwoincomingprincipalrays,asarrow.Objectwillbetotheleftofmirror.Objectarrowwillbeorientedoppositetoimagearrow,i.e.,invertedrelativetoimage.So,ifimagearrowwasdrawnupward,objectarrowmustpointdownward.Else,ifimagearrowwasdrawndownward,objectarrowmustpointupward.Eitherchoiceofimagearrowisacceptable.1P:State:objectisvirtual.Reasoning:i)objectisfoundtotherightofthemirroràobjectisnotontheincomingsideii)àbysignconvention:objectisvirtualandd<0.

SolutionsPHYS1252Exam#1,Spring2017,V.200205-1807hbsRaydiagramsforallexamversions,#1A,1B,1C,1D,1E,1F,areshownbelow.Inallversions,theobjectisvirtual.

.

NotOut Out

NotIn In

PHYS1252,Sp’2017,Exam#1,ProblemII:RayDiagram,Version#1A

F’-ray

F-ray

P-ray

Q

Q’

H

C-rayImage

Object

OpticalAxisA

P’-ray

C’-rayRESULTS:Objectislocatedon“NotIn”

à Objectisvirtual,d<0Objectandimagearrowshaveoppositeorientation

àImageisinvertedrelativetoobject

INPUTS:Imageisvirtualandtotheleftofmirrorà NotOut=Leftà Out=RightDeviceisamirror(notalens!)

à In=Out,NotIn=NotOutMirrorisdivergent,f<0

à Fon“NotIn”,F’on“NotOut”

F� F’

Out NotOut

In NotIn

PHYS1252,Sp’2017,Exam#1,ProblemII:RayDiagram,Version#1B

F’-ray

F-ray

P-ray

Q

Q’

H

C-rayImage

Object

OpticalAxisA

P’-ray

C’-rayRESULTS:Objectislocatedon“NotIn”

à Objectisvirtual,d<0Objectandimagearrowshaveoppositeorientation

àImageisinvertedrelativetoobject

INPUTS:Imageisvirtualandtotherightofmirrorà NotOut=Rightà Out=LeftDeviceisamirror(notalens!)

à In=Out,NotIn=NotOutMirrorisdivergent,f<0

à Fon“NotIn”,F’on“NotOut”

F� F’

SolutionsPHYS1252Exam#1,Spring2017,V.200205-1807hbs

.

.

NotOut OutNotIn In

F’-rayF-ray

P-ray

Q

Q’H

C’-ray

Image

ObjectOpticalAxisA

P’-ray

C-ray

F� F’

PHYS1252,Sp’2017,Exam#1,ProblemII:RayDiagram,Version#1C

RESULTS:Objectislocatedon“NotIn”

à Objectisvirtual,d<0Objectandimagearrowshavesameorientation

àImageiserectrelativetoobject

INPUTS:Imageisrealandtotherightofmirrorà Out=Rightà NotOut=LeftDeviceisamirror(notalens!)à In=Out,NotIn=NotOutMirrorisdivergent,f<0à Fon“NotIn”,F’on“NotOut”

Out NotOutIn NotIn

F’-rayF-ray

P-ray

Q

Q’H

C’-ray

Image

Object OpticalAxisA

P’-ray

C-ray

F� F’

PHYS1252,Sp’2017,Exam#1,ProblemII:RayDiagram,Version#1D

RESULTS:Objectislocatedon“NotIn”

à Objectisvirtual,d<0Objectandimagearrowshavesameorientation

àImageiserectrelativetoobject

INPUTS:Imageisrealandtotheleftofmirrorà Out=Leftà NotOut=RightDeviceisamirror(notalens!)à In=Out,NotIn=NotOutMirrorisdivergent,f<0à Fon“NotIn”,F’on“NotOut”

SolutionsPHYS1252Exam#1,Spring2017,V.200205-1807hbs.

.

.

NotOut OutNotIn In

F’-rayF-ray

P-ray

Q

Q’H

C’-ray

Image

ObjectOpticalAxisA

P’-ray

C-ray

F� F’

PHYS1252,Sp’2017,Exam#1,ProblemII:RayDiagram,Version#1E

RESULTS:Objectislocatedon“NotIn”

à Objectisvirtual,d<0Objectandimagearrowshavesameorientation

àImageiserectrelativetoobject

INPUTS:Imageisrealandtotherightofmirrorà Out=Rightà NotOut=LeftDeviceisamirror(notalens!)à In=Out,NotIn=NotOutMirrorisdivergent,f<0à Fon“NotIn”,F’on“NotOut”

NotOut OutNotIn In

PHYS1252,Sp’2017,Exam#1,ProblemII:RayDiagram,Version#1F

F’-ray

F-ray

P-ray

Q

Q’

H

C-rayImage

Object

OpticalAxisA

P’-ray

C’-rayRESULTS:Objectislocatedon“NotIn”

à Objectisvirtual,d<0Objectandimagearrowshaveoppositeorientation

àImageisinvertedrelativetoobject

INPUTS:Imageisvirtualandtotheleftofmirrorà NotOut=Leftà Out=RightDeviceisamirror(notalens!)à In=Out,NotIn=NotOutMirrorisdivergent,f<0à Fon“NotIn”,F’on“NotOut”

F� F’

SolutionsPHYS1252Exam#1,Spring2017,V.200205-1807hbsProblemIII:RefractionataPrismImmersedinAir, 35Ptotal=15P(a)+10P(b)+10P(c)DetailedSolutionforExamVersion#1A:

PrismsurfaceBandtheincidentray(outsidetheprism)arehorizontal.TheanglebetweenprismsurfacesLandBisα=15o.TheindexofrefractionoftheglassisnG=2.10.TherefractedrayfromsurfaceLinsidetheprismstrikessurfaceB.

a) Sinceincidentray(IR)isparalleltosurfaceB(SB),theanglebetweenIRandsurfaceL(SL)isa=15o.Thatangleaandangleofincidence,Q1,addupto90o,sinceQ1ismeasuredfromIRtothenormaltosurfaceL(NSL).Hence,theangleofincidence,Q1,is[7P] Q1=90o-a=90o-15o=75oBySnell’slawofrefraction:

nAirsin(Q1)=nGsin(Q2) andwecanusenAir»1.0.

à sin(Q2)=(nAir/nG)sin(Q1)=(1.0/2.10)sin(75o)=0.459965Hence,theangleofrefraction,Q2,is[8P] à Q2=arcsin(0.459965)=27.385o

b) SumofinterioranglesintriangleOPQis180o.Hence,seedrawingabove:[8P] 180o=(90o+Q2)+f+a[2P] à f=180o-(90o+Q2+a)=90o-(27.385o+15o)=47.615o

c) Thatanglefandangleofincidence,Q3,addupto90o,sinceQ3ismeasuredfromrayinsidetheprism(IP)tothenormalofsurfaceB(NSB).Hence:[5P] Q3=90o-f=90o-47.615o=42.385oThecriticalangleofincidenceatsurfaceBis:

sin(Qcrit)=(nAir/nG)=(1.0/2.10)=0.47619 [5P] à Qcrit=arsin(nAir/nG)=arcsin(0.47619)=28.437o à Q3.>Qcrit

àActualangleofincidenceatSBexceedscriticalangleofincidence.à TherayincidentatSBwillundergototalinternalreflectionandnorefractedray

emergesthroughSBfromtheprismintotheair.Alternatively,wecouldalsouseSnell’slawtocalculatethesineoftheangleofrefractionatSB(callitQ4,notshownindrawing),assumingtherewasarefractedray:[10P-Alt] sin(Q4)=(nG/nAir)sin(Q3)=(2.1/1.0)sin(42.385o)=1.416>1.Sincesin(Q4)exceeds1,theangleofrefraction,Q4,doesnotexistandnorefractedrayexitsatSB.

GlassnG

φ

SurfaceR

Refractionataprismimmersedinair:

PrismsurfaceBandtheincidentray(outsidetheprism)arehorizontal.TheanglebetweenprismsurfacesLandBisα=25o.TheindexofrefractionoftheglassisnG =1.80.TherefractedrayfromsurfaceLinsidetheprismstrikessurfaceB.

A) FindtheangleofincidenceatsurfaceLB) FindtheangleofrefractionatsurfaceLC) Findtheangleφ atsurfaceBD) FindtheangleofincidenceatsurfaceBE) DoestherayexittheprismthrusurfaceB?Ifso,atwhatangleofrefraction?Ifnot,whynot?

α

Exercise1SurfaceL

SurfaceB

α

ϴ1

ϴ2

QP

O

ϴ3

SolutionsPHYS1252Exam#1,Spring2017,V.200205-1807hbsDetailedSolutionforExamVersion#1D:

PrismsurfaceBandtheincidentray(outsidetheprism)arehorizontal.TheanglebetweenprismsurfacesLandBisα=35o.The”apex”anglebetweenprismsurfacesLandRisγ=97o.TheindexofrefractionoftheglassisnG=1.60.TherefractedrayfromsurfaceLinsidetheprismstrikessurfaceB.

a) Sinceincidentray(IR)isparalleltosurfaceB(SB),theanglebetweenIRandsurfaceL(SL)isa=19o.Thatangleaandangleofincidence,Q1,addupto90o,sinceQ1ismeasuredfromIRtothenormaltosurfaceL(NSL).Hence:[7P] Q1=90o-a=90o-35o=55oBySnell’slawofrefraction:

nAirsin(Q1)=nGsin(Q2) andwecanusenAir»1.0.

à sin(Q2)=(nAir/nG)sin(Q1)=(1.0/1.60)sin(55o)=0.511970[8P] à Q2=arcsin(0.511970)=30.795o

b) SumofinterioranglesintriangleFGHis180o.Hence,seedrawingabove:[8P] 180o=(90o-Q2)+y+g à y=180o-(90o-Q2+g)=90o+Q2-g[2P] y=90o+30.975o-97o=23.795o

c) Thatangleyandangleofincidence,Q5,addupto90o,sinceQ5ismeasuredfromrayinsidetheprism(IP)tothenormalofsurfaceR(NSR).Hence:

Q5=90o-y[5P] Q5=90o-23.795o=66.205oThecriticalangleofincidenceatsurfaceR(SR)is:

sin(Qcrit)=(nAir/nG)=(1.0/1.6)=0.625000 [5P] à Qcrit=arcsin(0.625000)=38.682o à Q5.>Qcrit

à ActualangleofincidenceatSRexceedscriticalangleofincidence.à TherayincidentatSRwillundergototalinternalreflectionandnorefractedray

emergesthroughSRfromtheprismintotheair. Alternatively,wecouldalsouseSnell’slawtocalculatethesineoftheangleofrefractionatSR(callitQ6,notshownindrawing),assumingtherewasarefractedray:[10P-Alt] sin(Q6)=(nG/nAir)sin(Q5)=(1.6/1.0)sin(66.205o)=1.464>1.Sincesin(Q6)exceeds1,theangleofrefraction,Q6,doesnotexistandnorefractedrayexitsatSR.

Refractionataprismimmersedinair:

PrismsurfaceBandtheincidentray(outsidetheprism)arehorizontal.TheanglebetweenprismsurfacesLandBisα=25o.TheindexofrefractionoftheglassisnG =1.80. Theapexangleatthetopofprism(enclosedbetweenLandR)isƔ = 100o.TherefractedrayfromsurfaceLinsidetheprismstrikessurfaceR.

A) FindtheangleofincidenceatsurfaceLB) FindtheangleofrefractionatsurfaceLC) Findtheangleψ atsurfaceRD) FindtheangleofincidenceatsurfaceRE) DoestherayexittheprismthrusurfaceR?Ifso,atwhatangleofrefraction?Ifnot,whynot?F) Whatisthemaximumvaluewhichtheapexangle,Ɣ,mustnotexceedifarefractedrayistoemerge

throughsurfaceRintoair,givenα=25o andnG =1.80.

Exercise2

α GlassnGSurfaceL

SurfaceR

ψ

Ɣ

SurfaceB

ϴ1

ϴ2α

ϴ5

F

G

H

SolutionsPHYS1252Exam#1,Spring2017,V.200205-1807hbsNumericalSolutionsforExamVersions#1A–#1F:

.

PHYS1252, Sp'17, Exam#1, Problems III Manual Inputs:Final Results:

Part Exam#: 1A 1B 1C 1D 1E 1FProblem III Inputs:

(a), (b) α [deg] 15.000000 19.000000 22.000000 35.000000 32.000000 22.000000(b), (c) ɣ [deg] N/A N/A N/A 97.000000 104.000000 102.000000(a), (c) n_G 2.100000 1.900000 1.700000 1.600000 1.750000 1.950000

Problem III Results:(a) θ_1 [deg] 75.000000 71.000000 68.000000 55.000000 58.000000 68.000000(a) sin(θ_2) 0.459965 0.497641 0.545402 0.511970 0.484599 0.475479(a) θ_2 [deg] 27.384828 29.844076 33.052157 30.795142 28.986198 28.390536(b) ϕ [deg] 47.615172 41.155924 34.947843 N/A N/A N/A(b) ψ [deg] N/A N/A N/A 23.795142 14.986198 16.390536(c) θ_crit [deg] 28.436890 31.756864 36.031879 38.682187 34.849905 30.851886(c) θ_3 [deg] 42.384828 48.844076 55.052157 N/A N/A N/A(c) sin(θ_4) 1.415624 1.430551 1.393446 N/A N/A N/A(c) θ_4 [deg] -- -- -- N/A N/A N/A(c) Refracted ray thru B? No No No N/A N/A N/A(c) θ_5 [deg] N/A N/A N/A 66.204858 75.013802 73.609464(c) sin(θ_6) N/A N/A N/A 1.463990 1.690479 1.870753(c) θ_6 [deg] N/A N/A N/A -- -- --(c) Refracted ray thru R? N/A N/A N/A No No No

SolutionsPHYS1252Exam#1,Spring2017,V.200205-1807hbsProblemIV:MicroscopeorTelescope 35Ptotal=15P(a)+10P(b)+10P(c)DetailedSolutionforExamVersion#1A:CompoundMicroscope.

.a)FindimageofLens1:Img1[1P] Given|d1|=1.15cmandObj1istotheleft=incomingsideofLens1àd1=+1.15cm>0.

Alsogiven:f1=+1.10cm.[8P] àd1’=(1/f1-1/d1)-1=(1/1.10-1/1.15)-1cm=+25.30cm=distancefromImg1toLens1[2P] d1’>0àImg1isonoutgoingsideofLens1àImg1istotherightofLens1[2P] d1’>0àImg1isreal[2P] m1=-d1’/d1=-22.00<0àImg1isinvertedrel.toObj1 (Note:onlythecorrectsignofm1isrequiredheretogetfullcredit).

1:in1:notout

1:notin1:out

2:in2:notout

2:notin2:out

h2’

h1’=h2

h1 F2F1

d2

d2’

d1 d1’L

Eye

CompoundMicroscope:InitialDrawing

H2H1

SolutionsPHYS1252Exam#1,Spring2017,V.200205-1807hbsb)FindObjectofLens2:Obj2.FindLens-to-LensDistance.[4P] Given|d2’|=29cmandImg2istotheleft=notoutgoingsideofLens2

àImg2isvirtualandd2’=-29cm<0.

Alsogiven:f2=+1.90cm.[4P] àd2=(1/f2-1/d2’)-1=(1/1.90-1/(-29))-1cm=+1.7832cm=distancefromObj2=Img1toLens2[2P] L=d1’+d2=(25.30+1.7832)cm=27.083cm=distanceLens2fromLens1. Noticealso(notrequired,seedrawing): d2>0àObj2isonincomingsideofLens1àObj2istotheleftofLens2 d2>0àObj2isreal m2=-d2’/d2=+16.26>0àImg2iserectrel.toObj2andImg2isinvertedrel.toObj1.c)FindSizeoffinalimage,Img2,anditsorientationrel.tooriginalobject,Obj1.[2P] m1=-d1’/d1=-(25.30/1.15)=-22.00

Given:h1=0.095mm

[2P] àh1‘=m1h1=(-22.00)(0.095mm)=-2.09mm[2P] m2=-d2’/d2=-((-29.00)/1.7832)=+16.26

h2=h1’=-2.09mm

[2P] àh2‘=m2h2=(+16.26)(-2.09mm)=-34.0mm;finalimagediameteris34.0mm. m1<0àImg1invertedrel.toObj1 m2>0àImg2erectrel.toObj2=Img1[2P] àImg2invertedrel.toObj1Alternativesolution:[6Palt] mtot=m1m2=(d1’/d1)(d2’/d2)=(25.30/1.15)((-29.00)/1.7832)=-357.8

Given:h1=0.095mm[2Palt]àh2‘=mtoth1=(-357.8)(0.095mm)=-34.0mm;finalimagediameteris34.0mm.[2Palt] mtot<0àImg2invertedrel.toObj1

SolutionsPHYS1252Exam#1,Spring2017,V.200205-1807hbsDetailedSolutionforExamVersion#1D:GalileanTelescope.

.a)FindimageofLens1:Img1[1P] Given|d1|=738×106km=738×1011cmismuch,muchlargerthangivenf1=+112.0cm.andObj1istothe

left=incomingsideofLens1à1/d1isvery,verysmallcomparedto1/f1à1/d1canbeneglectedinthe

imageformationeq.ford1’:

[8P] àd1’=(1/f1-1/d1)-1≈(1/f1)-1=f1=+112.0cm=distancefromImg1toLens1[2P] d1’>0àImg1isonoutgoingsideofLens1àImg1istotherightofLens1[2P] d1’>0àImg1isreal

GivenObj1istotheleft=incomingsideofLens1àd1=+738×1011cm>0.[2P] àm1=-d1’/d1=-1.518×10-12<0àImg1isinvertedrel.toObj1 (Note:onlythecorrectsignofm1isrequiredheretogetfullcredit).

Galilean Telescope: Initial Drawing

EyeF2� F2

Obj.1

Img.2 Img.1 = Obj.2

d1 d1�

d2d2�

In,NotOut

Out,NotIn

Divergent Lens 2: f2<0Convergent Lens 1: f1>0

H2H1

In,NotOut

Out,NotIn

Lh1

F1�

h1’=h2

h2'

SolutionsPHYS1252Exam#1,Spring2017,V.200205-1807hbsb)FindImageofLens2:Img2.

Givenlens-to-lensdistanceL=108.7cmandL=d1’+d2[2P] àd2=L-d1’=(108.7-112.0)cm=-3.3cm.

Alsogiven:f2=-3.0cm.[4P] àd2‘=(1/f2-1/d2)-1=(1/(-3.0)-1/(-3.3))-1cm=-33.00cm;distancefromImg2toLens2is33.0cm[2P] d2‘<0àImg2isonnotonoutgoingsideofLens2àImg2istotheleftofLens2[2P] d2‘<0àImg2isvirtual Noticealso(notrequired,seedrawing): d2<0àObj2isnotonincomingsideofLens1àObj2istotherightofLens2 d2<0àObj2isvirtual m2=-d2’/d2=-11.00<0àImg2isinvertedrel.toObj2andImg2iserectrel.toObj1.c)FindSizeoffinalimage,Img2,anditsorientationrel.tooriginalobject,Obj1.[2P] àm1=-d1’/d1=-(+112/+738×1011) =-1.518×10-12<0

Given:h1=140×103km=140×109mm

[2P] àh1‘=m1h1=(-1.518×10-12)(140×109mm)=-0.2125mm[2P] m2=-d2’/d2=-((-33.00)/(-3.3))=-10.00

h2=h1’=-0.2125mm

[2P] àh2‘=m2h2=(-10.00)(-0.2125mm)=+2.13mm;finalimagediameteris2.13mm. m1<0àImg1invertedrel.toObj1 m2<0àImg2invertedrel.toObj2=Img1[2P] àImg2erectrel.toObj1Alternativesolution:[6Palt] mtot=m1m2=(d1’/d1)(d2’/d2)=(+112/+738×1011)((-33.00)/(-3.3))=+1.518×10-11>0

Given:h1=140×103km=140×109mm

[2Palt]àh2‘=mtoth1=(+1.518×10-11)(140×109mm)=+2.12mm;finalimagediameteris2.12mm.[2Palt] mtot>0àImg2erectrel.toObj1

SolutionsPHYS1252Exam#1,Spring2017,V.200205-1807hbsProblemV:AngularMagnificationbyMicroscopeorTelescope 10P(Bonus)DetailedSolutionforExamVersion#1A:AngularMagnificationbyCompoundMicroscopeUsinginputsandresultsfromProblemIV,Part(c).

.

FromProblemIV,input: de=|d2’|=29.0cmFromProblemIV,Part(c): he=|h2’|=34.0mm=3.40cm

[4P] àθe≈tan(θe)=he/de=3.40/29.0=0.1172radians=6.72o Or:θe=arctan(he/de)=arctan(3.40/29.0)=0.1167radians=6.69o

Given: dref=dnear=25.0cmFromProblemIV,input: href=|h1|=0.095mm=0.0095cm

[4P] àθref≈tan(θref)=href/dref=0.0095/25.0=0.0003800radians=0.02177o Or:θref=arctan(href/dref)=arctan(3.40/29.0)=0.0003800radians=0.02177oItissufficientifθeandθrefarebothstatedonlyinradiansoronlyindegrees.[2P] àM'=θe/θref=(0.1172)/(0.0003800)=308 Or:M'=θe/θref=(0.1167)/(0.0003800)=307WillalsoacceptM'=−308orM'=−307ascorrectsolutions,withM'<0indicatingthatthefinalimageseenbytheeye(Img2)isinvertedrelativetotheoriginalobject(Obj1).

CompoundMicroscope:AngularMagnification

θe

he =|h2’|

|d2’|=de

θrefhref =h1

dref =dnear

MΘ =θe /θref

Img.2

Obj.1

SolutionsPHYS1252Exam#1,Spring2017,V.200205-1807hbsDetailedSolutionforExamVersion#1D:AngularMagnificationbyGalileanTelescopeUsinginputsandresultsfromProblemIV,Parts(b)and(c).

.

FromProblemIV,Part(b): de=|d2’|=33.0cmFromProblemIV,Part(c): he=|h2’|=2.12mm=0.212cm

[4P] àθe≈tan(θe)=he/de=0.212/33.0=0.006424radians=0.3681o Or:θe=arctan(he/de)=arctan(0.212/33.0)=0.6424radians=0.3681o

FromProblemIV,input: dref=d1=738×106kmFromProblemIV,input: href=h1=140×103km

[4P] àθref≈tan(θref)=href/dref=(140×103km)/(738×106)=0.0001897radians=0.01087o Or:θref=arctan(href/dref)=arctan((140×103km)/(738×106))=0.0001897radians=0.01087oItissufficientifθeandθrefarebothstatedonlyinradiansoronlyindegrees.[2P] àM'=θe/θref=(0.006424)/(0.0001897)=33.9

GalileanTelescope:AngularMagnification MΘ =θe /θref

θrefhref =h1

dref =d1Obj.1

θehe =|h2’|

|d2’|=de

Img.2

SolutionsPHYS1252Exam#1,Spring2017,V.200205-1807hbs.NumericalSolutionsforExamVersions#1A–#1F:

.

PHYS1252, Sp'17, Exam#1, Problems IV and V Manual Inputs:Final Results:

Part Exam#: 1A 1B 1C 1D 1E 1FProblem IV Inputs:

(a) f_1 [cm] 1.100000 1.300000 1.250000 112.000000 120.000000 105.000000(b) f_2 [cm] 1.900000 1.800000 1.700000 -3.000000 -2.700000 -2.600000(a),(c) d_1 [cm] 1.150000 1.400000 1.300000 7.380000E+13 7.380000E+13 7.380000E+13(b) L [cm] N/A N/A N/A 108.700000 117.100000 102.200000(b),(c) |d_2'| [cm] 29.000000 31.000000 29.000000 N/A N/A N/A(b) Img2 rel. to Lens2 left of left of left of N/A N/A N/A(c) h_1 [cm] 0.009500 0.008500 0.008900 1.400000E+10 1.400000E+10 1.400000E+10

Problem V Inputs:d_near [cm] 25.000000 25.000000 25.000000 N/A N/A N/A

Problem IV Results:(a) d_1' [cm] 25.300000 18.200000 32.500000 112.000000 120.000000 105.000000(a) Img1 rel. to Lens1 right of right of right of right of right of right of(a) Img1 real or virtual real real real real real real(a) sign(m_1) -1.000000 -1.000000 -1.000000 -1.000000 -1.000000 -1.000000(a) Img1 orientn. rel. to Obj1 inverted inverted inverted inverted inverted inverted(b) d_2 [cm] 1.783172 1.701220 1.605863 -3.300000 -2.900000 -2.800000(b) d_2' [cm] -29.000000 -31.000000 -29.000000 -33.000000 -39.150000 -36.400000(b) Img2 rel. to Lens2 N/A N/A N/A left of left of left of(b) Img2 real or virtual virtual virtual virtual virtual virtual virtual(b) L [cm] 27.083172 19.901220 34.105863 N/A N/A N/A(c) m_1 -22.000000 -13.000000 -25.000000 -1.517615E-12 -1.626016E-12 -1.422764E-12(c) m_2 16.263158 18.222222 18.058824 -10.000000 -13.500000 -13.000000(c) m_tot -357.789474 -236.888889 -451.470588 1.517615E-11 2.195122E-11 1.849593E-11(c) h_1'=h_2 [cm] -0.209000 -0.110500 -0.222500 -0.021247 -0.022764 -0.019919(c) h_2' [cm] -3.399000 -2.013556 -4.018088 0.212466 0.307317 0.258943(c) Img2 orientn. rel. to Obj1 inverted inverted inverted erect erect erect

Problem V Results:h_e [cm] 3.399000 2.013556 4.018088 2.1247E-01 3.0732E-01 2.5894E-01d_e [cm] 29.000000 31.000000 29.000000 3.3000E+01 3.9150E+01 3.6400E+01

θ_e [radians] 0.116675 0.064862 0.137678 6.4383E-03 7.8496E-03 7.1137E-03θ_e [degrees] 6.684960 3.716336 7.888380 3.6889E-01 4.4975E-01 4.0759E-01

h_ref [cm] 0.009500 0.008500 0.008900 1.4000E+10 1.4000E+10 1.4000E+10d_ref [cm] 25.000000 25.000000 25.000000 7.3800E+13 7.3800E+13 7.3800E+13

θ_ref [radians] 3.800000E-04 3.400000E-04 3.560000E-04 1.8970E-04 1.8970E-04 1.8970E-04θ_ref [degrees] 2.177240E-02 1.948056E-02 2.039730E-02 1.0869E-02 1.0869E-02 1.0869E-02

M_θ 307.0 190.8 386.7 33.9 41.4 37.5