Mastering Physics Assg9

11/1/2015 Assignment9

https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=4032264 1/43

Assignment9Due:8:30pmonSunday,November1,2015

Youwillreceivenocreditforitemsyoucompleteaftertheassignmentisdue.GradingPolicy

ImagesProducedbyDifferentConcaveMirrorsConceptualQuestion

Thesameobjectisplacedatdifferentdistances infrontofsixdifferentconcavesphericalmirrors.Eachmirrorhasthefocallength listedbelow.

PartA

Which,ifany,ofthesescenariosproducearealimage?Which,ifany,ofthesescenariosproduceavirtualimage?

Sortthefollowingscenariosintotheappropriatebins.

Hint1.Howtodrawaprincipalraydiagram

Thepropertiesoftheimagecanbedeterminedgraphicallybydrawingaprincipalraydiagram.Startbydrawingtworaysfromtheobject,comingfromthesamepoint.Drawonerayparalleltotheopticalaxisitsreflectedraywillgothroughthefocalpoint.Drawtheotherraygoingthroughthefocalpointitsreflectedraywillbeparalleltotheopticalaxis.Theimagewillbeformedatthepointwherethetworaysintersect.(Youmightneedtoextendthereflectedraysstraightbackthroughthemirrortofindtheintersectionpoint.)

Hint2.Definitionofarealimage

Arealimageisanimagethatisformedwhenlightraysoriginatingfromanobjectarereflectedoffthemirrorandintersect.Ifthelightraysfromtheobjectintersectonlywhentheyareextendedthroughthemirror,thentheimageformedisvirtual.

Hint3.Principalraysforobjectswithinthefocallength

"



Hint4.Principalraysforobjectsoutsidethefocallength

ANSWER:



Correct

PartB

Which,ifany,ofthesescenariosproduceaninvertedimage?Which,ifany,ofthesescenariosproduceanuprightimage?

Sortthefollowingscenariosintotheappropriatebins.

ANSWER:



Correct

PartC

Ranktheimagesonthebasisofthemagnitudeoftheirmagnification.

Rankthesefromlargesttosmallest.Torankitemsasequivalent,overlapthem.

Hint1.Ratiooffocallengthtoobjectdistancefromfocalpoint

Youcandeterminethemagnificationoftheimageusingtheobjectdistance, ,andfocallength, .Theratioofthefocallengthtoobjectdistancefromthefocalpointisallyouneed.Recallthattheobjectdistance, ,ismeasuredfromtheplaneofthemirror.Asketchofthemirrorwillhelpyoutovisualizethedistancesgivenandratioyourequire.

Hint2.Objectatthefocalpoint

"

"



Ifanobjectislocatedexactlyatthefocalpointofthemirror, ,theimageis"infinitely"large.Astheobjectmovesawayfromthefocalpoint,ineitherdirection,theimagesizedecreases.

ANSWER:

Correct

Problem24.10

Examiningyourimageinaconvexmirrorwhoseradiusofcurvatureis24.4 ,youstandwiththetipofyournose11.0 fromthesurfaceofthemirror.

PartA

Whereistheimageofyournoselocated?

Expressyouranswerwiththeappropriateunits.

"

DN DN



ANSWER:

Correct

PartB

Whatisitsmagnification?

ANSWER:

Correct

PartC

Yourearis10.0 behindthetipofyournosewhereistheimageofyourearlocated?

Expressyouranswerwiththeappropriateunits.

ANSWER:

Correct

PartD

Whatisitsmagnification?

ANSWER:

= 5.78/ DN

= 0.526 ) g

DN

= 7.72/ DN



Correct

PhETTutorial:GeometricOptics

LearningGoal:

Tounderstandhowthepropertiesofaconvexlensandthedistancebetweentheobjectandthelensaffecttheimage.

Forthistutorial,usethePhETsimulationGeometricOptics.Thissimulationallowsyoutoexperimentwithpropertiesofaconvexlensandseehowtheresultingimagechanges.

Startthesimulation.Whenyouclickthesimulationlink,youmaybeaskedwhethertorun,open,orsavethefile.Choosetorunoropenit.

Youcandragthelensandtheobjecttoanylocationandseetheresultingimage.Youcanadjustthecurvatureofthelens,therefractiveindexofthelensmaterial,andthediameterofthelens,usingthesliderbarsatthetop.

Feelfreetoplayaroundwiththesimulation.Whenyouaredone,beginpartA.

PartA

Aconvexlensisatransparentinstrumentthatusesrefractiontobendandfocuslightfromanobject,formingasharpimage.We'llfirstinvestigatehow

= 0.367 ) g



thelensproducesanimagefromapointsourceoflight(alamp).

SelectScreenatupperright,whichmakestheobjectalampandgivesablackscreenthatcanbedraggedaround.

SelectManyraysatupperlefttoseehowthelensbendssomeoftheraysoflightfromthelamp.

Withthelamppositionedfartotheleftofthelens,youshouldseethattheraysthatgothroughthelensconvergetoapoint.Ifthescreenisplacedwherethebeamsconverge,theimageonthescreenwillbeinfocus(itwillbeasmalldotoflightsincethatiswhattheobjectlookslikeinthiscase).Thescreenisthenatthefocalplane.Asthelampismovedclosertothelens,thedistancebetweenthefocalplaneandthelens_____.

ANSWER:

CorrectWhenthelampisclosertothelens,theraysgoingintothelensaredivergingmorequicklythanwhenthelampisfurtheraway.Thus,aftergoingthroughthelens,theyaren'tconvergingasquickly,soittakesalongerdistanceforthemtofocus.

PartB

Thehorizontalbluelinethroughthemiddleofthelensiscalledtheopticalaxis.Asthelampismovedaboveorbelowtheaxis,keepingthehorizontaldistancetothecenterofthelensconstant,howdoesthehorizontaldistancefromthelenstotheimagechange?

ANSWER:

decreases.remainsthesame.increases.

Itincreases.Itdoesnotchange.Itdecreases.



Correct

Thisexplainswhywerefertoafocalplane.Allobjectswiththesamehorizontaldistancetothelenswillbefocusedbythelensonthesameplane.Youcanseethisbyselecting2ndPointandplacingoneofthelampsaboveoriginallamp.

PartC

Dotheraysthatgothroughthelensalwaysconvergeontherightsideofthelens(formingarealimage),regardlessofthepositionofthelamp?

ANSWER:

Correct

ThetwoyellowX's(oneithersideofthelens)arecalledfocalpoints.Thedistancebetweenafocalpointandthelensiscalledthefocallengthofthelens.Iftheobjectisclosertothelensthanthisdistance,thelenscannotbendtheraysenoughforthemtoconverge.

Noifthelampisveryclosetothelens,theraysdon'tconverge.Noifthelampisveryfaraway,theraysdon'tconverge.Yes



PartD

Placethelampsothatthepointlikesourceisdirectlyontheleftfocalpoint.Whathappenstotheraysthatpassthroughthelensontherightsideofthelens?

ANSWER:

Correct

Whentheobjectisatthefocalpoint,theraysneitherdivergenorconverge,butareperfectlyparallel.Conversely,iftheobjectwereatinfinity,theraysgoingfromtheobjecttothelenswouldbeparallel,andafterpassingthroughthelens,theywouldconvergeatthefocalpoint.

PartE

Thefocallength____________whentherefractiveindexofthelensisincreasedand__________whenthecurvatureradiusofthelensisincreased.

ANSWER:

Correct

Ahigherindexofrefractioncausestheraysoflighttobendmore(goingintoandcomingoutofthelens).Theraysthereforefocusmorequickly,andthefocallengthisshorter.Ifthesurfaceofthelenswheretherayentersisparalleltothesurfacewheretherayexits,theraywillnotbebentatall.Thegreatertheanglebetweenthetwosurfaceswheretherayentersandleavesthelens(asisthecaseforasmallerradiusofcurvature),themoretheoutgoingraywillbebentrelativetotheincomingray.

Theyareparallel.Theydiverge.Theyconverge.

decreases/increasesincreases/decreasesdecreases/decreasesincreases/increases



PartF

Thefocallength__________whenthediameterofthelensisincreased.

ANSWER:

Correct

Thefocallengthdoesnotdependonthelens'sdiameteragreaterdiametersimplyallowsmorelighttobefocused.

PartG

Makethecurvatureradius0.6 ,therefractiveindex1.5,andthediameter0.6 .Placethelampsothatthesourceoflightis120 fromthemiddleofthelens(usetheruler).

Thefocallengthofthelensis____________,andthefocalplaneis___________fromthelens.

Hint1.Howtoapproachtheproblem

Usetherulertomeasurethedistancefromthecenterofthelenstoeitherofthefocalpoints.Thisisthefocallength.Then,usetherulertomeasurethedistancefromthecenterofthelenstowheretheraysconverge(thelocationoftheimage).

ANSWER:

doesnotchangeincreasesdecreases

N N DN

30 /12060 /18060 /6060 /120120 /120

DN DN

DN DN

DN DN

DN DN

DN DN



Correct

Forthislens,whentheobjectis120 away,theimageisthesamedistanceontheothersideofthelens.

PartH

Now,movethelampsothatitis90 fromthecenterofthelens.

Howfarfromthelensistheimage?

ANSWER:

DN

DN

1601209018060

DN

DN

DN

DN

DN



Correct

Thedistancestotheobject, theimage, andthefocallength, ,arerelatedbythelensequation:

.

If decreases,then mustincrease(asyoufoundinPartA).Noticethat,asyoufoundinPartF, if istwicethefocallengthofthelens(1/120+1/120=1/60).

PartI

Now,let'slookattheimagesofextendedobjects.DeselectScreen.Youshouldseeapencilastheobjectanditsimage.Everypointonthepencilemitsrayslikeapointsource.SelectingManyraysshowsraysfromthepencilstip.

Movethepencilaround,andlookattheresultingimage.Howdoesthesizeoftheimagedependonthepositionofthepencil(keepthedistancegreaterthanthefocallengthofthelens)?

ANSWER:

/ /

"

/

/

"

/ /

/

/ /



Correct

Theimagegetsverybigastheobjectapproachesthefocalpoint.

PartJ

Themagnificationofanobjectisdefinedas ,where istheheightoftheimageand istheheightoftheobject.Iftheimageisinverted,thenisnegative.

Placethepencil90 fromthelens.Whatisthemagnificationoftheimage(besurethecurvatureradiusisstill0.6 andtherefractiveindexis1.5)?

ANSWER:

Correct

Noticethattheimageisalsotwiceasfarfromthelensastheobject.

PartK

Howfarfromthelensdoesthepencilneedtobeforthemagnification tobe1?

ANSWER:

Thesizeoftheimagedoesn'tchange.Thesizeoftheimagedecreasesasthedistancefromthepenciltothelensdecreases.Thesizeoftheimageincreasesasthedistancefromthepenciltothelensdecreases.

$

$

$

$

$

DN N

2.01.00.51.5



Correct

Noticethat,asyousawearlierforthislens,whentheobjectis120 away,theimageisalso120 ,sotheratioofthedistances and isequaltoone.Infact,asyoumighthaveguessedbynow,themagnificationcanalsobeexpressedas .

PhETInteractiveSimulationsUniversityofColoradohttp://phet.colorado.edu

RayTracingforaConcaveMirror

AnobjectOisplacedatthelocationshowninfrontofaconcavesphericalmirror.Useraytracingtodeterminethelocationandsizeofthereflectedimage.Asyouwork,keepinmindthefollowingpropertiesofprincipalrays:

1. Arayparalleltotheaxis,afterreflection,passesthroughthefocalpointFofaconcavemirrororappearstocomefromthe(virtual)focalpointofaconvexmirror.

2. Araythrough(orproceedingtoward)thefocalpointFisreflectedparalleltotheaxis.3. ArayalongtheradiusthroughorawayfromthecenterofcurvatureCintersectsthesurfacenormallyandisreflectedbackalongitsoriginalpath.4. AraytothevertexVisreflected,formingequalangleswiththeopticaxis.

PartA

Tracethepathofarayemittedfromthetipoftheobjectthroughthefocalpointofthemirrorandthenthereflectedraythatresults.Startbyextendingtheexistingrayemittedfromthetipoftheobject.Thencreatethereflectedray.

Drawthevectorforthereflectedraystartingfromthepointwheretheincidentfocalrayhitsthemirror.Thelocationandorientationofthevectorwillbegraded.Thelengthofthevectorwillnotbegraded.

ANSWER:

180It'simpossible!60120

DN

DN

DN

DN DN / /

/

/



Correct

PartB

Nowtracethepathofarayemittedfromthetipoftheobjectparalleltotheaxisofthemirrorandthenthereflectedraythatresults.Startbyextendingtheexistingrayemittedfromthetipoftheobject.Thencreatethereflectedray.


ANSWER:



Correct

PartC

Now,usingthetworaysyouhavetraced,drawtheimageoftheobjectatthecorrectlocation,withthecorrectorientationandsize.

Drawthevectorendingattheintersectionofthereflectedrays.Thelocation,orientation,andlengthofthevectorwillbegraded.

Hint1.Imageformation(tipofthearrow)

Sinceboththefocalrayandparallelrayreflectfromthetipoftheobject,wheretheycrossisthelocationoftheimageofthetipoftheobject.

Hint2.Imageformation(baseofthearrow)



Ifyouweretodrawthefocalrayandparallelrayfromthebaseoftheobject,whichliesonthecentralaxisofthemirror,theserayswouldbothtraveltowardandreflectfromthemirroralongthecentralaxis.Thisimpliesthatifthebaseoftheobjectliesonthecentralaxis,thebaseoftheimageliesonthecentralaxis.

ANSWER:

Correct

Youcouldhavedrawnadditionalrayssuchastherayincidentfromthetipoftheobjectthatstrikesthevertexofthemirroranditscorrespondingreflectedray.However,tworaysareoftenenoughtodeterminetheimagelocation.

RayTracingforaConvexMirror



AnobjectOisplacedatthelocationshowninfrontofaconvexsphericalmirror.Useraytracingtodeterminethelocationandsizeoftheimageinthemirror.Asyouwork,keepinmindthefollowingpropertiesofprincipalrays:

1. Arayparalleltotheaxis,afterreflection,passesthroughthefocalpointFofaconcavemirrororappearstocomefromthe(virtual)focalpointofaconvexmirror.

2. Araythrough(orproceedingtoward)thefocalpointFisreflectedparalleltotheaxis.3. ArayalongtheradiusthroughorawayfromthecenterofcurvatureCintersectsthesurfacenormallyandisreflectedbackalongitsoriginalpath.4. AraytothevertexVisreflected,formingequalangleswiththeopticaxis.

PartA

Tracethepathofarayemittedfromthetipoftheobjecttowardthefocalpointofthemirrorandthenthereflectedraythatresults.Startbyextendingtheexistingrayemittedfromthetipoftheobject.Thencreatethereflectedray.


ANSWER:



Correct

PartB

Nowtracethepathofarayemittedfromthetipoftheobjectparalleltotheaxisofthemirrorandthenthereflectedraythatresults.Startbyextendingtheexistingrayemittedfromthetipoftheobject.Thencreatethereflectedray.


Hint1.Constructingthereflectedray

Accordingtopoint#1intheproblemintroduction,thereflectedraywillappeartohaveoriginatedfromthefocalpoint.Trydrawingavectorfromthefocalpointthroughthepointwheretheincidentrayhitsthemirror.Thenshortenthisvectorsothatitsstartpointisonthemirror,withoutchanging



itsangle.Alternatively,youcandrawanunlabeledvectorextendingfromthefocalpointandpassingthroughthepointwheretheincidentrayhitsthemirror.Thenyoucandrawthereflectedrayontopoftheunlabeledvector.Keepinmindthattheunlabeledvectorshouldnotbepartofyoursubmission.

ANSWER:

Correct

PartC

Whattypeofimageoftheobjectwilltheconvexmirrorcreate?

ANSWER:



Correct

PartD

Now,usingthetworaysyouhavetraced,drawtheimageoftheobjectatthecorrectlocation,withthecorrectorientationandsize.Startbyextendingtheexistingvirtualraysfromthesurfaceofthemirrorwherethereflectedraystouchthemirror.

Drawthevectorendingattheintersectionofthevirtualrays.Thelocation,orientation,andlengthofthevectorwillbegraded.

Hint1.Virtualextension

Avirtualextensionisarayonthevirtualsideofthemirrorthatrepresentsthepaththatthereflectedrayappearstohavefollowedinsidethemirror.Todrawavirtualextension,simplyextendthevirtualrayonthevirtualsideofthemirrorwithoutchangingitsdirection.

Hint2.Imageformation(tipofthearrow)

Sinceboththefocalrayandparallelrayreflectfromthetipoftheobject,wheretheycrossisthelocationoftheimageofthetipoftheobject.

Hint3.Imageformation(baseofthearrow)

Ifyouweretodrawthefocalrayandparallelrayfromthebaseoftheobject,whichliesonthecentralaxisofthemirror,theserayswouldbothtraveltowardandreflectfromthemirroralongthecentralaxis.Thisimpliesthatifthebaseoftheobjectliesonthecentralaxis,thebaseoftheimageliesonthecentralaxis.

ANSWER:

Avirtualimagebehindthemirror.Arealimageinfrontofthemirror.



Correct

Thesimpleraytracingyouworkedthroughinthisproblemisthestartingpointforthephotorealisticsceneryandlightingeffectsthatarecommonincomputergames.

HowDeepIstheGoldfish?

Atankwhosebottomisamirrorisfilledwithwatertoadepthof20.9 .Asmallfishfloatsmotionless8.00 underthesurfaceofthewater.

PartA

Whatistheapparentdepthofthefishwhenviewedatnormalincidencetothewater?

Expressyouranswerincentimeters.Use1.33fortheindexofrefractionofwater.

DN DN



Hint1.Usetheobjectimagerelation

Recallthattheobjectimagerelationforasphericalrefractingsurfaceisgivenbytheequation

,

where istheobjectdistanceinthematerialwithindexofrefraction , istheimagedistanceinthematerialwithindexofrefraction ,andistheradiusofcurvatureofthesurface.

Hint2.Radiusofcurvatureofthesurface

Noticethatthesurfaceofthewaterinthefishbowlisessentiallyflat.Asaresult,theradiusofcurvatureofthissurfaceisinfinite.

ANSWER:

Correct

PartB

Whatistheapparentdepthofthereflectionofthefishinthebottomofthetankwhenviewedatnormalincidence?

Expressyouranswerincentimeters.Use1.33fortheindexofrefractionofwater.

Hint1.Followthepathoflight

Keepinmindthatthelighthastobouncefromthebottomofthefish,downtothemirror,andthenbackupandoutofthewatertotheviewer.Firstthinkofhowfardownthefishwouldappeariftheindexofrefractionwerethesameinthewaterasout.Youcanusethelawofreflectionforaplanemirrortodeterminewherethefishwouldappearwithoutthewater.Thenapplythecorrectionfortherealrefractionduetothewater.

ANSWER:

*

B

/

B

*

C

/

C

*

C

*

B

/

B

*

B

/

C

*

C

6.02 DN

25.4 DN



Correct

Problem24.28

Toapersonswimming0.770 beneaththesurfaceofthewaterinaswimmingpool,thedivingboarddirectlyoverheadappearstobeaheightof5.35abovetheswimmer.

PartA

Whatistheactualheightofthedivingboardabovethesurfaceofthewater?

ANSWER:

Correct

RayTracinginLensesandMirrors

Weusuallylocatetheimageformedbyanobjectbylookingfortheintersectionofjusttwooftherayscomingfromthatobject,andweusuallychoosethesetworaystobeassimpletodrawaspossible.Infact,alltherayscomingfromtheobjectthatpassthroughthelensmustconvergeatthelocationoftheimage.Inthisproblemyouwillbeaskedtoidentifysomeoftheseotherrays,whicharenotsosimpletodraw.

PartA

N N

= 3.44 $ N



Thefirstfigureshowsanobjecttotheleftofaconverginglens.ThefocalpointofthelensislabeledF.Fiveraysaredrawnfromthetipoftheobjecttothelens,butonlyoneofthemistracedthroughthelenscorrectly.Choosetheraythatistracedcorrectly.

Hint1.Howtoapproachthisproblem

Alloftheraysshownaresimilarlytothestandardraysusedtodrawraydiagrams(i.e.,paralleltothelens'axis,passingthroughthefocalpoint,orpassingthroughthecenterofthelens).Tofigureoutifarayiscorrectlydrawn,thinkaboutthestandardraythatisclosesttotherayyouarelookingat.Considerwhatwouldhappentothestandardrayasyoumovedtheleftsideslightlytoalignitwiththerayyouarelookingat.

ANSWER:

Correct

PartB

ray1ray2ray3ray4ray5



Thesecondfigureshowsanobjecttotherightofaconcavemirror.ThefocalpointofthemirrorislabeledF.Fourraysareshownreflectingoffthemirror,butonlyoneofthemisdrawncorrectly.Choosetheraythatistracedcorrectly.

Hint1.Howtoapproachthisproblem

Alloftheraysshownaresimilartothestandardraysusedtodrawraydiagrams(i.e.,paralleltothemirror'saxis,passingthroughthefocalpoint,orpassingthroughthecenterofthemirror).Tofigureoutifarayiscorrectlydrawn,thinkaboutthestandardraythatisclosesttotherayyouarelookingat.Considerwhatwouldhappentothestandardrayasyoumovedtheleftsideslightlytoalignitwiththerayyouarelookingat.

ANSWER:

Correct

Problem24.44

Aconverginglenswithafocallengthof12.6 formsavirtualimage8.10 tall,16.8 totherightofthelens.

ray1ray2ray3ray4

DN NN DN



PartA

Determinethepositionoftheobject.

ANSWER:

Correct

PartB

Determinethesizeoftheobject.

ANSWER:

Correct

PartC

Istheimageerectorinverted?

ANSWER:

Correct

PartD

= 7.20 / DN

= 0.347 ]5] DN

Theimageiserect.Theimageisinverted.



Aretheobjectandimageonthesamesideoroppositesidesofthelens?

ANSWER:

Correct

MultipleChoiceProblem24.03

PartA

Ifasinglelensformsavirtualimage,wecanconcludethat

ANSWER:

Correct

UnderstandingLenses

LearningGoal:

Tolearnthequantitativeuseofthelensequation,aswellashowtodeterminequalitativepropertiesofsolutions.

Inworkingwithlenses,therearethreeimportantquantitiestoconsider:Theobjectdistance isthedistancealongtheaxisofthelenstotheobject.Theimagedistance isthedistancealongtheaxisofthelenstotheimage.Thefocallength isanintrinsicpropertyofthelens.Thesethreequantitiesarerelatedthroughtheequation

onthesamesideonoppositesides

Itisaconverginglens.Itisadiverginglens.Itcouldbeeithertypeoflens.

/

/

"



.

Notethatthisequationisvalidonlyforthin,sphericallenses.Unlessotherwisespecified,alensproblemalwaysassumesthatyouareusingthin,sphericallenses.

Theequationaboveallowsyoutocalculatethelocationsofimagesandobjects.Frequently,youwillalsobeinterestedinthesizeoftheimageorobject,particularlyifyouareconsideringamagnifyingglassormicroscope.Theratioofthesizeofanimagetothesizeoftheobjectiscalledthemagnification.Itisgivenby

,

where istheheightoftheimageand istheheightoftheobject.Thesecondequalityallowsyoutofindthesizeoftheimage(orobject)withtheinformationprovidedbythethinlensequation.

Allofthequantitiesintheaboveequationscantakebothpositiveandnegativevalues.Positivedistancescorrespondtorealimagesorobjects,whilenegativedistancescorrespondtovirtualimagesorobjects.Positiveheightscorrespondtouprightimagesorobjects,whilenegativeheightscorrespondtoinvertedimagesorobjects.Thefollowingtablesummarizestheseproperties:

real virtual

real virtual

upright inverted

upright inverted

Thefocallength canalsobepositiveornegative.Apositivefocallengthcorrespondstoaconverginglens,whileanegativefocallengthcorrespondstoadiverginglens.

Consideranobjectwith thatproducesanimagewith .Notethatwheneveryouareworkingwithaphysicalobject,theobjectdistancewillbepositive(inmultipleopticssetups,youwillencounter"objects"thatareactuallyimages,butthatisnotapossibilityinthisproblem).Apositiveimagedistancemeansthattheimageisformedonthesideofthelensfromwhichthelightemerges.

PartA

Findthefocallengthofthelensthatproducestheimagedescribedintheproblemintroductionusingthethinlensequation.

Expressyouranswerincentimeters,asafractionortothreesignificantfigures.

ANSWER:

/

/

"

)

5

5

/

/

5

5

QPTJUJWF OFHBUJWF

/

/

5

5

"

/ DN DN/



Correct

PartB

Consideringthesignof ,isthelensconvergingordiverging?

ANSWER:

Correct

PartC

Whatisthemagnification ofthelens?

Expressyouranswerasafractionortothreesignificantfigures.

ANSWER:

Correct

PartD

Thinkaboutthesignof andthesignof ,whichyoucanfindfromthemagnificationequation,knowingthataphysicalobjectisalwaysconsideredupright.Whichofthefollowingdescribesthenatureandorientationoftheimage?

ANSWER:

= 6.67 " DN

"

convergingdiverging

)

= 1.25)

/

5



Correct

Nowconsideradiverginglenswithfocallength ,producinganuprightimagethatis5/9astallastheobject.

PartE

Istheimagerealorvirtual?Thinkaboutthemagnificationandhowitrelatestothesignof .

ANSWER:

Correct

PartF

Whatistheobjectdistance?Youwillneedtousethemagnificationequationtofindarelationshipbetween and .Thensubstituteintothethinlensequationtosolvefor .


ANSWER:

realanduprightrealandinvertedvirtualanduprightvirtualandinverted

" DN

/

realvirtual

/ /

/

= 12.0 / DN



Correct

PartG

Whatistheimagedistance?


ANSWER:

Correct

Alensplacedattheoriginwithitsaxispointingalongthexaxisproducesarealinvertedimageat thatistwiceastallastheobject.

PartH

Whatistheimagedistance?


ANSWER:

Correct

PartI

Whatisthexcoordinateoftheobject?Keepinmindthatarealimageandarealobjectshouldbeonoppositesidesofthelens.


ANSWER:

= 6.67 / DN

4 DN

= 24.0 / DN



Correct

PartJ

Isthelensconvergingordiverging?

ANSWER:

Correct

Youcansolvethelensequationfor intermsof and .Ifyoudothisandthensubstituteyourresultintothemagnificationequation,youwillseethattheonlywaytoobtainanimageofarealobjectthatislargerthantheobjectitselfiswithaconverginglens.

PartK

Findthefocallengthofthelens.


ANSWER:

Correct

RayTracingandImageFormationwithaConcaveLens

= 12.0 4 DN

convergingdiverging

/

/ "

= 8.00 " DN



Aconcavelensrefractsparallelraysinsuchawaythattheyarebentawayfromtheaxisofthelens.Forthisreason,aconcavelensisreferredtoasadiverginglens.

PartA

Considerthefollowingdiagrams,whereFrepresentsthefocalpointofaconcavelens.Inthesediagrams,theimageformedbythelensisobtainedusingtheraytracingtechnique.Whichdiagramsareaccurate?

TypeAifyouthinkthatonlydiagramAiscorrect,typeABifyouthinkthatonlydiagramsAandBarecorrect,andsoon.

Hint1.Arayparalleltothelensaxis

Arayparalleltotheaxisofaconcavelensisrefractedalongalinethatextendsbackthroughthefocalpointonthesamesideofthelens.

Hint2.Araythatpassesthroughthefocalpoint

Araythatisdirectedtowardthefocalpointontheothersideofthelensisrefractedparalleltothelensaxis.

Hint3.Araythatpassesthroughthemiddleofthelens

Araythatpassesthroughthemiddleofaconcavelenscontinuesonitsoriginaldirectionwithessentiallynodisplacementafterpassingthroughthelens.

ANSWER:

AC



Correct

Aconcavelensalwaysformsanimagethatisonthesamesideofthelensastheobject.

PartB

Ifthefocallengthoftheconcavelensis7.50 ,atwhatdistance fromthelensshouldanobjectbeplacedsothatitsimageisformed3.70fromthelens?

Expressyouranswerincentimeters.

Hint1.Howtoapproachtheproblem

Todeterminetheobjectdistanceyoucanusethethinlensequation,butbecarefultoassignthecorrectsigntoeachquantityinvolvedintheequation.

Hint2.Thethinlensequation

Thethinlensequationforalenswithafocallength is

,

where and aretheobjectdistanceandtheimagedistance,respectively.

Hint3.Findtheimagedistance

Whatistheimagedistance foraconcavelensthatformsanimage3.70 fromthelens?

Expressyouranswerincentimeters.

Hint1.Signconventionforimagedistances

Conventionallytheimagedistancehasapositivesignwhentheimageisontheoppositesideofthelensfromtheobject(arealimage),andanegativesignwhentheimageisonthesamesideofthelensastheobject(avirtualimage).

ANSWER:

DN

P

DN

"

P

J

"

P

J

J

DN



ANSWER:

Correct

PartC

Whatisthemagnification producedbytheconcavelensdescribedinPartB?

Expressyouranswernumerically.

Hint1.Magnification

Themagnification producedbyalensisgivenbytheformula

,

where and aretheobjectdistanceandtheimagedistance,respectively.

ANSWER:

Correct

PartD

Whereshouldtheobjectbemovedtohavealargermagnification?

= 3.70 J

DN

= 7.30 P

DN

)

)

)

J

P

P

J

= 0.507)



Hint1.Magnificationandimagesize.

Recallthatalargermagnificationcorrespondstoalargerimage.YoumayfindtheraydiagramsfoundinPartAhelpfulindetermininghowthesizeoftheimagevariesastheobjectismovedcloserorfartherfromthelens.

ANSWER:

Correct

Problem24.54

Thefigureshowsasmallplantnearathinlens.Therayshownisoneoftheprincipalraysforthelens.Eachsquareis1.5 alongthehorizontaldirection,buttheverticaldirectionisnottothesamescale.Useinformationfromthediagramtoanswerthefollowingquestions:

PartA

Usingonlytherayshown,decidewhattypeoflensthisis.

ANSWER:

Theobjectshouldbemovedclosertothelens.Theobjectshouldbemovedfartherfromthelens.Theobjectshouldbemovedtothefocalpointofthelens.Theobjectshouldnotbemovedclosertothelensthanthefocalpoint.

DN



Correct

PartB

Whatisthefocallengthofthelens?

Expressyouranswerusingtwosignificantfigures.

ANSWER:

Correct

PartC

Calculatewheretheimageshouldbe.

Expressyouranswerusingtwosignificantfigures.

ANSWER:

Correct

VideoTutor:PartiallyCoveringaLens

First,launchthevideobelow.Youwillbeaskedtouseyourknowledgeofphysicstopredicttheoutcomeofanexperiment.Then,closethevideowindowand

convergingdiverging

= 4.5 " DN

= 2.3 / DN



answerthequestionatright.Youcanwatchthevideoagainatanypoint.

PartA

Supposewereplacethelensinthevideowithonethathasalongerfocallength.Tokeeptheimagesharp,howshouldwemovethescreen?

Hint1.Howtoapproachtheproblem.

Ifyouhadtosolvethisproblemforanumericalresult,youcouldusethethinlensequation,whichrelatesthedistancesbetweenthebulb,lens,andscreentothefocallengthofthelens.However,sincethequestionisqualitative,letsusearaydiagram.Alenswithalongerfocallengthisonethatrefractsrayslessstrongly:



Hereisaprincipalraydiagramforthelenswiththeshorterfocallength.Theobjectrepresentsthebulb,andtheimagedistance iswhereyouneedtoplacethescreentogetasharpimage:

Now,constructanequivalentdiagramforthelenswiththelongerfocallength.Willtheimagedistance (thedistancetothescreen)begreater,smaller,orthesame?



ANSWER:

Correct

Ifweincrease ,thenwemustincreasethedistancebetweenthelensandthescreeninordertogetasharpimage.

Problem24.62

An3.85 tallobjectis23.0 fromthecenterofasilveredsphericalglassChristmastreeornament5.80 indiameter.

PartA

Whatisthepositionofitsimage(measuredfromthecenteroftheornament)?

ANSWER:

Correct

TowardthelensAwayfromthelensThereisnoneedtomovethescreen.

"

NN DN DN

= 1.55 DN



PartB

Whatistheheightofitsimage?

ANSWER:

AnswerRequested

ScoreSummary:Yourscoreonthisassignmentis90.0%.Youreceived135.07outofapossibletotalof150points.

= 0.259 ] ]5 NN

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Mastering Physics Assg9