第46届国际物理奥林匹克竞赛(IPhO2015)理论试题与解答(3)

Q T-2 You may use: ∫ constant, where ? or ∫ √ Page 2 of 2 ( √ ) B4 Obtain the value of , the point where the beam meets the bottom of the tank. Take cm, 0.8 , cm (1 cm = 10-2 m). C The Extremum Principle and the Wave Nature of Matter

We now explore the connection between the PLA and the wave nature of a moving particle. For this we assume that a particle moving from to can take all possible trajectories and we will seek a trajectory that depends on the constructive interference of de Broglie waves. As the particle moves along its trajectory by an infinitesimal distance , relate the change in the phase C1 0.6 of its de Broglie wave to the change in the action and the Planck constant. Recall the problem from part A where the particle traverses from to (see Fig. 4). Let an opaque partition be placed at the boundary AB between the two regions. There is a small opening CD of width in AB such that and . C2 Consider two extreme paths C and D such that C lies on the classical trajectory discussed in part A. Obtain the phase difference between the two paths to first order. D Matter Wave Interference

Consider an electron gun at which directs a collimated beam of electrons to a narrow slit at in the opaque partition A B at such that is a straight line. is a point on the screen at (see Fig. 5). The speed in I is × m s and . The potential in II is such that speed × m s . The distance is ( ). Ignore electron-electron interaction.

?? I ?? D C A II ?? ?? 1.2 ?? ?? ?? Figure 4 ?? B CD d ?? I A II 215.00 nm

B ?? Figure 5 ?? 250 mm ?? D1 If the electrons at have been accelerated from rest, calculate the accelerating potential . Another identical slit is made in the partition A B at a distance of nm ( nm m) below slit D2 (Fig. 5). If the phase difference between de Broglie waves arriving at P through the slits F and G is , calculate . What is the smallest distance from P at which null (zero) electron detection maybe expected on the D3 screen? [Note: you may find the approximation useful] The beam has a square cross section of × and the setup is 2 m long. What should be the D4 minimum flux density Imin (number of electrons per unit normal area per unit time) if, on an average, there is at least one electron in the setup at a given time? 0.3 0.8 1.2 0.4 TheoreticalTask2(T-2):Solutions

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TheExtremumPrinciple1

A.TheExtremumPrincipleinMechanicsConsiderahorizontalfrictionlessx-yplaneshowninFig.1.Itisdividedintotworegions,IandII,byalineABsatisfyingtheequationx=x1.Thepotentialenergyofapointparti-cleofmassminregionIisV=0whileitisV=V0inregionII.TheparticleissentfromtheoriginOwithspeedv1alongalinemakinganangleθ1withthex-axis.ItreachespointPinregionIItravelingwithspeedv2alongalinethatmakesanangleθ2withthex?axis.Ignoregravityandrelativistice?ectsinthisentiretaskT-2(allparts).

Figure1

Obtainanexpressionforv2intermsofm,v1andV0.

Solution:

FromtheprincipleofConservationofMechanicalEnergy

12112

2mv=2mv2

+V0v2V02=(v2

1?

1/2

m

)Expressv2intermsofv1,θ1andθ2.

Solution:

Attheboundarythereisanimpulsiveforce(∝dV/dx)inthe?xdirection.Henceonlythevelocitycomponentinthex?directionv1xsu?erschange.Thecomponentinthey?directionremainsunchanged.Thereforev1y=v2y

v1sinθ1=v2sinθ2

Wede?neaquantitycalledactionA=m??

v(s)ds,wheredsisthein?nitesimallengthalongthetrajectoryofaparticleofmassmmovingwithspeedv(s).Theintegralistakenoverthepath.Asanexample.foraparticlemovingwithconstantspeedvonacircularpathofradiusR,theactionAforonerevolutionwillbe2πmRv.ForaparticlewithconstantenergyE,itcanbeshownthatofallthepossibletrajectoriesbetweentwo?xedpoints,theactualtrajectoryistheoneonwhichAde?nedaboveisanextremum(minimumormaximum).HistoricallythisisknownasthePrincipleofLeastAction(PLA).

1

ManojHarbola(IIT-Kanpur)andVijayA.Singh(ex-NationalCoordinator,ScienceOlympiads)weretheprincipalauthorsofthisproblem.ThecontributionsoftheAcademicCommittee,AcademicDevelopmentGroupandtheInternationalBoardaregratefullyacknowledged.

(A1)[0.2]

[0.3]

(A2)TheoreticalTask2(T-2):Solutions

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(A3)PLAimpliesthatthetrajectoryofaparticlemovingbetweentwo?xedpointsinaregion

ofconstantpotentialwillbeastraightline.Letthetwo?xedpointsOandPinFig.1havecoordinates(0,0)and(x0,y0)respectivelyandtheboundarypointwheretheparticletransitsfromregionItoregionIIhavecoordinates(x1,α).Notex1is?xedandtheactiondependsonthecoordinateαonly.StatetheexpressionfortheactionA(α).UsePLAtoobtainthetherelationshipbetweenv1/v2andthesecoordinates.

Solution:

Byde?nitionA(α)fromOtoPis

A(α)=mv??1x21+α2+mv2??

(x0?x1)2+(y0?α)2Di?erentiatingw.r.t.αandsettingthederivativeofA(α)tozero

v1αv2(y0?(x21

+α2)1/2?α)[(x=00?x1)2+(y0?α)2]1/2∴

v1(y0?α)(x21+α21/2

v=)2α[(x0?x1)2+(y0?α)2]1/2

NotethisisthesameasA2,namelyv1sinθ1=v2sinθ2.

B.TheExtremumPrincipleinOpticsAlightraytravelsfrommediumItomedium

IIwithrefractiveindicesn1andn2respectively.Thetwomediaareseparatedbyalineparalleltothex-axis.Thelightraymakesananglei1withthey-axisinmediumIandi2inmediumII(seeFig.2).Toobtainthetrajectoryoftheray,wemakeuseofanotherextremum(minimumormaximum)principleknownasFermat’sprinci-pleofleasttime.

Figure2

(B1)Theprinciplestatesthatbetweentwo?xedpoints,alightraymovesalongapathsuch

thatthetimetakenbetweenthetwopointsisanextremum.Derivetherelationbetweensini1andsini2onthebasisofFermat’sprinciple.

Solution:

ThespeedoflightinmediumIisc/n1andinmediumIIisc/n2,

wherecisthespeedoflightinvacuum.Letthetwomediabeseparatedbythe?xedliney=y1.ThentimeT(α)forlighttotravelfromorigin(0,0)and(x0,y0)is

T(α)=n??1(y21+α2)/c+n2(??

(x0?α)2+(y0?y1)2)/c

[1.0]

[0.5]

TheoreticalTask2(T-2):Solutions

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Di?erentiatingw.r.t.αandsettingthederivativeofT(α)tozero

n1αn2(y0?(y21

+α2)1/2?α)

[(x)2+(y=00?α0?y1)2]1/2∴n1sini1=n2sini2

[Note:DerivationissimilartoA3.ThisisSnell’slaw.]ShowninFig.3isaschematicsketchofthe

pathofalaserbeamincidenthorizontallyonasolutionofsugarinwhichtheconcentrationofsugardecreaseswithheight.Asaconse-quence,therefractiveindexofthesolutionalsodecreaseswithheight.

Figure3

Assumethattherefractiveindexn(y)dependsonlyony.UsetheequationobtainedinB1

toobtaintheexpresssionfortheslopedy/dxofthebeam’spathintermsofn0aty=0andn(y).

Solution:

FromSnell’slawn0sini0=n(y)sini

Then,dy

dx

=?coti

nn(y)

0sini0=??1+(

dy)2

dy??dx

dx=???n(y)??2

nsini?100

Thelaserbeamisdirectedhorizontallyfromtheorigin(0,0)intothesugarsolutionata

heighty0fromthebottomofthetankasshown.Taken(y)=n0?kywheren0andkarepositiveconstants.ObtainanexpressionforxintermsofyandrelatedYoumayuse:??tanθ)+constantsecθ=1/cosθor??quantities.secθdθ=ln(secθ+√dxx2?ln(x+√

1=x2?1)+constant.

Solution:

??

??dy

=??dx

(n?ky?

02nsini)?100

Notei0=90ososini0=1.

(B2)[1.5]

(B3)[1.2]

TheoreticalTask2(T-2):Solutions

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MethodIWeemploythesubstitution

ξ=

n0?kyn0

??dξ(?n0k)????ξ2?1

=?dxLetξ=secθ.Then

n0

k

ln(secθ+tanθ)=x+cOrMETHODII

Weemploythesubstition

ξ=

n0?kyn0

??dξ(?n0)??kξ2?1=???dx??

?n??

0

n0?ky

??k

lnn+(

n0?ky0

n)2

?1=?x+c

0

Nowcontinuing

Consideringthesubstitutionsandboundarycondition,x=0fory=0weobtainthattheconstantc=0.

Henceweobtainthefollowingtrajectory:

??x=n0n0?ky??n??0klnn+(?ky)2?10n0

Obtainthevalueofx0,thepointwherethebeammeetsthebottomofthetank.Takey0

=10.0cm,n0=1.50,k=0.050cm?1(1cm=10?2m).

Solution:

Giveny0=10.0cm.n0=1.50

k=0.050cm?1

From(B3)

???1/2?xn00=kln???n0?ky??????n0?ky??2

n+?10n0

?Herey=?y0

(B4)[0.8]

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