Centrale Physique 2 PC 2017

Thme de l'preuve Interfromtrie atomique
Principaux outils utiliss thermodynamique, mcanique quantique, optique ondulatoire, propagation d'onde
Mots clefs tuyre, nombre de Mach, onde stationnaire, polarisabilit, diffraction, interfromtre de Mach-Zehnder, effet Sagnac

Corrig

(c'est payant, sauf le dbut): - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

Extrait gratuit du corrig

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nonc complet

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Rapport du jury

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nonc obtenu par reconnaissance optique des caractres


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Extrait du corrig obtenu par reconnaissance optique des caractres


 Centrale Physique 2 PC 2017 -- Corrig Ce corrig est propos par Tom Morel (professeur en CPGE) ; il a t relu par Guillaume Maimbourg (agrg de physique) et Louis Salkin (professeur en CPGE). Ce sujet comporte trois parties consacres respectivement  la dynamique de particules dans une tuyre,  la diffraction d'ondes de matire et  l'interfromtre de Mach-Zehnder. La premire partie est entirement indpendante des deux autres.  Dans la premire partie, aprs avoir dmontr plusieurs formules de thermodynamique et de dynamique des fluides, on cherche  estimer les ordres de grandeur de la temprature et de la vitesse des atomes en sortie de la tuyre.  La diffraction de particules  travers un rseau optique modul est tudie dans la deuxime partie. Aprs avoir analys l'interaction entre un atome et une onde lectromagntique stationnaire, on s'intresse  la propagation d'une onde de matire dans ce potentiel. Cette partie repose essentiellement sur des notions du cours d'lectromagntisme et de mcanique quantique.  Enfin, la troisime partie s'intresse  la mesure de la vitesse de rotation de la Terre sur elle-mme grce  l'interfromtre de Mach-Zehnder. Russir cette partie suppose d'avoir bien compris la prcdente. L'preuve, relativement complique, fait appel  de nombreuses notions relatives aux ondes : propagation dans un milieu matriel, interaction avec une onde stationnaire, interfrences, dualit onde-corpuscule. De longueur raisonnable pour cette banque de concours, le sujet alterne des questions trs calculatoires et d'autres o le raisonnement physique est primordial. Cette preuve peut servir de problme de rvision une fois que les cours sur les ondes et la mcanique quantique sont matriss. Indications Partie I I.A.3 crire l'identit thermodynamique sur l'enthalpie. I.A.5 Utiliser le rsultat de la question I.A.3. I.A.6 Prendre le logarithme du dbit massique puis diffrentier cette expression. I.B.1 L'enthalpie massique s'crit, d'aprs la question I.A.2, r T(z) h(z) = -1 I.B.4 Calculer numriquement la grandeur col P0 T0 -4/3 . Partie II II.A Dans un conducteur parfait, le champ lectrique est nul. crire alors la continuit du champ lectrique tangentiel en x = xM . II.C.4 crire l'quation de d'Alembert dans l'espace des frquences. - II.D.3 Projeter l'expression de k d sur l'axe (Oz). Partie III III.A.2 Faire un schma simple en faisant apparatre les ordres 0, -1 et 1 pour connatre l'orientation de chaque rayon transmis et les vecteurs d'onde correspondants. III.A.3 L'intensit est proportionnelle  |h + b |2 . III.B.3 Utiliser le rsultat de la question II.D.3. III.B.4 Les rseaux 1 et 3 sont espacs d'une distance de 2L. Pour une rotation autour de l'axe (Oy), v3x (t2 ) - v1x (t2 ) = 2L y Injecter ensuite les rsultats des questions III.B.4, II.C.1 et II.D.1. Interfromtrie atomique I. Caractrisation de la source atomique I.A.1 L'coulement tant stationnaire, le dbit massique est conserv, d'o Dm = (z) v(z) A(z) I.A.2 La loi des gaz parfaits s'crit P(z) V(z) = n R T(z). Comme n = m/M, avec P(z) = r (z) T(z) r= R M Pour un gaz parfait, la variation d'enthalpie massique s'crit dh = cP dT avec cP la capacit thermique massique  pression constante. Or, d'o cP = R M( - 1) dh = r dT -1 I.A.3 Pour une transformation adiabatique et rversible d'un gaz parfait, la loi de Laplace s'crit P = Cte La deuxime identit thermodynamique pour une transformation adiabatique rversible donne galement dh = Tds + Ainsi V(z) V(z) dP = dP m m dh = dP (z) Le programme de PC stipule que les identits thermodynamiques ne doivent pas tre connues mais rappeles par l'nonc. Cette question est donc  moiti hors-programme. I.A.4 Les transferts thermiques entre le gaz et les parois sont ngligeables. De mme, les forces de pesanteur sont ngliges. Le premier principe appliqu  l'coulement entre le four et la position z devient h + ec = 0 avec ec l'nergie cintique macroscopique massique. Avec 1 2 1 ec = v (z) - v0 2 v 2 (z) 2 2 il vient 1 2 v (z) + h(z) = h0 2 I.A.5 D'aprs la question I.A.3, P(z) = Cte (z). Exprimons le carr de la clrit : P 2 cs = S = Cte -1 (z) = P(z) (z) avec P(z) = Cte (z) d'aprs la question I.A.2 cs 2 = r T(z) Par consquent, cs (z) = p r T(z) I.A.6 Prenons le logarithme de l'expression du dbit massique : ln Dm = ln + ln v + ln A Il est toujours intressant d'utiliser le logarithme et de diffrentier l'expression pour obtenir des termes en df /f . D'aprs la question I.A.1, le dbit massique est conserv, donc d(ln Dm ) = 0, d'o d dv dA + + =0 (z) v(z) A(z) Exprimons le rapport d/ en fonction de dv/v. Or, avec la loi de Laplace, d = d dP = S dP = 2 (z) cs (z) d'o D'aprs la question I.A.3, Avec la question I.A.4, Il vient dP = (z) S dP P d dh = 2 (z) cs 2 v = -v(z) dv dh = -d 2 d v(z) dv dv =- = -M2 (z) (z) cs 2 v(z) Combinons ces relations. Finalement, dA dv + 1 - M2 (z) = 0 A(z) v(z) I.A.7 Le long de la tuyre, le gaz est acclr donc dv > 0. Pour v < cs (resp. v > cs ), M < 1 (resp. M > 1). Il vient dA(v < cs ) < 0 et dA(v > cs ) > 0 La section est dcroissante pour v < cs et croissante pour v > cs . La section doit donc possder un minimum.  la transition, v = cs c'est--dire M = 1.