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.