[User]
\begin{wrapfigure}{r}{0.33\textwidth}%
\vspace{-10pt}
\includegraphics[width=\linewidth]{2012-v3g-09-laeng_ing}%
\end{wrapfigure}
A positively charged particle is accelerated from the origin with the help of a voltage $4U$ (where $U>0$) to a certain velocity which lies on the $x-y$ plane under the angle of $30$ degrees with respect to the $x$-axis (see figure). The electric potential $V(x,y)\equiv V(x)$ only depends on the $x$-coordinate: if $3a<x<4a$ then $V(x)=U$ and if not then $V(x)=0$; besides the electrostatic force no other forces are applied to the particle and $a>0$.\\
\osa Sketch the trajectory of the charge on the $x-y$ plane (you do not have to show the geometrical measurements or angles).\\
\osa The source of the charged particles is now a coaxial vacuum diode at the origin. Because of that the particles accelerated with the voltage $4U$ are moving isotropically (at equal quantities) to all directions of the $x-y$ plane; the $z$-directional component of the velocity for all particles is 0. What part of all the emitted particles reaches the region of space $x>4a$?

Solution:
\osa Kuivõrd potentsiaal sõltub ainult $x$-koordinaadist, siis on elektriväli kõikjal $x$-telje sihiline ning
impulsi $y$-komponent säilib. Seega, kui osake siseneb kõrgema potentsiaaliga piirkonda, siis
väheneb vaid kiiruse $x$-komponent ning
trajektoor on selline, nagu oleks valguskiirel optiliselt hõredamat kihti läbides.\\
\osa Täielik \enquote{sisepeegeldus} toimub siis, kui positiivse $x$-suunaga
seotud kineetiline energia on suurem kui $qU$, mis vastab kiirusvektori
ja $x$-telje vahelisele nugale
\[
\arccos \sqrt{\frac {qU}{4qU}}=\frac \pi 3.
\]
Ainult sellest nurgast väiksemate nurkade puhul saab osake läbida kõrgema
potentsiaaliga kihti ning selliset osakeste suhteline hulk on
\[
\xi = \frac \pi 3 \div \pi = \frac 13.
\]

Laengud

\begin{wrapfigure}{r}{0.33\textwidth}%
\vspace{-10pt}
\includegraphics[width=\linewidth]{2012-v3g-09-laeng}%
\end{wrapfigure}
Positiivselt laetud osake kiirendatakse koordinaatide alguspunktis pinge $4U$ (kus $U>0$) abil teatud kiiruseni, mis
lebab $x-y$ tasandis $30$-kraadise nurga all $x$-telje suhtes, vt joonist. Elektriline potentsiaal $V(x,y)\equiv V(x)$
sõltub ainult $x$-koordinaadist: kui
$3a<x<4a$, siis $V(x)=U$ ning vastasel korral $V(x)=0$; peale elektrostaatilise jõu osakesele mingeid muid jõude ei mõju ning $a>0$.

\osa Visandage laengu trajektoor $x-y$-tasandis (geomeetrilisi mõõtmeid ja nurki pole vaja märkida).

\osa Nüüd on laetud osakeste allikaks koordinaatide alguspunktis asuv koaksiaalne vaakumdiood, mistõttu pinge $4U$
abil kiirendatud osakesi liigub isotroopselt (võrdsel hulgal) kõigis $x-y$-tasandi suundades;
$z$-suunaline kiiruskomponent on kõigil osakestel 0. Milline osa kõigist kiiratud osakestes jõuab ruumipiirkonda $x>4a$?

Kuivõrd potentsiaal sõltub ainult $x$-koordinaadist, on elektriväli kõikjal $x$-telje sihiline ning
impulsi $y$-komponent säilib. Täielik \enquote{sisepeegeldus} toimub siis, kui positiivse $x-$suunaga
seotud kineetiline energia on suurem kui $qU$.


[User]
01@   IPhO 2010, problem 2. A neat, tricky problem about how chimneys work.

    3      Statistical Mechanics
    There are fundamentally two approaches to describing systems of many interacting particles: bottom-
    up and top-down. In the top-down approach of thermodynamics, we try to roughly describe the
    behavior of the whole system in terms of a few macroscopically measurable observables, such as
    pressure and temperature, and hope this is enough information to extract what we want. In
    the bottom-up approach, we start by analyzing the behavior of individual molecules, governed by
    Newtonian mechanics. Of course, we can’t do this exactly, but it turns out to be possible to make
    probabilistic statements about individual molecules. This is the approach of statistical mechanics.

         Idea 5: Boltzmann Distribution
         The probability distribution for the states of a particle in a system of temperature T is
         proportional to e−E/kB T . Specifically:

           • For quantum systems, where the energy levels are discrete, the probability of being in a
             state n with energy En is proportional to e−En /kB T .

           • For a single classical particle, the state is instead specified by (x, p), the position and the
             momentum, and the probability density in this space, called phase space, is proportional

           to e−E(x,p)/kB T .

       It isn’t possible to derive the Boltzmann distribution from anything we’ve already covered,
       but you’ll see in T2 how it emerges from a simpler postulate.

       Example 5: Isothermal Atmosphere

       Do problem 7 using statistical mechanics.

       Solution
       The energy of each particle is
                                                              p2
                                           E(x, p) = mgz +       .
                                                              2m
       The probability distribution for height z is found by integrating over all the other quantities,
                     Z     Z     Z                                 Z     Z     Z
                                         −E(x,p)/kB T    −mgz/kB T                       2
              p(z) ∝ dx dy d p e     3
                                                      =e              dx dy d3 p e−p /2m .

       Here, all integrals are implicitly from −∞ to ∞. Now, the remaining integrals are just
       constants independent of z, so we just get

                                             p(z) ∝ e−mgz/kB T .

       Since the particles are assumed independent (since we have an ideal gas), the probability for
       a particle to be at a point is proportional to the density of gas at that point. We see the
       density falls exponentially with height, so by the ideal gas law, the pressure does too.

Solution:
01@   IPhO 2010, problem 2. A neat, tricky problem about how chimneys work.

   3     Statistical Mechanics
   There are fundamentally two approaches to describing systems of many interacting particles: bottom-
   up and top-down. In the top-down approach of thermodynamics, we try to roughly describe the
   behavior of the whole system in terms of a few macroscopically measurable observables, such as
   pressure and temperature, and hope this is enough information to extract what we want. In
   the bottom-up approach, we start by analyzing the behavior of individual molecules, governed by
   Newtonian mechanics. Of course, we can’t do this exactly, but it turns out to be possible to make
   probabilistic statements about individual molecules. This is the approach of statistical mechanics.

       Idea 5: Boltzmann Distribution
       The probability distribution for the states of a particle in a system of temperature T is
       proportional to e−E/kB T . Specifically:

         • For quantum systems, where the energy levels are discrete, the probability of being in a
           state n with energy En is proportional to e−En /kB T .

         • For a single classical particle, the state is instead specified by (x, p), the position and the

           momentum, and the probability density in this space, called phase space, is proportional
           to e−E(x,p)/kB T .

       It isn’t possible to derive the Boltzmann distribution from anything we’ve already covered,
       but you’ll see in T2 how it emerges from a simpler postulate.

       Example 5: Isothermal Atmosphere

       Do problem 7 using statistical mechanics.

       Solution
       The energy of each particle is
                                                              p2
                                           E(x, p) = mgz +       .
                                                              2m
       The probability distribution for height z is found by integrating over all the other quantities,
                     Z     Z     Z                                 Z     Z     Z
                                         −E(x,p)/kB T    −mgz/kB T                       2
              p(z) ∝ dx dy d p e     3
                                                      =e              dx dy d3 p e−p /2m .

       Here, all integrals are implicitly from −∞ to ∞. Now, the remaining integrals are just
       constants independent of z, so we just get

                                             p(z) ∝ e−mgz/kB T .

       Since the particles are assumed independent (since we have an ideal gas), the probability for
       a particle to be at a point is proportional to the density of gas at that point. We see the
       density falls exponentially with height, so by the ideal gas law, the pressure does too.

Kevin Zhou Physics Olympiad Handouts — educational use, attribute to Kevin Zhou (knzhou.github.io). LICENSE FLAG: must reconfirm permissions before public release.


[User]
1. FLYıNG DUMBBELL (10 points)

— Jaan
Kalda.
In this problem, we shall study the dy‐
namics of a dumbbell consisting of two steel
balls, each with radius r = 1cm, connected
by a steel rod with diameter d = 1mm and
length l = 10cm, in the absence of gravity.
Unless instructed otherwise, assume steel is
perfectly elastic. You may simplify your cal‐
culations by assuming l ≫ r.
i) (2 points) Given that the Young’s modulus
of steel is Y = 2 × 1011
Pa and the density
of steel is ρ = 7800kgm−3
, determine the
period T of free longitudinal oscillations of
the dumbbell. (Do not consider oscillations
with standing waves in the rod where the
balls remain almost at rest.) Young’s modu‐
lus is the ratio of a material’s stress (force per
unit area) to its strain (relative deformation).
ii) (2points)Estimate the impact time τ when
a steel ball bounces off a steel wall.
iii) (2 points) The dumbbell moves toward a
steel wall with velocity⃗ v = −vx̂, with its axis
perpendiculartothewall,andbouncesback.
Here, x̂ denotes a unit vector along the axis
perpendicular to the wall. Sketch how the x‐
component vx of the front ball’s velocity (the
ball that collides with the wall) depends on
time.
iv) (2 points) Now, the dumbbell moves to‐
ward a steel wall with velocity ⃗ v = −vx̂ as
before, but its axis forms an angle α with the
x‐axis. The interaction of the front ball with
thewalldependsqualitativelyontheangleα,
withatransitionfromonetypeofinteraction
to another occurring at α = α0. Determine
the value of α0. Hint: min
(sinx
x
)
≈ −0.217.
v) (2 points) Under the assumptions of the
previous task, let α > α0. Additionally, as‐
sume that while steel is highly elastic, it is
not infinitely so: any oscillations excited in
the rod will decay by the time the rear ball
collides with the wall. Determine the speed
with which the centre of mass of the dumb‐
bell departs from the wall.

Solution:
1. FLYıNG DUMBBELL

— Solution
by Jaan Kalda, grading schemes by Author 2.
i) (2 points)Free oscillations of the dumbbell
take place around the centre of mass, i.e. the
centre of the rod. Therefore, we need the
stiffness of a half of the rod. This stiffness
is expressed as k = Y π
2 d2
/l. We also need
the mass of the ball m = 4
3πr3
ρ. The oscil‐
lation angular frequency ω =
√
k/m, hence
the period
T = 2π
√
m
k
= 4π
r
d
√
2ρrl
3Y
≈ 0.64ms.
ii) (2 points)The easiest way to estimate is to
notice that a compressed ball is essentially a
compression wave in steel, so the period is
on the order of a wave with wave length 2r.
Knowing that the sound speed cs =
√
Y /ρ,
we obtain τ ∼ 2r/cs = 2r
√
ρ/Y = 4μs. Al‐
ternatively,onecanapproximatetheballasa
springofstiffnessκ ∼ Y r andmass∼ m,and
obtain a similar result with τ ∼ 2π
√
m/κ.
iii) (2 points) When the dumbbell with axis
perpendicular to the wall approaches with
velocity ⃗ v = −vx̂, the front ball impacts
the wall first. Since the impact time (τ ≈
4 μs) is much shorter than the oscillation
period (T ≈ 0.64 ms), the front ball’s velocity
changes almost instantaneously from −v to
+v, while the rear ball continues with velo‐
city −v. Since the balls have equal masses,
the centre of mass remains stationary. The
dumbbell then oscillates about this station‐
ary centre of mass, with the front ball’s velo‐
city following a half‐period sinusoidal oscil‐
lation, changing from +v to −v over a time
interval of T/2. When the velocity reaches
−v,thefrontballimpactsthewallagain,and
its velocity changes instantaneously from −v
to +v. After this second impact, both balls
move away from the wall, with the same ve‐
locity+v,sothedumbbellasawholedeparts
with velocity +v.
t
v
iv) (2 points) During the impact, the front
ball velocity becomes opposite, so the centre
of mass stops (as the rear ball moves still
with its old speed). After the collision, the
front ball obtains a component v cosα along
its axis, and v sinα perpendicular to it. The
former initiates oscillations of period T, and
the latter — a rotation at angular speed
Ω = v sinα/(l/2) = 2v sinα/l
. The ball will hit the wall twice if the rota‐
tion is slow, and only once if the rotation is
fast enough; let us study this in more details.
By time t ≪ 1/Ω, the rotation angle is Ωt,
and the distance of the farthest point of the
ballfromtherotationcentreisl/2−asin(ωt),
where the oscillation amplitude can be ob‐
tainedfromtheenergyconservationlaw,a =
v cosα
√
m/k = v cosα/ω. So, the distance
from the wall of the closest point of the ball
is
l
2
cosα −
[
l
2
− asin(ωt)
]
cos(α + Ωt) ≈
≈
l
2
Ωtsinα + acosαsinωt =
= vtsin2
α +
v
ω
cos2
αsinωt =
=
v
ω
ssin2
α +
v
ω
cos2
αsins, s ≡ ωt.
If this expression becomes negative, there
will be a second collision. So, the cross‐over
value of α = α0 is such that the expression
becomes never negative, hence
tan2
α0 = −min
sins
s
≈ 0.217,
hence
α0 = arctan
√
0.217 ≈ 25°.
Ifwedividethisexpressionbycorresponding
v) (2 points)Using the results of the previous
task, the angular speed after the initial col‐
lision is Ω = 2v sinα/l. The dumbbell ro‐
tates around its centre of mass, longitudinal
oscillations decay by the time of the second
collision. It rotates until the other ball will
hit the wall. At the moment of the second
collision, the velocity of the ball is v sinα,
and its projection to the surface normal of
thewallis−v sin2
α. Duringthesecondcolli‐
sion,thatcomponentsreversessign,andasa
result, both balls have now x‐directional ve‐
locity component v sin2
α. Hence, this is also
the speed of the centre of mass —‐ the speed
with which the dumbbell departs from the
wall.
Grading: (preliminary)
• i)
• explaining that oscillation is symmetric
around centre of the rod (invoking New‐
ton’s third law suffices as well) 0.5pts
• expressing stiffness of half‐rod 0.5pts
• minor mistake made in stiffness expres‐
sion −0.2pts
• mass of the ball m = 4
3πr3
ρ 0.2pts
• Realising that the system can be treated as
a spring 0.3pts
• oscillation period T = 2π
√m
k 0.3pts
• final answer 0.2pts
• ii)
• Solution 1:
• Realise compressed ball is essentially a
compression wave 0.5pts
• Formula for speed of sound 0.5pts
• Relation between time, radius and speed
0.5pts
• Final answer 0.5pts
• Solution 2:
• Realise the ball can be thought of as a
spring 0.5pts
• Estimate spring constant 0.5pts
• Relation between spring constant and time
or frequency 0.5pts
• Final answer 0.5pts
• iii)
• 2 hits 0.4pts
• velocity of front ball flips almost instantan‐
eously 0.4pts
• centre of mass stays at rest 0.4pts
• sinusoidal movement of front ball 0.4pts
• constant velocity −v of front ball after
second hit 0.4pts
• iv)
• Realise it behaves as in previous question
(balls at velocity ‐v and v, CM at rest), but
it now also rotates and oscillates around
centre of mass 0.5pts
• Expression for the angular speed of rota‐
tion 0.3pts
• Expression of the amplitude of oscillations
0.2pts
• Realise the difference in interaction is that
if the first ball bounces once or twice
0.2pts
• Formulaforthedistanceofthefrontballto
the wall over time 0.3pts
• Realise that if the distance is over 0 for all
t > 0 the first ball does not hit the wall
twice 0.2pts
• Finding the critical angle given this condi‐
tion 0.3pts
• v)
• Realise that the dumbbell rotates around
its centre of mass (after first collision)
0.2pts
• Realise that the longitudinal oscillations
have decayed by the time of the second
collision 0.4pts
• Expression for velocity of ball v sinα0.5pts
• Expression for the component of velo‐
city of ball in direction of surface normal
v sin2
α 0.5pts
• Realise the component of velocity of
second ball in direction of surface nor‐
mal is also v sin2
α 0.2pts
• Realise the speed of the centre of mass
v sin2
α 0.2pts

FLYıNG DUMBBELL


[User]
See pilt pöörlevast lennukipropellerist on tehtud telefoni kaameraga, mis salvestab korraga ühe vertikaalse veeru pikselid. Pilt tekib vasakult paremale veergude kaupa skaneerides.\\
\osa Mis suunas pöörleb propeller fotograafi poolt vaadatuna (päripäeva või vastupäeva)?\\
\osa Mitu laba on propelleril?\\
\osa Mitu pööret teeb propeller ühes minutis, kui kogu pildi tegemiseks kulunud aeg on 1/8 sekundit?\\

\begin{center}
	\includegraphics[width=90mm]{2010-v3g-10-Propeller.jpg}
\end{center}

Solution:
\osa Propeller pöörleb vastupäeva, sest pildi ülaosas liiguvad labad vastu parajasti salvestatavale pikseliveerule ja seetõttu paiknevad seal labade kujutised tihedamalt.\\
\osa Vasakpoolsel joonisel on ülalt alla tõmmatud üks veerg millel on korraga peal maksimaalset 2 laba. Kui labasid oleks 2, peaks veerus paistma korraga vaid üks laba. Labad ise on kantud joonisele mustaga. Näha on, et labade vaheline nurk on suurem kui 90 kraadi ja seega on propelleril kolm laba.

\begin{center}
	\includegraphics[width=50mm]{2010-v3g-10-Propeller2.jpg}
	\qquad
	\includegraphics{2010-v3g-10-proplah}
\end{center}

\textit{Alternatiivne lahendus} \\
Tähistame labade tekitatud jooned numbritega 1 kuni 7, nii nagu näidatud parempoolsel joonisel.
Joonise alumises servas eelneb joon 2 joonele 6.
See tähendab, et joonele 2 vastav laba peab eelnema joonele 6 vastavale labale.
Joonise ülemise serva põhjal võime analoogselt väita, et joonele 5 vastav laba peab eelnema joonele 6 vastavale labale.
Järelikult peavad jooned 5 ja 2 vastama samale labale. Ülemises servas jääb joonte 5 ja 2 vahele veel 2
joont, st sellele labale vastavad jooned korduvad perioodiga 3 joont.
See periood peab olema propelleri labade arvu $n$ kordne. Et 3 on algarv, siis ainus variant on $n=3$.\\
\osa Iga kolmas triip pildil kujutab sama propellerilaba. Järgneval joonisel on valgega nummerdatud labad;
propelleri telje kõrgusel on tõmmatud joon mille kogupikkus moodustus pildistamise aja jooksul ehk kogupikkus on $1/8\;$s.
Punasega on märgitud aeg millega laba number üks jõudis liikuda 1,5 pööret.
Punase osa pikkus moodustab ligikaudu 4/5 pildi kogulaiusest. Seetõttu moodustab ka nende punktide ajaline intervall 4/5 pildi tegemise koguajast. Selle aja jooksul teeb propeller poolteist pööret. Ühes sekundis teeb propeller $\num{1,5} /(1/8\cdot 4/5)=15$ pööret ja ühes minutis $15\cdot 60=900$ pööret.

\begin{center}
	\includegraphics[width=50mm]{2010-v3g-10-Propeller3.jpg}
\end{center}

Propeller

See pilt pöörlevast lennukipropellerist on tehtud telefoni kaameraga, mis salvestab korraga ühe vertikaalse veeru pikselid. Pilt tekib vasakult paremale veergude kaupa skaneerides.\\
\osa Mis suunas pöörleb propeller fotograafi poolt vaadatuna (päripäeva või vastupäeva)?\\
\osa Mitu laba on propelleril?\\
\osa Mitu pööret teeb propeller ühes minutis, kui kogu pildi tegemiseks kulunud aeg on 1/8 sekundit?\\

\begin{center}
	\includegraphics[width=90mm]{2010-v3g-10-Propeller.jpg}
\end{center}

\osa
Vastavalt sellele, kas labad liiguvad salvestatavatele pikslite veergudele vastu või eemale, on labade kujutiste tihedus vastavalt suurem või väiksem.\\
\osa
Vaadeldes pildil ühte vertikaalset pikslite veergu ei ole moonutusi näha. Seega tasub uurida, kui palju labasid erinevatel pikslite veergudel näha on.\\
\osa
Teades labade arvu on võimalik vaadelda täpselt, kui palju üks laba salvestamise käigus liigub.


[User]
Optiline süsteem koosneb kumerläätsest ja nõguspeeglist, mille optilised peateljed ühtivad. Kumerpeegli asukohta pole joonisel märgitud. On teada, et objektist $A$ tekib teisele poole läätse kaks kujutist $K_1$ ja $K_2$. Konstrueerige kumerpeegli kõveruskeskpunkt $O$ ja kumerpeeglis objektist $A$ tekkinud näiv kujutis $A'$. Eeldada, et optilises süsteemis on nurgad piisavalt väiksed, et sfäärilisi aberratsioone ei teki.

\begin{center}
	\includegraphics[width=\linewidth]{2007-v3g-09-yl}
\end{center}

Solution:
Paneme tähele, et kõverustsenter $O$, objekt $A$ ning kujutis $A'$ kumerpeeglis asuvad samal sirgel. Seega, kui meil õnnestub leida sirge $AA'$, siis selle lõikepunkt optilise teljega annaks punkti $O$. Teisest küljest, kiir, mis läbib punkti $A$ kujutist läätses, $K_2$, ja punkti $A'$ kujutist, $K_1$, jätkub peale murdumist läätses sirgena $s'$ ja läbib nii punkti $A$ kui $A'$. Tänu sellele leiamegi sirge $AA'$ ja punkti $O$.

Punkti $A'$ leidmiseks konstrueerime esmalt fokaaltasandi $F$. Selleks leiame fokaaltasandis lebava punkti --- sirge $s$ lõikepunkti sirgega $t$, mis on paralleelne sirgega $s'$ ja läbib läätse keskpunkti. Kujutisi $K_1$ ja $K_2$ ühendav sirge murdub nii, et selle pikenduse lõikepunkt optilise peateljega ongi otsitav kõverustsenter. Punkti $A'$ leiame kui sirge $s'$ lõikepunkti sirgega $q'$, mis peale murdumist läbib fookuse ja punkti $K_1$ (sest $K_1$ on $A'$ kujutiseks).

\begin{center}
	\includegraphics[width=\linewidth]{2007-v3g-09-lah}
\end{center}

Nõguspeegel

Optiline süsteem koosneb kumerläätsest ja nõguspeeglist, mille optilised peateljed ühtivad. Kumerpeegli asukohta pole joonisel märgitud. On teada, et objektist $A$ tekib teisele poole läätse kaks kujutist $K_1$ ja $K_2$. Konstrueerige kumerpeegli kõveruskeskpunkt $O$ ja kumerpeeglis objektist $A$ tekkinud näiv kujutis $A'$. Eeldada, et optilises süsteemis on nurgad piisavalt väiksed, et sfäärilisi aberratsioone ei teki.

\begin{center}
	\includegraphics[width=\linewidth]{2007-v3g-09-yl}
\end{center}

Kahe kujutamise tekkeks peab kumerpeegel asuma objektist $A$ vasakul. $A'$ leidmiseks peab määrama läätse fookuse, aga selleks peab silmas pidama asjaolu, et pole teada, kumb kujutistest $K_1$ ja $K_2$ kuulub objektile $A$.


[User]
Comparing the single and double slit.

     (a) Show that the minima for the single slit occur when

                                                a sin θ = nλ,   n ̸= 0.

     (b) Note that this looks almost identical to the result for the maxima of a double slit with
         separation a. Explain the difference using phasors.

        (c) In our analysis of the double slit, we didn’t account for the small but nonzero width of each
            slit. Using the idea of problem 4, sketch the diffraction pattern accounting for this.

    For tips on accurately measuring diffraction patterns, see the handout on experimental methods.

[3] Problem 12 (MPPP 123). Some imperfect diffraction gratings. For this problem, you can ignore
    secondary maxima. Use the small angle approximation throughout, and neglect any diffraction
    effects from the finite widths of the slits.

     (a) In an imperfect diffraction grating, the slits have equal widths, but the distances between the
         slits are alternately d and 3d. Sketch the resulting diffraction pattern, indicating the relative
         heights of the maxima.

     (b) In another imperfect diffraction grating, the slits are evenly spaced, but their widths are
         alternatively a and b, where a ≈ b. Sketch the resulting diffraction pattern, indicating the
         relative heights of the maxima.

    It may be useful to refer to problem 4.

    4      Higher Dimensions
    In these problems we tackle interference and diffraction effects in more than one dimension, which
    can be used to infer the structure of molecules and crystals.

         Idea 3
         We’ve shown that in a single narrow, tall slit, the central maximum is a tall band, bounded
         by minima at θ = ±λ/d. If we instead had a circular slit, the central maximum is a
         circle, bounded by minima at θ ≈ 1.22λ/d. The radii of the higher-order minima then
         get closer and closer spaced as one moves outward. The resulting pattern is called an Airy disc.

  You can straightforwardly write down an integral that gives the intensity I(r), but the integral
  can only be performed in terms of special functions, called Bessel functions.

  Idea 4: Babinet’s Principle

  Consider all of the rays R that strike a point P on the screen. If the intensity at P is zero,
  then the rays must completely destructively interfere. That means that if we split R into
  two sets of rays R1 and R2 in any way, then the amplitudes due to the rays R1 and the rays
  R2 must be equal and opposite, which means either set of rays alone would produce the
  same intensity at P . This is Babinet’s principle.

  As a concrete example, consider shining a laser pointer at a wall. There will be a bright spot
  on the wall at the exact location the laser hits, and darkness everywhere else. Consider some
  dark point P . If we had instead passed the laser through two slits, we would only get the
  rays R1 going through the slits, and we would generally get some nonzero intensity at P , due
  to the double slit interference pattern. Babinet’s principle tells us we would get the exact
  same intensity at P if we put two slit-shaped obstacles in the way, because then we would
  get precisely the rays R2 which don’t hit the slits. In other words, the diffraction pattern
  from an obstacle is precisely the same as the diffraction pattern from a complementary slit.

  Example 6: BPhO 2016.5

  When a laser pointer hits a spring, the following pattern is produced on a screen behind it.

  Explain why this happens, and what we can learn about the spring.

  Solution
  If we look at the spring along the direction the laser pointer is going, it’s essentially two sets
  of obstructions, one going up and to the right (the front of the spring, in the picture), and
  one going up and to the left (the back of the spring). By Babinet’s principle, the resulting
  diffraction pattern should be the same as if we had two sets of openings instead. Thus, we
  expect to see two independent diffraction patterns, one due to each of these obstructions.
  The angle between these two patterns is twice the angle that the spiral path of the spring
  makes with the horizontal.

  Now focus on one interference pattern. By Babinet’s principle, it’s basically a single slit
  pattern, which is indeed what we see. However, from the reflection of the laser in the picture,
  we see that the laser beam is wide enough to hit two separate turns of the spring. The result
  is a double slit pattern multiplied by a single slit pattern, where the latter yields a minimum
  at approximately every 5 double slit minima. The spacing between the single slit minima
  tells us the thickness of the wire, and the spacing between the double slit minima tells us
  the spacing between the turns in the spring. Combining this with what we know about the
  angle of the spring tells us about the radius of the spring. Thus, we can figure out essentially
  everything about its 3D shape.

  The coolest thing about this problem is that all the logic here applies to diffraction off any
  helical structure. For example, below is the X-ray diffraction pattern of DNA, taken by
  Rosalind Franklin in 1952.

  As discussed in detail here, such data was used to discover the structure of DNA.

  Example 7: MPPP 125

  An opaque sheet is perforated by many small holes arranged in a square grid of side length
  d. It is illuminated by light of wavelength λ, and a screen lies a distance D behind it.

  Assuming D ≫ d ≫ λ, find the locations of the primary diffraction maxima on the screen.

  Solution
  Let (x, y) denote coordinates on the sheet, and (x′ , y ′ ) denote coordinates on the screen, with
  the same center. When we considered one-dimensional diffraction gratings, we found that
  light which originates from y and hits point y ′ on the screen has a path length difference
  yy ′ /D relative to light that came from point y = 0. A similar argument shows light which
  comes from (x′ , y ′ ) gets a path length difference

                                                 xx′ + yy ′
                                          ∆ℓ =
                                                     D
  relative to light coming from x = y = 0.

        For a square grid, (x, y) = (nd, md) for integers n and m, giving a path length difference

                                                     d
                                           ∆ℓn,m =     (nx′ + my ′ ).
                                                     D
        We get a diffraction maximum at (x′ , y ′ ) when the light from each hole arrives in phase, which
        means this quantity must be a multiple of λ for all n and m. This occurs precisely when
                                                  λD ′                λD ′
                                           x′ =      n,        y′ =      m
                                                   d                   d
        for integers n′ and m′ . That is, the diffraction maxima also form a square grid of side length
        λD/d. Notice again that the diffraction pattern is “inverse” to the pattern on the sheet. It
        gets bigger when the sheet gets smaller; for instance, if the sheet is compressed horizontally,
        the maxima on the screen are stretched horizontally.

[3] Problem 13 (MPPP 126). Continuing on the previous example, suppose the holes are instead
    arranged in a triangular grid with side length d.

    Find the primary diffraction maxima on the screen. What kind of grid do they form?

Solution:
(a) Using the formula derived above, the minima occur when ka sin θ/2 = πn, and
         with k = 2π/λ, this immediately yields a sin θ = nλ, n ̸= 0.

     (b) For the double slit, when d sin θ = nλ, there are only 2 phasors, which point in the same
         direction. In a single slit with width a, we’re summing infinitesimal phasors that form a circle
         arc. Since the points in the single slit that provide the first and last phasors are separated
         by a distance a, that indicates that when a sin θ = nλ, the first and last phasors point in the
         same direction. That means that the phasors went in a full circle (or multiple), so the net
         amplitude is zero.

     (c) Since the slits are small, by the idea of problem 4, we should multiply a single slit intensity
         pattern with a much wider double slit intensity pattern, giving the following result.

[3] Problem 12 (MPPP 123). Some imperfect diffraction gratings. For this problem, you can ignore
    secondary maxima. Use the small angle approximation throughout, and neglect any diffraction
    effects from the finite widths of the slits.

     (a) In an imperfect diffraction grating, the slits have equal widths, but the distances between the
         slits are alternately d and 3d. Sketch the resulting diffraction pattern, indicating the relative
         heights of the maxima.

     (b) In another imperfect diffraction grating, the slits are evenly spaced, but their widths are
         alternatively a and b, where a ≈ b. Sketch the resulting diffraction pattern, indicating the
         relative heights of the maxima.

    It may be useful to refer to problem 4.

    Solution. (a) By blocking every other slit, we can see that this pattern is the sum of two gratings
         with spacing 4d and a displacement of d. Thus we can add the amplitudes of what we would
         get with a grating of spacing 4d,
                                                   sin(N k(4d sin θ)/2)
                                              A∼                        ,
                                                    sin(k(4d sin θ)/2)

      except one of them has a phase shift of eikd sin θ . This results in a net amplitude of A(1+eikd sin θ ).
      Factoring out a common factor and squaring the amplitude gets

                                          sin2 (2N kd sin θ)
                                     I∝                      cos2 (kd sin θ/2).
                                           sin2 (2kd sin θ)

 (b) Without loss of generality, we can assume that a > b. Then the pattern is the sum of a grating
     with spacing d and slit width b, and a grating with spacing 2d and slit width a − b. Since the
     amplitude of the second component is much smaller, the overall pattern looks like a diffraction
     grating with spacing d and peaks with intensity proportional to (a + b)2 , except there will be
     a small peak with intensity (a − b)2 in between the primary peaks instead of a minimum.

4    Higher Dimensions
In these problems we tackle interference and diffraction effects in more than one dimension, which
can be used to infer the structure of molecules and crystals.

    Idea 3
    We’ve shown that in a single narrow, tall slit, the central maximum is a tall band, bounded
    by minima at θ = ±λ/d. If we instead had a circular slit, the central maximum is a
    circle, bounded by minima at θ ≈ 1.22λ/d. The radii of the higher-order minima then
    get closer and closer spaced as one moves outward. The resulting pattern is called an Airy disc.

  You can straightforwardly write down an integral that gives the intensity I(r), but the integral
  can only be performed in terms of special functions, called Bessel functions.

  Idea 4: Babinet’s Principle

  Consider all of the rays R that strike a point P on the screen. If the intensity at P is zero,
  then the rays must completely destructively interfere. That means that if we split R into
  two sets of rays R1 and R2 in any way, then the amplitudes due to the rays R1 and the rays
  R2 must be equal and opposite, which means either set of rays alone would produce the
  same intensity at P . This is Babinet’s principle.

  As a concrete example, consider shining a laser pointer at a wall. There will be a bright spot
  on the wall at the exact location the laser hits, and darkness everywhere else. Consider some
  dark point P . If we had instead passed the laser through two slits, we would only get the
  rays R1 going through the slits, and we would generally get some nonzero intensity at P , due
  to the double slit interference pattern. Babinet’s principle tells us we would get the exact
  same intensity at P if we put two slit-shaped obstacles in the way, because then we would
  get precisely the rays R2 which don’t hit the slits. In other words, the diffraction pattern
  from an obstacle is precisely the same as the diffraction pattern from a complementary slit.

  Example 6: BPhO 2016.5

  When a laser pointer hits a spring, the following pattern is produced on a screen behind it.

  Explain why this happens, and what we can learn about the spring.

  Solution
  If we look at the spring along the direction the laser pointer is going, it’s essentially two sets
  of obstructions, one going up and to the right (the front of the spring, in the picture), and

  one going up and to the left (the back of the spring). By Babinet’s principle, the resulting
  diffraction pattern should be the same as if we had two sets of openings instead. Thus, we
  expect to see two independent diffraction patterns, one due to each of these obstructions.
  The angle between these two patterns is twice the angle that the spiral path of the spring
  makes with the horizontal.

  Now focus on one interference pattern. By Babinet’s principle, it’s basically a single slit
  pattern, which is indeed what we see. However, from the reflection of the laser in the picture,
  we see that the laser beam is wide enough to hit two separate turns of the spring. The result
  is a double slit pattern multiplied by a single slit pattern, where the latter yields a minimum
  at approximately every 5 double slit minima. The spacing between the single slit minima
  tells us the thickness of the wire, and the spacing between the double slit minima tells us
  the spacing between the turns in the spring. Combining this with what we know about the
  angle of the spring tells us about the radius of the spring. Thus, we can figure out essentially
  everything about its 3D shape.

  The coolest thing about this problem is that all the logic here applies to diffraction off any
  helical structure. For example, below is the X-ray diffraction pattern of DNA, taken by
  Rosalind Franklin in 1952.

  As discussed in detail here, such data was used to discover the structure of DNA.

  Example 7: MPPP 125

  An opaque sheet is perforated by many small holes arranged in a square grid of side length
  d. It is illuminated by light of wavelength λ, and a screen lies a distance D behind it.

      Assuming D ≫ d ≫ λ, find the locations of the primary diffraction maxima on the screen.

      Solution
      Let (x, y) denote coordinates on the sheet, and (x′ , y ′ ) denote coordinates on the screen, with
      the same center. When we considered one-dimensional diffraction gratings, we found that
      light which originates from y and hits point y ′ on the screen has a path length difference
      yy ′ /D relative to light that came from point y = 0. A similar argument shows light which
      comes from (x′ , y ′ ) gets a path length difference

                                                       xx′ + yy ′
                                                ∆ℓ =
                                                           D
      relative to light coming from x = y = 0.

      For a square grid, (x, y) = (nd, md) for integers n and m, giving a path length difference

                                                    d
                                          ∆ℓn,m =     (nx′ + my ′ ).
                                                    D
      We get a diffraction maximum at (x′ , y ′ ) when the light from each hole arrives in phase, which
      means this quantity must be a multiple of λ for all n and m. This occurs precisely when
                                                λD ′                λD ′
                                         x′ =      n,        y′ =      m
                                                 d                   d
      for integers n′ and m′ . That is, the diffraction maxima also form a square grid of side length
      λD/d. Notice again that the diffraction pattern is “inverse” to the pattern on the sheet. It
      gets bigger when the sheet gets smaller; for instance, if the sheet is compressed horizontally,
      the maxima on the screen are stretched horizontally.

[3] Problem 13 (MPPP 126). Continuing on the previous example, suppose the holes are instead
    arranged in a triangular grid with side length d.

    Find the primary diffraction maxima on the screen. What kind of grid do they form?
    Solution. Note that the points on a triangular grid can be written as
                                               (x, y) = nr1 + mr2
    where the “lattice basis vectors” are
                                                                 √
                                        r1 = (d, 0),   r2 = (d/2, 3d/2).
    Again, we need to find the points (x′ , y ′ ) on the screen where the light from all holes arrives in phase.
    This is a bit less obvious than in the above example, so let’s think about it more systematically.
    First, we can find a peak (x′ , y ′ ) = r′1 where changing n by one changes the path length by λ, and
    changing m by one doesn’t change the path length at all. Similarly, we can find a peak (x′ , y ′ ) = r′2
    where changing m by one changes the path length by λ, while changing n by one doesn’t change
    the path length. The general solution will then take the form
                                                 r′ = n′ r′1 + m′ r′2
    for integers n′ and m′ , where r′1 and r′2 are called the “reciprocal lattice” basis vectors.
       In equations, the criterion we have stated above are
                           r′1 · r1 = λD,   r′1 · r2 = 0,        r′2 · r1 = 0,   r′2 · r2 = λD.
    Plugging in the lattice basis vectors and solving the equations gives

                                  ′   λD          1       ′   λD       2
                                r1 =        1, − √     , r2 =      0, √     .
                                       d           3           d        3
                                                                                √
    In other words, the diffraction maxima also form a triangular grid, but the side length is (2/ 3)(λD/d),
    and the whole thing is rotated by 90◦ .


[User]
4. Questions consist of sub-questions. Write your detailed answer in the blank
   space provided below the sub-question and final answer to the sub-question in
   the smaller box which follows the blank space. Note that your detailed answer
   will be considered in the evaluation.

Solution:
4. The realization towards the latter half of nineteenth century that blackbody radiation
   in a cavity can be considered to be a gas in equilibrium became an important milestone
   in the development of modern physics. The equation of state of such a gas with
   pressure P and volume V is P = u/3. Here u is the internal energy per unit volume.
   Also, u = βT 4 , where β is a constant and T is absolute temperature at which radiation
   is in equilibrium with the walls of the cavity.
   (a) Using the first law of thermodynamics obtain an expression for dQ in terms of         [3]
       volume, temperature and other related quantities.

         Solution: From first law of thermodynamics we have

                                                   6
                                dQ = dU + dW = d(uV ) + P dV

         Using u = βT 4 , du = 4βT 3 dT

                                  C1
                                  dQ =
                                          4β 4
                                           3
                                             T dV + 3T 3 V dT

                                                                                (7)

   (b) For a reversible process the entropy is defined as dS = dQ/T . Obtain an expres-      [2]

             HB
       sion for S(V, T ) for the radiation gas.

         Solution:

         At T = 0, S = 0. Hence
                                    dQ
                                     T
                                       = dS = d

                                                4β 3
                                                       4β 3
                                                        3
                                                          T V

                                                                                (8)

                                           S=      T V                               (9)
                                                 3

   (c) Imagine the universe to be a perfectly spherical adiabatic enclosure of radius r      [3]
       containing pure radiation. Rate of expansion of the universe is governed by a
       law
                                           dr
                                              = αr
                                           dt
       where α is a constant. At t = 0, the universe has size r0 and temperature T0 .
       Determine its temperature T (t) for t > 0.

         Solution: As universe expands adiabatically, dQ = 0. Equation (7) gives

                                          dV    dT
                                             +3    =0
                                           V    T

                                          Page 5

                      INPhO - 2016 - Final Solutions

         Using V = 4πr3 /3

                                    dr dT
                                       +      = 0                                 (10)
                                     r    T
                                 ⇒ rT = r0 T0 = constant

         Using rate of expansion law

                                            r = r0 eαt                            (11)
                                        ⇒ T = T0 e−αt                             (12)

   (d) Here α is 72 km·s−1 ·Mpc−1 where 1 Mpc= 3.26×106 light years. State the               [1]
       dimensions of α. Obtain its numerical value in SI units as per the dimensions
       stated.

         Solution:
                                            [α] = T−1
                                       α = 2.33 × 10−18 s−1

                                    C1                   6

INPhO 2016 Problem 4


[User]
Heast soojusjuhist plaadile asetatakse kuivast jääst (st tahkest
süsihappegaasist) kerge seib raadiusega $r=\SI{1}{cm}$; seibi surutakse
pealt jõuga $F=\SI{10}{N}$. Millise minimaalse aluse
temperatuuri juures hõljub seib sublimeeruva süsihappegaasi tekitatud
gaasipadjal? Aluse temperatuur lugege ühtlaseks ja samaks temaga
vahetus kontaktis oleva ainekihiga. Õhurõhk $p_{0}=\SI{100}{kPa}$
ja kuiva jää aururõhu sõltuvus temperatuurist on kujutatud graafikul.

\begin{center}
	\includegraphics{2010-v3g-06-Aururohk}
\end{center}

Solution:
Olukorras, kus aluse temperatuur on minimaalne, on rõhk seibi alumise külje vahetus läheduses võrdne
süsihappegaasi aururõhuga. Seibi surutakse alla jõuga $F$ ja kuna seibi
pindala on $\pi r^2$, peab surumist tasakaalustav rõhk olema $p={F\over \pi r^2}$.
Vaadeldes rõhkude tasakaalu seibi ülemise ja alumise pinna läheduses, saame et kuiva jää aururõhk on
\[
p\idx{kuiv} = p + p_0= p_0+{F\over \pi r^2} = \SI{131.8}{kPa}.
\]
Sellele vastab graafiku põhjal temperatuur $\sim \SI{195}{K}$.

Õhuhoki

Heast soojusjuhist plaadile asetatakse kuivast jääst (st tahkest
süsihappegaasist) kerge seib raadiusega $r=\SI{1}{cm}$; seibi surutakse
pealt jõuga $F=\SI{10}{N}$. Millise minimaalse aluse
temperatuuri juures hõljub seib sublimeeruva süsihappegaasi tekitatud
gaasipadjal? Aluse temperatuur lugege ühtlaseks ja samaks temaga
vahetus kontaktis oleva ainekihiga. Õhurõhk $p_{0}=\SI{100}{kPa}$
ja kuiva jää aururõhu sõltuvus temperatuurist on kujutatud graafikul.

\begin{center}
	\includegraphics{2010-v3g-06-Aururohk}
\end{center}

Seibi alumise külje läheduses surub süsihappegaas teatud rõhuga seibi üles. Vastav jõud on ülemiselt küljelt tasakaalustatud nii õhurõhu, kui ka surumisjõu poolt.


[User]
USAPhO 2014, problem A4.
[4] Problem 27 (MPPP 150). A solid metal sphere of radius R is divided into two parts by a planar
    cut, so that the surface area of the curved part of the smaller piece is πR2 . The cut surfaces are
    coated with a negligibly thin insulating layer, and the two parts are put together again, so that the
    original shape of the sphere is restored. Initially the sphere is electrically neutral.

    The smaller part of the sphere is now given a small positive electric charge Q, while the larger
    part of the sphere remains neutral. Find the charge distribution throughout the sphere, and the
    electrostatic interaction force between the two pieces of the sphere.

Solution:
We know that distributing charge uniformly on the outer surface of the entire sphere will
    give a valid configuration, in the sense that the field is everywhere perpendicular to the conductors.
    Similarly, distributing equal and opposite charges uniformly on the two flat faces will give a valid
    configuration, since it acts like a parallel plate capacitor, making the field vanish everywhere outside.
       Neither of these solutions have the right total charge on each piece, but we can fix this by
    superposing the two. By solving a system of two equations, we find the charge distribution is

      • total charge Q distributed uniformly on the sphere,

      • charges ±(3/4)Q distributed uniformly on the flat surfaces.

    The two flat faces attract each other and the two curved faces repel each other; there are no other
    forces by the shell theorem. The pressure on the flat faces is σ 2 /2ϵ0 . With a little trigonometry, we
    find the area of the flat faces is (3/4)πR2 , giving a force

                                                        3Q2
                                                F1 =           .
                                                       8πϵ0 R2
    As for the repulsive force, using the result of problem 11 we get
                                        3Q2                            45 Q2
                             F2 = −             ,   Ftot = F1 + F2 =              .
                                      128πϵ0 R2                        128 πϵ0 R2


[User]
This subtle problem will expose a hidden assumption we’ve made in the previous two
    examples. Suppose that all of space is filled with uniform charge density ρ.
     (a) Show that E = (ρ/ϵ0 )xx̂ obeys the differential form of Gauss’s law.

     (b) Show that E = (ρ/3ϵ0 )rr̂ also obeys Gauss’s law.

     (c) Argue that by symmetry, E = 0. Show that this does not obey Gauss’s law.

     (d) ⋆ What’s going on? Which, if any, is the actual field? If you think there’s more than one
         possible field, how could that be consistent with Coulomb’s law, which gives the answer
         explicitly? For that matter, what does Coulomb’s law say about this setup, anyway?

       Idea 4
       A tricky, occasionally useful idea is to use Newton’s third law: it may be easier to calculate
       the force of A on B than the force of B on A.

       Example 5: Purcell 1.28

       Consider a point charge q. Draw any imaginary sphere of radius R around the charge. Show
       that the average of the electric field over the surface of the sphere is zero.

       Solution
       Imagine placing a uniform surface charge σ on the sphere. Then the average of the point
       charge’s electric field over the sphere times 4πR2 σ is the total force of the point charge on
       the charged sphere. But this is equal in magnitude to the force of the charged sphere on
       the point charge, which must be zero by the shell theorem. Thus the average field over the
       sphere has to vanish.

       Example 6

       Consider two spherical uniformly charged balls of charge q and radii ai , with their centers
       separated by a distance r > a1 + a2 . What is the net force of the first on the second?

       Solution

       It might seem obvious that the answer is q 2 /4πϵ0 r2 , with no dependence on a1 and a2 . In
       fact, if you’ve done any orbital mechanics, you’ve almost certainly assumed that the force
       between two spherical bodies (such as the Earth and Sun) is Gm1 m2 /r2 , which is equivalent.

       This has a simple but slightly tricky proof. By the shell theorem, we can set a1 = 0, replacing
       the first ball with a point charge, because this produces the same field at the second ball.

       But the force on the second ball depends on the electric field at every point on it, which
       seems to require doing an integral. To avoid this, we use Newton’s third law, which tells us
       it’s equivalent to compute the force on the first ball. To compute that, we may set a2 = 0 by
       the shell theorem again. This reduces us to the case of two point charges, giving the answer.

[3] Problem 7 (Purcell 1.28). Some extensions of the previous example.

      (a) Show that if the charge q is instead outside the sphere, a distance r > R from its center, the
          average electric field over the surface of the sphere is the same as the electric field at the center
          of the sphere.

      (b) Show that for any overall neutral charge distribution contained within a sphere of radius R,
          the average electric field over the interior of the sphere is −p/4πϵ0 R3 where p is the total
          dipole moment.

Solution:
(a) We see that ∇ · E = ∂x ((ρ/ϵ0 )x) = ρ/ϵ0 , as desired.

      (b) In Cartesian coordinates, this field is
                                                     ρ
                                              E=        (xx̂ + yŷ + zẑ)
                                                    3ϵ0
          whose divergence is ρ/ϵ0 , as desired.

      (c) This has to hold by symmetry because the electric field can’t point in any particular direction,
          by rotational symmetry. It also can’t just point radially, because that breaks translational
          symmetry; the center is a special point. So the only option is E = 0.
          But then clearly ∇ · E = 0, so Gauss’s law is not obeyed.

 (d) The issue is boundary conditions. Just like any differential equation, the solution for the
     electric field is not defined without boundary conditions, just like how a solution for Newton’s
     second law, F = ma, is not defined without specifying an initial position and velocity.
     One common way to resolve the ambiguity, used in example 3, is to assume the field goes to zero
     at infinity. This is equivalent to assuming there’s no “extra” field produced by other charges
     outside the ones we’ve considering. (Of course, in real experiments, that’s not guaranteed,
     and we always have to worry about “shielding” external fields, such as from other pieces of
     equipment or static charges. Even the Earth makes a sizable vertical electric field, which
     is occasionally discharged by lightning.) However, we can’t do this here because the charge
     density extends out to infinity too.
     Another common way to resolve the ambiguity, used in example 4, is to assume that the
     field shares the same symmetries as the charge distribution. But in this case, that’s not
     even possible, because the charge distribution has too much symmetry. (It is translationally
     symmetric, and rotationally symmetric about every point!) It is impossible to pick a set of
     boundary conditions that maintains all these symmetries. That’s why there are many equally
     good answers, depending on which symmetries you break.
     At first glance, it could seem that Coulomb’s law could give us a unique answer. Coulomb’s
     law for a point charge is itself derived by implicitly assuming that there are no “extra” fields
     flying around, just the spherically symmetric field of the point charge itself. This looks very
     reasonable, so what stops us from just saying that each charge in this problem has such a field,
     and then integrating over the charges? R ∞Well, if you write down the integral, you’ll find that it’s
     divergent, analogous to the integral −∞ x dx. By itself, the integral is not even well-defined.
     In order to get an answer, you have to “regulate” the integral (i.e. change it in a way that
     makes it well-defined). One possible
                                       R L regulator, for example, is to just chop off the limits of
     integration at finite values, like −L x dx. But that particular regulator is equivalent to just
     replacing the charge distribution with a finite one centered at the origin! In other words,
     Coulomb’s law also fails to give a unique answer, because it requires a regulator to give a well-
     defined answer, and there are many possible regulators. If you treat the charge distribution
     as a giant ball with center at the origin, you get the result of part (b). If you treat it as a
     thick, huge slab along the yz plane centered at the origin, you get the result of part (a). The
     symmetry argument fails once again, because all the regulators break some symmetry. This
     is a simple example of “anomalous symmetry”, an important idea in theoretical physics.
     The exact same problem appears in Newtonian cosmology, where charge density is replaced
     with mass density, and this problem confused Newton himself, who incorrectly thought that
     g = 0 by symmetry. In this context, all regulators/boundary conditions are unsatisfactory.
     Of course, we want a rotationally symmetric universe to match experiment, so we have to put
     that in by hand. But then every solution has a center towards which everything collapses, so
     to keep the solar system an inertial frame, we’d have to put it at the center of the universe!
     Surely, this would make Copernicus roll in his grave.
     Some of these problems are fixed in general relativity. You still have to postulate rotational
     symmetry (again, on the basis of experimental data), but once you do that, there are no
     further problems. That’s because in general relativity, acceleration is not absolute in the
     way it is in Newtonian mechanics. Instead, there is no center; everything just gets closer to
     everything else. For further discussion and references, see this paper.

       Idea 4
       A tricky, occasionally useful idea is to use Newton’s third law: it may be easier to calculate
       the force of A on B than the force of B on A.

       Example 5: Purcell 1.28

       Consider a point charge q. Draw any imaginary sphere of radius R around the charge. Show
       that the average of the electric field over the surface of the sphere is zero.

       Solution
       Imagine placing a uniform surface charge σ on the sphere. Then the average of the point
       charge’s electric field over the sphere times 4πR2 σ is the total force of the point charge on
       the charged sphere. But this is equal in magnitude to the force of the charged sphere on
       the point charge, which must be zero by the shell theorem. Thus the average field over the
       sphere has to vanish.

       Example 6

       Consider two spherical uniformly charged balls of charge q and radii ai , with their centers
       separated by a distance r > a1 + a2 . What is the net force of the first on the second?

       Solution

       It might seem obvious that the answer is q 2 /4πϵ0 r2 , with no dependence on a1 and a2 . In
       fact, if you’ve done any orbital mechanics, you’ve almost certainly assumed that the force
       between two spherical bodies (such as the Earth and Sun) is Gm1 m2 /r2 , which is equivalent.

       This has a simple but slightly tricky proof. By the shell theorem, we can set a1 = 0, replacing
       the first ball with a point charge, because this produces the same field at the second ball.
       But the force on the second ball depends on the electric field at every point on it, which
       seems to require doing an integral. To avoid this, we use Newton’s third law, which tells us
       it’s equivalent to compute the force on the first ball. To compute that, we may set a2 = 0 by
       the shell theorem again. This reduces us to the case of two point charges, giving the answer.

[3] Problem 7 (Purcell 1.28). Some extensions of the previous example.
     (a) Show that if the charge q is instead outside the sphere, a distance r > R from its center, the
         average electric field over the surface of the sphere is the same as the electric field at the center
         of the sphere.

     (b) Show that for any overall neutral charge distribution contained within a sphere of radius R,
         the average electric field over the interior of the sphere is −p/4πϵ0 R3 where p is the total
         dipole moment.
    Solution. The same Newton’s third law trick will work for both parts.
     (a) Let the desired answer be Eavg and let the charge q be at r. Now imagine a charge Q
         is uniformly distributed over the surface of the sphere. The force of the charge q on the

          distributed charge Q is precisely FqQ = QEavg . But we also know that
                                                                  kQq
                                                 FqQ = −FQq = −       r̂
                                                                   r2
          by Newton’s third law and the shell theorem. Therefore we have
                                                              kq
                                                    Eavg = − 2 r̂
                                                              r
          which is precisely the electric field at the center of the sphere due to q. (Note that r̂ points
          from the center of the sphere to the charge q.)
      (b) Let the desired answer be Eavg . Now imagine a charge Q is uniformly distributed over the
          volume of the sphere. The force of the charge distribution (with charge density ρ(x)) on the
          distributed charge Q is precisely FqQ = QEavg . But we also know that
                                                         Z
                                       FqQ = −FQq = − ρ(r)EQ (r) d3 r

          where EQ is the field due to Q. Now, this field is easy to find, as it is just the field of a
          uniformly charged sphere, so
                                                         kQ
                                                   EQ = 3 r
                                                          R
          as shown in problem 1. Putting this in the integral, we have
                                                          Z
                                                       kQ
                                            QEavg = − 3     ρ(r)r d3 r
                                                       R
          so by the definition of the dipole moment,
                                                     Z
                                                  k                   kp
                                         Eavg = − 3 ρ(r)r d3 r = − 3
                                                  R                   R
          as desired.


[User] IPhO 2012, problem 1C. A neat and clean problem about superconductors.

Solution: IPhO 2012, problem 1C. A neat and clean problem about superconductors.


[User] Juku is living in a cylindrical space station that rotates so that artificial gravitational force is created. The radius of the station is $R$, its rotation’s angular velocity is $\omega$. Juku throws a ball straight up with an initial speed $v=\frac{\sqrt{3}}{3}\omega R$. From how far away from Juku along the surface of the station will the ball land?

Solution:
Vaatleme palli lendu jaama teljega seotud inertsiaalses taustsüsteemis. Olgu Juku koordinaadid palli viskamise hetkel $(0,-R)$. Sellisel juhul liigub pall pärast viset ühtlaselt ning sirgjooneliselt, kusjuures selle kiiruse vertikaalsihiline komponent on $v$ ning horisontaalsihiline komponent $\omega R$. Järelikult $\tan\alpha=\frac{\omega R}{v}$, millest saame $\alpha=\frac{\pi}{3}$. Kuna kehtib $\varphi=\pi-2\alpha$, siis saame järeldada, et ka $\varphi=\frac{\pi}{3}$ ning pall läbib enne jaama pinnani jõudmist teepikkuse $R$. Palli kiirus on
\[
\sqrt{v^2+\omega^2R^2}=\frac{2\sqrt{3}}{3}\omega R,
\]
seega on pall õhus aja $t=\frac{\sqrt{3}}{2\omega}$, mille jooksul jõuab jaam pöörduda nurga $\theta=\omega t=\frac{\sqrt{3}}{2}$ võrra. Järelikult näeb Juku otse üles visatud palli maanduvat enda ees kaugusel
\[
\left(\varphi-\theta\right)R=\left(\frac{\pi}{3}-\frac{\sqrt{3}}{2}\right)R.
\]

\begin{center}
\begin{tikzpicture}[scale=1]
\draw (2.5,0) arc (0:360:2.5);
\draw[->,thick] (-3,0) -- (3,0);
\node[below] at (2.9,0) {x};
\draw[->,thick] (0,-3) -- (0,3);
\node[left] at (0,2.9) {y};
\node[left] at (0,-2.8) {(0,-R)};
\draw (0,-2.5) -- (2.17,-1.25) -- (0,0);
\node[right] at (-0.125,-2.25) {$\alpha$};
\draw (0,-2) arc (90:30:0.5);
\node[left] at (2.17,-1.25) {$\alpha$};
\draw (1.75, -1.5) arc (210:150:0.5);
\node[right] at (-0.125,-0.3) {$\varphi$};
\draw (0,-0.5) arc (270:330:0.5);
\node[left] at (0,-1.25) {R};
\node[right] at (1.09,-0.55) {R};
\end{tikzpicture}
\end{center}

Pallivise

Juku elab silindrikujulises kosmosejaamas, mille pöörlemine tekitab kunstliku raskusjõu. Jaama raadius on $R$, selle pöörlemise nurkkiirus $\omega$. Juku viskab palli otse üles algkiirusega $v=\frac{\sqrt{3}}{3}\omega R$. Kui kaugele Jukust mööda jaama pinda pall maandub?

Palli lendu on mugav vaadelda laboratoorses taustsüsteemis. Sellisel juhul liigub pall pärast viset ühtlaselt ning sirgjooneliselt, kusjuures selle kiiruse vertikaalsihiline komponent on $v$ ning horisontaalsihiline komponent $\omega R$.


[User]
Question B2
In this problem, use a particle-like model of photons: they propagate in straight lines and obey the
law of reflection, but are subject to the quantum uncertainty principle. You may use small-angle
approximations throughout the problem.
    A photon with wavelength λ has traveled from a distant star to a telescope mirror, which has
a circular cross-section with radius R and a focal length f  R. The path of the photon is nearly
aligned to the axis of the mirror, but has some slight uncertainty ∆θ. The photon reflects off the
mirror and travels to a detector, where it is absorbed by a particular pixel on a charge-coupled
device (CCD).
    Suppose the telescope mirror is manufactured so that photons coming in parallel to each other
are focused to the same pixel on the CCD, regardless of where they hit the mirror. Then all small
cross-sectional areas of the mirror are equally likely to include the point of reflection for a photon.

  a. Find the standard deviation ∆r of the distribution for r, the distance from the center of the
     telescope mirror to the point of reflection of the photon.

  b. Use the uncertainty principle, ∆r∆pr ≥ h̄/2, to place a bound on how accurately we can
     know the angle of the photon from the axis of the telescope. Give your answer in terms of R
     and λ. If you were unable to solve part a, you may also give your answer in terms of ∆r.

   c. Suppose we want to build a telescope that can tell with high probability whether a photon it
      detected from Alpha Centauri A came the left half or right half of the star. Approximately
      how large would a telescope have to be to achieve this? Alpha Centauri A is approximately
      4 × 1016 m from Earth and has a radius approximately 7 × 108 m. Assume visible light with
      λ = 500 nm.

                   Copyright c 2018 American Association of Physics Teachers

2018 USA Physics Olympiad Exam                  Part B                                              9

Solution:
Question B2
In this problem, use a particle-like model of photons: they propagate in straight lines and obey the
law of reflection, but are subject to the quantum uncertainty principle. You may use small-angle
approximations throughout the problem.
    A photon with wavelength λ has traveled from a distant star to a telescope mirror, which has
a circular cross-section with radius R and a focal length f  R. The path of the photon is nearly
aligned to the axis of the mirror, but has some slight uncertainty ∆θ. The photon reflects off the
mirror and travels to a detector, where it is absorbed by a particular pixel on a charge-coupled
device (CCD).
    Suppose the telescope mirror is manufactured so that photons coming in parallel to each other
are focused to the same pixel on the CCD, regardless of where they hit the mirror. Then all small
cross-sectional areas of the mirror are equally likely to include the point of reflection for a photon.

  a. Find the standard deviation ∆r of the distribution for r, the distance from the center of the
     telescope mirror to the point of reflection of the photon.

                                                 Solution

      The square of the standard deviation is the variance, so

                                             (∆r)2 = hr2 i − hri2 .

      Computing the average value of r2 is mathematically the exact same thing as computing the
      moment of inertia of a uniform disk; we have
                                                  Z R
                                      2     1                       1
                                    hr i =              r2 (2πrdr) = R2 .
                                           πR2     0                2

      Similarly, the average value of r is
                                                  Z R
                                             1                     2
                                      hri =             r (2πrdr) = R.
                                            πR2     0              3

      Then we have
                                                       R
                                                 ∆r = √ .
                                                       18

  b. Use the uncertainty principle, ∆r∆pr ≥ h̄/2, to place a bound on how accurately we can
     know the angle of the photon from the axis of the telescope. Give your answer in terms of R
     and λ. If you were unable to solve part a, you may also give your answer in terms of ∆r.

                                                 Solution

      We have
                                                             h
                                             ∆pr ≈ p∆θ =       ∆θ.
                                                             λ

                   Copyright c 2018 American Association of Physics Teachers

2018 USA Physics Olympiad Exam                 Part B                                                 15

    Applying the uncertainty principle, we have
                                                       √
                                                         18λ
                                               ∆θ ≥          .
                                                        4πR
    Since the factors of h̄ canceled out, this is really a classical calculation; one could get a similar
    result by considering classical diffraction from a circular aperture.

  c. Suppose we want to build a telescope that can tell with high probability whether a photon it
     detected from Alpha Centauri A came the left half or right half of the star. Approximately
     how large would a telescope have to be to achieve this? Alpha Centauri A is approximately
     4 × 1016 m from Earth and has a radius approximately 7 × 108 m. Assume visible light with
     λ = 500 nm.

                                               Solution

    We need ∆θ to be much smaller than the actual angular separation, or

                                             7 × 108 m
                                      ∆θ               ≈ 2 × 10−8 .
                                             4 × 1016 m
    This means that                   √         √
                                     18λ            18(5 × 10−7 m)
                                R≈                                = 8.4 m
                                   4π∆θ             4π(2 × 10−8 )

                  Copyright c 2018 American Association of Physics Teachers

2018 USA Physics Olympiad Exam                  Part B                                              16

USAPhO 2018 Question B2


[User]
Theory
                                               English (Official)                       Q2-1
Nonlinear Dynamics in Electric Circuits (10 points)
Please read the general instructions in the separate envelope before you start this problem.

Introduction

Bistable non-linear semiconducting elements (e.g. thyristors) are widely used in electronics as switches
and generators of electromagnetic oscillations. The primary field of applications of thyristors is con-
trolling alternating currents in power electronics, for instance rectification of AC current to DC at the
megawatt scale. Bistable elements may also serve as model systems for self-organization phenomena
in physics (this topic is covered in part B of the problem), biology (see part C) and other fields of modern
nonlinear science.

Goals

To study instabilities and nontrivial dynamics of circuits including elements with non-linear I − V charac-
teristics. To discover possible applications of such circuits in engineering and in modeling of biological
systems.

Part A. Stationary states and instabilities (3 points)
Fig. 1 shows the so-called S-shaped I − V characteristics of a non-linear element X. In the voltage
range between Uh = 4.00 V (the holding voltage) and Uth = 10.0 V (the threshold voltage) this I − V
characteristics is multivalued. For simplicity, the graph on Fig. 1 is chosen to be piece-wise linear (each
branch is a segment of a straight line). In particular, the line in the upper branch touches the origin if it
is extended. This approximation gives a good description of real thyristors.

                                 10

                                  9

                                  8

                                  7

                             I [A] 6

                                  5

                                  4

                                  3
                                                     Uh
                                  2
                                                                                Uth
                                  1

                                  0
                                   0   1   2     3        4    5      6   7   8   9   10
                                                              U [V]

                       Figure 1: I − V characteristics of the non-linear element X.

                                                 Theory
                                              English (Official)                  Q2-2
     A.1     Using the graph, determine the resistance Ron of the element X on the upper            0.4pt
             branch of the I − V characteristics, and Roff on the lower branch, respectively.
             The middle branch is described by the equation
                                                            U
                                                I = I0 −        .                             (1)
                                                           Rint
             Find the values of the parameters I0 and Rint .

The element X is connected in series (see Fig.2) with a resistor R, an inductor L and an ideal voltage
source E. One says that the circuit is in a stationary state if the current is constant in time, I(t) = const.

                                                    R               L

                                                        E

              Figure 2: Circuit with element X, resistor R, inductor L and voltage source E.

     A.2     What are the possible numbers of stationary states that the circuit of Fig. 2 may      1pt
             have for a fixed value of E and for R = 3.00 Ω? How does the answer change for
             R = 1.00 Ω?

     A.3     Let R = 3.00 Ω, L = 1.00 μH and E = 15.0 V in the circuit shown in Fig. 2.             0.6pt
             Determine the values of the current Istationary and the voltage Vstationary on the
             non-linear element X in the stationary state.

The circuit in Fig. 2 is in the stationary state with I(t) = Istationary . This stationary state is said to be
stable if after a small displacement (increase or decrease in the current), the current returns towards the
stationary state. And if the system keeps moving away from the stationary state, it is said to be unstable.

     A.4     Use numerical values of the question A.3 and study the stability of the stationary     1pt
             state with I(t) = Istationary . Is it stable or unstable?

Part B. Bistable non-linear elements in physics: radio transmitter (5 points)
We now investigate a new circuit configuration (see Fig. 3). This time, the non-linear element X is con-
nected in parallel to a capacitor of capacitance C = 1.00 μF. This block is then connected in series to a
resistor of resistance R = 3.00 Ω and an ideal constant voltage source of voltage E = 15.0 V. It turns out
that this circuit undergoes oscillations with the non-linear element X jumping from one branch of the
I − V characteristics to another over the course of one cycle.

                                                   Theory
                                                English (Official)                    Q2-3
                                                                R

                                                 C             E

              Figure 3: Circuit with element X, capacitor C, resistor R and voltage source E.

     B.1     Draw the oscillation cycle on the I − V graph, including its direction (clockwise          1.8pt
             or anticlockwise). Justify your answer with equations and sketches.

     B.2     Find expressions for the times t1 and t2 that the system spends on each branch             1.9pt
             of the I − V graph during the oscillation cycle. Determine their numerical val-
             ues. Find the numerical value of the oscillation period T assuming that the time
             needed for jumps between the branches of the I − V graph is negligible.

     B.3     Estimate the average power P dissipated by the non-linear element over the                 0.7pt
             course of one oscillation. An order of magnitude is sufficient.

The circuit in Fig. 3 is used to build a radio transmitter. For this purpose, the element X is attached to
one end of a linear antenna (a long straight wire) of length s. The other end of the wire is free. In the
antenna, an electromagnetic standing wave is formed. The speed of electromagnetic waves along the
antenna is the same as in vacuum. The transmitter is using the main harmonic of the system, which has
period T of question B.2.

     B.4     What is the optimal value of s assuming that it cannot exceed 1 km?                        0.6pt

Part C. Bistable non-linear elements in biology: neuristor (2 points)
In this part of the problem, we consider an application of bistable non-linear elements to modeling of
biological processes. A neuron in a human brain has the following property: when excited by an external
signal, it makes one single oscillation and then returns to its initial state. This feature is called excitability.
Due to this property, pulses can propagate in the network of coupled neurons constituting the nerve
systems. A semiconductor chip designed to mimic excitability and pulse propagation is called a neuristor
(from neuron and transistor).
We attempt to model a simple neuristor using a circuit that includes the non-linear element X that we
investigated previously. To this end, the voltage E in the circuit of Fig. 3 is decreased to the value E′ =
12.0 V. The oscillations stop, and the system reaches its stationary state. Then, the voltage is rapidly
increased back to the value E = 15.0 V, and after a period of time τ (with τ < T ) is set again to the value
E′ (see Fig. 4). It turns out that there is a certain critical value τcrit. , and the system shows qualitatively
different behavior for τ < τcrit and for τ > τcrit .

                                           Theory
                                       English (Official)               Q2-4
                 16

                 15

         E [V] 14

                 13

                 12
                                          t0                      t0 + τ
                 11
                                                       t
               Figure 4: Voltage of the voltage source as a function of time.

C.1   Sketch the graphs of the time dependence of the current IX (t) on the non-linear   1.2pt
      element X for τ < τcrit and for τ > τcrit .

C.2   Find the expression and the numerical value of the critical time τcrit for which   0.6pt
      the scenario switches.

C.3   Is the circuit with τ = 1.00 × 10−6 s a neuristor?                                 0.2pt

Solution:
Problem 2 : Solution/marking scheme – Nonlinear Dynamics in Electric Circuits
(10 points)
Part A. Stationary states and instabilities (3 points)

  Solution A1:                                                                                [0.4]

  By looking at the I − V graph, we obtain

                                          Roff = 10.0 Ω,                                           0.1
                                          Ron = 1.00 Ω,                                            0.1
                                          Rint = 2.00 Ω,                                           0.1
                                             I0 = 6.00 A.                                          0.1
  Note: No penalty for the number of digits in this question

  Solution A2:                                                                                [1]

  Kirchoff law for the circuit (U is the voltage of the bistable element):

                                             E = IR + U                                            0.1

  This yields
                                                E −U
                                             I=                                                    0.1
                                                  R
  Hence, stationary states of the circuit are intersections of the line defined by this equation
  and the I − V graph of X .                                                                       0.2

  For R = 3.00 Ω, one always gets exactly one intersection.                                        0.2
  For R = 1.00 Ω, one gets 1, 2 or 3 intersections depending on the value of E.                    0.4
  The following table summarizes the number of points granted for possible answers to the
  last subquestion with R = 1.00 Ω:

                     Possible answer     1    2     3    1,3   1,2   2,3   1,2,3
                     Points              0    0   0.2    0.3     0   0.2     0.4

  Solution A3:                                                                                [0.6]

  The stationary state is on the intermediate branch, one can thus use the corresponding
  equation:                                                                                0.2

                                                  1

                                                E − Rint I0
                                   Istationary =                                                0.1
                                                 R − Rint
                                              = 3.00 A                                          0.1

                                   Ustationary = Rint (I0 − I)                                  0.1
                                              = 6.00 V                                          0.1

 Extra (non-physical) stationary states on the switched on and/or switched off branches
 lead to a penalty of 0.2 point.

 Solution A4:                                                                                [1]

 Any correct modeling such as the following:                                                    0.5
 The Kirchoff law for the circuit reads
                                          dI                         dI
                      E = IR + UX + L        = IR + (I0 − I)Rint + L
                                          dt                         dt
 This implies
                                  dI
                               L     = E − I0 Rint − (R − Rint )I
                                  dt
 The separation between two cases is of importance, especially because of the relative sign
 of dI/dt:
 If I > Istationary , we have dI/dt < 0 and I decreases.                                        0.2
 If I < Istationary , we have dI/dt > 0 and I increases.                                        0.2
 Note: Formulas with time derivatives are not essential, any other correct justification is
 accepted.
 We conclude that the stationary state is stable.                                               0.1

 Note: The checkbox gives 0.1 points if “stable” is checked, regardless of the previous rea-
 soning (also if there is nothing). A wrong reasoning leading to check the “unstable” option
 doesn’t however give any point for the checkbox.

Part B. Bistable non-linear elements in physics and engineering: radio transmitter (5 points)

 Solution B1:                                                                                [1.8]

 A correctly drawn cycle gives 1.2 points, distributed as follows:

     • Switched on branch is part of the cycle                                                  0.2

     • Switched off branch is part of the cycle                                                 0.2

     • Jumps are vertical (constant U )                                                         0.2

     • Jumps are positioned at Uh and Uth                                                       0.2

     • The system moves to the left on the switched on branch                                   0.2

                                                2

   • The system moves to the right on the switched off branch                                 0.2

Each of the following observations individually gives up to 0.2 points, but their total
cannot exceed 0.6:

   • U constant during jumps because the charge on the capacitor cannot change in-
     stantaneously                                                                            0.2

   • The intermediate branch cannot be part of the cycle because there is a stationary
     state on it                                                                              0.2

   • Jumps occur at corners of the IV graph because at those points the system has
     nowhere else to go                                                                       0.2

   • The system moves moves to the left on the switched on branch because it approaches
     the stable stationary state (which is located outside the IV graph), or argument with
     the Kirchoff law                                                                         0.2

   • The system moves moves to the right on the switched off branch because it ap-
     proaches the stable stationary state (which is located outside the IV graph), or
     argument with the Kirchoff law                                                           0.2

                                      In
                                      ⇒
                                      =
                                      ÷:÷
                            ^   IIA   )

                        "

                                       l¥
                        :
                                                      oscillation
                                                        cycle
                        8

                        :

                                                                        stationary
                                                  I     it
                                                                            state
                                                  I
                        3

                                0         2   3         45
                                                             =      6   7           85

                                                                                th

Solution B2:                                                                                [1.9]

Since the non-linear element is oscillating between the switched on and switched
off branches we can put UX = Ron/off IX . On either of the branches, the circuit behaves
as a standard RC-circuit with conductance C and resistance Ron/off R/(Ron/off + R) (the
resistor and the element X being connected in parallel). Another way to express it is to      0.5

                                                             3

write the Kirchhoff law for the switched on and switched off branches
                                       dIX
                          Ron/off RC       = E − (Ron/off + R)IX
                                        dt

The time constant of the circuit is
                                         Ron/off R
                                                    C.
                                        Ron/off + R

If the branch in question (switched on or switched off) extended indefinitely, after a long
time the system would have landed in a stationary state with the voltage
                                                  Ron/off
                                    Uon/off =               E.
                                                Ron/off + R

Then, the time dependence of the voltage drop on the non-linear element is a sum of the
constant term Uon/off and of the exponentially decaying term:
                                                               Ron/off +R
                       Ron/off                   Ron/off

                                                              −R          t
            UX (t) =             E + Uon/off −             E e on/off RC
                     Ron/off + R               Ron/off + R

There are 0.5 points distributed as follow for UX (t):

   • Correct exponential                                                                      0.2

   • Correct constant term (t → ∞)                                                            0.1

   • Correct coefficient in front of the exponential                                          0.1

   • Correct equation for UX (t)                                                              0.1

Time spent by the system on the switched on branch during one cycle:

                         Ron R          Uth − Uon
                  ton =         C log               = 2.41 · 10−6 s,                          0.4
                        Ron + R         Uh − Uon

Time spent by the system on the switched off branch during one cycle:

                         Roff R          Uoff − Uh
                 toff =          C log                = 3.71 · 10−6 s.                        0.4
                        Roff + R         Uoff − Uth

The total period of oscillations:

                               T = ton + toff = 6.12 · 10−6 s                                 0.1

Note: Correct final answers give full points. One may earn points for intermediate steps
(see above) for partial answers.

Solution B3:                                                                                [0.7]

Neglect the energy consumed on the switched off branch.            The energy consumed

                                                 4

 on the switched on branch during the cycle is estimated by
                                                  2
                              1         Uh + Uth
                          E=                            ton = 1.18 · 10−4 J.                  0.4
                             Ron           2

 For the power, this gives an estimate of
                                              E
                                        P ∼     = 19.3 W.                                     0.3
                                              T
 Note:

     • Formula + answer inside 5 W ≤ P ≤ 50 W give full points

     • Formula + answer outside the range above but inside 1 W ≤ P ≤ 100 W give 0.5
       points

     • answer outside range but good formula gives 0.4 points

 Also, the proposed formula is only an example, any other reasonable approximation of the
 integral of the upper branch should be accepted.

 Solution B4:                                                                                [0.6]

 The wave length of the radio signal is given by λ = cT = 1.82 · 103 m.                       0.2
 The optimal length of the antenna is λ/4 (or 3λ/4, 5λ/4 etc.)                                0.3
 The only choice which is below 1 km is s = λ/4 = 459 m.                                      0.1
 Note: The correct answer s = λ/4 = 459 m gives full points, and the mistake s = λ/2 =
 918 m only 0.4 pts.

Part C. Bistable non-linear elements in biology: neuristor (2 points)

 Solution C1:                                                                                [1.2]

 For Ẽ = 12.0 V, the steady state of the system is located on the switched off
 branch:
                                   Roff
                            Ũ =          Ẽ = 9.23 V.
                                 R + Roff
 When the voltage is increased to E = 15.0 V, the system starts moving to the right along
 the switched off branch (in the same way it did in task B).
 If the voltage drops again before the system reaches the threshold voltage, it will simply
 return to the stationary state.
 If system reaches the threshold voltage, it will jump to the switched on branch, and it
 will make one oscillations (since τ < T ) before the voltage drops again and it returns to
 the stationary state.

                                                   5

                        I

                                                                 ÷
           ^
                                x
                                                                                T   <

                                                     to
                                                                                0                 return              to               the

                                                     ⇐:#
                                                                                original                  stationary
                                                                                state

                                                          :h÷
                                                        osier                                                                                .

                                                                                >
                                                                                t
                                to                               tote

1. Approach to the new stationary state                                                                                0.2

2. Return to the old stationary state                                                                                0.2

          ' '
                 T
                            X
                    -

                                                                          5    evolution           on
                                                                                T            >
                                                                                Tent
                                                                      f       upper
                                                                                branch

                                                             Ji\!#            i
                                                to
                            jump
                4                                            i

                                                                                6                              to
                                         branch                                                                        jump
                upper

                                                                                )
                                                             ,                                                        lower               branch
                                                             :
                                                                                ,

                                         apprach                                                        L                7                    return      to
                                3         new        state   ,

                                                                                Stationary

                                                                                (
                                                                                :

                                          ¥
                                                                                state
                                                                                ,

                            .        -
                                                                                '
                                                                                ✓

           .

           .                    i                                 I
                                                                                )

                            to                                   tot      T                                                                   t

3. Approach to the new stationary state                                                                                0.1

4. Jump to the upper branche before t0 + τ                                                                                0.2

5. Evolution on the upper branch                                                                                0.2

6. Jump to the lower branche below the old stationary state                                                                                0.1

7. Return to the old stationary state (from below)                                                                                0.2

                                                                               6

Solution C2:                                                                               [0.6]

The time needed to reach the threshold voltage is given by
                                                    !
                          Roff R          Uoff − Ũ
                 τcrit =          C log                = 9.36 · 10−7 s.
                         Roff + R        Uoff − Uth

Note: This is the same formula as for toff in task B2, with Uh replaced by Ũ .

   • Correct time constant                                                                 0.2

   • Correct choice of voltages                                                            0.2

   • Correct final formula                                                                 0.1

   • Correct numerical value                                                               0.1

Note: Correct final answers give full points. One may earn points for intermediate steps
(see above) for partial answers.

Solution C3:                                                                               [0.2]

Since τ > τcrit , the system will make one oscillation.      We conclude that the sys-
tem is a neuristor.                                                                        0.2

Note: 0.2 are given only if “Yes” is checked, regardless of the development of the other
tasks.

                                            7

English (Official)                       Q2-1