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2014 年第 45 届国际物理学奥林匹克竞赛理论题和实验题及其
解答(英文)
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竞赛题及答案完整版
Problem 1
Solution Part A
Consider the forces acting on the puck and the cylinder and depicted in the figure
on the right. The puck is subject to the gravity force ???? and the reaction force
from the cylinder ??. The cylinder is subject to the gravity force ????, the reaction
force from the plane ??1, the friction force ?????? and the pressure force from the
puck ??′=???. The idea is to write the horizontal projections of the equations of
motion. It is written for the puck as follows
??????=??sin??,
(A.1)
where ???? is the horizontal projection of the puck acceleration.
For the cylinder the equation of motion with the acceleration ?? is found as
????=??sin?????????.
(A.2)
Since the cylinder moves along the plane without sliding its angular acceleration
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is obtained as
??=??/??
(A.3) Then the equation of rotational motion around the center of
mass
form
of
the
cylinder
takes
the
????=??http://doc.guandang.net/bb316ba8befa958cb491190ff.html??????,
(A.4) where the inertia moment of the hollow cylinder is given by
??=????2.
(A.5) Solving (A.2)-(A.5) yields
2????=??sin??.
From equations (A.1) and (A.6) it is easily concluded that
(A.6)
??????=2????.
(A.7)
Since the initial velocities of the puck and of the cylinder are both equal to zero,
then, it follows from (A.7) after integrating that
????=2????.
written as
(A.8) It is obvious that the conservation law for the system is
??????=2 2 2,
(A.9) where the angular velocity of the cylinder
is found to be
??
??=??,
(A.10) since it does not slide over the plane.
Solving (A.8)-(A.10) results in velocities at the lowest point of the puck trajectory
written as
??=2 (2?? ??)??=?? (2?? ??)??
????????????????2
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????2
????2
(A.12) (A.13)
In
the
refhttp://doc.guandang.net/bb316ba8befa958cb491190ff.htmlerence
frame
sliding progressively along with the cylinder axis, the puck moves in a circle of
radius ?? and, at the lowest point of its trajectory, have the velocity
????????=?? ??
??rel=??
(A.14) and the acceleration
(A.15)
At the lowest point of the puck trajectory the acceleration of the cylinder axis is
equal to zero, therefore, the puck acceleration in the laboratory reference frame
is also given by (A.15).
???????=??
puck and the cylinder is finally found as
??
??=3???? 1 3?? .
2
??????????
then the interaction force between the
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2??
(A.16) (A.17)
Part B
1) According to the first law of thermodynamics, the amount of heat transmitted ????
to the gas in the bubble is found as
????=?????????? ??????,
(B.1) where the molar heat capacity at arbitrary
process is as follows
1??????????://doc.guandang.net/bb316ba8befa958cb491190ff.html
??=??????=???? ??????
(B.2)
Here ???? stands for the molar heat capacity of the gas at constant volume, ??
designates its pressure, ?? is the total amount of moles of gas in the bubble, ??
and ?? denote the volume and temperature of the gas, respectively.
Evaluate the derivative standing on the right hand side of (B.2). According to the
Laplace formula, the gas pressure inside the bubble is defined by
4??
??=??,
(B.3) thus, the equation of any equilibrium process with the gas in
the bubble is a polytrope of the form
??3??=const.
form
????=??????,
(B.4) The equation of state of an ideal gas has the
(B.5) and hence equation (B.4) can be rewritten as
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??3???2=const.
(B.6)
Differentiating (B.6) the derivative with respect to
temperature sought is found as
????3??
=
(B.7)
????2??
Taking into accounhttp://doc.guandang.net/bb316ba8befa958cb491190ff.htmlt that the
molar heat capacity of a diatomic gas at constant volume is
5
????=2??,
(B.8) and using (B.5) it is finally obtained that
3J
??=???? 2??=4??=33.2 mole?K.
(B.9)
2) Since the heat capacity of the gas is much smaller than the heat capacity of the
soap film, and there is heat exchange between them, the gas can be considered as
isothermal since the soap film plays the role of thermostat. Consider the fragment
of soap film, limited by the angle ?? as shown in the figure. It's area is found as
??=??(????)2.
??=??????.
(B.10) and the corresponding mass is obtained as
(B.11)
Let ?? be an increase in the radius of the bubble, then
the Newton second law for the fragment of the soap film mentioned above takes the
form
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???? =??′??′???????????,
resultant
(B.12) where ?????????? denotes the projection of the
surface
tension
actihttp://doc.guandang.net/bb316ba8befa958cb491190ff.htmlng
force
in
the
radial
direction, ??′ stands for the gas pressure beneath the surface of the soap film and
??
??′=?? 1 2?? .
?????????? is easily found as
??????????=????????=???2?2??[ ?? ?? ??]???. (B.13)
Since the gaseous process can be considered isothermal, it is written that
??′??′=????.
(B.14)
Assuming that the volume increase is quite small, (B.14) yields
113??
??′=????≈??≈?? 1??? .
(B.15)
1
??
1 ??
Thus, from (B.10) - (B.16) and (B.3) the equation of small oscillations of the soap
film is derived as
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8??
?????? =???2??
(B.16)
with the frequency
??= ??????2=108 s?1.
8??
(B.17)
Part C
The problem can be solved in different ways. Herein several possible solutions are
considered.
At
the
moment
when
the
current
in
the
coils
is
a
mhttp://doc.guandang.net/bb316ba8befa958cb491190ff.htmlaximum, the total voltage
across the coils is equal to zero, so the capacitor voltages must be equal in magnitude
and opposite in polarity. Let ?? be a voltage on the capacitors at the time moment
just mentioned and ??0 be that maximum current. According to the law of charge
conservation
??0=2???? ????,
(C1.1) thus,
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??0
??=3??.
(C1.2)
Then, from the energy conservation law
2?2??
the maximum current is found as
??0=2??0
2????0
22????0
2
????22
2????22
(C1.3) (C1.4)
=2
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After the key ?? is shortened there will be independent oscillations in both circuits
with the frequency
??=
(C1.5)
and their amplitudes are obtained from the corresponding energy conservation laws
written as
2????22????2
://doc.guandang.net/bb316ba8befa958cb491190ff.html2????0
2=2
(C1.7)
2
Hence, the corresponding amplitudes are found as
0.
??1= ??0,
(C1.8)
??2=
(C1.9)
Choose the positive directions of the currents in the circuits as shown in the figure
on the right. Then, the current flowing through the key is written as
follows
??=??1???2.
(C1.10)
The currents depend on time as
=??cos???? ??sin????,
(C1.12)
??1 ?? =??cos???? ??sin????,
(C1.11)
??2 ??
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The constants ??,??,??,?? can be determined from the initial values of the currents
and their amplitudes by putting down the following set of equations
??1 0 =??=??0,
(C1.13)
2
??2 ??2=??1,
(C1.14)
??2 0 =??=??0,
(C1.15)
2
??2 ??2=??2.
(C1.16)
Solving (C1.13)-(C1.16) it is found that
??=2??0,
(C1.17)
??=???0,
(C1.18)
The
sign
in
??
is
chosen
nhttp://doc.guandang.net/bb316ba8befa958cb491190ff.htmlegative, since at the time
moment of the key shortening the current in the coil 2?? decreases.
Thus, the dependence of the currents on time takes the following form
=??0(cos???? 2sin????),
(C1.19)
??1 ??
??2 ?? =??0(cos?????sin????).
(C1.20)
In accordance with (C.10), the current in the key is dependent on time according
to
?? ?? =??1 ?? ???2 ?? =3??0sin????.
(C1.21)
Hence, the amplitude of the current in the key is obtained as
??max=3??0=????0=
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(C1.22)
222????0
=
2????1
2
22????2
(C1.6)
Instead of determining the coefficients ??,??,??,?? the vector diagram shown in the
figure on the right can be used. The segment ???? represents the current sought and
its projection on the current axis is zero at the time of the key shortening. The
current
??1
in
the
coil
of
inductance
??
grows
at
momenthttp://doc.guandang.net/bb316ba8befa958cb491190ff.html
the
same
because
time
the
capacitor 2?? continues to discharge, thus, this current is depicted in the figure
by the segment ????. The current ??2 in the coil of inductance 2?? decreases at the
time of the key shortening since it continues to charge the capacitor 2??, that is
why this current is depicted in the figure by the segment ????.
It is known for above that ????=??0,????= ??0,????= 0. Hence, it is found from the
Pythagorean theorem that
????= ?????????=2??0,
(C2.2) Thus, the current sought is found as
??max=????=???? ????=3??0=????0=
(C2.3)
(C2.1)
????==??0,
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It is clear that the current through the key performs harmonic oscillations with the
frequency
??=
(C3.1)
and it is equal to zero at the time of the key shortening, i.e.
?? ?? =??maxsin????.
(C3.2)
Since the current is equal to zero at the time of the key shortening, then the current
amplithttp://doc.guandang.net/bb316ba8befa958cb491190ff.htmlude is equal to the
current derivative at this time moment divided by the oscillation frequency. Let us
find that current derivative. Let the capacitor of capacitance 2?? have the charge ??1.
Then the charge on the capacitor of capacitance ?? is found from the charge
conservation law as
??2=??0???1.
(C3.3)
After shortening the key the rate of current change in the coil of inductance ?? is
obtained as
=??1
??1
(C3.4)
2????
whereas in the coil of inductance 2?? it is equal to
=???0???1
2????
??2
(C3.5)
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Since the voltage polarity on the capacitors are opposite, then the current derivative
with respect to time finally takes the form
???2=??0=??2??0.
??=??1
(C3.6)
2????
Note that this derivative is independent of the time of the key shortening!
Hence, the maximum current is found as
??max=??=???http://doc.guandang.net/bb316ba8befa958cb491190ff.html?0=
and it is independent of the time of the key shortening!
??
(C3.7)
Problem 2.Van der Waals equation of state
Solution
Part А. Non-ideal gas equation of state
A1.If ??=??is substituted into the equation of state, then the gas pressure turns
infinite. It is obvious that this is the moment when all the molecules are tightly
packed. Therefore, the parameter ?? is approximately equal to the volume of all
molecules, i.e.
??=??????3
(A1.1) A2.In the most general case thevan der
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Waals equation of state can be rewritten as
??????3? ?????? ?????? ??2 ?????????=0
(A2.1).
Since at the critical values of the gas parameters the straight line disappears, then,
the solution of (A2.1) must have one real triple root, i.e. it can be rewritten as
follows
????(???????)3=0
Comparing
(A2.2).
the
coefficients
of
expression
(A2.1)
and
(http://doc.guandang.net/bb316ba8befa958cb491190ff.htmlA2.2), the following set of
equations is obtained
3????????=?????? ??????
3??????2
(A2.3).
??=??
??????3??=????
Solution to the set (A2.3) is the following formulas for the van der Waals coefficients
??=
27??2????264??????????
??
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(A2.4), (A2.5).
??=8??
The critical parameters are achieved in the presence of an inflection point in the
isotherm, at which the first and second derivatives are both zero. Therefore, they
are defined by thefollowingconditions
????
=0
(A2.6), ????and
??2??
????????
=0
(A2.7).
Thus, the following set of equations is obtained
??????2??? =0
??
????? 2
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which has the same solution (A2.4) and (A2.5).
A3.Numericalcalculationsforwaterproduce the following result
://doc.guandang.net/bb316ba8befa958cb491190ff.html
????=0.56
m6?Pa
?? ?? ????? =????
???? ??????
2??????
??????? 3
?
????6??????
=0
(A2.8),
????=
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A4.From equations (A1.4) and (A3.2) it is found that
3
mole?5m3
3.1?10mole ??
(A3.1). (A3.2). (A4.1).
????= ??=3.7?10?10m≈4?10?10m
??
Part B. Properties of gas and liquid
B1.Usingtheinequality???????, the van der Waals equation of state can be written as
??0
??????????
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????=????
4????
(B1.1), (B1.2).
which has the following solutions
????=2?? 1± 1?
????
Smaller root in (B1.2) gives the volume in an unstable state on the rising branch
of
thevan
der
Waals
isotherm.
The
volume
of
gas
is
givehttp://doc.guandang.net/bb316ba8befa958cb491190ff.htmln by the larger root,
since at ??=0an expression for the volume of an ideal gasshould be obtained, i.e.
????
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????=2?? 1
1?
????
????4????0
????
(B1.3).
?3
For given values of the parameters the value ????=5.8?10. It can therefore be
assumed
????0????
1? ??0??????????0
??????
? ??0????
that ?????1, then (B1.3)takes the form
????≈
2
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B2. For an ideal gas
hence,
???
??0
=
4????0
≈
????0
(B1.4). (B2.1), (B2.2) (B3.1)
????0=
1
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???? =
????0?????
????0
=2 1? 1?
????????=0.58%.
B3.Mechanical
stability
of
a
thermodynamic
system
is
inpower
provided
that://doc.guandang.net/bb316ba8befa958cb491190ff.html
????
??
The minimum volume, in which the mattercan still exist in the gaseous state,
corresponds to a point in which
????
??????????→ ???? =0
(B3.2).
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Using the van der Waals equation of state (B3.2) is written as
????????2??
????=?
=0
??
(?????)
??
??
(B3.3). (B3.4). (B3.5). (B4.1) (B4.2).
From (B3.2) and (B3.3), and with the help of?????????????, it is found that
2??
??????????=????
??????????????
2????
=
Thus,
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B4. Usingtheinequality?????/??, the van der Waals equation of state is written as
??
??????? =????, whose solution is
????=2???? 1± 1?
??
4??????
??
??2??2
2????0
=86
In
this
case,
the
smaller
http://doc.guandang.net/bb316ba8befa958cb491190ff.htmltaken,
root
since
liquid volume????=?? must be obtained according to (B4.1), i.e.
shouldbe
at??→0the
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????=2???? 1? 1?
??
4??????
??
≈?? 1
??????
. ??
(B4.3). (B5.1). (B6.1).
B5. Since (B4.3) givesthevolumeoftheonemoleofwaterits mass density is easily found
as
??????2kg
????=??=≈=5.8?10
??m3
??
B6. Inaccordancewith (B4.3) the volume thermal expansion coefficient is derived as
1???????????
??=???????=?? ??????≈??=4.6?10?4К?1
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??
?? 1 ??
B7.The heat, required to convert the liquid to gas, is used to overcome the
intermolecular forces that create negative pressure ??/??2, therefore,
????11
??=????≈ ??????2????=?? ?????
(B7.1),
??
and using?????????, (B7.1) yields
??=
??=????
??
????
??://doc.guandang.net/bb316ba8befa958cb491190ff.html
???? 1 ??
≈????=1.0?106kg??J
(B7.2).
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B8.Consider some water of volume??. To make a monolayer of thickness ?? out of it,
the following work must be done
??=2????
(B8.1).
Fabrication of the monomolecular layer may be interpreted as the evaporation of an
equivalent volume of water which requires the following amount of heat
(B8.2), where the mass is given by
??=??????
??=????
(B8.3).
Using (A4.1a), (B5.1)and(B7.2), one finally gets
??N
??=2??2????=0.12?10?2m
(B8.4).
Part С. Liquid-gas systems
C1.At equilibrium, the pressure in the liquid and gas should be equalat all depths.
The pressure??in the fluid at the depth ?is related to the pressure of saturated vapor
above the flat surface by
??=??0 ???????
(C1.1). The surface tension creates additional pressure defined
by the Laplace formula ashttp://doc.guandang.net/bb316ba8befa958cb491190ff.html
2??
?????=??
(C1.2).
The same pressure??inthefluidatthedepth? depends on the vapor pressure ??? over the
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curved liquid surface and its radiusofcurvature as
2??
??=??? ??
related by
(C1.3). Furthermore, the vapor pressure at different heights are
???=??0 ???????
(C1.4). Solving (C1.1)-(C1.4), it is found that
2??
?=(?????)????
(C1.5).
Hence,the pressure difference sought is obtained as
2??????2????
?????=??????0=???????=???????≈??????.
??
??
??
??
??
(C1.6).
Note that the vapor pressure over the convex surface of the liquid is larger than
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the pressure above the flat surface.
C2.Let ????be vapor pressure at a temperature ????, and ??????????be vapor pressure
at a temperature ??????????.
In accordance with equation (3) from problem statement,
whhttp://doc.guandang.net/bb316ba8befa958cb491190ff.htmlentheambient temperature
falls by an amount of ?????the saturated vapor pressure changes by an amount
??
?????=??
(C2.1). ????????????? In accordance with the Thomson formula obtained
in part C1, the pressure of saturated vapor above the droplet increases by the
amountof ?????. While a droplet is small in size, the vapor above its surface remains
unsaturated.
Whena
droplet
hasgrownuptoacertainminimumsize,
thevaporaboveitssurface turns saturated.
Since
the
pressure
remains
hold
?????????? ?????=????
unchanged,
the
following
must
(C2.2). Assuming the vapor is almost ideal gas,
its density can be found as
????
????=???????????
condition
(C2.3).
From equations (C2.1)-(C2.3), (B5.1) and (C1.6) one finds
2???????????????
=???? ??Thus, it is finally obtained that
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??=
2????2????????????????? ??
????????
http://doc.guandang.net/bb316ba8befa958cb491190ff.html??
??
(C2.4). (C2.5).
=1.5?10?8m
Problem 3. Simplest model of gas discharge
Solution
Part А. Non-self-sustained gas discharge
A1.Let us derive an equation describing the change of the electron number density
with time. It is determined by the two processes; the generation of ion pairs by
external ionizer and the recombination of electrons with ions. At ionization process
electrons and ions are generated in pairs, and at recombination processthey disappear
in pairs as well.Thus, their concentrations are alwaysequal at any given time, i.e.
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?? ?? =???? ?? =????(??)
(A1.1).
Then the equation describing the numberdensityevolution of electrons and ions in time
can be written as
????(??)2
=???????(??)
(A1.2). ??????????
It is easy to show that at??→0 the function tanh????→0, therefore, by virtue of
thttp://doc.guandang.net/bb316ba8befa958cb491190ff.htmlhe initial condition ?? 0
=0,one finds
??0=0
(A1.3).
Substituting ???? ?? =??tanh???? in (A1.2) and separating it in the independent
functions (hyperbolic, or 1 and????), one gets
??=
??????????
(A1.4),
??= ??????????
(A1.5).
A2.According to equation (A1.4) the number density of electronsat steady-state is
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expressed in terms of the external ionizer activity as
????1= ????2= ????=
????????1??????????2????
(A2.1), (A2.2), (A2.3).
????????1 ????????2
Thus,the following analogue of the Pythagorean theorem is obtained as
10?3
????= ????
(A2.4) 1 ????2=20.0?10cm.
A3.In the steady state, the balance equations of electrons and ions in the tube volume
take the form
??
????????????=?????????????? ??
(A3.1),
????????????=???????http://doc.guandang.net/bb316ba8befa958cb491190ff.html??????
? ??
(A3.2).
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It follows from equations (A3.1) and (A3.2) that the ion and electron currents are
equal, i.e.
????=????
(A3.3). At the same time the total current in each
tube section is the sum of the electron and ion currents
??=???? ????
(A3.4). By definition ofthe current density the following relations hold
??
????=2=??????????=????????????
(A3.5),
??
??
????
????=2=??????????=????????????
(A3.6).
Substituting (A3.5) and (A3.6) into (A3.1) and (A3.2), the following quadratic
equation for the current is derived
????????????=?????? 2????????
??
2
??
2??
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(A3.7). (A3.8). (A3.9).
The electric field strength in the gas is equal to
??
??=??
and solution to the quadratic equation (A3.7) takes the form
??=http://doc.guandang.net/bb316ba8befa958cb491190ff.html
????2??2??????3
?1± 1
4????????????4??2??2
It is obvious that only positive root does make sense, i.e.
??=
????2??2??????3
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1
????????????
??
4????????????4??2??2
?1
(A3.10).
A4.At low voltages (A3.10) simplifies and gives the following expression
??=2??????
.
(A4.1)
which is actually the Ohm law.
Using the well-known relation
??
??=??
together with
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??
??=????
one gets
??=2???? ??
??????
(A4.2) (A4.3), (A4.4).
1??
Part B. Self-sustained gas discharge
B1.Consider a gas layer located between ??and?? ????.The rate of change in the
electron number inside the layer due to the electric current is givenfor a small time
intervhttp://doc.guandang.net/bb316ba8befa958cb491190ff.htmlal ???? by
??????
??????=??????=????????.
(B1.1).
????????
This change is due to the effect of the external ionization and the electron avalanche
formation. The external ionizer creates the following number of electrons in the
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volume??????
??????
??????=??????????????????
(B1.2). whereas the electron avalanche produces
the number of electrons found as
??(??)??
??????=??????????=????????????????=??????????????
(B1.3).
The
balance
equationfor the number of electrons is written as
??????????
??????=?????? ??????
(B1.4), whichresults in the following differential
equation for the electron current
??????(??)
=???????????? ??????(??)
(B1.5).
????
On substituting???? ?? =??1????1?? ??2,one derives
????????????
??2=?
(B1.7).
??://doc.guandang.net/bb316ba8befa958cb491190ff.html
??1=??
(B1.6),
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B2.Given the fact that the ions flow in the direction opposite to the electron
motion,the balance equationfor the number of ionsis written as
????????=???????????? ????????
(B2.1), where
????
????????=??????=?????????
(B2.2).
????????
????????????=??????????????????
(B2.3).
??(??)
????????=??????????????
(B2.4). Hence, the following differential equation
for the ion current is obtained
??????(??)
?????=???????????? ??????(??).
(B2.5)
Onsubstituting the previouslyfound electron current together with the ion
current,???? ?? =??2 ??1????2??,yields
??1=???1
(B2.6),
??2=??
(B2.7).
B3.Sincetheionsstartstomovefrom the anode located at??=??, the following condition
holds
???? ?? =0
coefficient
(B3.1). B4.By definition of secondary electron emission
the
following
shouldhttp://doc.guandang.net/bb316ba8befa958cb491190ff.html be imposed
condition
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???? 0 =?????? 0
(B4.1).
?? ?? ???? ???(??)1???? ??
?? ?? ???(?? ????)1???? ??
B5.Total current in each tube section is the sum of the electron and ion currents:
??????
??=???? ????=??2???????
??
Aftersubstituting the boundary conditions (B3.1) and (B4.1):
??2???1??????=0
and
??????
??1???????=??(??2???1)
??
Solving (B5.2) and (B5.3) one can obtain:
??????1
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??2=??????
?????????(1 ??)???So the total current:
??=
??????????????
(B5.1). (B5.2) (B5.3). (B5.4). (B5.5).
???????(1 ??)????1
1
B6.When the discharge gap length is increased, the denominator in formula (B5.1)
decreases. At that
moment, when it turns zero, the electric current in the gas becomes self-sustaining
and external ihttp://doc.guandang.net/bb316ba8befa958cb491190ff.htmlonizer can be
turned off. Thus,
11
??????=??ln 1 ??
(B6.1).
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Experimental problem. To see invisible!
Part 1. Qualitative observations!
Section 1.1. Polarizers
For determination of the transmission plane of the polarizer, one can use a glaring
effect from any shining surface. It is known that the reflected light is polarized
in the plane of the reflecting surface. The corresponding transmission planes are
shown in the figure on the right.
Section 1.2. Rulers
1.2.1. In case of the incident light being polarized along the optical axis or
perpendicular to it, there is only one kind of waves generated in the medium. This
means that no change in light polarization is to occur. Thus, it is possible to
determine either the direction of the optical axis or the direction which is
perpendicular
to
it.
Those
possible
alternatives
are
shown
in
the
fhttp://doc.guandang.net/bb316ba8befa958cb491190ff.htmligure above (either along
the ruler, or perpendicular to it).
1.2.2. One can see at what parts of the rulers similar colors are observed, mainly
with blue hue.
The distance between those bands for the ruler No.1 is
rules stacked together it is ~ 8 cm.
~12 cm, while for the two
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Section 1.3. Strip
1.3.1 Possible directions of the optical axis of the strip can be determined in a
similar way. As shown in the figure on the right, those directions make a small
angle ?10? with the sides of the strip.
1.3.2
The
coordinates
of
the
dark
bands
are
approximately
found
as
follows ????=3,5????, ????=7,5????.
Section 1.4. Liquid crystal cell
1.4.1 In case of zero voltage, the directions of the optical axis can be determined
in the same way: It is either horizontal or vertical.
At the maximum voltage applied
the optical axis orients along the electric field, which means it turns
perphttp://doc.guandang.net/bb316ba8befa958cb491190ff.htmlendicular to the cell
plane.
1.4.2 The voltage at which suach a sharp transition in orientation of
molecules of the liquid crystal occurs is approximately equal to
??????=2 V
.
Part 2. Measure!
Section 2.1. Investigating a photodiode
2.1.1 In the figure below a position for a circuit switch is shown. During measurements
of the resistance, the circuit switch should be unshorted.
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2.1.2 In table 1 the results are presented of the measurements of the voltage U as
a function of the resistance.
Those data are plotted in the corresponding graph.
Note that the optimal resistance should be within the range 5-15 ???????, which
corresponds to the largest variation in the voltage.
2.1.3 In table 2 the results are shown of the measurements for the voltage as a function
of the number of light filters at different values of resistance.
://doc.guandang.net/bb316ba8befa958cb491190ff.htmlr
Intensity of the light ???? that has passed through the filter decreases as a geometric
progression when increasing the number of filters ??:
????=??0????.
(1)
In case if the measured voltage is proportional to the intensity of the incident light,
it obeys a similar law:
????=??0????.
(2)
To verify equation (2), one needs to use a semi-logariphmic scale. In other words,
it is necessary to plot ln?? as a function of
ln????=ln??0 ??ln??.
(3)
??:
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That plot is shown in the following figure.
According to the graph above, by decreasing the resistance the dependence turns a
linear function and further measurements should be made at the lowest resistance among
given values, i.е. at ??=10 ???????.
2.1.4. According to equation (3) , the slope is ??=ln??. Using the Method of Least
Squares,
we
can
obtain
its
value
??=?0.53±0.03.
Thuhttp://doc.guandang.net/bb316ba8befa958cb491190ff.htmls, the coefficient of
transmission turns to be equal to ??=exp??=0.59 with an error, which can be calculated
by applying the following formula ???=exp(??)???=0.02. Finally we obtain
??=0.59±0.02.
Note that values for ??=10 ??????? produce the following result:
??=0.59±0.02
.
Часть 2.2 Light transition through a plastic ruler
2.2.1 Results of measurements of the light intensity as a function of coordinates
of transmission points through ruler #1, #2 and both rulers, are shown in table 3
and in the graph below.
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2.2.2 To calculate a phase shift, we use equation (1), mentioned in the problems
formulation, which can be represented as
???
??=????????sin22,
(1)
where ???????? is the largest value of voltage. But we have to be sure that this value
actually corresponds to the maximum of function (1), not just another boundary point.
http://doc.guandang.net/bb316ba8befa958cb491190ff.htmlAccording to measurements,
(see
the
graph)
for
each
ruler
the
most
suitable
value
for
????????
is ????????=160 ????.
The following equation
???
??0=sin22
(1)
has multiple roots and it is not easy to find actual values of the phase shift, even
if it’s possible to calculate certain value for Umax, roots of the equation mentioned
above are shown in the figure below
Formally we can represent the roots in different forms, for example,
???=±2(arcsin 0 ????),
??=0,1,2….
???=±2(π?arcsin 0 ????),
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Choosing a correct root should depend on a function obtained experimentally.
Values
for the phase shifts calculated by the equation
???=2arcsin ??
??
??????
(2)
(3)
are shown in table 3
It is clear that the function ???(??) has to be monotonous, that is why the signs
of roots for the fihttp://doc.guandang.net/bb316ba8befa958cb491190ff.htmlrst two
points must be changed, which is a mathematically correct operation (just reflecting
the graph). Note that the phase shifts are calculated with uncertainty of ±2????.
2.2.3 Obtained functions are close to linear, using MLS we get
???1=0.059???0.94, ???2=0.028?? 0.52.
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Graphs of those functions are shown below.
2.2.4 If two rulers are stacked together, then phase shifts simply add, and,
theoretically, the intensity as function of phase shifts can be written as
??? ???
??=????????sin2122.
(5)
Here ???????? is the largest value of voltage at the light transition through both
rulers and can be obtained from experimental data.
Results of the calculations are shown in table 3 and in the graph above. Consistency
of theoretical calculations and experimental data can clearly be seen.
Part 2.3 Liqhttp://doc.guandang.net/bb316ba8befa958cb491190ff.htmluid crystal cell
2.3.2 Light transmission through LCC
2.3.1 Results of the intensity measurements as a functions of voltage ?????? are shown
in table 51. Graph of the obtained function is drawn in the figure below.
1
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We do not expect that participants can take the same number of measurements, 15-20
points are enough. It is principally important to find the dip in the graph.
It is important to choose correct roots of equation (2) in order to adequately
calculate phase shifts. In this case it is rather obvious because at large values
of ?????? the voltage difference tends to zero, ???→0. Other solutions and
corresponding equations are shown in the figure below.
Results of calculation of
??? ′=arcsin ??
??
??????
and correct values of phase shift ??? are shown in the figure
below.
This
figure
is
drawn
for
ohttp://doc.guandang.net/bb316ba8befa958cb491190ff.htmlf
the
understanding.
sake
(It’s
not required to draw it for participants).
2.3.2 The value of the phase shift at zero voltage is ???0≈10.6.
2.3.3 In order to check applicability of the power function ???=?????? it is
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recommended to redraw the last graph logarithmically, as shown in the figure below.
It can be seen from the graph that in the range of 1 V to 5 V the function is almost
linear, which justifies the applicability of the power law. The power in that equation
is equal to the slope of the graph, its numerical value is ??≈1.75.
Section 2.4 Light transmission through a curved strip
2.4.1 Results of the measurements of the light intensity as a function of coordinate
z of the point of light penetration into the strip are presented in table 6 and plotted
below.
2.4.2
The
shape
of
the
curve
indicates
that
???0
lies
at
the
ahttp://doc.guandang.net/bb316ba8befa958cb491190ff.htmlscending
part
of
the
relation between the intensity and the phase shift, which can be calculated as
???0=10?? 2arcsin ??
??0
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??????
≈33.9.
Graph of the phase shift is drawn in the figure on right. Because we are interested
in the central part of the graph, “reflection” parts are not shown. (This graph
is not required from participants)
2.4.3 The graph shows that the central part is approximately parabolic function of
z
???=????2 ??.
(10) In order to determine the coefficients of the function we can
draw graph of ??? as a function of ??2 (see figure on the right). Using MLS, we can
determine the parameters
??=0.0104 мм?1, ??=2.45.
It is necessary to add 10?? to the obtained value of b.
Comparing equations (9) and (10), we conclude that the parameters can be represented
by the strip characteristics as:
???0
??=2??2??http://doc.guandang.net/bb316ba8befa958cb491190ff.html (11) 2??=???0.
From those equations we get the radius of curvature of the strip
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??=?? 2??.
1
??
(12)
Substitution of the obtained results leads us to ??=29 ????. Note that the obtained
result is quite rough, due to uncertainties in measuring.
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