1 PHYSICS EXAMINATION PROBLEMS SOLUTIONS AND HINTS FOR STUDENT SELF-STUDY Module Code PHY1024 Name of module Properties of Matter Date of examination May / June 2014 1. (i) The pressure increases as V decreases until the dew-point is reached and condensate forms. As V is reduced further more molecules enter the liquid state until the sample is all liquid. At this point the pressure starts to rise again at a rate governed by the compressibility. d dv f v (v) ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ v p = 0 ⇒ Mv 2 = 2k B T ⇒ v p = 2 k B TM U = 3 2 nRT = 3 2 × 0.75mol × 8.314 J K −1 mol −2 × 350 K=3.3kJ M = 2k B T v p 2 = 2 × 1.38 × 10 −23 JK −1 × 350 K 390ms −1 ( ) 2 = 64 × 10 −27 kg 2 (a) Y = F ⊥ A 0 Δll 0 = 60 MPa 0.04% = 150 GPa (b) Δl max = 0.09% × 250 mm = 225 μm (c) π d 2 4 ( ) × 60 MPa = 7.3kN ⇒ d = 4 × 7.3kN π × 60 MPa ( ) = 12.4 mm (d) E el Δl ( ) = Fx () 0 Δl ∫ dx = YA l 0 x 2 2 # $ % % & ' ( ( 0 Δl = AY Δl 2 2l 0 (e) P max = max(stress × area) × rateof extension = 90 MPa × π 12.4 mm ( ) 2 4 × 250 mm × 0.09% 3 × 60s = 14 mW P V T < T C vapour liquid 2-phase region
49
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1
PHYSICS EXAMINATION PROBLEMS SOLUTIONS AND HINTS FOR STUDENT SELF-STUDY
Module Code PHY1024
Name of module Properties of Matter
Date of examination May / June 2014
1. (i) The pressure increases as V decreases until
the dew-point is reached and condensate forms. As V is reduced further more molecules enter the liquid state until the sample is all liquid. At this point the pressure starts to rise again at a rate governed by the compressibility.
measured (relative) parallax = true parallax of target - true parallax of reference measured parallax = 1/20 - 1/500 = 0.048 measured distance = 20.83 pc (overestimate by 0.83 pc)
GAIA will use absolute parallaxes (bookwork)
CMD: de-redden the MS, then find distance modulus, convert to distance distance = 1 kpc
approx 3 mag difference between tip of upper MS and Sun luminosity is approx 16 solar luminosities L ~ M^3.5 , so T_MS ~ L/M ~ M^-2.5 MS lifetime approx 0.14 solar MS lifetimes (approx 1.4 Gyr)
2. eccentricity > 0 (not sinusoidal) symmetry: peri- and ap-astron orbital velocities oriented along line of sight at extrema, subject to sin(i) projection Vp sin (i) / Va sin(i) = Vp / Va = (1+e)/(1-e) = 180/25 re-arrange to find e = 0.75
sketch: bookwork photometric monitoring for transits to test whether inclination is approx 90 degrees.
semiamplitude K1 approx 102.5 km/s (measure from plot)
mass function equation, with appropriate approximation to find secondary mass approx 2.1 MJ
KIII to find a approx 0.95 a.u.
max separation approx a(1+e) = 1.66 a.u. at 46 pc distance, this is 0.036 arcseconds Rayleigh criterion, min diameter 3.85 m longer wavelength PRO star/companion flux contrast ratio smaller CON larger telescope required
3. specific intensity, Planck function, flux density: bookwork Bnu(T) = 3.14 x 10^-16 W m^-2 Hz^-1 sr^-1 RJ approx: not valid (require hnu/kT << 1) mass calculation (bookwork) approx 0.5 solar masses core mean density approx 3.8 x 10^-16 kg m^-3 free-fall time approx 0.1 Myr
PHYSICS EXAMINATION PROBLEMS SOLUTIONS AND HINTS FOR STUDENT SELF-STUDY
Module Code PHY2031
Name of module Lasers, Materials and Nanoscale Probes for Quantum Applications
Date of examination May 2014
2.
4. Number of longitudinal HeNe modes lasing = 3
1 PHY3052
PHYSICS EXAMINATION PROBLEMS SOLUTIONS AND HINTS FOR STUDENT SELF-STUDY
Module Code PHY3052
Name of module NUCLEAR & HIGH ENERGY PHYSICS
Date of examination June 2014
1. (i) (a) 6- o /0 + e- + ve (d) /0 + p o + e+ + ve
(b) W o S- + S0 + vW (e) ve + p o n + e+
(c) S + e- o S+ ve (f) W o P + vP + vW
Weak interaction
Lorentz factor = J = 14.5 / 1.784 = 8.13
From this, speed = 0.992 c
Length travelled = 0.992 × c × J × W = 6.29 × 10-2 cm
(ii) (a) 1/2, - (b) 3/2, + (c) 0, + (d) eQo
-2517
8177 e )( O N
(e) eQo e)( Cl Ar 2337
173718
(f) eQo -4821
4820 e)(4 *Sc Ca
Bookwork
2. (i) 3H o 3He + e- + ve Bookwork
(ii) Bookwork is a magic nucleus. This means product state has high BE, so that reaction will give out greater than expected energy. Released as KE of alpha particle due to light mass.
Q – mvc2
W
vW vP
P
W
2 PHY3052
vD = 0.0518 c = 1.55 x 107 m/s
Aparent = 210, so R= 9.09 fm
Zdaughter = 82
24 8Zc mc RZG v c
SD D = 145.2 – 81.10 = 64.10
O = 1.23x 10-7 s-1
t1/2 = ln(2)/O = 5.6 x 106 s = 64.8 days
(iii) Neutral pion has quark – antiquark structure:
First need to calculate energy of pion particle in lab frame, assume stationary
proton. From energy and momentum conservation:
ES = (Mp2 + MS
2 – Me2) c2 / (2 Mp)
From this, Lorentz factor = JS = 3.548
Speed = vS = 0.9595 c
Energy of photons in pion frame = MS c2 / 2
Then need to calculate Lorentz/Doppler shift for moving pion.
Maximum/minimum energy will be for photon travelling towards/away from
observer: Emax = 469.3 MeV, Emin = 9.7 MeV
3 (i) Bookwork
P = Fcas / A
0 12
uu ddS
3 PHY3052
0.0013 Pa for 1Pm.
Bookwork
(ii) (a) weak
(b) strong
(c) weak
(d) EM and weak
(e) weak
(f) weak
(g) EM and weak
(h) weak
Bookwork
4. (i)
Bookwork
(ii) Bookwork
2H + 3H o 4He + n
Q = BE (4-He) – BE(2-H) – BE (3-H) = 17.6 MeV
Bookwork
Using Q value from above, total nuclear density = 6.8 x 1018 m-3.
Density of deuterium = 3.4 x 1018 m-3. 7 orders smaller than density of air.
1s
1p1/2
2s 1d5/2 1d3/2
1p3/2
1
PHYSICS EXAMINATION PROBLEMSSOLUTIONS AND HINTS FOR STUDENT SELF-STUDY
Module Code PHY3065Name of module Quantum optics and photonicsDate of examination May/June 2014
1. Solution for problem plus optional comment/hint as to method if appropriate
2.
3.
4.
5.
1. (i)
a, a†
= 1, [a, a] = 0,a†, a†
= 0.
Also acceptable:a, a†
= 1
Action on number states:
a|ni =
pn|n 1i, a†|ni =
pn + 1|n + 1i.
Matrix elements of a† and a:
hm|a|ni =
pnhm|n 1i =
pn m,n1, hm|a†|ni =
pn + 1hm|n + 1i =
pn + 1 m,n+1,
(where m,n1 is the Kronecker delta). Hence a† and a have the following expressions in terms of their matrix elements in thenumber-state basis:
a =
1X
n,m=0
hm|a|ni|mihn| =
1X
n=0
pn|n 1ihn|, a†
=
1X
n,m=0
hm|a†|ni|mihn| =
1X
n=0
pn + 1|n + 1ihn|.
(ii)Hamiltonian of single-mode quantum light:
ˆH = ~!
a†a +
1
2
.
Schrodinger equation:
i~ d
dt| (t)i =
ˆH| (t)i
with solution:
| (t)i = eiHt/~| (0)i.
| (0)i =
1p2
(|4i + |1i).
Applying time evolution equation above, and the single-mode Hamiltonian:
| (t)i = ei!(a†a+1/2)t 1p2
(|4i + |1i) = ei!t/2 1p2
(e4i!t|4i + ei!t|1i).
Thus
| (/!)i = ei/2 1p2
(e4i|4i + ei|1i) = ip2
(|4i |1i).
2
Inner product of | (0)i and | (/!)i is
h (/!)| (0)i =
ip2
(h4| h1|)
1p2
(|4i + |1i)
=
i
2
(1 1) = 0,
so | (0)i and | (/!)i are orthogonal.(iii)
Calculate h↵|i and find condition for it to vanish.
h↵|i = e|↵|2/21X
n=0
(↵)
n
pn!
hn| (c1|1i + c2|2i) = e|↵|2/21X
n=0
(↵)
n
pn!
(c1n1 + c2n2)
= e|↵|2/2
c1↵p
1!
+ c2(↵
)
2
p2!
= e|↵|2/2↵
c1 + c2
↵p
2
We thus have h↵|i = 0 if
c1 = c2↵p
2
.
(iv)A density operator given by
=
X
i
pi| iih i|
describes a system that has probability pi of being in the state | ii.For any vector | i,
h | ˆO†ˆO| i =
ˆO| i
† ˆO| i
0
because this is the norm squared of the vector ˆO| i, which is a positive number. Thus ˆO†ˆO is positive.
2. (i)Energy of state |gi is ~!/2. Energy of state |ei is ~!/2.In the coupling term, a|eihg| takes the atom from the ground to the excited state and annihilates a photon; thus absorption of a
photon by the atom. a†|gihe| takes the atom from the excited to the ground state and creates a photon; thus emission of a photonby the atom.
ˆH| +i =
ˆH1p2
(|1, gi + |0, ei) =
1p2
~! 3
2
|1, gi 1
2
~!|1, gi +
1
2
~|0, ei + ~! 1
2
|0, ei +
1
2
~!|0, ei +
1
2
~|1, gi
=
1p2
~! +
2
|1, gi + ~
! +
2
|0, ei
= ~
! +
2
1p2
(|1, gi + |0, ei)
= ~! +
2
| +i
Thus | +i is an eigenstate of ˆH with eigenvalue ~! +
2
.
ˆH| i =
ˆH1p2
(|1, gi + |0, ei) =
1p2
~! 3
2
|1, gi +
1
2
~!|1, gi 1
2
~|0, ei + ~! 1
2
|0, ei +
1
2
~!|0, ei +
1
2
~|1, gi
=
1p2
~! +
2
|1, gi + ~
!
2
|0, ei
= ~
!
2
1p2
(|1, gi + |0, ei)
= ~!
2
| i
Thus | i is an eigenstate of ˆH with eigenvalue ~!
2
.
3
Time evolution is given by the operator eiHt/~, which gives the time evolution
ei(!+/2)t| +i and ei(!/2)t| i.
Initial state at t = 0 is
1p2
(| +i | i) = |1, gi
so the light mode has one photon and the atom is in the ground state.State at time t is
1p2
(ei(!+/2)t| +i ei(!/2)t| i) = ei!t 1p2
(eit/2| +i eit/2| i)
At t = / this is
ei!/ 1p2
(ei/2| +i ei/2| i) = ei!/ ip2
(| +i | i) = iei!/|0, ei
so the mode is in the vacuum state and the atom is in its excited state.At t = 2/ state is
e2i!/ 1p2
(ei| +i ei| i) = e2i!/ 1p2
(| +i + | i) = e2i!/|1, gi
so the mode has one photon and the atom is in the ground state.Probability amplitude of finding atom in state |gi is coefficient of |1, gi in total state at time t, which is
ei!t 1
2
(eit/2+ eit/2
) = ei!tcos(t/2),
so probability is cos
2(t/2).
(ii)Photons of frequency ! have energy ~! so the probability P (n) of having n photons is
P (n) = A exp
n~!
kBT
The normalization constant is such thatP
n P (n) = 1, so
A1=
1X
n=0
exp
n~!
kBT
=
1X
n=0
exp( ~!
kBT)
n=
1
1 exp
~!
kBT
Thus
P (n) =
1 exp
~!
kBT
exp
n~!
kBT
.
3. (i)In linear media the polarization (electric dipole moment per unit volume) due to an applied electric field is taken to be
proportional to the applied field.In nonlinear media the polarization of the medium depends nonlinearly on the applied electric field; the polarization is gener-
ally taken as a series in the electric field.
PNLi(r, t) = "0X
j,k
(2)ijkEj(r, t)Ek(r, t)
Also acceptable:
PNL(z, t) = "0(2)
[E(z, t)]2
4
In parametric down-conversion a pump photon of frequency !p is converted to two photons of frequencies ! and !p!. Thegenerated photon with the higher frequency is usually called the signal, the one with the lower frequency is called the idler.
The sum of the signal and idler frequencies is equal to the pump frequency. Pump wavelength p = 800 nm, signal wavelength = 1400 nm. Idler frequency is
c
p c
so idler wavelength is
1
p 1
1
=
1
800
1
1400
1
nm =
5600
3
nm = 1867 nm.
(ii)
[aL, aL] =
"p5
2
a0 +
1
2
e5i a†0,
p5
2
a0 +
1
2
e5i a†0
#=
p5
2
1
2
e5i[a0, a
†0] +
1
2
e5i
p5
2
[a†0, a0] =
p5
2
1
2
e5i 1
2
e5i
p5
2
= 0,
[a†L, a†
L] =
"p5
2
a†0 +
1
2
e5i a0,
p5
2
a†0 +
1
2
e5i a0
#=
p5
2
1
2
e5i[a†
0, a0] +
1
2
e5i
p5
2
[a0, a†0] = 0,
[aL, a†L] =
"p5
2
a0 +
1
2
e5i a†0,
p5
2
a†0 +
1
2
e5i a0
#=
5
4
[a0, a†0] +
1
4
[a†0, a0] =
5
4
1
4
= 1.
State is input vacuum state, given by a0|0i. Average obtained is the expectation value of the output number operator a†LaL in
the state |0i:
h0|a†LaL|0i =
1
4
h0|a0a†0|0i =
1
4
h1|1i =
1
4
.
Average of 1/4 photons is measured.For uncertainty, need expectation value of square of number operator a†
LaL:
h0|a†LaLa†
LaL|0i..
Only terms with same number of a†0 and a0, and a†
0 at the end, and a0 at the beginning, contribute:
h0|a†LaLa†
LaL|0i = h0|12
e5i a0
5
4
a0a†0 +
1
4
a†0a0
1
2
e5i a†0|0i =
1
16
h1|5a0a†0 + a†
0a0|1i =
1
16
h1|5 + 6a†0a0|1i =
11
16
.
Square of uncertainty in photon number is
11
16
1
4
2
=
5
8
so uncertainty isp
5/8.Parametric down-conversion creates photons in pairs, so number measurements will always find an even number of photons
in the output beam (assuming perfect detection efficiency).In a squeezed state the uncertainty in one of the quadratures is less than the vacuum value (1/2). The uncertainty in the
conjugate quadrature is increased compared to the vacuum value so that the quadrature uncertainty relation is satisfied.4. (i)
a†3a3 =
1p2
a†1
ip2
a†2
1p2
a1 +
ip2
a2
=
1
2
a†1a1 + a†
2a2
a†4a4 =
1p2
a†2
ip2
a†1
1p2
a2 +
ip2
a1
=
1
2
a†2a2 + a†
1a1
5
a4
a3
a2
a1
Thus
a†3a3 + a†
4a4 = a†1a1 + a†
2a2
so number of photons in output beams equals number of photons in input beams.We have
a1 =
1p2
a3 ip2
a4, a2 =
1p2
a4 ip2
a3.
Input state is
|1i1 |1i2 = a†1a
†2|0i =
1
2
a†3 + ia†
4
a†4 + ia†
3
|0i =
i
2
a†3a
†3 + a†
4a†4
|0i =
i
2
(|2i3 + |2i4)
So output state is i((|2i3 + |2i4)/2.Two photons will be found in arm 3, with probability 1/2, OR two photons will be found in arm 4, with probability 1/2. One
photon in each of arms 3 and 4 has probability 0.Photons with a narrow frequency spread are wave packets of finite length. For output arms of equal length, the photon wave
packets in each input arm must exactly overlap at the beam splitter in order for the probability of one photon in each output armto vanish. If the arrival time of the photons at the beam splitter is varied then there are zero coincidences in the output detectorswhen the arrival times are the same (zero probability of one photon in each output arm), whereas when the arrival times aredifferent there are coincidences (non-zero probability of one photon in each output arm).
The decrease in the coincidence rate when the photon wave packets overlap at the beam splitter is called the Hong-Ou-Mandeldip.
For classical particles with probability 1/2 of reflection and probability 1/2 of transmission, the probability of detecting bothphotons in output arm 4 is (1/2)(1/2) = 1/4, which is also the probability of detecting both photons in output arm 3 . Thus theprobability of detecting both photons in same output arm is 1/2. The probability of detecting one photon in each output arm is(1/2)(1/2) + (1/2)(1/2) = 1/2.
2.The polarization measurement on photon 1 yields either H with probability 1/2, after which photon 2 is in the state |Hi2, OR
it yields V with probability 1/2, after which photon 2 is in the state |V i2. Photon 2 thus has probability 1/2 of being in state|Hi2 and probability 1/2 of being in state |V i2 .
Density operator for state of photon 2 is
=
1
2
|Hi2 2hH| + 1
2
|V i2 2hV |.
PHY3068: Principles of Theoretical Physics
Hints and tips for the 2013/14 exam
1. Least action principle states that the actual trajectory of a mechanical system isa stationary point of the action functional. This stationary point can be foundby computing the variation of the action and setting it to zero to the first orderin variation.
The Lagrange function is the difference of potential and kinetic energies. Inthe moving reference frame, it is sufficient to change the velocity as explainedin the question. (The potential energy depends only upon the absolute valueof r, which is the same in both reference frames.) Expanding the kinetic energyresults in two terms, mv · Ω × r and m
2(Ω × r)2. The variation can be easily
computed; the respective contributions of the two terms in the action to theforce are the Coriolis force, 2mΩ×v, and the centrifugal one: −mΩ× (Ω×r).
One can notice that the cross-term in the action, mv · Ω × r has the formof vector potential term, ev · A(r). The magnetic field can be found fromthe standard expression, B = curlA, which gives B = 2mΩ/e. This analogycan be employed to write Schrodinger equation, by introducing the covariantderivative, D = ∇ + ie
hA, with eA(r) = mΩ × r. The gradient term in the
Schrodinger equation then takes the form D2ψ.
2. Spontaneous symmetry breaking occurs when the Lagrangian possesses certainsymmetry, but its individual minima break the symmetry. Goldstone theoremstates that the spectrum of such a system must include a massless (or gapless)mode.
Obviously, this Lagrangian is invariant with respect to rotations in the internalspace of fields φ1,2, as the quantity φ2
1 + φ22 is invariant. Steady-state config-
urations are provided by minima of the Lagrangian. Broken-symmetry statesoccur if a > 0, so that the potential energy is minimal when φ2
1 + φ22 = a/b.
When the Lagrangian is expanded to the second order in χ, it takes the form ofLagrangian describing two decoupled scalar fields. The field χ1 represents themassive radial mode, with the mass m1 = 2a, while the field χ2 represents themassless Goldstone mode.
3. The rate of quantum tunneling is proportional to the exponential of imaginary-time action. The turning points can be found from the condition U(x∗) =E = 0; the central domain, U(x) > 0, is classically forbidden. The imaginary-time action can be obtained by changing the sign of U(x), which results inthe action of a quantum harmonic oscillator with frequency Ω. The trajectorythat connects the two classical turning points is a simple harmonic motion
x(τ) = x∗ sinΩτ , with the amplitude x∗ =!
2ϵmΩ2 . Thus, the travel time is
1
given by half-period of the oscillations, τ∗ = π/Ω. The two terms in the action,x2 and x2, cancel each other after averaging over a half-period, so that the actionis ϵτ∗. This gives the tunneling action πϵ/Ω, and tunneling rate ∼ exp
"
−2πϵhΩ
#
.The analysis is applicable if ϵ≫ hΩ.
4. In Feynman’s approach, the contribution of an individual trajectory to the totaltransition amplitude is given by a complex exponential of the classical action forthat trajectory. In the presence of an external electromagnetic field, this actionis modified due to the term ev ·A in the Lagrangian. The integral of this termover the full trajectory is proportional to the line integral
$
A · dr. For a loop,this integral can be related to the flux of the magnetic field through an arbitrarysurface bounded by the loop. The orientation of the normal to the loop is takenso that the loop is traversed clockwise, as seen along the normal. The flux ofa monopole through the full sphere is 4πg, each hemisphere contributing 2πg.The loop shown in the figure can be viewed as a boundary of either northernor southern hemisphere. In the first case, the flux is 2πg, in the second casethe flux is −2πg, due to different orientation of the loop with respect to thesurface. In quantum theory, this would result in ambiguity in phase factor dueto the monopole: the phase factor is equal to exp
"
±2πigeh
#
, depending on thechoice of the hemisphere. However, when the phase is a multiple of π, the twoexpressions for the phase factor give identical values. This gives eg = 1
2hn,
where n is an integer.
2
1 PHYM007
PHYM007
UNIVERSITY OF EXETER
PHYSICS
MAY / JUNE 2014
ULTRAFAST PHYSICS
SOLUTIONS AND HINTS
2 PHYM007
1. (i) Sketch a diagram of an all-optical two-colour pump-probe experiment for
measurements of the intensity and polarisation response of samples deposited on
opaque substrates. [5]
On the sketch (diagram) indicate clearly:
(a) the ultrafast laser; [1]
(b) the optical paths of the pump and probe beams; [2, 2]
(c) the focusing and polarisation optics; [2, 2]
(d) the optical bridge detector; [1]
(e) the sample. [1]
HINT: See the lecture notes
Describe the factors limiting the temporal resolution of the experiment. [5]
HINT: (a) durations of the pump and probe pulses; (b) the minimal step,
accuracy and repeatability of the motion of the mechanical delay line.
Explain why time-resolved pump-probe experiments are limited to measurements of
repeatable phenomena? [4]
HINT: Each point in a time-resolved signal is an average of millions of
measurements of the sample's response for the specific pump-probe delay. In
addition, the signals measured for different pump-probe delay values must be
consistent among themselves in order to constitute a meaningful time-resolved
signal.
(ii) State the Nyquist-Shannon sampling theorem. [3]
HINT: See the lecture notes
3 PHYM007
Consider a time resolved experiment designed to study coherent phonons with
frequencies in the range 100 GHz – 1 THz and linewidths in the range 1 – 5 GHz.
Estimate the maximum time step and the minimum length of the time resolved
signals that need to be acquired in order to resolve peaks associated with the
phonons in the Fourier spectra of the time resolved signals. [3, 3]
HINT: Applying the Nyquist-Shannon theorem, obtain 0.5 ps and 1 ns for the
maximum time step and minimum signal length, respectively.
4 PHYM007
2. (i) Consider a solid state sample pumped by a high frequency electromagnetic wave
with the electric field rkEE tt Zcos, 0r .
Using a Taylor series expansion with respect to small atomic displacements, define
the Raman contribution to the polarisation of the sample induced by the incident
electromagnetic wave and label the Raman tensor and terms corresponding to the
Stokes and anti-Stokes scattering processes. [4, 1, 1, 1]
Name the other two main classes of optical processes that contribute to the optically
induced electrical polarisation of the sample. [1, 1]
HINT: See the lecture notes
(ii) Describe the mechanism of displacive optical excitation of coherent phonons in
solids and the conditions in which the mechanism could be expected to be active. [4]
HINT: See the lecture notes
The contribution to the reflectivity signal induced by the optical phonon, δR(td),
measured in a pump-probe experiment has been successfully fitted by the following
where td is the pump-probe time delay, ω and τ are the phonon's frequency and life
time, erf (x) is the error function, and σ and c are parameters.
(a) Identify the likely mechanism of excitation of the phonon and justify your
answer in terms of the temporal character of the signal. [1, 2]
(b) Name parameter b and describe the physical implications of its non-zero value
for interpretation of the signal. [1, 2]
HINT: (a) Impulsive: note the phase at td = 0; (b) See the lecture notes
5 PHYM007
Is it possible to observe excitation of acoustic phonons in an all-optical pump-probe
experiment? Justify your answer. [1, 3]
HINT: Yes. Indeed, although acoustic phonons do not couple to the optical
field directly, they could still be excited if the optical pump action is non-
uniform (e.g. due to finite skin depth or tight focusing) or if the sample is itself
non-uniform (e.g. a multlayer)
(iii) Describe the concept of meta-materials. [3]
Explain the significance of resonant phenomena for creation of negative refractive
index electromagnetic meta-materials. [3]
Sketch the ray paths from vacuum through a perfect lens made of a meta-material
with refractive index of -1. [3]
HINT: See the lecture notes
(iv) Find the frequency separation between longitudinal cavity modes in a laser with
cavity length of 1 m. [2]
Numerical answer: 0.15 GHz
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3. (i) What polarisation of the pump beam results in a specular optical Kerr effect (SOKE)
of maximal strength? [2]
Sketch on the same graph the intensity envelope of an ultrashort optical pump pulse
and the temporal dependence of the pump-induced specular inverse Faraday effect
(SIFE) signal, for the case when the characteristic life time of the pump induced
angular polarisation is comparable to the pump pulse duration. [1, 2]
HINT: See the lecture notes
(ii) Consider a sample excited by an optical pump pulse with duration much shorter than
the electronic thermalisation time. The electronic density of states in the sample can
be assumed to be constant over the energy range affected by the excitation. The
central angular frequency of the pump pulse is ω0.
Sketch on the same graph the electron distribution functions, f (E), where E is the
electron energy,
(a) at equilibrium before the excitation; [2]
(b) just after the excitation when the electrons are still "hot"; [3]
(c) shortly after the excitation when the electron sub-system has thermalised. [2]
Neglecting the temperature dependence of the Fermi energy, clearly label on the
sketch
(d) the Fermi energy level, EF; [1]
(e) the energies EF±ħω0, where ħ is the Planck constant; [1]
(f) the energies EF±kBT0 and EF±kBT, where kB is the Boltzmann constant, T0 is the
equilibrium temperature and T is the temperature of thermalised electrons. [2]
HINT: See the lecture notes
(iii) Write the equations of the two-temperature model in one dimension, naming and
7 PHYM007
explaining physical meaning of all variables and material parameters in the
equations. [10]
State the main assumption made when the two-temperature model is applied to the
interpretation of ultrafast pump-probe measurements of metals. [2]
HINT: See the lecture notes
(iv) In order to generate ultrashort electrical pulses, some pump-probe experiments use
GaAs-based Auston photoconductive switches gated by optical pulses that are
shorter than the electron-hole recombination time. Consider an Auston switch made
from intrinsic GaAs, which has a band gap of 1.5 eV and room temperature carrier
density of 2.5∙1014 m-3.
Estimate the maximum value of the resistivity ratio ρ(T = 300 K) / ρ photo-induced
induced in the Auston switch by 50% absorption of an ultrashort optical pulse tuned
to the GaAs interband absorption edge. Assume an excitation density of 1 pJ / µm2
and an optical skin depth of 100 nm. [6]
Numerical answer: 1011
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4. (i) In the interpretation of pump-probe measurements of ultrafast demagnetisation, it is
sometimes assumed that the ferromagnetic ordering is partly preserved even when
the transient electron temperature exceeds the Curie temperature of the material that
is being studied. Explain how this assumption could be justified. [4]
HINT: Note that the standard value of the Curie temperature applies to
equilibrium. However, upon ultrafast excitation, when the electron subsystem
is heated above Tc, the full demagnetisation might not occur if there has been
not enough time for the angular momentum associated with the magnetic
ordering to be transferred to the environment, e.g. the lattice sub-system.
Estimate the value of the exchange integral of a ferromagnetic material with Curie
temperature of 1000 K. [4]
Numerical answer: ≈ 1.4·10-21 J
Using a sketch, explain how and under what conditions ultrafast demagnetisation of
a thin ferromagnetic film could initiate precession of its magnetisation. [6]
HINT: See the lecture notes
(ii) Write the Landau-Lifshitz equation of motion of the magnetisation in the absence of
dissipation. [3]
Write the Landau and Gilbert magnetic damping terms and demonstrate their
equivalence in the limit of small dissipation. [3, 3, 4]
HINT: See the lecture notes
(iii) A meta-material exploits resonant coupling of the high frequency magnetic field of
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an electromagnetic wave to standing spin waves in a magnetic film that has its
magnetisation pinned at one or both surfaces.
Assuming perfect pinning at both film surfaces, calculate the ratio of the overlap
integrals between the high frequency magnetic field and the first two standing spin
wave modes that can couple to the electromagnetic field. [7]
Numerical answer: 3
1
PHYSICS EXAMINATION PROBLEMS SOLUTIONS AND HINTS FOR STUDENT SELF-STUDY
Module Code PHYM008
Name of module Physical Methods in Biology and Medicine Date of examination May/June 2014
1. (a) see notes on muscle structure, (b) e.g., confocal for nucleus, optical super-res for
mitochondria, using respective resolution formulae from notes, (c) e.g. x-ray crystallography, single particle cryo electron microscopy, (d) L10ms = 375 nm, L1s = 3.75 um
2. (a) see, for example, typical dual colour filter sets from Semrock; (b) thin film interference;
manufacturing processes include evaporative deposition, ion-assisted deposition, ion beam sputtering; show schematics, advantages (c) OD ~ 6 for emitter in excitation bands; (d) Iscatter/Iem ~ 1785
3. (a) NA 1.3, mag 63, variable immersion, plan/flat field correction, coverslip 0.15-0.19 mm,
DIC; (b) rlateral = 235 nm with water immersion; (c) Mtotal should be 68.1 for sampling, requires coupling magnification of 1.08; total field 117.5x117.5 um^2; (d) change mag by 500/400, otherwise may result in aliasing; (e) solid angle fraction based on NA: 0.24, Ndet = 1800 photons, SNR = 42.4
4. live cell conditions, contrast method (e.g. DIC, phase contrast or fluorescence), movement with
spatial sensor (camera), shape may require high-resolution (optical super-res, EM); temporal sampling based on characteristic time from mito size and velocity, determines required frame rate, spatial sampling ½ of smallest detail and requires appropriate magnification on camera; mechanisms: ATP, acto-myosin, microtubule & motors, test by specific inhibition of possible mechanisms