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THE UNIVERSITY OF SYDNEY
FACULTY OF SCIENCE
INTERMEDIATE PHYSICS
PHYS 2213 PHYSICS 2EE
NOVEMBER 2007 TIME ALLOWED: 3 HOURS
ALL QUESTIONS HAVE THE VALUE SHOWN
INSTRUCTIONS:
This paper consists of 3 sections.
Section A Optics 40 marks Section B Electromagnetic Properties of Matter 40 marks Section C Solid State and Device Physics 40 marks
Candidates should attempt all questions.
USE A SEPARATE ANSWER BOOK FOR EACH SECTION.
In answering the questions in this paper, it is particularly important to give rea- sons for your answer. Only partial marks will be awarded for correct answers with inadequate reasons.
No written material of any kind may be taken into the examination room. Calcu- lators are permitted.
Table of constants
Avogadro’s number NA = 6. 022 × 1023 mole−^1
speed of light c = 2. 998 × 108 m.s−^1
electronic charge e = 1. 602 × 10 −^19 C
electron rest mass me = 9. 110 × 10 −^31 kg
electron rest energy energy = 511 keV
electron volt 1 eV = 1. 602 × 10 −^19 J
protron rest mass mp = 1. 673 × 10 −^27 kg
neutron rest mass mn = 1. 675 × 10 −^27 kg
Planck’s constant h = 6. 626 × 10 −^34 J.s
Planck’s constant (reduced) ¯h = 1. 055 × 10 −^34 J.s
Boltzmann’s constant kB = 1. 380 × 10 −^23 J.K−^1
Stefan’s constant σ = 5. 670 × 10 −^8 W.m−^2 .K−^4
Coulomb constant (^4) π^10 = 8. 988 × 109 N.m^2 .C−^2
permittivity of free space 0 = 8. 854 × 10 −^12 C^2 .N−^1 .m−^2
permeability of free space μ 0 = 4 π × 10 −^7 kg.m.C−^2
gravitational constant G = 6. 673 × 10 −^11 N.m^2 .kg−^2
atomic mass constant u = 1. 660 × 10 −^27 kg
refractive index of air nair = 1.
degrees/radian 180 /π ' 57.
P (z) P (0)
= 10 −^0.^1 αz
sin(θC ) =
n 2 n 1
N A = sin(θ^12 ) = (n^21 − n^22 )
(^12)
∆τ = D∆λBW L
V = 2 π
a λ
N A
M =
π^2
V 2
n = n 0 +
∆n 2
1 + cos
2 π
x Λ
2 nΛ = λB
∆λ λ
∆n n
κac = π
∆n λ
R = tanh^2 (κacL)
∆λ =
λ^2 nL
∆λ =
λ^2 π
κac
∆τ =
OP L
c
2 nL c I(∆φ) Imax
= cos^2
∆φ 2
∆Lcoh =
c ∆νBW
∆n(t) = n^30 KpE(t)
F <
c 2 nL
Please use a separate book for this section.
Answer ALL QUESTIONS in this section.
Keep explanations concise. Be careful to use the appropriate units.
- Below is a schematic of an optical disc storage system.
(a) Name any three of the optical components (1) to (9) in the diagram of the CD ROM optical head shown above.
(b) Name 3 functions that the detector has to perform.
(c) Why is a polarizing beam splitter used rather than a non-polarizing beam splitter?
(d) Name 2 places in the CD reader where diffraction plays a role.
(e) Assuming the reflected light intensity obeys a simple interferometric form I = Io cos^2 (∆φ/2) where ∆φ is the phase difference between the two beams, what bump height (at λ = 780 nm) will give 100% contrast between 1 ′s and 0 ′s? The refractive index of the coating film is 1.4.
- (a) What does WDM stand for?
(b) What are the two main advantages of WDM systems over older, single wavelength, systems?
(c) What two critical functions did erbium doped fibre amplifiers enable when they were introduced into fibre optic systems in the mid 1990s.
(d) Name one advantage of fibre Bragg gratings over thin film filters.
(e) Name one advantage of thin film filters over fibre Bragg gratings.
(f) A fibre Bragg grating is written in a fibre (n=1.5) with a strength κ = 2 cm−^1 and length L= 2 cm at λ=1550 nm.
(i) What is the bandwidth of the grating in nm?
(ii) What is the reflectivity (in %) and transmission (in dB)?
(iii) If this grating were written in birefringent fibre where the differ- ence in refractive index for the two polarizations, ∆n = 10−^3 , what is the separation in grating wavelengths for the two polarizations?
(iv) Will this grating be useful for filtering unpolarized light?
(15 marks)
SECTION B
ELECTROMAGNETIC PROPERTIES OF MATTER
FORMULAS
Ex = −
∂V
∂x
Ey = −
∂V
∂y
Ez = −
∂V
∂z
FE = qE
E =
4 π 0
q r^2
ˆr ΦE =
E · dA =
qenclosed 0
E = σ/ 0
p = qd τ = p × E U = −p · E V =
4 π 0
q r
Q = CV C =
K 0 A
d
U =
CV 2
P = np P = σbound D = K 0 E = E D = σf ree D = 0 E + P P = χe 0 E
p = αE α =
n
K − 1
K + 2
i =
dq dt
J = I/A i = nqvdrif tA VH =
Bi nel J = nqvdrif t R = V /I ρ = E/J
ΦB =
B · dA FB = qv × B μ = N iAnˆ
ωc = qB/m B = μ 0 (H + M) τ (^) B = μ × B
dFB = idL × B B = μH W = −μ · B M = χH
W = V
HdB B = μrμ 0 H S =
μ 0
E × B dB =
μ 0 4 π
ids × r r^3
prad = I/c or prad = 2I/c
B · dA = 0 B(z) =
μ 0 ia^2 2(z^2 + a^2 )^3 /^2 ∂^2 Ey ∂x^2
− μ 0 0
∂^2 Ey ∂t^2
E · ds = −
dΦB dt
B(r) =
μ 0 i 2 πr
H = N i/L
Ey = Ey 0 cos(ωt − kx) Bz = Ey/c
B · ds = μ 0 i + μ 0 0
dΦE dt
B = μ 0 ni ∮ D · dA = qf ree U/V = B^2 / 2 μ 0
H · ds = if ree +
dΦD dt
- (a) Draw a hysteresis curve for a typical ferromagnetic material and label “coercivity” and “remanence” on your diagram.
(b) Briefly define remanence and coercivity.
(c) An iron permanent magnet is in the form of a circular disk with a small gap in it, as shown in the diagram below.
The average radius of the circular disk is r and the length of the gap is `.
(i) Draw three sketches of this iron permanent magnet in your answer book. On one show the field lines of B in the iron and in the gap. On another show the field lines of the magnetization. On the third show the field lines of H in the iron and in the gap.
(ii) Use Ampere’s Law to show that the field in the iron is is given by
Hiron = −
`
(2πr − `) μ 0
B
(17 marks)
SECTION C
SOLID STATE AND DEVICE PHYSICS
FORMULAS
E =
p^2 2 m
F (E) =
1 + e
E−EF kT
n = NC e
−(Ec−EF ) kT
p = NV e
−(EF −EV ) kT
ni =
NC NV e
−Eg 2 kT
np = n^2 i
n = nie
EF −Ei kT
p = nie
Ei−EF kT
ρ =
q(nμn + pμp)
μ =
vd ε Dn/p μn/p
kT q
Jn = qμnnε + qDn
dn dx
Jp = qμppε − qDp
dp dx
W =
2 s q
NA + ND
NAND
Vbi
J = Js(e
qV kT (^) − 1)
Js =
qDppn 0 Lp
qDnnp 0 Ln
- An n − type germanium semiconductor sample is brought into contact with a p − type silicon sample. The germanium sample has a carrier concentra- tion of 4. 5 × 1016 cm−^3 and the silicon sample has a carrier concentration of
- 0 × 1016 cm−^3. At 300K the intrinsic carrier concentration of germanium is 2. 4 × 1013 cm−^3 and its band gap is 0.66 eV. At 300K the intrinsic carrier concentration of silicon is 1. 45 × 1010 cm−^3 and its band gap is 1.12 eV.
(a) Describe the differences between an n − type semiconductor and a p − type semiconductor.
(b) Calculate the energy difference between the centre of the band gap and the Fermi level in both cases.
(c) Draw a labeled diagram showing the edges of the bands of the semicon- ductors and the Fermi level(s) when the semiconductors are brought into contact. Show the region near the junction on both sides of the junction.
(d) Name the type of junction and briefly discuss one application.
(15 marks)
THERE ARE NO MORE QUESTIONS.
