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THE UNIVERSITY OF SYDNEY
FACULTY OF SCIENCE
INTERMEDIATE PHYSICS
PHYS 2011 PHYSICS 2A
JUNE 2008 TIME ALLOWED: 2 HOURS
ALL QUESTIONS HAVE THE VALUE SHOWN
INSTRUCTIONS:
This paper consists of 2 sections.
Section A Optics 50 marks Section B Nuclear 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 mec^2 = 511 keV
electron volt 1 eV = 1. 602 × 10 −^19 J
proton 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
Universal gas constant R = 8. 315 J.mol−^1 K−^1
Stefan’s constant σ = 5. 670 × 10 −^8 W.m−^2 .K−^4
permittivity of free space ε 0 = 8. 854 × 10 −^12 C^2 .N−^1 .m−^2
gravitational constant G = 6. 673 × 10 −^11 N.m^2 .kg−^2
atomic mass constant u = 1. 660 × 10 −^27 kg
degrees/radian 180 /π ' 57.
Itrans = Iincid cos^2 φ
tan θp =
nb na
∆θ =
- 22 λ D
∆φ = 2 π
( 2 n 2 t λvac
)
∆φ = 2 π
( 2 n 2 t cos θ 2 λvac
)
It Ii
1 + F sin^2 (δ/2)
where δ = 2π
( 2 t cos θ λ
)
F =
4 R
(1 − R)^2
F =
π
F
2 t cos θ = mλ
R = m
π 2
F
∆λF SR =
λ m
λ^2 2 t
Please use a separate book for this section.
Answer ALL QUESTIONS in this section.
- (a) Consider a telescope with a perfect circular objective (focussing) lens, operating in space above the Earth’s atmosphere. When we make an image of a distant star (considered as a ‘point’ source) using this tele- scope, what do we see on a CCD detector at the focal plane of the telescope? Briefly explain the reason for this.
(b) A light source illuminates two closely-spaced slits in an otherwise opaque plate. The light that passes through the slits falls on a white screen well away from the slits. Two light sources are used in turn:
(i) a household electric light bulb with a red filter (bulb and filter lo- cated close to the slit assembly), and a
(ii) a red laser.
What difference will there be in the patterns cast on the screen in the two cases? Explain briefly the reason for this difference. (10 marks)
- (a) When it rains on an area of roadway where cars have dropped oil, one may see coloured patterns. Explain briefly the reason for this. You may use a diagram to illustrate your answer.
(b) How is spectral (chromatic) resolving power defined? Under what cir- cumstances would a large value of this quantity be useful? Give rea- sons for your answer.
(c) Polaroid sunglasses can greatly reduce the glare of sunlight reflected from a wet road. With the help of a diagram, explain why this is so.
(d) Make a labelled sketch of a setup that could be used to examine stresses in a transparent model of a mechanical component such as a supporting beam. Explain briefly how the system is able to reveal stresses in the material. (16 marks)
SECTION B
NUCLEAR PHYSICS
USEFUL INFORMATION
Selected formulae
R = R 0 A^1 /^3 N = N 0 e−λt
λ =
ln 2 t 1 / 2 A = N λ A = nϕσ(1 − e−λt) R = φnσ EB = [ZMH + N Mn − M (AZ X)] × 931 .494 MeV
EB = C 1 A − C 2 A^2 /^3 − C 3
Z(Z − 1)
A^1 /^3
− C 4
(A − 2 Z)^2
A
± C 5 A−^4 /^3
U = k
Z 1 Z 2 e^2 r
Constants
R 0 = 1. 2 × 10 −^15 m C 1 = 15 .75 MeV C 2 = 18 .80 MeV C 3 = 0 .710 MeV C 4 = 23 .69 MeV C 5 = 39 MeV NA = 6. 02 × 1023 mole−^1
k =
4 π 0
= 8. 98755178 × 109 N.m^2 /C^2
Selected conversion factors
1u = 1. 66053873 × 10 −^27 kg = 931.5 MeV/c^2 1 Ci = 3. 70 × 1010 Bq = 3. 70 × 1010 decays/s 1 year = 3. 16 × 107 s 1 barn = 10 −^24 cm^2
Selected particle masses
mp = 1 .007276u = 1. 672623 × 10 −^27 kg mn = 1 .008665u = 1. 674929 × 10 −^27 kg me = 0 .000548580u = 9. 10939 × 10 −^31 kg
Selected atomic masses
M
( 1 1 H
) = 1 .007825u M
( 2 1 H
) = 2 .0141024u M
( 3 1 H
) = 3 .016049u M
( 3 2 He
) = 3 .016029u M
( 4 2 He
) = 4 .002603u M
( 7 3 Li
) = 7 .016004u M
( 7 4 Be
) = 7 .016929u M
( 16 8 0
) = 15 .994915u M
( 23 10 Ne
) = 22 .994467u M
( 23 11 Na
) = 22 .989770u M
( 56 26 Fe
) = 55 .934939u M
( 238 92 U
) = 238 .050784u
- Give a brief description of alpha, beta (β− and β+) and gamma decay. For each case write the corresponding decay equation for nucleus AZ X. (10 marks)
THERE ARE NO MORE QUESTIONS.
