# Search in the document preview

9322

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.

9322 Semester 1 2008 Page 2 of 10

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 mec2 = 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) h̄ = 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 ε0 = 8.854× 10−12 C2.N−1.m−2

gravitational constant G = 6.673× 10−11 N.m2.kg−2

atomic mass constant u = 1.660× 10−27 kg degrees/radian 180/π ' 57.2958

9322 Semester 1 2008 Page 3 of 10

SECTION A

OPTICS

FORMULAS

n = c/v v = fλ

na sin θa = nb sin θb

Iav = 1

2 0cE

2 max

d sin θ = mλ

d sin θ = (m+ 1

2 )λ

E2tot = E 2 1 + E

2 2 + 2E1E2 cos(φ2 − φ1)

I = I0 cos 2 ( πd sin θ

λ

)

β = 2πa sin θ

λ

I = I0

sin(β/2) (β/2)

2

sin θ = mλ

a

φ = 2πd sin θ

λ

I = I0 cos 2(φ/2)

sin(β/2) (β/2)

2

` = λ2

∆λ

I = I0

sin(β/2) (β/2)

2 sin(Nφ/2) sin(φ/2)

2

sin θ − sin θi = mλ

d sin θ =

mλ

d

R = λ

∆λ R = mN

9322 Semester 1 2008 Page 4 of 10

Itrans = Iincid cos 2 φ

tan θp = nb na

∆θ = 1.22λ

D

∆φ = 2π

( 2n2t

λvac

) + φ12 + φ23

∆φ = 2π

( 2n2t cos θ2

λvac

) + φ12 + φ23

It Ii

= 1

1 + F sin2 (δ/2) where δ = 2π

( 2t cos θ

λ

)

F = 4R

(1−R)2

F = π √ F

2 2t cos θ = mλ

R = m π

2

√ F

∆λFSR = λ

m ' λ

2

2t

9322 Semester 1 2008 Page 5 of 10

Please use a separate book for this section. Answer ALL QUESTIONS in this section.

1. (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)

2. (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)

9322 Semester 1 2008 Page 6 of 10

3. A transmission diffraction grating is illuminated at normal incidence by light from a sodium lamp (wavelength 589.0 nm). It is observed that the first order diffracted light leaves the grating at an angle of 23.1◦ from the axis.

(a) Find the spacing of lines in the grating.

(b) At what angle will the second order light of the same wavelength leave the grating?

(c) What advantage would there be to observing the second order light rather than the first order?

(d) Explain why it is not physically possible for a third order diffracted beam to exist in this case.

(12 marks)

4. (a) Two flat rectangular glass plates are placed one on top of the other, touching along one edge but separated by the thickness of a human hair along the opposite edge. The system is illuminated by monochromatic light from a sodium vapour lamp, and is viewed from above.

(i) Describe what will be seen by an experimenter viewing the plates. Include a diagram to illustrate your answer.

(ii) Will a bright or dark region be seen along the line where the plates touch? Give reasons for your answer.

(b) A certain Fabry-Perot interferometer has plates with a separation of 3.00 mm, and intensity reflectance of 0.80. It is being used to study light with wavelength near 450 nm.

(i) Calculate the resolving power of the interferometer. State any as- sumptions or approximations you need to make.

(ii) Calculate the smallest detectable wavelength interval between close spectral lines at 450 nm.

(12 marks)

9322 Semester 1 2008 Page 7 of 10

SECTION B

NUCLEAR PHYSICS

USEFUL INFORMATION

Selected formulae

R = R0A 1/3

N = N0e −λt

λ = ln 2

t1/2 A = Nλ A = nϕσ(1− e−λt) R = φnσ

EB = [ZMH +NMn −M(AZX)]× 931.494 MeV

EB = C1A− C2A2/3 − C3 Z(Z − 1) A1/3

− C4 (A− 2Z)2

A ± C5A−4/3

U = k Z1Z2e

2

r

Constants

R0 = 1.2× 10−15m C1 = 15.75 MeV C2 = 18.80 MeV C3 = 0.710 MeV C4 = 23.69 MeV C5 = 39 MeV NA = 6.02× 1023 mole−1

k = 1

4π0 = 8.98755178× 109N.m2/C2

9322 Semester 1 2008 Page 8 of 10

Selected conversion factors

1u = 1.66053873× 10−27kg = 931.5 MeV/c2

1 Ci = 3.70× 1010 Bq = 3.70× 1010 decays/s 1 year = 3.16× 107 s 1 barn = 10−24 cm2

Selected particle masses

mp = 1.007276u = 1.672623× 10−27kg mn = 1.008665u = 1.674929× 10−27kg me = 0.000548580u = 9.10939× 10−31kg

Selected atomic masses

M ( 1 1H

) = 1.007825u

M ( 2 1H

) = 2.0141024u

M ( 3 1H

) = 3.016049u

M ( 3 2He

) = 3.016029u

M ( 4 2He

) = 4.002603u

M ( 7 3Li

) = 7.016004u

M ( 7 4Be

) = 7.016929u

M ( 16 8 0 )

= 15.994915u

M ( 23 10Ne

) = 22.994467u

M ( 23 11Na

) = 22.989770u

M ( 56 26Fe

) = 55.934939u

M ( 238 92 U

) = 238.050784u

9322 Semester 1 2008 Page 9 of 10

Please use a separate book for this section. Answer ALL QUESTIONS in this section.

5. (a) In an experiment, aluminium foil of thickness 0.3 mm and area 1 cm2 was bombarded with neutrons with a flux of 5.0×1012 neutrons. cm−2.s−1. Calculate the 27Al(n, γ)28Al reaction cross-section if 1.8×107 neutrons. cm−2.s−1are captured. Density of aluminium is 2.7 g. cm−3.

(b) In a subsequent experiment, the activity of the aluminium foil con- taining radioactive 28Al was measured. At times 2 and 4 minutes af- ter irradiation, the aluminium foil had activities of 9.85 × 106 Bq and 5.39× 106 Bq respectively. What is the half-life of 28Al?

(15 marks)

6. (a) Define binding energy and binding energy per nucleon.

(b) Calculate the binding energy of 5626Fe. Compare your result with the figure shown below and comment.

(c) Calculate the binding energy per nucleon for 21H (deuterium) and 3 1H

(tritium). Compare your results with the figure shown below and com- ment.

(15 marks)

9322 Semester 1 2008 Page 10 of 10

7. Give a brief description of alpha, beta (β− and β+) and gamma decay. For each case write the corresponding decay equation for nucleus AZX.

(10 marks)

THERE ARE NO MORE QUESTIONS.