SolnsHints 2010-11 Term-2 - Physics and Astronomy - - Physics …€¦ ·  · 2014-04-15PHYSICS'EXAMINATION'PROBLEMS' ......

PHYSICS'EXAMINATION'PROBLEMS'SOLUTIONS'AND'HINTS'FOR'STUDENT'SELF5STUDY'

Module'Code' PHY1023'Name'of'Module' Waves'and'Optics'Date'of'Examination' June'2011'

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1.# Deduced#power#R#and#T#coefficients:##derivations#available#in#lecture#notes.#

# Reflection#and#transmission#of#triangular#wave#pulses:##follow#examples#from#lecture#notes.#

# Fraction#of#incident#energy#transmitted#across#junction#=#0.89.#

Modifications#to#system#for#zero#reflection:##insert#a#length#of#new#string#between#two#existing#strings#

which#has#length#λ/4#and#impedance#=# 21zz .#

To#propagate#500#Hz#waves#without#reflection#at#the#interface,#the#length#of#string#to#insert#would#be#0.5#m#and#its#impedance#would#be#14.14#kg#m−1#s.#(ie.#the#tension#will#be#continuous#across#all#three#strings).#

#

2.# Equations#of#motion,#graphs#/#sketches#of#motion#vs#frequency#etc#and#statement#of#conditions#for#specific#motions#are#all#available#in#high#resolution#from#lecture#notes.#

# (Case#3):##Velocity#AmplitudebF0= =#2#ms−1.#

#

3.# Huygens’#principle#and#intensity#pattern#available#from#lecture#notes.#

# Three#angles#of#diffraction#maxima;##θ1#=#2.3°,#θ2#=#4.6°,#θ3#=#6.9°.#

# Highest#diffraction#order#n#=#24##(the#n=25#order#propagates#along#the#interface#at#90°#and#so#is#not#visible).#

# At#an#angle#of#incidence#of#30°,#the#highest#order#visible#is#n=37.#

At#400#nm,#3rd#diffracted#order#occurs#at#θ#=#46°;#while#at#700#nm,#2nd#diffracted#order#occurs#at#θ#=#56°.#

Using#the#Rayleigh#Criterion,#separation#of#two#points#by#50#m#can#just#be#resolved#using#the#Mount#Palomar#telescope.##

Also#using#the#Rayleigh#Criterion,#a#mirror#of#diameter#128#km#is#needed#to#resolve#newsprint#approx#2mm#in#size.#

#

#

#

4.# Use#lecture#notes#for#impedance#expressions.#

# Current#in#C#circuit#=#0.06#A;##current#in#L#circuit#=#1.06#A.#

# IC#and#IL#will#be#equal#at#the#circuit’s#resonant#frequency:##f#=#205#Hz.#

Expression#for#parallel#circuit#impedance#available#from#lecture#notes.#The#difference#in#frequency#response#between# parallel# and# series# circuits# relates# to# high# f# and# low# f# filtering# properties# of# individual# branch#components#(inductors#filter#high#f;#capacitors#filter#low#f#current).##

Power#transferred#in#a#series#circuit#is#maximum#at#the#same#resonant#frequency#(i.e.#f#=#205#Hz).#

At#resonance,#this#power#transferred#=#2.5#×#10−3#W.#

#

5.# Definition#and#examples#of#Doppler#effect#from#lecture#notes.#

# Maximum#and#minimum#frequencies#are#1097#Hz#and#919#Hz.#

Standard# relations# for# wavelength,# wavevector,# phase# and# group# velocities# and# definitions# for# types# of#dispersion#are#available#from#lecture#notes.#

Derivation#of#group#velocity#of#water#waves#relation#is#completed#using#standard#mathematics#and##group#velocity#equation.#

Estimate#wavelength#of#waves#at#Saunton#Beach#to#be#20#m.##Subsequent#phase#velocity#is#6#m#s−1.##

'

6.# Standard#relations#for#phase#velocity#in#terms#of#string#tension#and#linear#density#is#available#from#lecture#notes.###

# Combine#this#relation#and#normal#expression#for#velocity#(dist#/#time)#to#prove#L#=#9.8#m.#

# Relation#for#frequencies#of#normal#modes#on#this#string;#f#=#5n##(where#n#represents#the#harmonic#number).#

# Derivation#of#total#energy#of#vibrating#string#available#from#lecture#notes.#

# The#stated#equation#represents#the#harmonic#mode#with#k#=#2.#

# Total#energy#(Pav)#=#5#×#10−3#W.#

#

#

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PHYSICS EXAMINATION PROBLEMS SOLUTIONS AND HINTS FOR STUDENT SELF-STUDY

Module Code PHY1026

Name of module Mathematics for Physicists

Date of examination May 2011

1. (i) x, y, z( ) = 1.5, 3 2,1( ) , ρ,φ, z( ) = 3,π 6,1( ) (ii) i (iii) 11.592 (write the cosine in terms of complex exponentials)

2. (i) dydx

= −

∂ f∂x y∂ f∂y x

=y x + y( )−11− x x + y( )

(ii) ∂f∂x y

= 3x2; ∂f∂y x

= 3y2; dfdx

= 3x2 + 6x5

3. (i) M = m03

1+ e2( )3 2 − 23 2⎛⎝⎜

⎞⎠⎟ = 7.156m0

(ii) I = πa4

4

4. (i) Directional gradient = 12

π2−1⎛

⎝⎜⎞⎠⎟

(ii) see notes (iii) Both integrals equal 2πa2h

5. (i) y = −x 1+ 1ln x a( )

⎛

⎝⎜⎞

⎠⎟ (arbitrary constant written in the form ln a( ) )

(ii) x = exp −t( ) Aexp 3 it( )+ Bexp − 3 it( )⎡⎣

⎤⎦ (or sine and cosine form)

x

t

period

PHY 2007. Hints 2011

Q1: Required equations: L = I!, T = 1/2 I!.!, N = I" (" = angular acceleration)

For rod, I = 1/3Ml2.

a) ! = 3g/2l b) f = 1/4Mg c) "f = (3g/l)1/2 d) f= 5Mg/2 e) " = (3g/l)1/2

Q2: Identify contibutiions ofrom alinear and angular acceleration of frame, coriolos and centrifugal terms.

Bullet deflected to right of path

Pendulum is sufficiently long that effects of coriolis force, leading to rotation of plane of oscillation are observable.

Tides are sensitive to coriolis force which enhances movement up beach in France as tide is coming in and vice-versa

Q3: Should recover the Newtonian result F = -dU/dx = mx..

# is a generalised coordinate and the equation is: $.. = -g/l(sin$ +cos$)

Q4: All answers can be found in notes

Q5: Measured on earth the trip takes 26.7 y. For the astronaut it takes 25.4y.

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1

PHYSICS EXAMINATION PROBLEMS

SOLUTIONS AND HINTS FOR STUDENT SELF-STUDY

Module Code PHY2020

Name of module Lasers and Materials for Quantum Applications

Date of examination June 2011

1. Pmax = 0.064 Pa

3. v = 1.8 × 107 ms

-1

EKE = 1590 eV

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PHYSICS EXAMINATION PROBLEMS SOLUTIONS AND HINTS FOR STUDENT SELF-STUDY

Module Code PHY3135

Name of module NUCLEAR & HIGH ENERGY PARTICLE PHYSICS

Date of examination June 2011

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HINTS AND TIPSfor PHYM422 exam (2011)

1. According to selection rules, the only non-zero matrix elements of thedipole moment is the one between l = 0 and l = 1, m = 0 states. Thematrix element can be found directly, and is equal to eER/

√3. The analysis

is applicable when δE ≪ ∆E = h̄2/mR2.2. The density of states in three dimensions is proportional to the finalmomentum; the latter is determined by the energy conservation, Ef = Ei +h̄ω. The threshold is determined by the condition Ef > 0. The transitionmatrix element vanishes at p = 0 by symmetry x → −x. Therefore, thetransition rate is proportional to p2fg(pf) ∝ p3f ∝ (ω − ωthr)3/2.3.(i) The states can be separated into two non-interacting pairs, (14) and(23). For each pair, the characteristic equation is solved trivially, which givesenergy levels −4A, 2A for (14) pair, and 0, 2A for the second pair. Note thatthe 2A level is doubly degenerate.3.(ii) For the S=0 state, both particles can be placed into single-particleground state, n = 1, which gives the energy 2E1. For the S = 1 state, theminimal energy is E1 + E2. The wave functions have the standard form.For S = 1 state, there are three different wave functions, correspondingto S1 = −1, 0, 1, all being symmetric with respect to a permutation of twospins.4. The integration is achieved by introducing the change of variable u = qr.The angular distribution of the scattered particles is isotropic at low energies,and exhibits forward scattering peak at high energies. Therefore, angularintegration in the low-energy case is trivial, and amounts to multiplicationby 4π.

1