Individual Study Plan: Unit 2 Waves and Light
Subject:PhysicsTeacher:N.Lad
Student:UnitUnit 2
Study Task outline:Waves ISPThis ISP covers the Waves and Light
part of Physics Unit 2: Physics at Work. While much of the topic
contains mathematical concepts which you will be asked
quantitatively (using numbers) you may be asked to explain these
ideas qualitatively (using words). This ISP will help you build the
skills required to answer both styles of questions and aid you to
build your knowledge for your assessments. It is of imperative
importance that you can answer both quantitative and qualitative
Physics questions; without this skill you will struggle to do any
better than a D grade. To achieve a grade A-C you will need to be
able to bring together different areas of Physics you have learnt
(e.g. electrical energy anf the photoelectric effect). You must
also be able to use an equation in a qualitative manner, explaining
the implications of changing a variable in the equation. The
specification points in this ISP are all of the specification
points within the Waves module (see your specification or the
Revision checklist in your revision guide).
The A* grade is awarded if the candidate meets two requirements:
Grade A for the overall Advanced GCE 90% of the total available
uniform marks for the A2 unitsThis means that you can get a B grade
in AS Physics and still achieve an A*in A2 PhysicsA* students
should be able to: apply principles and concepts in familiar and
new contexts involving only a few steps in the argument describe
significant trends and patterns shown by data presented in tabular
or graphical form and interpret phenomena with few errors and
present arguments and evaluations clearly explain and interpret
phenomena with few errors and present arguments and evaluations
clearly carry out structured calculations with few errors and
demonstrate good understanding of the underlying relationships
between physical quantities.
Exam questions where you are asked to describe experiments are
often worth 6 marks, which can help you reach the next grade up of
an exam paper. In these questions you need to:Clearly state the
equipment required, explain and justify the equipment you have
chosen, the measurement you make, and calculations you will do and
also how you will make the experiment reliable to satisfy the
criteria for full marks.Past papers and other examboard information
is available on the following website:
http://www.edexcel.com/quals/gce/gce08/physics/Pages/default.aspx
TaskTask detailResources neededCompleted/comments
1
Spec points covered: 63,64,65,66,67, 69,70,71,72You should read
through the notes from lessons and p146-155. The areas to focus on
are:1. Being able to explain why we believe light behaves like a
particle and a wave2. Explaining the importance of the
photoelectric effect3. Define and use radiation flux as power per
unit areaComplete Explaining Everything 1Complete questions for
Task 1 from your Waves ISP booklet and mark using mark schemeA*/A
link the photoelectric effect to electrical energy
Edexcel AS Physics textbook
Waves Booklet and mark scheme
Explaining everything 1Q E MWWW
EBI
MRI
2
Specification points covered : 68You should read through the
notes from lessons and p156-163. The areas to focus on are:1.
Explaining atomic line spectra in terms of transitions between
discrete energy levelsComplete Explaining Everything 2Complete
questions for Task 2 from your Waves ISP booklet and mark using
mark schemeA/A* - link the photoelectric effect to atomic line
spectraEdexcel AS Physics textbook
Waves Booklet and mark scheme
Explaining everything 2Q E MWWW
EBI
MRI
3
Specification points covered : 28, 30, 32, 31, 32, 35You should
read through the notes from lessons and p76-91. The areas to focus
on are:1. Stating the types of waves2. Describing the key features
of a wave3. Describe standing and progressive wavesComplete
Explaining Everything 2Complete questions for Task 2 from your
Waves ISP booklet and mark using mark schemeA*/A research how
electron behave as standing wavesEdexcel AS Physics textbook
Waves Booklet and mark scheme
Explaining everything 3Q E MWWW
EBI
MRI
4
Spec points covered : 33,34, 36, 37, 38, 41,42, 43, 44, 45You
should read through the notes from lessons and 92-99. The areas to
focus on are:1. Describing the process of superposition2.
Describing the formation of stationary waves and comparing them to
progressive wave3. Describing reflection , refraction and how to
find the refractive index4. Describing diffractionComplete
Explaining Everything 4Complete questions for Task 4 from your
Waves ISP booklet and mark using mark schemeA*/A research de
Broglie diffractionEdexcel AS Physics textbook
Waves Booklet and mark scheme
Explaining everything 4Q E MWWW
EBI
MRI
5
Specification points covered : 29, 39, 40, 46, 47, 48, 49You
should read through the notes from lessons and p100-113. The areas
to focus on are:1. Describing the process of polarisation2.
Describing the EM spectrum3. Describe the process of the Doppler
shift, and pulse echo techniques such as ultrasoundComplete
Explaining Everything 5Complete questions for Task 5 from your
Waves ISP booklet and mark using mark schemeA*/A link the EM
spectrum to the photoelectric effectEdexcel AS Physics textbook
Waves Booklet and mark scheme
Explaining everything 5Q E MWWW
EBI
MRI
Q - Questions completedE - Explaining everything completedM -
Questions markedWWW want went wellEBI even better ifMRI My response
is (student response)Explaining Everything 1The Photoelectric
effectExplain why the wave model of light cannot explain emission
of photoelectrons, but how the particle model of light does.
Describe the experiment performed to prove the particle model of
light, and include a labelled diagram.
Keywords to include are:Photon Wave Energy Work function
Threshold frequency Photoelectron Emission Interactions Kinetic
energyInclude any equations related to the photoelectric effect,
and any graphs to help you determine constants/variable involved n
the photoelectric effect equations.
Draw V vs I (or I vs V) graphs for a diode, a filament lamp and
a resistor. Explain the shape of the graphs.Terms/Words to
include:Resistancetemperaturecurrentpotential differenceconstantone
direction
Explaining everything 2Energy LevelsUsing a diagram, show how
absorption or emission of photons can be shown using an energy
level diagram. Explain how we can calculate the wavelength or
frequency of the photon. Include any equations.
Keywords to include are:Electron Volts
(eV)AbsorptionEmission
Explaining Everything 3In the space below complete the following
tasks:Draw a wave labelling the key features (wavelength,
frequency, amplitude)Include the wave equation and how to calculate
the periodCompare and contrast longitudinal waves and transverse
wavesCompare and contrast standing and progressive waves
Explaining Everything 4: Light as a WaveExplain the evidence
that supports the theory that light acts as a wave.
Keywords to include are:Interference constructive destructive
path difference superposition
fdfds
Explaining everything 5Describe how light can become
polarised:
Describe an experiment that you could use to test how the much
sugar solution rotates the plane of polarisation of light:
Describe the Doppler Effect:
Describe pulse echo techniques:
List of data, formulae and relationshipsDataAcceleration of free
fallg = 9.81 m s2(close to Earths surface)Boltzmann constant k =
1.38 1023 J K1Coulombs law constantk = 1/4 = 8.99 109 N m2
C2Electron chargee = 1.60 1019- CElectron massme = 9.11 1031
kgElectronvolt1 eV = 1.60 1019 JGravitational constantG = 6.67 1011
N m2 kg2Gravitational field strengthg = 9.81 N kg1(close to Earths
surface)Permittivity of free space= 8.85 1012 F m1Planck constant h
= 6.63 1034 J sProton massmp = 1.67 1027 kgSpeed of light in a
vacuumc = 3.00 108 m s1Stefan Boltzmann constant5.67 108 W m2
K4Unified atomic mass unitu = 1.66 1027 kg
2.MechanicsKinematic equations of motionv = u + ats = ut + at2v2
= u2 + 2asForcesF = mag = F/mW = mgWork and energyW = FsEk =
mv2Egrav = mgh
MaterialsStokes lawF = 6rvHookes lawF = kxDensity = m/VPressurep
= F/AYoungs modulusE = / whereStress = F/AStrain = x/xElastic
strain energyEel = FxAppendix 8 Formulae
3.WavesWave speed v = fRefractive index12 = sin i/sin r = v1/v2
ElectricityPotential differenceV = W/QResistanceR = V/IElectrical
power, energy andP = VIefficiencyP = I2RP = V2/RW = VIt% efficiency
= [useful energy (or power) output/total energy (or power) input]
100%ResistivityR = l/ACurrentI = Q/tI = nqvA
Quantum physicsPhoton modelE = hfEinsteins photoelectrichf = +
mv2maxequation
Waves ISP BookletTask 1
1.(a)The maximum wavelength of electromagnetic radiation which
can release photoelectrons from the surface of caesium is 6.5 107
m.(i)State the part of the electromagnetic spectrum to which this
radiation
belongs............................................................................................................................(1)(ii)Show
that caesium has a work function of about 3 1019
J........................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................(2)
(b)The caesium cathode of a photocell is illuminated by radiation
of frequency f. The circuit shown is used to measure the stopping
potential Vs for a range of frequencies.
(i)Explain what is meant by the term stopping
potential..................................................................................................................................................................................................................................................................................................................................................................................
(2)
(ii)The experiment is repeated, using different photocells, to
measure the stopping potentials of calcium and beryllium. The graph
shows how the stopping potentials Vs for all three metals vary with
frequency f.
Use the relationshiphf = eVs + to explain why all three graphs
are
parallel..................................................................................................................................................................................................................................................................................................................................................................................
(2)(Total 7 marks)2.(a)Magnesium has a work function of 5.89 1019
J. Explain the meaning of this
statement................................................................................................................................................................................................................................................................................................................................................................................................................(2)
(b)Ultraviolet radiation from an extremely faint source is incident
normally on a magnesium plate. The intensity of the radiation is
0.035 W m2. A single magnesium atom occupies an area of about 8
1020 m2 on the surface of the plate.(i)Show that, if the radiation
is regarded as a wave motion, it should take at least 200 s for a
magnesium atom to absorb 5.89 1019 J of
energy.............................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................(3)
(ii)In practice, it is found that photoemission from the plate
begins as soon as the radiation source is switched on. Explain how
the photon model of electromagnetic radiation accounts for this.
You may be awarded a mark for the clarity of your
answer...................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................
(4)(Total 9 marks)3.A leaf of a plant tilts towards the Sun to
receive solar radiation of intensity 1.1 kW m 2, which is incident
at 50 to the surface of the leaf.
(a)The leaf is almost circular with an average radius of 29 mm.
Show that the power of the radiation perpendicular to the leaf is
approximately 2
W.....................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................(3)
(b)Calculate an approximate value for the amount of solar energy
received by the leaf during 2.5 hours of
sunlight...........................................................................................................................................................................................................................................................................Energy
= ..........................................(2)(Total 5 marks)
4.The photoelectric effect supports a particle theory of light
but not a wave theory of light. Below are two features of the
photoelectric effect. For each feature explain why it supports the
particle theory and not the wave theory.(a)Feature 1: The emission
of photoelectrons from a metal surface can take place
instantaneously.Explanation
.......................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................
(2)(b)Feature 2: Incident light with a frequency below a certain
threshold frequency cannot release electrons from a metal
surface.Explanation
.......................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................
(2)(Total 4 marks)5.The table gives information about two beams of
monochromatic light.Intensity /Wm2Colour
Beam A6.0red
Beam B0.2blue
Each beam is shone in turn onto a barium plate. Beam B causes
photoemission but beam A does not. A student says that this is
because the blue beam is more energetic than the red beam.(a)In one
sense the students statement is correct. In another sense the
statement is incorrect. Explain how it
is:correct...............................................................................................................................................................................................................................................................................................................................................................................................................incorrect...........................................................................................................................................................................................................................................................................(3)
(b)Barium has a work function of 3.98 1019 J.(i)Explain the
meaning of the term work
function..................................................................................................................................................................................................................................................................................................................................................................................(2)(ii)Calculate
the photoelectric threshold frequency for the barium
plate..................................................................................................................................................................................................................................................................................................................................................................................Threshold
frequency = ......................................(2)(Total 7
marks)
Task 26.The diagram shows some of the energy levels of a
tungsten atom.
(a)An excited electron falls from the 1.8 keV level to the 69.6
keV level. Show that the wavelength of the emitted radiation is
approximately 0.02
nm...............................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................
(4) (b)To which part of the electromagnetic spectrum does this
radiation
belong?.....................................................................................................................................(1)(Total
5 marks)7.Astronomers can identify different gases present in the
outer parts of stars by analysing the line spectra of the
starlight.Explain the meaning of line spectra.(2)Explain how line
spectra provide evidence for the existence of energy levels in
atoms.(3)(Total 5 marks)8.The following is a simplified energy
level diagram for atomic hydrogen.
State the ionisation energy of atomic
hydrogen.............................................................................................................................................................................................................................................................................................
Account for the labelling of the energy levels with negative
numbers.......................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................(3)Calculate
the wavelength of the photon emitted when an electron moves from
the1.51 eV energy level to the 3.40 eV energy
level.........................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................Wavelength
= ........................................(3)Describe how you would
produce a line spectrum of atomic hydrogen in a
laboratory.........................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................(2)Sketch
what you would expect to see.
(1)(Total 9 marks)
9.Below is a simplified energy level diagram for atomic
hydrogen.__________________0 eVfirst excited
state__________________3.4 eVground state__________________13.6 eV
(a)A free electron with 12 eV of kinetic energy collides with an
atom of hydrogen. As a result the atom is raised to its first
excited state. Calculate the kinetic energy of the free electron,
in eV, after the collision.
Kinetic energy = ...................... eV(2)(b)Calculate the
wavelength of the photon emitted when the atom returns to its
ground state.
Wavelength =................................(3)(Total 5
marks)10.The diagram shows the lowest three energy levels of atomic
hydrogen.(a)Excited hydrogen atoms can emit light of wavelength 656
nm. By means of a suitable calculation, determine which transition
between energy levels is responsible for this
emission..........................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................Transition:
from level ............ to level .............(4)
(b)The spectrum of light from the Sun contains a dark line at a
wavelength of 656 nm. With reference to the energy level diagram,
explain how this line is produced. You may be awarded a mark for
the clarity of your
answer.........................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................(4)(c)In
the spectrum of light received from another galaxy, the same line
appears at a wavelength of 695 nm.How can we deduce from this that
the galaxy is receding from the
Earth?..........................................................................................................................................................................................................................................................................(1)(Total
9 marks)
Task 311.A source of light emits a train of waves lasting 0.04
s. The light has a wavelength of 600 nm and the speed of light is 3
108 m s1. How many complete waves are sent out?A2.0 107B4.5 107C2.0
1010D4.5 1013(Total 1 mark)12.Electromagnetic waves are produced by
oscillating charges. Sound waves are produced by oscillating tuning
forks. How are these waves similar?Athey are both longitudinal
waves.Bthey are both transverse waves.Cthey both have the same
frequency as their respective sources.Dthey both require a medium
to travel through.(Total 1 mark)13.It has been suggested that
tigers use infrasound low frequency sounds inaudible to humans to
keep rivals away from their territory and to attract mates.Sound is
a longitudinal wave. Describe how sound travels through the
air...........................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................
(3)State what is meant by
frequency..............................................................................................................................................................................................................................................................................................................................................................................................................................................(1)The
frequency range of the sound produced by the tigers extends down to
18 Hz. Calculate the wavelength in air for sounds of this
frequency. Speed of sound in air = 330 m
s1..............................................................................................................................................................................................................................................................................................................................................................................................................................................Wavelength
= ....................................(2)(Total 6 marks)
14.The photograph shows a laboratory machine for illustrating a
sinusoidal transverse wave.Beside the machine is a rule. (You may
make marks on the photograph if you wish.)
Find a value for the wavelength of the
waves...............................................................................................................................................................................................................................................................................................Wavelength
= .......................................(2)Find a value for the
amplitude of the
waves...............................................................................................................................................................................................................................................................................................Amplitude
= ..........................................(2) The machine is
operated so that every rod along the wave moves up and down through
a complete cycle every 2 s. Calculate the frequency of the
waves...............................................................................................................................................................................................................................................................................................Frequency
= ............................................(1) What is meant by
a transverse
wave?.............................................................................................................................................................................................................................................................................................................................................................................................................................................(2)Discuss
whether this machine would be helpful in illustrating how a sound
wave
travels............................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................(3)(Total
10 marks)15.A loudspeaker emits a sound wave of wavelength 0.66 m.
The diagram shows how displacement varies with distance from the
loudspeaker at one instant of time.
(a)Which letter indicates the wavelength of the sound
wave?.....................................................................................................................................(1)
(b)Sound travels at 330 m s1 in air. Calculate the period of the
wave...........................................................................................................................................................................................................................................................................Period
= ..........................(3)(Total 4 marks)16.Two points on a
progressive wave differ in phase by radian. The distance between
them is 0.50 m. The frequency of the oscillations is 10 Hz. The
maximum speed of the wave isA2.50 m s1B5.00 m s1C12.5 m s1D40.0 m
s1(Total 1 mark)17.Which of the following statements about standing
waves is true?Aparticles between adjacent nodes all have the same
amplitude.Bparticles undergo no disturbance at an
antinode.Cparticles immediately either side of a node are moving in
opposite directions.Dparticles between adjacent nodes are out of
phase with each other.(Total 1 mark)
18.The diagram shows a wave on a rope. The wave is travelling
from left to right.
At the instant shown, point L is at a maximum displacement and
point M has zero displacement. Which row in the table correctly
describes the motion of points L and M during the next half cycle
of the wave?Point LPoint M
ARisesfalls
BRisesfalls then rises
Crises then fallsrises
Drises then fallsfalls then rises
(Total 1 mark)19.The following apparatus is set up. When the
frequency of the vibrator is 60 Hz, the standing wave shown in the
diagram is produced.
(a)What is the wavelength of this standing wave?Wavelength =
...................(1)(b)The frequency of the vibrator is altered
until the standing wave has two more nodes.Calculate the new
frequency.
Frequency = ................(2)(Total 3 marks)
Task 420.(a)The diagram shows three possible stationary waves on
a string of length 1.20 m stretched between fixed points X and Y.
(i)Wave A has a frequency of 110 Hz.Complete the table below to
show the wavelengths and frequencies of the three
waves.WaveWavelength / mFrequency / Hz
A110
B
C
(3) (ii)Each of the waves has nodes at X and Y. Explain why
these points must be
nodes.......................................................................................................................................................................................................................................................(1)(b)There
is a similarity between the behaviour of the string in part (a) and
that of the electron in a hydrogen atom. Electron states can be
represented by stationary waves which have to fit inside the atom.
Stationary waves with greater numbers of nodes represent electrons
in higher energy levels. Explain why this is the
case................................................................................................................................................................................................................................................................................................................................................................................................................
(2)(Total 6 marks)21.(a)Explain what is meant by the term
transverse wave. You may wish to illustrate your answer with the
help of a simple diagram.
....................................................................................................................................................................................................................................................................................................................................................................................................................(3)(b)State
two differences between a stationary wave and a progressive
wave.Difference 1
........................................................................................................................................................................................................................................................Difference
2
.......................................................................................................................................................................................................................................................(2)(c)Spiders
are almost completely dependent on vibrations transmitted through
their webs for receiving information about the location of their
prey. The threads of the web are under tension. When the threads
are disturbed by trapped prey, progressive transverse waves are
transmitted along the sections of thread and stationary waves are
formed.Early in the morning droplets of moisture are seen evenly
spaced along the thread when prey has been trapped.
(i)Explain why droplets form only at these
points.............................................................................................................................
(1) (ii)The speed of a progressive transverse wave sent by trapped
prey along a thread is 9.8 cm s1. Use the diagram to help you
determine the frequency of the stationary
wave.................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................Frequency
= ................................................(4)(Total 10
marks)
22.State two conditions necessary for total internal reflection
to occur at an interface between air and water.Condition 1
...........................................................................................................................................................................................................................................................................Condition
2
...........................................................................................................................................................................................................................................................................(Total
2 marks)23.(a)Light changes direction when it passes from air to
water.(i)Name the process of light changing direction in this
way............................................................................................................................(1)(ii)Explain
why this process takes
place.......................................................................................................................................................................................................................................................(1)
(b)The diagram represents some fish under water and a butterfly
above the water. (i)Draw a ray to show the path of light travelling
from the butterfly to the eye of fish B.(2)(ii)Explain what is
meant by critical
angle........................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................(2)(iii)Explain
how rays from fish A could reach the eye of fish B along two
different paths. Add rays to the diagram to illustrate your
answer..................................................................................................................................................................................................................................................................................................................................................................................(4)(Total
10 marks)24.(a)Explain what is meant by superposition of
waves.........................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................(2)(b)The
electron micrograph shows a small area of the surface of a copper
sample.
Electrons move over the surface and behave like waves. When a
wave reaches an edge or an atom, it reflects, and a standing wave
is formed. (i)Explain how a standing wave is formed by the
reflection of a
wave.............................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................
(3) (ii)A standing wave is visible on the micrograph between atoms
A and B. There are nodes at X and Y. The dark lines between X and Y
show antinodes. The distance between points X and Y on the
micrograph is 4.2 109 m. Use this information to calculate the
wavelength of the electron
waves.............................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................Wavelength
=
.................................................................(2)(iii)Explain
the meaning of the terms amplitude and
antinode.Amplitude.................................................................................................................................................................................................................................................................................................................................................................................(1)Antinode.................................................................................................................................................................................................................................................................................................................................................................................(1)(Total
9 marks)25.A ray of light travelling in air, strikes the middle of
one face of an equilateral glass prism as shown.State what happens
to the following properties as the light goes from the air into the
glass.Frequency
..............................................................................................................................Wavelength
............................................................................................................................Speed
.....................................................................................................................................(Total
3 marks)
Task 526.(a)Ultrasound images of the body are a useful
diagnostic tool for doctors. A single transducer can be used both
to send and receive pulses of ultrasound.The diagram shows a
lateral cross-section through part of the abdomen. The diagram is
not to scale.(i)Calculate the time interval between sending out a
single pulse and receiving its echo from interface B. The speed of
ultrasound in the abdominal wall is1500 m s1.
Time interval = .......................(3) (ii)The time between
pulses being emitted by the transducer is 200 s. At what frequency
are the pulses emitted?
Frequency = ................(2) (iii)The time interval before
the echo returns from interface D is 250 s. Suggest why this time
interval will make reflections from D difficult to interpret and
what could be done to overcome this
problem.............................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................(3)(iv)State
one reason why ultrasound rather than X-rays is now used to scan
expectant
mothers..................................................................................................................................................................................................................................................................................................................................................................................(1)(b)Ultrasound
is also used to measure blood flow in the body. It uses the Doppler
shift of the reflected pulse to measure the speed of blood through
the arteries of the body.Describe the principle of this method and
how it can be used to determine the speed of
blood....................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................(4)(Total
13 marks)27.(a)Explain with the aid of diagrams why transverse
waves can be polarised but longitudinal ones cannot be
polarised.
.........................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................(3)(b)Describe
with the aid of a diagram how you could demonstrate that light can
be polarised.
.........................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................(3)(Total
6 marks)28.When the Moon is full, bright moonlight makes it
difficult for astronomers to study the stars. Moonlight is
scattered by atoms in the atmosphere causing it to become plane
polarised.Draw labelled diagrams to show how the polarised light
differs from unpolarised light.
Polarised lightUnpolarised light(2)Explain how an astronomers
telescope could be adapted to overcome the problem of the bright
moonlight..............................................................................................................................................................................................................................................................................................................................................................................................................................................(2)(Total
4 marks)
29.The Raman effect can be used to identify the chemical
composition of solid materials in a non-destructive manner, e.g. to
identify fake diamonds. When laser light is shone on a material
some of the light is scattered. The scattered light will be plane
polarised and some of it will have been frequency shifted (that is,
it will have undergone a change in frequency). This happens when
the light interacts with the vibrating atoms in the materials.
Analysis of these Raman frequency shifts reveals the chemical
composition of the material.What is meant by plane polarised
light?.............................................................................................................................................................................................................................................................................................................................................................................................................................................(2)Some
of the light will have been frequency shifted by the moving atoms.
Name the phenomenon which caused the frequency shifting and briefly
explain how it
arises..........................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................(4)
The Raman frequency shifts are often measured in wavenumbers.
Wavenumber is defined as .When laser light of wavelength 1064 nm is
shone onto diamond, the scattered light has a Raman wavenumber
shift of +1300 cm1. Show that the frequency of this scattered light
is about 3 1014
Hz.........................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................(5)
A student correctly remarks that if the frequency of the laser
light has changed then its energy must have changed.State which
model of light is being used by this student, and explain what must
have happened to the vibrating atoms in the diamond during this
interaction...........................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................(3)(Total
14 marks)
30.Explain what is meant by the term polarisation when referring
to
light.............................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................(2)Sugar
is produced from plants such as sugar cane. The stems are crushed
and the juice extracted. The concentration of sugar in the juice is
used to value the crop.The concentration can be determined using
polarised light.Explain how to measure the angle of rotation of
polarised light when it passes through a sugar
solution..........................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................
(4)A student has carried out this experiment and obtains three
results. He has plotted them on the graph below. He takes three
more results and tabulates them.Angle of
rotation/degreesConcentration of solution/kg per litre
170.25
330.50
500.75
Add these results to the graph.(3)
Use your graph to determine the concentration of an unknown
sample which gives a rotation of38.Concentration:
....................................................................................................
kg per litre(1)(Total 10 marks)31.A photographer uses a polarising
filter over the camera lens. She notices that the intensity of the
light received from the blue sky changes as she rotates the
filter.What does this suggest about light from the
sky?...............................................................................................................................................
(1)Explain the change in intensity as the filter is
rotated...............................................................................................................................................................................................................................................................................................
(2) The use of a polarising filter makes a blue sky appear darker,
but the clouds remain bright.Suggest why there is little change in
the intensity of the light from the
clouds..............................................................................................................................................................................................................................................................................................................................................................................................................................................(1)
Astronomers notice the same effect with the radio waves emitted by
some galaxies.What does this suggest about these radio
waves?..............................................................................................................................................................................................................................................................................................(1)State
why radio waves should behave in the same way as
light...............................................................................................................................................................................................................................................................................................(1)(Total
6 marks)32.A food packaging factory is moving soup through a 0.075
m diameter pipe when an obstruction occurs in the pipe. An
ultrasound probe, connected to an oscilloscope, is moved along the
pipe to find the obstruction (figure 1). The oscilloscope trace is
shown below(figure 2).Figure 1 Figure 2
Oscilloscope time base = 20 106 s cm1.
On figure 2, pulse A is the outgoing signal from the probe and
pulse B is the reflected signal from the other side of the
pipeCalculate the speed of the ultrasound in the liquid in the
pipe................................................................................................................................................Speed
= .............................................................(2)
State one way in which the oscilloscope trace will change when the
ultrasound probe is above the
obstruction..............................................................................................................................................................................................................................................................................................................................................................................................................................................(1)After
the obstruction has been cleared, a Doppler ultrasound probe is
used to measure the speed of the soup in the pipe. Describe the
principle of this
method.........................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................(3)What
must be measured to determine the speed of the
soup?..............................................................................................................................................................................................................................................................................................(1)
Someone says that this would be easier if the soup contained lumps
like vegetables. Comment on this
suggestion..............................................................................................................................................................................................................................................................................................................................................................................................................................................(1)(Total
8 marks)
1.(a)Part of spectrumLight / Visible / red (1)1Calculation of
work functionUse of = hc/ (1)3.06 1019 [2 sig fig minimum]
(1)2(6.63 1034 J s)(3.00 108 m s1)/(6.5 107 m)= 3.06 1019
J(b)(i)Meaning of stopping potentialMinimum potential difference
between C and A / across thephotocell (1)Which reduces current to
zero OR stops electrons reaching A /crossing the gap / crossing
photocell (1)2(ii)Why the graphs are parallelCorrect rearrangement
giving Vs = hf/e /e (1)Gradient is h/e which is constant / same for
each metal (1)[Second mark can be awarded without the first if
norearrangement is given, or if rearranged formula is wrong butdoes
represent a linear graph with gradient h/e]2[7]2.(a)Meaning of
statement(5.89 1019 J / work function) is the energy needed to
remove anelectron [allow electrons] from the (magnesium)
surface/plateConsequent markMinimum energy stated or indicated in
some way [e.g. at least /or more] (1)2 (b)(i)Calculation of timeUse
of P = IA (1)Use of E = Pt (1)[use of E = IAt scores both
marks]Correct answer [210 (s), 2 sig fig minimum, no u.e.]
(1)[Reverse argument for calculation leading to either
intensity,energy or area gets maximum 2 marks]Example calculation:t
= (5.89 1019 J)/(0.035 W m2 8 1020 m2)3(ii)How wave-particle
duality explains immediate photoemissionQOWC (1)Photon energy is hf
/ depends on frequency / depends on wavelength (1)One/Each photon
ejects one/an electron (1)The (photo)electron is ejected at
once/immediately (1)[not just photoemission is
immediate]4[9]3.(a)Solar PowerUse of P = Ir2 [no component needed
for this mark] (1)Use of cos 40 or sin 50 (with I or A) (1)2.2 [2
sf minimum. No ue] (1)3e.g. P=1.1 103 W m2 cos 40 (29 103 m)2= 2.2
W(b)EnergyUse of E = Pt (1)1.8 104 J/2.0 104J (1)2e.g. E = 2.2W
(2.5 3600 s)= 2.0 104 J[5]
4.Photoelectric effect(a)Explanation:Particle theory: one photon
(interacts with) one electron (1)Wave theory allows energy to build
up, i.e. time delay (1)2(b)Explanation:Particle theory: f too low
then not enough energy (is released byphoton to knock out an
electron) (1)Wave theory: Any frequency beam will produce enough
energy (to release an electron, i.e. should emit whatever the
frequency) (1)2[4]5.(a)Why statement correctBlue photon has more
energy than red photon (1)Why statement incorrectBlue beam carries
less energy per unit area per second / Blue beamcarries less energy
per second / Blue beam carries less energy perunit area / Blue beam
has lower intensity and intensity = energy per unitarea per
secondAdditional explanation[Under correct] Blue has a higher
frequency (OR shorter wavelength) /[Under incorrect] Blue beam has
fewer photons (1)[Allow reverse statements about Red throughout
part a]3(b)(i)Meaning of work functionEnergy to remove an electron
from the surface (ORmetal OR substance) (1)[Dont accept from the
atom. Dont accept electrons.]Minimum energy / Least energy / Energy
to just/ without giving the electron any kinetic energy
(1)2(ii)Calculation of threshold frequencyUse of = hf0 (1)Correct
answer [6.00 1014 Hz] (1)e.g.(3.98 1019 J)/(6.63 1034 J s) = 6.00
1014 Hz2[7]6.(a)WavelengtheV to J (1)Use of E = hf (1)Use of c = f
(1)1.8 1011 [2 sf minimum. No ue] (1)4e.g. f =(1.8 keV ( 69.6 keV))
(103 1.6 1019 J keV1) / 6.6 1034 J s= 1.64 1019 Hz = 3.00 108 m
s1/1.64 1019Hz= 1.8 1011 m(b)TypeX rays [Accept gamma rays]
(1)1[5]7.Explanation of line spectra:Specific frequencies or
wavelengths (1)Detail, e.g. absorption/emission (1)OR within narrow
band of wavelengths2Explanation how line spectra provide evidence
for existence or energy levels in atoms:Photons (1)Associated with
particular energies (1)Electron transitions (1)Discrete levels (to
provide line spectra) (1)3[5]
8.Ionisation energy of atomic hydrogen:13.6 eV OR 2.18 1018 J [
sign, X](1)1Why energy levels are labelled with negative
numbers:Work/energy is needed to raise the electrons/atoms to an
energy of 0 eV, somust start negative(1)(1)ORWork/energy is given
out when the electrons/atoms move to the groundstate, so energy now
less than 0, i.e. negative(1)(1)ORthe ground state is the most
stable/lowest energy level of theelectrons/atoms and must be less
than 0, i.e. negative(1)(1)2[1st mark essential: e
highest/maximum/surface/ionised/free has energy = 0eV2nd mark:
raising levels means energy in OR falling levels means energy out
negative levels]Wavelength of photon:E = 1.89 (eV)(1)Convert E to
joules, i.e. (1.6 1019)OR = [Their E](1)= 6.6 107 (m) [6.5
6.7](1)3Production of line spectrum of atomic hydrogen in a
laboratory:Source hydrogen discharge tube/hydrogen lamp/low p
hydrogen with high V across(1)(view through) diffraction
grating/prism/spectrometer/spectroscope(1)2 Sketch:
A few vertical lines on a blank background OR sharp bandsDark on
light/light on dark NOT equally spaced(1)1Absorption spectrum:White
light through gas in container (1)Diffraction
grating/prism/spectrometer (1)Must be dark lines on bright
background (1)[9]9.(a)Calculation of energy required by atom
(1)Answer [1.8 (eV)] (1)Example of answer:Energy gained by atom =
13.6 eV 3.4 eV = 10.2 eVKE of electron after collision = 12 eV 10.2
eV = 1.8 eV2(b)Use of E = hf and c = f (1)Conversion of eV to
Joules (1)Answer = [1.22 107 m] (1)Example of answerE = hf and c =
f E = hc/ = (6.63 1034 J s 3 108 m s2) (10.2 eV 1.6 1019 C) = 1.21
107 m3[5]
10.(a)Which transitionUse of ()E = hc/ OR ()E = hf and f = c/
(1)Use of 1.6 1019 (1)Correct answer [1.9 eV] (1)C to B / 1.5 to
3.4 (1)[Accept reverse calculations to find wavelengths]e.g.(6.63
1034 J s)(3.00 108 m s1)/(656 109 m)(1.6 1019 J eV1)= 1.9
eV4(b)Explanation of absorption lineQOWC (1)Light of this
wavelength is absorbed by hydrogen (1)In the outer part of the Sun
(OR Suns atmosphere) (1)Absorbed radiation is reemitted in all
directions (1)Transition from B to C (OR 3.4 to 1.5) (1)Max 4(c)Why
galaxy recedingWavelength increased (OR stretched) / red shift
/frequency decreased1[9]11.A[1]12.C[1]13.Description of
soundParticles/molecules/atoms oscillate/vibrate (1)(Oscillations)
parallel to/in direction of wave propagation / wavetravel / wave
movement [Accept sound for wave] (1)Rarefactions and compressions
formed [Accept areas of high and low pressure] (1)3 Meaning of
frequencyNumber of oscillations/cycles/waves per second / per unit
time (1)1 Calculation of wavelengthRecall v = f (1)Correct answer
[18 m] (1)2Example of calculationv = f = 330 m s1 18 Hz= 18.3
m[6]14.Value of wavelength = 13.9 cm 0.5 cm (using interpolated
sine curve) (1)= 13.4 cm [accept 13.2 to 13.6 cm] (1)2[12.3 to 12.5
cm for distance using rods (1) ]Value of amplitudePeak to peak =
4.5 cm [Accept 4.3 cm to 4.7 cm] (1)Amplitude = peak to peak= 2.25
cm [Accept 2.15 cm to 2.35 cm] [Allow ecf for 2nd mark if (1)2first
part shown]Calculation of frequencyf = 1/T= 1 2 s = 0.5 Hz
(1)1Explanation of why waves are
transverseOscillation/vibration/displacement/disturbance at right
angle (1)to direction of propagation/travel of wave
(1)2[Oscillation not in direction of wave (1)] Description of use
of machine to illustrate sound waveSound is longitudinal/not
transverse (1)with oscillation along the direction of propagation /
compressions and rarefactions (1)so model not helpful
(1)3[10]15.(a)D1 (b)WavelengthUse of v = f (1)Use of f = 1/T
(1)Answer T = [0.002 s] (1)[give full credit for candidates who do
this in 1 stage T = /v]Example of answerv = ff = 330 / 0.66T = 1/f
= 0.66 / 330T = 0.002 s3[4]16.D[1]17.C[1]18.B[1]19.(a)[1.0 m] (1)1
(b)Ratio of (5 or 6 / 3 ) 60 (1)Answer [f = 100 Hz]
(1)2[3]20.(a)(i)Tablef2.4(110)1.22200.8330All wavelengths correct
(2)[One or two wavelengths correct gets 1]Both frequencies correct
(1)[Accept extra zero following wavelength figure, e.g. 2.40.Accept
units written into table, e.g. 2.4 m, 220 Hz]3(ii)Why nodesString
cannot move / no displacement / zero amplitude /no oscillation /
phase change of on reflection / two wavescancel out / two waves are
exactly out of phase (1)(OR have phase difference of OR half a
cycle) /destructive interference1 (b)Why waves with more nodes
represent higher energiesMore nodes means shorter wavelength
(1)Momentum will be larger (1)[OR Allow 1 mark for More nodes means
higher frequency and E = hf]2[6]
21.(a)Transverse wave(Line along which) particles/em field
vectors oscillate/vibrate (1)Perpendicular to (1)Direction of
travel or of propagation or of energy flow or velocity
(1)3(b)DifferencesAny two:Standing wavesProgressive waves1. store
energy1. transfer energy (1)2. only AN points have max2. all have
the max ampl/displ (1) ampl/displ3. constant (relative) phase3.
variable (relative) phaserelationshiprelationship (1)Max
2(c)(i)DropletsFormed at nodes / no net displacement at these
points (1)1(ii)SpeedUse of = f (1)Evidence that wavelength is twice
nodenode distance (1)Wavelength = 1.2 (cm) (1)Frequency = 8.0 [8.2
/ 8.16] Hz or s1 only (1)4[10]22.Direction of travel of light is
water air (1)Angle of incidence is greater than the critical angle
(1)2[2]23.(a)(i)Name processRefraction (1)1(ii)Explanation of
refraction taking placechange in speed / density / wavelength
(1)1(b)(i)Draw ray from butterfly to fishrefraction shown
(1)refraction correct (1)2(ii)Explain what is meant by critical
angleIdentify the angle as that in the denser medium (1)Indicate
that this is max angle for refraction OR total internalreflection
occurs beyond this (1)[angles may be described in terms of relevant
media]2(iii)Explain two paths for rays from fish A to fish Bdirect
path because no change of medium/refractive index/density (1)(total
internal) reflection along other path /angle of incidence >
critical angle (1)direct ray correctly drawn with arrow (1)total
internal reflection path correctly drawn with arrow (1)[lack of
ruler not penalised directly] [arrow penalised once
only]4[10]24.(a)Meaning of superpositionWhen
vibrations/disturbances/waves from 2 or more sources coincideat
same position (1)resultant displacement = sum of displacements due
to individual waves (1)2(b)(i)Explanation of formation of standing
wavedescription of combination of incident and reflected
waves/waves in opposite directions (1)described as superposition or
interference (1)where in phase, constructive interference /
antinodesOR where antiphase, destructive interference / nodesOR
causes points of constructive and destructive interferenceOR causes
nodes and antinodes (1)3(ii)Calculate wavelengthIdentify 2
wavelengths (1)Correct answer [2.1 109 m] (1)2Example of
calculation:(NANANANAN) X to Y is 2 = 4.2 109 m 2= 2.1 109
m(iii)Explain termsamplitude maximum displacement (from mean
position)(can use diagram with labelled displacement axis)
(1)antinode position of maximum amplitudeOR position where waves
(always) in phase (1)2[9]25.Frequency unaltered (1)Wavelength
decreases (1)Speed decreases (1)3[3]26.(a)(i)Use of speed =
distance over time (1)Distance = 4 cm (1)Answer = [2.7 105 s]
(1)Example of answert = 4 cm 1500 m s1t = 2.7 105 s3(ii)Use of f =
1/T (1)Answer = [5000 Hz] (1)2(iii)Time for pulse to return greater
than pulse interval (1)All reflections need to reach transducer
before next pulse sent. (1)Will result in an inaccurate image. (1)
(Max 2)Need to decrease the frequency of the ultrasound. (1) (Max
3)Max 3(iv)X-rays damage cells/tissue/foetus/baby but ultrasound
doesnot (need reference to both X-rays and ultrasound) (1)1 (b)The
answer must be clear, use an appropriate style and be organisedin a
logical sequence (QWC)Doppler shift is the change in frequency of a
wave when the source or thereceiver is moving (1)Requirement for a
continuous set of waves (1)Two transducers required (one to
transmit and one to receive) (1)Change in frequency is directly
related to the speed of the blood (1)4[13]27.(a)Transverse waves
oscillate in any direction perpendicular to wave direction
(1)Longitudinal waves oscillate in one direction only OR parallel
to wavedirection. (1)Polarisation reduces wave intensity by
limiting oscillations and wavedirection to only one plane OR
limiting oscillations to one direction only. (1)(accept vibrations
and answers in terms of an example such as a ropepassing through
slits)3(b)Light source, 2 pieces of polaroid and detector e.g. eye,
screen, LED ORlaser, 1 polaroid and detector (1)Rotate one polaroid
(1)Intensity of light varies (1)3[6]28.Unpolarised and plane
polarised light:Correct diagrams showing vibrations in one plane
only and in all planes (1)
Vibrations/oscillations labelled on diagrams (1)2 Telescope
adaptation:Fit polarising filter / lens [must be lens not lenses]
(1)At 90 to polarisation direction to block the moonlight / rotate
until2cuts out moonlight (1)[4]29.Meaning of plane
polarisedOscillations/vibrations/field variations (1)Parallel to
one direction, in one plane [allow line with arrow at both ends]
(1)2Doppler effectDoppler (1)If source/observer have (relative)
movement [reflections off vibrating/moving atoms] (1)Waves would be
bunched/compressed/stretched or formula quoted [accept diagram]
(1)Thus frequency / wavelength changes [accept red /blue shift]
(1)4Frequency about 3 1014 HzEvidence of use of 1/wavelength =
wavenumber (1)laser wavenumber = 9400 or wavelength change =7.69104
(1)New wavenumber = 10700 [or 8100] or conversion of wavelength
change to m [7.69 106] (1)New wavelength = 935 nm [or 1240 nm] Use
of frequency = c / wavelength [in any calculation] (1)f = 3.2 1014
Hz [note answer of 2.8 1014 = 3 , 3.4 1014 = 4](1)5Model of
lightParticle/photon/quantum model (1)Photon energy must have
changed / quote E = hf (1)Energy of atoms must have changed [credit
vibrating less/more/faster/slower] (1)3[14]30.PolarisationThe
(wave) oscillations (1)occur only in one plane (1)2[OR shown with a
suitable diagram]How to measure angle of rotationAny four points
from:Polaroid filter at one/both endswith no sugar solution,
crossed Polaroids (top and bottom oftube) block out lightsugar
solution introduced between Polaroidsone Polaroid rotated to give
new dark viewdifference in angle between two positions read from
scale (1) (1) (1) (1)Max 4 GraphPoints plotted correctly [1 for
each incorrect; minimum mark 0] (1) (1)Good best fit line to enable
concentration at 38 to be found (1)3Concentration0.57 ( 0.01) kg
l11[10]31.Light from sky:Light is polarised (1)1Change in
intensity:Filter allows through polarised light in one direction
only (1)When polarised light from the sky is aligned with filter,
light is let through (1)When polarised light is at right angles
with polarising filter, less light passes (1)Turn filter so that
polarised light from blue sky isnot allowed through, so sky is
darker (1)Max 2 Clouds:Light from clouds must be unpolarised
(1)1Radio waves:Radio waves can be polarised OR transverse (1)1 Why
radio waves should behave in same way as light:Both are
electromagnetic waves/transverse (1)[Transverse only, credited for
1 answer]1[6]32.Speed of ultrasoundUse of = s/t (1) = 150 103 (m)
132 106 (s)= 1140 m s1 (1)2 Change of traceExtra
pulse(s)ORReflected pulse moves closer1Principle of Doppler probe3
points from: Arrange probe so that soup is approaching OR Soup
reflects ultrasound /with changed frequency/wavelength OR change in
frequency/wavelength depends on speed OR Probe detects frequency of
reflected ultrasound OR Use of diagrams showing waves3
Determination of speed1 point from:Frequency/wavelength change
/Angle between ultrasound direction and direction of flow of
soup1CommentLumps give larger reflections/ Lumps travel slower1
[8]