IB Physics Course Book Oxford IB Diploma Program

8.2 THERMAL ENERGY TRANSFER

√___T =
So power = 5.7 × 10-8 × 6.2 × 1018 × 58004 4 _________4_8_________ = 1200 K.
5.7 ×10-8 × 3.8 × 10-4
= 4.0 × 1026 W
3 A spherical black body has an absolute
In one hour there are 3600 s, so the energy temperature T1 and surface area A.
radiated in one hour is 1.4 × 1030 J. Its surroundings are kept at a lower
temperature T2.
2 A metal filament used as a pyrometer in a
kiln has a length of 0.050 m and a radius of Determine the net power lost by the body.
1.2 × 10–3 m. Determine the temperature of
the filament at which it radiates a power Solution
of 48 W.
The power emitted by the body is; σAT14
Solution
the power absorbed from the surroundings
The surface area of the filament is 2πrh = is σAT24.
(2π × 1.2 ×10-3) × 0.050 = 3.8 × 10–4 m2
So the net power lost is σA(T14 - T24).
So the power determines the temperature as
Note that this is not the same as
48 = 5.7 × 10-8 × 3.8 × 10–4 × T 4
σA(T1 - T2)4.

Sun and the solar constant

The Sun emits very large amounts of energy as a result of its nuclear
fusion reactions. Because the Earth is small and a long way from the Sun,
only a small fraction of this arrives at the top of the Earth’s atmosphere.
A black body at the temperature of the Sun has just under half of its
radiation in our visible region, roughly the same amount in the infra-
red, and 10% in the ultraviolet. It is the overall difference between this
incoming radiation and the radiation that is subsequently emitted from
the Earth that determines the energy gained by the Earth from the Sun.
This energy is used by plants in photosynthesis and it drives the changes
in the world’s oceans and atmospheres; it is crucial to life on this planet.

The amount of energy that arrives at the top of the atmosphere is known
as the solar constant. A precise definition is that the solar constant is the
amount of solar radiation across all wavelengths that is incident in one second
on one square metre at the mean distance of the Earth from the Sun on a plane
perpendicular to the line joining the centre of the Sun and the centre of the Earth.

The energy from the Sun is spread over an imaginary sphere that has a
radius equal to the Earth–Sun distance. The Earth is roughly 1.5 × 1011 m
from the Sun and so the surface area of this sphere is 2.8 × 1023 m2.

The Sun emits 4 × 1026 J in one second. The energy incident in one

second on one square metre at the distance of the Earth from the radiation
pSruenciissio_24n_..80_f××_o_r11_00_t22_h36 i=s from Sun
1400 J. The answer is quoted to 2 s.f., a reasonable
estimate and represents about 5 × 10-10 of the entire one square metre at
top of atmosphere
output of the Sun. ▲ Figure 11 Defining the solar constant.

The value of the solar constant varies periodically for a number of reasons: 341

● The output of the Sun varies by about 0.1% during its principal 11-year
sunspot cycle.

● The Earth’s orbit is elliptical with the Earth slightly closer to the Sun
in January compared to July; this accounts for a difference of about

8 ENERGY PRODUCTION

7% in the solar constant. (Note that this difference is not the reason
for summer in the Southern Hemisphere – the seasons occur because
the axis of rotation of the Earth is not perpendicular to the plane of its
orbit around the Sun.)

● Other longer-period cycles are believed to occur in the Sun with
periods ranging from roughly hundreds to thousands of years.

Energy balance in the Earth surface–
atmosphere system

The solar constant is the power incident on the top of the atmosphere.
It is not the power that arrives at ground level. As the radiation from
the Sun enters and travels through the atmosphere, it is subject to
losses that reduce the energy arriving at the Earth’s surface. Radiation
is absorbed and also scattered by the atmosphere. The degree to which
this absorption and scattering occurs depends on the position of the Sun
in the sky at a particular place. When the Sun is lower in the sky (as at
dawn and sunset), its radiation has to pass through a greater thickness
of atmosphere and thus more scattering and absorption takes place. This
gives rise to the colours in the sky at dawn and dusk.

Even when the energy arrives at ground level, it is not necessarily going
to remain there. The surface of the Earth is not a black body and therefore
it will reflect some of the energy back up towards the atmosphere. The
extent to which a particular surface can reflect energy is known as its
albedo (from the Latin word for “whiteness”). It is given the symbol a:

a = _energy re_flected by a_given sur_face in a gi_ven time
total energy incident on the surface in the same time

Like emissivity, albedo has no units, it varies from 0 for a surface that
does not reflect any energy (a black body) to 1 for a surface that absorbs
no radiation at all. Unless stated otherwise, the albedo in the Earth
system is normally quoted for visible light (which as we saw earlier
accounts for nearly a half of all radiation at the surface).

The average annual albedo for the whole Earth is about 0.35, meaning
that on average about 35% of the Sun’s rays that reach the ground are
reflected back into the atmosphere.

This figure of 0.35 is, however, very much an average because albedo
varies depending on a number of factors:

Surface Albedo ● It varies daily and with the seasons, depending on the amount and
Ocean 0.06 type of cloud cover (thin clouds have albedo values of 0.3–0.4, thick
Fresh snow 0.85 cumulo-nimbus cloud can approach values of 0.9).
Sea ice 0.60
Ice 0.90 ● It depends on the terrain and the material of the surface. The table
Urban areas 0.15 gives typical albedo values for some common land and water surfaces.
Desert soils 0.40
Pine forest 0.15 The importance of albedo will be familiar to anyone who lives where
Deciduous forest 0.25 snow is common in winter. Fresh snow has a high albedo and reflects
most of the radiation that is incident on it – the snow will stay in
place for a long time without melting if the temperature remains cold.
However, sprinkle some earth or soot on the snow and, when the sun
shines, the snow will soon disappear because the dark material on its

342

8.2 THERMAL ENERGY TRANSFER

surface absorbs energy. The radiation provides the latent heat needed to
melt the snow. Albedo effects help to explain why satellites (including
the International Space Station) in orbit around the Earth can take
pictures of the Earth’s cloud cover and surface in the visible spectrum.

Worked examples Data

1 Four habitats on the Earth are: forest, Incident intensity from the Sun = 340 W m–2
grassland (savannah), the sea, an ice cap.
Reflected intensity at surface = 100 W m–2
Discuss which of these have the greatest and
least albedo. Radiated intensity from surface = 240 W m–2

Solution Re-radiated intensity from = 2 W m–2
atmosphere back to surface
A material with a high albedo reflects the incident
visible radiation. Ice is a good reflector and Solution
consequently has a high albedo. On the other hand,
the sea is a good absorber and has a low albedo. The definition of albedo is clear.

2 The data give details of a model of the energy It is _p_o_w_e_r _re_fl_e_ct_e_d_b_y_a_g_iv_e_n_s_u_rf_a_ce_
balance of the Earth. Use the data to calculate
the albedo of the Earth that is predicted by total power incident on the surface
this model.
So in this case the value is _1_00_ = 0.29

340

The greenhouse effect and temperature balance

The Earth and the Moon are the same average distance from the Sun,
yet the average temperature of the Moon is 255 K, while that of the
Earth is about 290 K. The discrepancy is due to the Earth having an
atmosphere while the Moon has none.

The difference is due to a phenomenon known as the greenhouse
effect in which certain gases in the Earth’s atmosphere trap energy
within the Earth system and produce a consequent rise in the average
temperature of the Earth. The most important gases that cause the effect
include carbon dioxide (CO2), water vapour (H2O), methane (CH4), and
nitrous oxide (dinitrogen monoxide; N2O), all of which occur naturally
in the atmosphere. Ozone (O3), which has natural and man-made
sources, makes a contribution to the greenhouse effect.

It is important to distinguish between:

● the “natural” greenhouse effect that is due to the naturally occurring
levels of the gases responsible, and

● the enhanced greenhouse effect in which increased concentrations of
the gases, possibly occurring as a result of human-derived processes,
lead to further increases in the Earth’s average temperature and
therefore to climate change.

The principal gases in the atmosphere are nitrogen, N2, and oxygen, O2,
(respectively, 70% and 20% by weight). Both of these gases are made up
of tightly bound molecules and, because of this, do not absorb energy
from sunlight. They make little contribution to the natural greenhouse
effect. The 1% of the atmosphere that is made up of the CO2, H2O, CH4
and N2O has a much greater effect.

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8 ENERGY PRODUCTION

Nature of The molecular structure of greenhouse-gas molecules means that they
science absorb ultraviolet and infra-red radiation from the Sun as it travels
through the atmosphere. Visible light on the other hand is not so
Other worlds, other readily absorbed by these gases and passes through the atmosphere
atmospheres to be absorbed by the land and water at the surface. As a result the
temperature of the surface rises. The Earth then re-radiates just like
The dynamic equilibrium in any other hot object. The temperature of the Earth’s surface is far lower
our climate has been very than that of the Sun, so the wavelengths radiated from the Earth will
important for the evolution peak in the long-wavelength infra-red. The absorbed radiation had,
of life on Earth. Venus of course, mostly been in the visible region of the electromagnetic
and Mars evolved very spectrum. So, just as gases in the atmosphere absorbed the Sun’s infra-
differently from Earth. red on the way in, now they absorb energy in the infra-red from the
Earth on the way out. The atmosphere then re-radiates the energy yet
Venus has similar again, this time in all directions meaning that some returns to Earth.
dimensions to the Earth This energy has been trapped in the system that consists of the surface
but is closer to the of the Earth and the atmosphere.
Sun with a very high
albedo at about 0.76. It’s The whole system is a dynamic equilibrium reaching a state where
atmosphere is almost the total energy incident on the system from the Sun equals the
entirely carbon dioxide. energy total being radiated away by the Earth. In order to reach this
In consequence, the surface state, the temperature of the Earth has to rise and, as it does so, the
temperature reaches a amount of energy it radiates must also rise by the Stefan–Boltzmann
730 K and a runaway law. Eventually, the Earth’s temperature will be such that the balance
greenhouse effect acts. of incoming and outgoing energies is attained. Of course, this balance
Venus and Mars are clear was established over billions of years and was steadily changing as the
reminders to us of the composition of the atmosphere and the albedo changed with changes in
fragility of a planetary vegetation, continental drift, and geological processes.
climate.
Why greenhouse gases absorb energy

Ultraviolet and long-wavelength infra-red radiations are absorbed by the
atmosphere.

Photons in the ultraviolet region of the electromagnetic spectrum
are energetic and have enough energy to break the bonds within the
gas molecules. This leads to the production of ionic materials in the
atmosphere. A good example is the reaction that leads to the production
of ozone from the oxygen atoms formed when oxygen molecules are
split apart by ultraviolet photons.

The energies of infra-red photons are much smaller than those of
ultraviolet and are not sufficient to break molecules apart. When the
frequency of a photon matches a vibrational state in a greenhouse gas
molecule then an effect called resonance occurs. We will look in detail
at the vibrational states and resonance in carbon dioxide, but similar
effects occur in all the greenhouse gas molecules.

In a carbon dioxide molecule, the oxygen atoms at each end are attached
by double bonds to the carbon in a linear arrangement. The bonds
resemble springs in their behaviour.

The molecule has four vibrational modes as shown in figure 12. The
first of these modes – a linear symmetric stretching does not cause
infrared absorption, but the remaining three motions do. Each one has

344

8.2 THERMAL ENERGY TRANSFER
O C O equilibrium position

symmetric stretching bending
anti-symmetric stretching modes

▲ Figure 12 Vibrational states in the carbon dioxide molecule.

a characteristic frequency. If the frequency of the radiation matches this, relative absorption
then the molecule will be stimulated into vibrating with the appropriate
mode and the energy of the vibration will come from the incident 23456
radiation. This leads to vibrational absorption at wavelengths of 2.7 µm, wavelength/µm
4.3 µm and 15 µm.
▲ Figure 13 Part of the absorption
These effects of these absorptions can be clearly seen in figure 13 spectrum for carbon dioxide.
which shows part of the absorption spectrum of carbon dioxide.
In this diagram a peak indicates a wavelength at which significant
absorption occurs.

Modelling the climate balance

We said earlier that about 1400 J falls on each square metre of the
upper atmosphere each second: the solar constant. We use the physics
introduced in this topic to see what the consequences of this are for the
Earth’s surface–atmospheric system.

The full 1400 J does not of course reach the surface. Of the total, about
25% of the incident energy is reflected by the clouds and by particles in
the atmosphere, about 25% is absorbed by the atmosphere, and about
6% is reflected at the surface.

The incoming radiation falls on the portion of the Earth’s surface which

is normal to the Sun’s radiation – i.e. a circle of area equal to π × (radius

of Earth)2, as only one side of the Earth faces the Sun at any one time.

However this radiation has to be averaged over the whole of the surface

which is 4π × (radius of Earth)2. So the mean power arriving at each
_1_4_00_
square metre is = 350 W.
4

The albedo now has to be taken into account to give an effective mean

power at one square metre of the surface of

(1 – a) × 350

For the average Earth value for a of about 0.3, the mean power absorbed
by the surface per square metre is 245 W.

345

8 ENERGY PRODUCTION

energy intensity The knowledge of this emitted power allows a prediction of the
(arbitrary units) 256 K temperature of a black body T that will emit 245 W m–2. Using the

Stefan–Boltzmann law:

245 = σT4

0 wavelength √So T = 4 ____2_4_5____ = 256 K
5.67 ×10-8
100%
transmittance This is very close to the value for the Moon, which has no atmosphere.
0 wavelength We need to investigate why the mean temperature of the Earth is about
▲ Figure 14 Intensity and 35 K higher than this.

transmittance for a completely We made the assumption that the Earth emits 245 W m-2 and that
transparent atmosphere. this energy leaves the surface and the atmosphere completely.
This would be true for an atmosphere that is completely transparent at
(a) 256 K all wavelengths, but Earth’s atmosphere is not transparent in this way.
energy intensity
(arbitrary units) energy Figure 14 shows the relative intensity–wavelength graph for a black
transmitted wavelength body at 256 K. As expected, the area under this curve is 245 W m–2
0 and represents the predicted emission from the Earth. It shows the
response of an atmosphere modelled as perfectly transparent at all
transmittance 100% energy not wavelengths. (Technically, this graph shows the transmittance of the
energy not transmitted atmosphere as a function of wavelength, a value of 100% means that
transmitted the particular wavelength is completely transmitted, 0 means that no
energy is transmitted at this wavelength.) Not surprisingly all the black-
0energy intensity wavelength body radiation leaves the Earth because the transmittance is 1 for all
(b) (arbitrary units) 290 K wavelengths in this model.
256 K
0 In fact the atmosphere absorbs energy in the infra-red and ultraviolet
wavelength regions. A simple, but slightly more realistic model for this absorption
will leave the transmittance at 100% for the visible wavelengths and
100%transmittance change the transmittance to 0 for the absorbed wavelengths. Figure 15
shows how this leads to an increased surface temperature.
0 wavelength
▲ Figure 15 Intensity and transmittance When the transmittance graph is merged with the graph for black-
for an atmosphere opaque to infra-red body radiation to give the overall emission from the Earth into space,
and ultraviolet radiation. the area under the overall emission curve will be less than 245 W m–2
because the infra-red and ultraviolet wavelengths are now absorbed
in the atmosphere and these energies are not lost (Figure 15(a)).
This deficit will be re-radiated in all directions; so some returns to
the surface.

In order to get the energy balance correct again, the temperature of the
emission curve must be raised so that the area under the curve returns to
245 W m–2. As the curve changes with the increase in temperature, the
area under the curve increases too. The calculation of the temperature
change required is difficult and not given here. However, for the emission
from the surface to equal the incoming energy from the Sun, allowing
for the absorption, the surface temperature must rise to about 290 K.
The net effect is shown in Figure 15(b) with a shifted and raised emission
curve compensating for the energy that cannot be transmitted through
the atmosphere.

The suggestion that the atmosphere completely removes wavelengths
above and below certain wavelengths is an over-simplification.

346

8.2 THERMAL ENERGY TRANSFER

Figure 16 shows the complicated transmittance pattern in the infrared
and indicates which absorbing molecules are responsible for which
regions of absorption.

transmittance (percent)100
80
60
40
20

0
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
wavelength/μm

O2 H2O CO2 H2O CO2 O3 H2O CO2 O3 H2O CO2 CO2
absorbing molecule

▲ Figure 16 Transmittance of the atmosphere in the infra-red.

The energy balance of the Earth

The surface–atmosphere energy balance system is very complex;
figure 17 is a recent diagram showing the basic interactions and you
should study it carefully.

global energy flows W m-2

102 reflected solar radiation 341 incoming solar 239 outgoing
101.9 W m-2 radiation longwave
341.3 W m-2 radiation
reflected by 238.5 W m-2
clouds and
atmosphere 79 40 atmospheric
window
79 emitted by 169 30
atmosphere
greenhouse
absorbed by gases
78 atmosphere

latent
17 80 heat

reflected by 356 40 333
surface back
396 radiation
23
161 therm1a7ls 80 surface 333
evapo- radiation
absorbed by absorbed by
surface transpiration surface

net absorbed
0.9 W m-2

▲ Figure 17 Factors that make up the energy balance of the Earth (after Stephens and others.
2012. An update on earth’s energy balance in light of the latest global observations.
Nature Geoscience.).

347

8 ENERGY PRODUCTION

Global warming

There is little doubt that climate change is occurring on the planet.
We are seeing a significant warming that may lead to many changes
to the sea level and in the climate in many parts of the world. The
fact that there is change should not surprise us. We have recently
(in geological terms) been through several Ice Ages and we are
thought to be in an interstadial phase at the moment (interstadial
means “between Ice Ages”). In the seventeenth century a “Little Ice
Age” covered much of northern Europe and North America. The
River Thames regularly froze and the citizens held fairs on the ice.
In 1608, the Dutch painter Hendrick Avercamp painted a winter
landscape showing the typical extent and thickness of the ice in
Holland (figure 18).

▲ Figure 18 Winter landscape with skaters (1608), Hendrick Avercamp.

Many models have been suggested to explain global warming, they include:

● changes in the composition of the atmosphere (and specifically the
greenhouse gases) leading to an enhanced greenhouse effect

● increased solar flare activity

● cyclic changes in the Earth’s orbit

● volcanic activity.

Most scientists now accept that this warming is due to the burning of
fossil fuels, which has gone on at increasing levels since the industrial
revolution in the eighteenth century. There is evidence for this. The
table below shows some of the changes in the principal greenhouse
gases over the past 250 years.

Gas Pre-1750 Recent % increase since
concentration concentration 1750
Carbon dioxide
Methane 280 ppm 390 ppm 40
Nitrous oxide 700 ppb 1800 ppb
Ozone 270 ppb 320 ppb 160
25 ppb
34 ppb 20

40

ppm = parts per million; ppb = parts per billion

348

8.2 THERMAL ENERGY TRANSFER

The recent values in this table have been collected directly in a number Nature of
of parts of the world (there is a well-respected long-term study of the science
variation of carbon dioxide in Hawaii where recently the carbon dioxide
levels exceeded 400 ppm for the first time for many thousands of An international
years). The values quoted for pre-1750 are determined from a number perspective
of sources:
There have been a number
● Analysis of Antarctic ice cores. Cores are extracted from the ice in the of international attempts
Antarctic and these yield data for the composition of the atmosphere to reach agreements over
during the era when the snow originally fell on the continent. Cores the ways forward for
can give data for times up to 400 000 years ago. the planet. These have
included:
● Analysis of tree rings. Tree rings yield data for the temperature and
length of the seasons and the rainfall going back sometimes hundreds ● The Kyoto Protocol
of years. was originally adopted
by many (but not all)
● Analysis of water levels in sedimentary records from lake beds can be countries in 1997 and
used to identify historical changes in water levels. later extended in 2012.

An enhanced greenhouse effect results from changes to the ● The Intergovernmental
concentration of the greenhouse gases: as the amounts of these Panel on Climate Change.
gases increase, more absorption occurs both when energy enters
the system and also when the surface re-radiates. For example, in ● Asia–Pacific Partnership
the transmittance–wavelength graph for a particular gas, when the on Clean Development
concentration of the gas rises, the absorption peaks will increase too. The and Climate.
surface will need to increase its temperature in order to emit sufficient
energy at sea level so that emission of energy by Earth from the top of ● The various other United
the atmosphere will equal the incoming energy from the Sun. Nations Conventions
on Climate Change, e.g.
Global warming is likely to lead to other mechanisms that will Cancùn, 2010.
themselves make global warming increase at a greater rate:
Do some research on the
● the ice and snow cover at the poles will melt, this will decrease Internet to find what is
the average albedo of Earth and increase the rate at which heat is presently agreed between
absorbed by the surface. governments.

● a higher water temperature in the oceans will reduce the extent to
which CO2 is dissolved in seawater leading to a further increase in
atmospheric concentration of the gas.

Other human-related mechanisms such as deforestation can also drive
global warming as the amount of carbon fixed in the plants is reduced.

This is a problem that has to be addressed at both an international
and an individual level. The world needs greater efficiency in power
production and a major review of fuel usage. We should encourage
the use of non-fossil-fuel methods. As individuals we need to be aware
of our personal impact on the planet, we should be conscious of our
carbon footprint. Nations can capture and store carbon dioxide, and
agree to increase the use of combined heating and power systems. What
everyone agrees is that doing nothing is not an option.

349

8 ENERGY PRODUCTION

Questions

1 (IB) b) The flow rate of water in the pipe is
400 m3 s–1. Calculate the power delivered
a) A reactor produces 24 MW of power. The by the falling water.
efficiency of the reactor is 32%. In the
fission of one uranium-235 nucleus 3 (IB)
3.2 × 10–11 J of energy is released.
The energy losses in a pumped storage power
Determine the mass of uranium-235 that station are shown in the following table.
undergoes fission in one year in this reactor.
Source of energy loss Percentage loss of energy
b) During its normal operation, the following
set of reactions takes place in the reactor. friction and turbulence of
water in pipe
01n + U238 → U239 27
92
92 _v

U239 → 29339Np + -01e + friction in turbine and ac
92 generator
29339Np → 29349Pu + -01e + _v
15

Comment on the international implications electrical heating losses 5
of the product of these reactions.

2 (IB) a) Calculate the overall efficiency of the
conversion of the gravitational potential
The diagram shows a pumped storage power energy of water in the tank into electrical
station used for the generation of electrical energy. energy.

b) Sketch a Sankey diagram to represent the
energy conversion in the power station.

310 m tank 4 (IB)
pipe
turbine A nuclear power station uses uranium-235
(U-235) as fuel.
lake
a) Outline:
Water stored in the tank falls through a pipe to (i) the processes and energy changes
a lake through a turbine that is connected to an that occur through which the internal
electricity generator. energy of the working fluid is increased
The tank is 50 m deep and has a uniform area (ii) the role of the heat exchanger of
of 5.0 × 104 m2. The height from the bottom of the reactor and the turbine in the
the tank to the turbine is 310 m. generation of electrical energy.
The density of water is 1.0 × 103 kg m–3.
a) Show that the maximum energy that can be b) Identify one process in the power station
where energy is degraded.
delivered to the turbine by the falling water
is about 8 × 1012 J. 5 (IB)

The intensity of the Sun’s radiation at the position
of the Earth is approximately 1400 W m–2.

Suggest why the average power received per
unit area of the Earth is 350 W m–2.

350

QUESTIONS

6 (IB) a) State the region of the electromagnetic
The diagram shows a radiation entering or spectrum to which the resonant frequency
leaving the Earth’s surface for a simplified model of nitrogen oxide belongs.
of the energy balance at the Earth’s surface.
b) Using your answer to (a), explain why
atmosphere TA = 242 K nitrogen oxide is classified as a greenhouse
gas.

transmitted through radiated by 9 (IB)
atmosphere 245 W m-2
atmosphere The diagram shows a simple energy balance
radiated by Earth’s 0.700 σT4A climate model in which the atmosphere and
surface = σT4E the surface of Earth are treated as two bodies
each at constant temperature. The surface
Earth’s surface TE of the Earth receives both solar radiation
and radiation emitted from the atmosphere.
a) State the emissivity of the atmosphere. Assume that the Earth’s surface and the
atmosphere behave as black bodies.
b) Determine the intensity of the radiation
radiated by the atmosphere towards the 344 W m-2
Earth’s surface.
e = 0.720 242 K
c) Calculate TE. a = 0.280

7 a) Outline a mechanism by which part of the atmosphere
radiation radiated by the Earth’s surface
is absorbed by greenhouse gases in the atmospheric radiation solar radiation
atmosphere. Go on to suggest why the
incoming solar radiation is not affected by Earth’s surface 288 K
the mechanism you outlined.
Data for this model:
b) Carbon dioxide (CO2) is a greenhouse
gas. State one source and one sink (that average temperature of the atmosphere of
removes CO2) of this gas. Earth = 242 K

8 (IB) emissivity e of the atmosphere of Earth = 0.720

The graph shows part of the absorption average albedo a of the atmosphere of
spectrum of nitrogen oxide (N2O) in which Earth = 0.280
the intensity of absorbed radiation A is plotted
against frequency f. solar intensity at top of atmosphere = 344 W m–2

A/arbitrary units average temperature of the surface of
Earth = 288 K

a) Use the data to determine:
(i) the power radiated per unit area of the
atmosphere
(ii) the solar power absorbed per unit area
at the surface of the Earth.

2345
f/×1013 Hz

351

8 ENERGY PRODUCTION

b) It is suggested that, if the production of 10 (IB)
greenhouse gases were to stay at its
present level, then the temperature of It has been estimated that doubling the amount
the Earth’s atmosphere would eventually of carbon dioxide in the Earth’s atmosphere
rise by 6 K. changes the albedo of the Earth by 0.01.
Estimate the change in the intensity being
Calculate the power per unit area that reflected by the Earth into space that will result
would then be from this doubling. State why your answer is
an estimate.
(i) radiated by the atmosphere
Average intensity received at Earth from the
(ii) absorbed by the Earth’s surface. Sun = 340 W m–2
Average albedo = 0.30
c) Estimate the increase in temperature of the
Earth’s surface.

352

9 WAVE PHENOMENA (AHL)

Introduction the usefulness of modelling using a spreadsheet –
both for graphing and developing relationships
In this topic we develop many of the concepts through iteration.
introduced in Topic 4. In general, a more
mathematical approach is taken and we consider

9.1 Simple harmonic motion

Understanding Applications and skills

➔ The defining equation of SHM ➔ Solving problems involving acceleration,
velocity and displacement during simple
➔ Energy changes harmonic motion, both graphically and
algebraicallyOBJ TEXT_UND
Nature of science
➔ Describing the interchange of kinetic and
The importance of SHM potential energy during simple harmonic
motion
The equation for simple harmonic motion (SHM)
can be solved analytically and numerically. ➔ Solving problems involving energy transfer
Physicists use such solutions to help them to during simple harmonic motion, both
visualize the behaviour of the oscillator. The use of graphically and algebraically
the equations is very powerful as any oscillation
can be described in terms of a combination of Equations _2π
harmonic oscillators using Fourier synthesis. T
The modelling of oscillators has applications in ➔ angular velocity–period equation: ω =
virtually all areas of physics including mechanics,
electricity, waves and quantum physics. In ➔ defining equation for shm: a = -ω2x
this sub-topic we will model SHM using a
simple spreadsheet and see how powerful this ➔ displacement–time equations: x = x0 sin ωt;
interpretation can be. x = x0 cos ωt

➔ velocity–time equations: v = ωx0 cos ωt;
v = -ωx0 sin ωt

➔ velocity-dis_p_la_c_e_m_e_nt equation:
v = ±ω √( x02 - x2 )

➔ kinetic energy equation: EK = _1__ mω2 (x02 - x2)

_1__ 2

➔ total energy equation: ET = 2 mω2 x02 ___l__
➔ period of simple pendulum:
T = √2π g

➔ period of mass–spring: T = 2π√__m_k___

353

9 WAVE PHENOMENA (AHL)

screen Introduction

metal sphere In this sub-topic we treat SHM more mathematically but restrict
turntable ourselves to two systems – the simple pendulum and of a mass
oscillating on a spring. Each of these systems is isochronous and is
light source usually lightly damped; this means that a large number of oscillations
drive belt occur before the energy in the system is transferred to the internal
energy of the system and the surrounding air.
Figure 1 Comparison of SHM and circular
motion. Angular speed or frequency (ω)

In Sub-topic 6.1 we considered the angular speed ω in relation to circular
motion; it is the rate of change of angle with time and is also called
angular frequency. It is measured in radians per second (rad s-1). This
quantity is important when we deal with simple harmonic motion because
there is a very close relationship between circular motion and SHM. This
relationship can be demonstrated using the apparatus shown in figure 1.

A metal sphere is mounted on a turntable that rotates at a constant angular
speed. A simple pendulum is arranged so that it is in line with the sphere
and oscillates with the same periodic time as that of the turntable – a little
trial and error should give a good result here. The pendulum and the
turntable are illuminated by a light that is projected onto a screen. The
shadows projected, onto the screen, of the circular motion of the sphere and
oscillatory motion of the pendulum show these motions to be identical.

Circular motion and SHM

In mathematical terms the demonstration in figure 1 is equivalent

to projecting the two-dimensional motion of a point onto the

single dimension of a line. Imagine a point P rotating around the

perimeter of a circle with a constant angular speed ω. The radius

of the circle r joins P with the centre of the circle O. At time t = 0

the radius is horizontal and at time t it has moved through an angle
θ radians. For constant angular speed ω = _θt , rearranging this gives θ =
ωt. Projecting P onto the y-axis gives the vertical component of r as

r sin θ. Projecting P onto the x-axis gives the horizontal component

of r as r cos θ. The variation of the vertical component with time or

angle is shown on the right of figure 2 and takes the form of a sine

curve. Because the rate of rotation is constant, the angle θ or ωt is

proportional to time. If we drew a graph of y against t the quantities
_T_
2π and π would be replaced by T and respectively.
2

yy

y = r sinθ r

P

r

θ = ωt x 0 π 2π ωt

0 x = r cosθ

-r

Figure 2 Projection of circular motion on to a vertical line.
The equations of the projections are y = y0 sin ωt and x = x0 cos ωt.

354

9.1 SIMPLE HARMONIC MOTION

Here y0 and x0 are the maximum values of y and x, which in this case are Note
identical to r. These are the amplitudes of the motion.
● You will meet differential
You may remember from Sub-topic 4.1 that in SHM the displacement, calculus in your Mathematics
velocity and acceleration all vary sinusoidally with time. Thus, the or Mathematics Studies
projection of circular motion on the vertical or horizontal takes the same course. This is not the
shape as SHM. This is very useful in analysing SHM. place to teach you to
differentiate – you will
The relationship between displacement, velocity, not be expected use
and acceleration calculus on your IB Physics
course. However, for
In Sub-topic 4.1 we saw that, starting from the displacement–time students studying physics,
curve, we could derive the velocity–time and acceleration–time curves engineering and allied
from the gradients. This can be done graphically, but another technique subjects at a higher level,
is to differentiate the equations with respect to time (differentiation is calculus will form a major
equivalent to finding the gradient). aspect of your course. In this
case, we will differentiate the
We have seen from the comparison with circular motion that the equations, but it is the results
that are important not the
displacement takes the form: x = x0 sin ωt or x = x0 cos ωt depending on method of obtaining them.
when we start timing. It really doesn’t matter whether the projection is onto
355
the x-axis or the y-axis – therefore we can use x and y interchangeably.

Let’s start with x = x0 sin ωt

tahnedvωeltohceityanisgugliavrenfrebqyuven=cy_.dd_xt
( )This means that = x0 ω cos ωt, where x0

is the amplitude

The maximum value that cosine can take is 1 so the maximum velocity

is v0 = x0 ω thus making the equation for the velocity at time t become
v = v0 cos ωt

We know that the acceleration will be the gradient of a velocity–time
graph so we have

a = _d_v = -v0ω sin ωt = -x0 ω2 sin ωt

dt

As for cosine, the maximum value that sine can take is 1 so

a0 = v0 ω = x0 ω2 giving a = -a0 sin ωt

Comparing the equations x = x0 sin ωt and a = -x0 ω2 sin ωt we can see a
common factor of x0 sin ωt

meaning that

a = -ω2 x

This may remind you, in Sub-topic 4.1, we saw that a = -kx for SHM.

So the constant k must actually be ω2 (the angular frequency squared).
We will look at the significance of ω2 very soon.

In examinations, you can be asked to find maximum values for velocity and

acceleration – this makes the calculations easier because the maximum of

sine and cosine are each 1 and, therefore, we don’t need to include the sine
or cosine term in our calculations. Thus the maximum velocity will be
v0 = x0 ω and the magnitude of the maximum acceleration will be a0 = x0 ω2
(the direction of a0 will be opposite that of x0).

Modelling SHM with a spreadsheet

Much can be learned about SHM by using a spreadsheet to graph it.
Figure 3 is a screen shot of part of a spreadsheet set up for this purpose

9 WAVE PHENOMENA (AHL)

356 AB C D E F GH I J K LMNO P Q
t1 = 0 x0 = 5
1 time/sdisplacement/m velocity/m s−1 acceleration/m s−2 Δt = 0.2 ω = 1.26 t/s
2 0.0 14.0
0.00 6.30 0.00
t/s
3 0.2 1.25 6.10 −1.98 14.0

4 0.4 2.41 5.52 −3.83 t/s
14.0
5 0.6 3.43 4.58 −5.45 SHM displacement-time graph
6.00
6 0.8 4.23 3.36 −6.71
4.00
7 1.0 4.76 1.93 −7.56
2.00
8 1.2 4.99 0.37 −7.92
0.00
9 1.4 4.91 −1.21 −7.79 0.0 2.0 4.0 6.0 8.0 10.0 12.0
10 1.6 4.51 −2.71 −7.16
11 1.8 3.83 −4.05 −6.09 −2.00
12 20. 2.91 −5.12 −4.62 x/m
13 2.2 1.81 −5.87 −2.87 −4.00
14 2.4 0.59 −6.26 −0.93
15 2.6 −0.67 −6.24 −6.00
1.06

16 2.8 −1.88 −5.84 2.99

17 3.0 −2.98 −5.06 4.73

18 3.2 −3.89 −3.96 6.17

19 3.4 −4.55 −2.62 7.22

20 3.6 −4.92 −1.11 7.81

21 3.8 −4.99 0.48 7.92 SHM velocity-time graph
8.00
22 4.0 −4.73 2.03 7.52 6.00
4.00
23 4.2 −4.18 3.45 6.64 2.00
0.00
24 4.4 −3.37 4.66 5.35 −2.00 0.0 2.0 4.0 6.0 8.0 10.0 12.0
−4.00
25 4.6 −2.34 5.57 3.72 −6.00
−8.00
26 4.8 −1.17 6.13 1.85

27 5.0 0.08 6.30 −0.13 v/m s−1

28 5.2 1.33 6.07 −2.11

29 5.4 2.49 5.46 −3.95

30 5.6 3.49 4.51 −5.54

31 5.8 4.27 3.27 −6.78

32 6.0 4.79 1.83 −7.60

33 6.2 5.0 0.26 −7.93

34 6.4 4.89 −1.31 −7.76
35 6.6 4.48 −2.81 −7.11
36 6.8 −4.13 −6.00
3.78
SHM acceleration-time graph
37 7.0 2.24 −5.18 −4.51 10.00 4.0 6.0 8.0 10.0 12.0
38 7.2 1.73 −5.91 −2.74 8.00
39 7.4 0.50 −6.27 −0.84 6.00
40 7.6 −0.75 −6.23 4.00
1.20 2.00
0.00
41 7.8 −1.96 −5.79 3.11 −2.00
−4.00
42 8.0 −3.05 −5.00 4.84 −6.00
−8.00
43 8.2 −3.94 −3.88 6.25 a/m s−2−10.00

44 8.4 −4.58 −2.52 7.28 0.0 2.0

45 8.6 −4.94 −1.00 7.84

46 8.8 −4.98 0.58 7.90

47 9.0 −4.71 2.13 7.47

48 9.2 −4.14 3.54 6.57

49 9.4 −3.31 4.73 5.25

50 9.6 −2.27 5.62 3.60

Figure 3 Spreadsheet for SHM.

(as before this uses Microsoft Excel but other spreadsheets will have
similar functions). Column A contains incremental times starting from
zero (cell H1 is copied into A2 – this means that the starting time can
be changed). Cell H2 determines the time increments (in this case
0.2 s) by adding the contents of H2 to each previous cell; the times can
be increased down the column. The formula in cell A3, in this case,
is =A2+$H$2 and the formula in cell A4 is =A3+$H$2, etc.

9.1 SIMPLE HARMONIC MOTION

The times generated, together with chosen values of x0 and ω, are used to Note
generate the displacement, velocity and time curves. The values for x0 and
ω are inserted into cells J1 and J2 respectively – these can then be changed We have used a spreadsheet to
at will. The equation for the displacement, x = x0 sin ωt is converted into show the solutions to the SHM
the formula =$J$1*SIN(A2*$J$2), which is copied into column B. The equation. Later we will discuss
equation for the velocity, v = x0 ω cos ωt is converted into the formula how we can actually solve the
=$J$1*$J$2*COS(A2*$J$2) which is copied into column C. Finally, the SHM equation using iteration.
equation for the acceleration, a = -x0 ω2 sin ωt, is converted into the
formula =(-1)*$J$1*($J$2)^2*SIN(A2*$J$2) and is copied into column D.

To produce the curves shown in figure 3, you now need to:

● insert a scatter chart

● choose the data series

● do a little formatting to improve the size and position of the chart
and label the axes, etc.

The SHM equation and ω2

Re-visiting the equation a = -ω2 x discussed earlier in this sub-topic,
we see that it fits the definition of SHM (motion in which the
acceleration is proportional to the displacement from a fixed
point and is always directed towards that fixed point). The
equation is the defining equation for SHM.

The sinusoidal graphs provide the solutions to this equation with respect

to time. This may seem strange because there is no apparent time factor

in a = -ω2 x. This is where ω2 comes in. We saw with circular motion

that ω represents the angular speed. Therefore ω2 is simply the square of
( )__1__ 2.
this measured in rad2 s−2. ω2 has dimensions equivalent to time

In circular motion we defined ω as being _∆_θ or simply _θ_ when it is

∆t t
constant. For a complete revolution the angle will be 2π radians and the
_2_π_
time will be the periodic time T. This means that ω = or, alternatively,
ω = 2πf; comparing this _1_ explains T
equation with f = why we can call ω
T
the angular frequency.

Worked example We are only asked to find the magnitude of
the velocity so it doesn’t matter which of the
An object performs SHM with a period of 0.40 s cosine curves the motion really takes.
and has amplitude of 0.20 m. The displacement is
zero at time zero. Calculate: Using v = x0 ω cos ωt

a) the maximum velocity ωt = _2πt = _(2π ×_0.10) ≈ _π rad
T 0.40 2
b) the magnitude of the velocity after 0.10 s
So v 0.20 _2π cos _π 0
c) the maximum acceleration of the object. 0.40 2

Solution = × × =

a) v0 x0 x0 _2π This could have been done without the full
T
= ω = × calculation once we had decided that it was a

v0 = 0.20 × _2π = 3.1 m s−1 cosine. 0.10 s represents the time for a quarter
0.40
of a period (0.40 s); the value of any cosine at
_π_
b) As the displacement is zero at time zero this a quarter of a period (or radian) is zero.
2
must be a sine or negative sine wave. Thus the
_2π 2
0.40
( )c)
velocity will be a cosine or negative cosine. a0 = x0 ω2 = 0.20 × ≈ 49 m s−1

357

9 WAVE PHENOMENA (AHL)

The velocity equation

The derivation of the SHM equation is for reference and you don’t need to
reproduce it – following it through, however, will help you to understand
what is going on.

We have already seen that v = x0 ω cos ωt and x = x0 sin ωt
and you may know that sin2 θ + cos2 θ = 1.

________

This means that cos θ = ±√1- sin2 θ (don’t forget that squaring either

the positive or negative cosine will give (+)cos2 θ. This means that ±

must be included in the square root of this.

________

So v = ±x0ω √1- sin2 ωt

But sin ωt = _x_ so sin2 ωt = ( )_x_ 2
x0
x0 _1_-___x_x0__2 = ±ω _x0_2_-__x_0_2 __x_x0__2

√ ( ) √ ( )v = ±x0 ω = ______

±ω √x02 - x2

______

The equation v = ±ω√x02 - x2 is useful for finding the velocity at a
particular position when you know the amplitude and period (or frequency

or angular frequency) – you don’t need to know the time being considered.

Worked example Note

An object oscillates simple harmonically with ● ± tells us that the object could be going in
frequency 60 Hz and amplitude 25 mm. Calculate either of the two opposite directions.
the velocity at a displacement of 8 mm.
● Don’t forget x02 - x2 ≠ ( x0 - x )2; it is a
Solution common mistake for students to equate these
two expressions.
ω = 2πf = 120π (there is no need to calculate the
value here but do not leave π in the answer in an
examination.

______

v = ±ω √x02 - x2

____________________

= ±120π √( 25 × 10-3 )2- ( 8 × 10-3 )2

= ±8.9 m s−1

θl Simple harmonic systems

Ft m 1. The simple pendulum
mg sin θ
The simple pendulum represents a straightforward system that oscillates
mg cos θ with SHM when its amplitude is small. When the pendulum bob (the
θ mass suspended on the string) is displaced from the rest position there
Figure 4 Restoring force for simple pendulum. is a component of the bob’s weight that tends to restore the bob to its
normal rest or equilibrium position. A condition of a system oscillating
358 simple harmonically is that there is a restoring force that is proportional
to the displacement from the equilibrium position – this is, in effect, tying
in with the equation defining SHM because F = ma and a = -ω2 x this
means that F = -mω2 x.

Figure 4 shows the forces acting on a pendulum bob. The bob is in
equilibrium along the radius when the tension in the string Ft equals
the component of the weight in line with the string (=mgcosθ). The
component of the weight perpendicular to this is not in equilibrium and
provides the restoring force.

9.1 SIMPLE HARMONIC MOTION

So the restoring force must be equal to the mass multiplied by the Note
acceleration according to Newton’s second law of motion:
● The period of a simple
mg sin θ = ma pendulum is independent
of the mass of the
For a small angle sin θ ≈ θ ≈ _x pendulum.
l
_g ● The period of a simple
( )rearranging gives -m l x = ma pendulum is independent
of the amplitude of the
(The minus sign is because the displacement (to the right) is in the pendulum.
opposite direction to the acceleration (to the left) in figure 4.
● l is the length of the
Cancelling m pendulum from the point of
suspension to the centre of
( )a = - _g x mass of the bob.
l
● The equation only applies to
this compares with the defining equation for SHM a = -ω2 x leading to small oscillations (swings
making angle of less than
ω2 = _g 10° with the rest position).
l
__
√As the periodic time for SHM is given by T = _2ωπ, this shows that T = 2π _l
g .

Investigate! clamp split cork
timer
1 Experimenting with the simple pendulum: ruler
l
● attach a piece of thread of length about
2 metres to a pendulum bob (a lump of G-clamp
modelling clay would do for this) bob

● suspend the thread through a split cork to θ < 10°
provide a stable point of suspension θ

● align a pin mounted in another piece AC
of modelling clay with the rest position B
of the bob – to act as a reference point
(sometimes called a fiducial marker)

● set the bob oscillating through a small
angle and start timing as the bob passes the
fiducial marker (you should consider why
your measurements are likely to be more
reliable when the bob is moving at its fastest)

● time thirty oscillations (remember that each
oscillation is from A to C and back to A)

● measure the length of the thread from the
point of suspension to the centre of mass
of the bob (take this to be the centre of
the bob)

● repeat the procedure and find an average
period for the pendulum

359

9 WAVE PHENOMENA (AHL)

● repeat your measurements for a range of increment the change in velocity will
lengths be the acceleration multiplied by the
time increment – when you add this
● first plot a graph of periodic time T against to your previous velocity you will
length l get the new velocity (actually it’s the
average velocity through the time
● referring to the simple pendulum equation increment)
you may see that this graph should not be
linear – what should you plot in order to ● because velocity is the change in
linearize your results? displacement divided by the time increment,
you can find the change in displacement by
● how can you use the graph to calculate a multiplying your current velocity value by
value for the acceleration of freefall g? your chosen time increment

● don’t forget to do an error analysis and ● the new value of the displacement will be
compare your result with the accepted the previous value added to the change in
value of g. displacement

2 Use of iteration

Making use of the powerful iterative ● This now feeds back into finding the next
functionality of a spreadsheet can be a value of the acceleration and the cycle
valuable learning aid. With iteration a repeats and you generate your data for
graphical investigation is possible from which you can plot displacement-time,
basic principles and basic equations without velocity-time and acceleration-time graphs
knowing the solution and without advanced (a example of the spreadsheet is included
mathematics. on the website).

● you will need to set up a worksheet The following flow chart illustrates the iteration:

with the headings time, displacement,

acceleration, change in velocity, velocity n=0
and change in displacement choose ∆t, tn, xn, k
υn = 0
● you now need to set up the initial

conditions:

1. choose a time increment for the an+1 = -k × xn

iteration (make it fairly large until your

spreadsheet is working well) ∆υ = an+1 × ∆t n n+1
2. start your time column at zero and υn+1 = υn + ∆υ

decide how long you want to run the

iteration for (make this, say, 10 time ∆x = υn+1 × ∆t
increments initially) xn+1 = xn + ∆x

3. choose an amplitude value for your enough no
oscillation increments ?

4. set the initial displacement value equal yes
to the amplitude

5. choose a constant (k or ω2) finish
6. set the initial velocity to be zero

● the acceleration will always be -k Subscripts n represent the current value of a
multiplied by the displacement variable and n + 1 the next value; at each loop
the next value always replaces the previous one.
● because acceleration is the change
in velocity divided by your time

360

9.1 SIMPLE HARMONIC MOTION

2. Mass–spring system Note

We have focused on a simple pendulum as being a very good ● The period of the
approximation to SHM. A second system which also behaves well and mass-spring system is
gives largely undamped oscillations is a mass–spring system. We will independent of amplitude
consider a mass being oscillated horizontally by a spring (see figure 5); (for small oscillations).
this is more straightforward than taking account of including the effects
of gravity experienced in vertical motion. We will assume that the ● The period of the
friction between the mass and the base is negligible. The mass, therefore, mass-spring system
exchanges elastic potential energy (when it is fully extended and is independent of the
compressed) with kinetic energy (as it passes through the equilibrium acceleration of gravity.
position).

relaxed spring mass restoring force
extended spring mass

base base

initial position x position of left edge
of left edge initial position when spring extended
of left edge

Figure 5 Restoring force for mass-spring system.

When a spring (having spring constant k) is extended by x from its
equilibrium position there will be a restoring force acting on the
mass given by F = -kx (the force is in the opposite direction to the
extension).

Using Newton’s second law

ma = -kx
which can be written as

a = - ( _k ) x
m

this compares with the defining equation for SHM

a = -ω2x

( )ω2 = _k
m

As the periodic time for SHM is given by

T = _2π
ω

this shows that

√T = 2π __m_
k

When the spring is compressed the quantity x represents the compression
of the spring. When the mass is to the left of the equilibrium position
the compression is positive but the restoring force will be negative. This,
therefore, leads to the same outcomes as for extensions.

361

9 WAVE PHENOMENA (AHL)

Worked example Energy in SHM systems

The mass–spring system is used We have seen that, in a simple pendulum, there is energy interchange
in many common accelerometer between gravitational potential and kinetic; in a horizontal mass–spring
designs. A mass is suspended by a system the interchange is between elastic potential and kinetic energy.
pair of springs which displaces when Because each system involves kinetic energy we will focus on this form
acceleration occurs. An accelerometer of energy and bear in mind that the potential energy will always be the
contains a mass of 0.080 kg coupled difference between the total energy and the kinetic energy at a particular
to a spring with spring constant of time. The total energy will be equal to the maximum kinetic energy.
4.0 kN m-1. The amplitude of the
mass is 20 mm. Calculate: From Topic 2 we know that the kinetic energy of an object of mass m,
moving at velocity v, is given by

a) the maximum acceleration EK = _1 mv2
2

b) the natural frequency of the mass. We also know that the equation for the velocity at a particular
position is
Solution _k
( )a) m ______
a= - x
v = ±ω√x02 - x2

_4.0 × 103
0.08
=( )a-  20 × 10-3 Combining these equations gives the kinetic energy at displacement x:

= 1000 m s−2 EK = _1 m ω2 (x02 - x2)
2
__m_
√b) T = 2πk so f = _1 This tells us that the maximum kinetic energy will be given by
T
E Kmax = _1 m ω2 x02
__ 2
_1 _k
√= 2π m and this must be the total energy (when the potential energy is zero) so
we can say
________
_1 _4.0 × 103
=√f2π 0.08 = 36 Hz ET = _1 m ω2 x02
2

The potential energy at any position will be the difference between the
total energy and the kinetic energy

so

EP = ET - EK = _1 m ω2 x02 - _1 m ω2 (x02 - x2) = _1 m ω2 x2
2 2 2

Figure 6 illustrates the variation with displacement of energy: the

green line shows the total energy, the red the potential energy and the
blue the kinetic energy. At any position the total energy is the sum of

the kinetic energy and the potential energy – as we would expect.

1.200

1.000

0.800

energy/J 0.600

0.400

0.200

0.000

−0.30 −0.20 −0.10 0.00 0.10 0.20 0.30

displacement/m

Figure 6 Variation of energy with displacement.

362

9.1 SIMPLE HARMONIC MOTION

The variation of the potential energy and kinetic energy with displacement
are both parabolas. With all the quantities in the total energy equation being
constant, the total energy is, of course, constant for an undamped system.

In addition to looking at the variation of energy with displacement we
should consider the variation of energy with time. Again, let us start
with the kinetic energy.

The velocity varies with time according to the equation

v = x0 ω cos ωt

so the kinetic energy will be

EK = _1 mv2 = _1 m(x0 ω cos ωt)2
2 2

or

EK = _1 mx02 ω2 cos2 ωt
2

When the cosine term equals 1, this gives the maximum kinetic energy.
The maximum kinetic energy occurs when the potential energy is zero
and so is numerically equal to the total energy at that instant.

ET = _1 mx 2 ω2
2 0

The potential energy will be the difference between the total energy and
the kinetic energy so

EP = ET - EK

= _1 mx02 ω2 - _1 mx02 ω2 cos2 ωt
2 2

= _1 mx02 ω2 sin2 ωt
2

This relationships are shown on figure 7.

The green line represents the total energy, the red curve the potential
energy and the blue curve the kinetic energy.

1.20 Note

1.00 ● The total energy is always
the sum of the kinetic and
0.80 potential energies.

energy/J 0.60 ● The graphs (unlike sine
and cosine) never become
0.40 negative.

0.20 ● The period of the energy
change is half that of the
0.00 10.0 12.0 variation with time of
0.0 2.0 4.0 6.0 8.0 displacement, velocity, or
acceleration.
time/s
● The frequency of the energy
Figure 7 Graph showing variation of energy with time. change is twice that of
the variation with time of
displacement, velocity, or
acceleration.

363

9 WAVE PHENOMENA (AHL)

9.2 Single-slit diffraction

Understanding Applications and skills

➔ The nature of single-slit diffraction ➔ Describing the effect of slit width on the
diffraction pattern
Nature of science
➔ Determining the position of first interference
Development of theories minimum

That rays “travel in straight lines” is one of the first ➔ Qualitatively describing single-slit diffraction
theories of optics that students encounter. It comes as a patterns produced from white light and from a
surprise when this theory cannot explain the diffraction range of monochromatic light frequencies
seen at the edges of shadows cast by small objects
illuminated using point sources. Although partial Equations
shadows can be explained by considering light sources
to be extended, this cannot account for the diffraction ➔ angle between first minimum and central
from a point source. The wave theory of diffraction and _λ__ _
how diffraction can be explained in terms of wavefront maximum θ =
propagation from secondary sources is a good example a
of how theories have been developed in order to explain
a wider variety of phenomena.

Io intensity Introduction

In Sub-topic 4.4 we introduced diffraction and saw that when a wave

passes through an aperture it spreads into the geometric shadow

region. We also saw that the diffraction pattern consists of a series of

bright and dark fringes. We will now consider the diffraction pattern

5% Io in more detail.

θ3 θ2 θ1 0 θ1 θ2 θ3 angle

Graph of intensity against angle

central Figure 1 shows a single-slit diffraction pattern together with a graph of
maximum the variation of the intensity of the diffraction pattern with the angle
measured from the straight-through position.

Figure 1 Variation of intensity with θ1, θ2, and θ3 are the angles with the straight-through position made by
angle for a diffraction pattern. the minima.

Note the second is about 2% and the third is about 1%
(figure 1 is not drawn to scale – the secondary maxima
● The central maximum has twice the angular width of are all larger than a scale-diagram would show).
the secondary maxima (each of these have the same
angular width). ● No light reaches the centre of the minima but, in going
towards the maxima, the intensity gradually increases
● The intensity falls off quite significantly from the – it is very difficult to decide the exact positions of the
principal maximum to the secondary maxima – maxima and minima.
the intensity of the first secondary maximum is
approximately 5% of that of the principal maximum, ● Theintensityisproportionaltothesquareoftheamplitude.

364

9.2 SINGLE-SLIT DIFFRACTION

The single-slit equation TOK

Returning to the work of Sub-topic 4.4 we saw how waves can Small angle approximation
superpose to give constructive interference (when they meet in phase)
or destructive interference (when they meet anti-phase). We say that diffraction is
most effective when the
(a) single slit (b) aperature is of the same order
of magnitude as the width of
towards first top of slit the slit; we can demonstrate
θ minimum this effectively using a ripple
incident aθ tank. In the equation θ = _λa___
plane waves a wavefront if we make λ ≈ a, then θ
= 1 radian (or 57.3°) or, if
bottom of θ you avoid the small angle
slit approximation, sin θ = 1
path difference = a sinθ and θ = 90°. Examine how
closely diffraction in a ripple
Figure 2 Deriving the single slit equation. tanks agrees with small angle
approximation prediction.
Figure 2(a) shows a single-slit of width a. Each point on the wavefront With poor agreement (as this
should show) why do we
within the slit behaves as a source of waves. These waves interfere when continue to use the equation
θ = _λa___?
they meet beyond the slit. For the two waves coming from the edges

of the slit making an angle θ with the straight through, there is a path

difference of a sin θ. Waves from a point halfway along the slit will have
_a_
a path difference of sin θ from the waves coming from each of the
2
edges. When this path difference equals half a wavelength, the waves

from halfway along the slit will interfere destructively with the waves

coming from the bottom edge of the slit. It follows that for each point in

the bottom half of the slit there will be a point in the top half of the slit
_a_ _a_ _λ2_  ,
at a distance sin θ from it. This means that when sin θ = there is
2 2
destructive interference between a wave coming from a point in the upper

half of the slit and an equivalent wave from the lower half of the slit.

Because we are dealing with small angles we can approximate sin θ to θ
and, by cancelling the factor of two, we arrive at the equation

θ = _λ 
a

The position of other minima (shown in figure 1) will be given by θ = _n_λ_

a
(where n = 2, 3, 4, ...) but you will not be examined on this relationship.

Note

● This is the equation for the angle of the first minimum.

● It follows from the equation θ = _n__λ__ that the minima are not actually separated

a
by equal distances. However, for values of θ that are less than about 10°, it is

a good approximation to consider the minima to be equally spaced.

● The principal or central maximum occurs because the pairs of waves from the
top and bottom halves of the slit will travel the same distance and have no
path difference.

● The angular width of the principal maximum (from first minimum on one side
to first minimum on the other) is 2θ.

365

9 WAVE PHENOMENA (AHL)

Worked example Solution

a) Explain, by reference to waves, the diffraction a) The wavefront within the slit behaves as a series
of light at a single slit. of point sources which spread circular wave fronts
from them. These secondary waves superpose
b) Light from a helium–neon laser passes through in front of the slit and the superposition of the
a narrow slit and is incident on a screen 3.5 m waves produces the diffraction pattern.
from the slit. The graph below shows the
variation with distance x along the screen of b) (i) In this case the graph is drawn for distance
intensity I of the light on the screen.
not angle, so we need to calculate the
I angle θ in order to be able to use θ = _λa_.

θ = _s_ = _3_.0_×__1_0_-_3

D 3.5

= 0.86 × 10−3 rad with s being the

distance from the centre of the principal

maximum to the first minimum. It is

more reliable to measure 2θ from the first

minimum on one side of the principal

maximum to the first minimum on the

other side of the principal maximum. D is

the distance from the slit to the screen.

a = _λ_ = _6_3_0_×__1_0_-_9

θ 0.86 × 10-3

−10 −5 0 5 10 = 0.73 × 10−3 m or 0.73 mm
x/mm

(i) The wavelength of the light emitted by the (ii) With a wider slit more light is able to pass
laser is 630 nm. Determine the width of
the slit. through. This will result in an increase in

(ii) State two changes to the intensity the intensity of the beam and so the peaks
distribution of the central maximum
when the single slit is replaced by one of will all be higher.
greater width.
a in the equation θ = _λ_ increases but λ

a
remains constant; the angle θ will decrease

which means the principal maximum will

become narrower and the minima will

move closer together.

Figure 3 Single slit with Single slit with monochromatic and white light
monochromatic and white light.
Figure 3 shows two images of the light emerging from a single-
slit. The upper image is obtained using monochromatic (one
frequency) green light and the lower image is obtained using white
light. Both the angular width of the central maximum and the
angular separation of successive secondary maxima depend on
the wavelength of the light – this is the reason why the secondary
maxima produced by the diffraction of white light are coloured. You
will see that, for the secondary maxima, the blue light is less deviated
than the other colours as it has the shortest wavelength. The edges
of the principal maximum are coloured rather than pure white. This
is because the principal maxima for the colours at the blue end of
the visible spectrum are less spread than the colours at the red end;
the edges are therefore a combination of red, orange, and yellow and
have an orange hue.

366

9.3 INTERFERENCE

Investigate!

Diffraction at a single slit

● Using a laser pointer, shine the light on the ● Measure the distance between the callipers
gap between the jaws of a pair of digital or and the screen, D, using a tape-measure.
vernier callipers (forming a slit) and project it
onto a white screen. The screen needs to be ● Calculate the wavelength of the light using
at least 3 m from the callipers. _s
λ ≈ aθ as we know that 2θ ≈ D
● Adjust the callipers so that the image is clear.
● Both the dependence of the width of the
● Measure the separation, s, of the first minima
on the screen using a metre ruler. central maximum and the separation of the

D maxima on the wavelength of light can be

investigated by using laser pointers with

different colours.

2θ s ● Research the wavelength ranges of the laser
laser light and then suggest how you could modify
the experiment to check the calibration of the
callipers callipers at small jaw separation.
screen
We will return to diffraction when we look at
Figure 4 Investigating laser light passing through slit. resolution in Sub-topic 9.4.

9.3 Interference

Understanding Applications and skills

➔ Young’s double-slit experiment ➔ Qualitatively describing two-slit interference
patterns, including modulation by one-slit
➔ Modulation of two-slit interference pattern by diffraction effect
one-slit diffraction effect
➔ Investigating Young’s double-slit experimentally
➔ Multiple slit and diffraction grating interference
patterns ➔ Sketching and interpreting intensity graphs of
double-slit interference patterns
➔ Thin film interference
➔ Solving problems involving the diffraction
Nature of science grating equation

Thin film interference ➔ Describing conditions necessary for constructive
and destructive interference from thin films,
The observation of colour is not simply a question including phase change at interface and effect of
of colour pigmentation. Certain mollusc and refractive index
beetle “shells”, butterfly wings, and the feathers
of hummingbirds and kingfishers can produce ➔ Solving problems involving interference from
beautiful colours as a result of thin film interference thin films
known as “iridescence”. The observed colour
changes depend on the angle of illumination or Equations _λ___D__
viewing, and are caused by multiple reflections from
the surfaces of films with thicknesses similar to the fringe separation for double slit: s = d
wavelength of light. These natural aesthetics require
the analysis of the physics of interference. diffraction grating equation: nλ = d sinθ

light reflection by parallel-sided thin film: 367
constructive interference 2dn = (m + _12__)λ
destructive interference 2dn = mλ

9 WAVE PHENOMENA (AHL)

Introduction

In Sub-topic 4.4 we considered interference of the waves emitted by two
coherent sources. We saw that interference is a property of all waves, and
we considered the equal fringe spacing in the Young double-slit experiment.
In this sub-topic we focus on the intensity variation in a double-slit
experiment and see how interference is achieved using multiple slits. We
then look at a second way of achieving interference, using division of
amplitude instead of division of wavefront as with double-slit interference.

Intensity variation with the double-slit

Figure 1 shows the image of the light from a helium–neon laser that
has passed through a double slit. The alternate red and dark fringes
are equally spaced as we saw in Sub-topic 4.4. However, looking at the
image closely we see that there are extra dark regions.

Figure 1 Double-slit diffraction pattern for light from He–Ne laser.

We know from Sub-topic 9.2 that a single slit produces a diffraction pattern
with a very intense principal maximum and much less intense secondary
maxima. A double slit is, of course, two single slits so each of the slits
produces a diffraction pattern and the waves from the two slits interfere.
The two effects mean that the intensity of the interference pattern is not
constant, but is modified by the diffraction pattern to produce the intensity.

Figure 2(a) below shows how the relative intensity would vary for a double-
slit interference pattern without any modification due to diffraction. By using
relative intensity we avoid the need to think about the actual intensity values
and units. Figure 2(b) shows the variation of relative intensity with angle
for a single slit. Figure 2(c) shows the superposition of the two effects so that
the single-slit diffraction pattern behaves as the envelope of the interference
pattern. Shaping a pattern in this way is called modulation and is important in
the theory of AM (amplitude modulation) radio. You will note that the fringe
spacing does not change between figures 2(a) and (c) but that the bright
fringes occurring between 11° and 14° in figure 2(a) are reduced to a much
lower intensity. An interference maximum coinciding with a diffraction
minimum is suppressed and does not appear in the overall pattern.

We saw in Sub-topic 4.4 that the fringe separation s for light of
wavelength λ is given by

s = _λD
d

where D is the distance from the double slit to the screen and d

is the separation of the slits. The value of s used in this equation is

for the interference pattern not the modulated pattern caused by

diffraction.

368

9.3 INTERFERENCE

relative intensity
(a)

20 15 10 5 0 5 10 15 20

angle from straight through position/°

relative intensity
(b)

20 15 10 5 0 5 10 15 20

angle from straight through position/°

relative intensity
(c)

20 15 10 5 0 5 10 15 20

angle from straight through position/°

Figure 2 Combination of diffraction and interference.

Multiple-slit interference

We have now seen the interference patterns produced both by a single
slit and a double slit. What happens if there are more than two slits?
The answer is that the bright fringes, which come from constructive
interference of the light waves from different slits, remain in the same
positions as for a double slit but the pattern becomes sharper. The bright
fringes are narrower and their intensity is proportional to the square of
the number of slits. Why is this?

At the centre of the principal maximum the waves reaching the screen
from all of the slits are in phase and so it is very bright here. By moving
to a position close to the centre of this maximum the path difference
between the light from two adjacent slits has changed by such a small
amount that it hardly affects the interference and so it is still bright. Slits
further away from each other will have more likelihood of a greater path
difference and therefore meeting out of phase.

369

9 WAVE PHENOMENA (AHL)

Imagine having three slits: at the principal maximum the path difference
between the waves will be small as their paths are nearly parallel. Each
of the three wave trains will meet nearly in phase at positions around
the principal maximum peak, which makes it fairly wide.
Now imagine having 100 slits: for every slit there will a second
slit somewhere that transmits a wave with a half wavelength path
difference from the first. When the waves interfere at a position
close to the centre of the maximum, destructive interference occurs
and reduces the overall intensity. This will be true for the waves
coming from the two slits adjacent to this pair, the two next to
those and so on. Increasing the number of slits will give destructive
interference close to the centre of the maxima – this reduces the
width of the central maximum (and the other maxima too). The
extra energy that is needed to increase the intensity of the maxima
must come from the regions that are now darker, so the maxima are
more intense.
The mathematics of modulation of waves is quite complex and for the
purpose of the IB Physics course you will simply need to recognize that
the modulation is happening and remember the effect of increasing the
number of slits.
Figure 3 shows the interference patterns obtained when red laser light
passes through various numbers of slits of identical width. Figures 3
and 4 show that the single-slit diffraction pattern always acts as an
envelope for the multiple-slit interference patterns.

Figure 3 The effect of increasing the number of slits on an interference pattern.
In figure 4 the relative intensities have been drawn the same size in
order to show the increased sharpness; however, for two, three, and five
slits the actual intensities are in the ratio 1 : 9 : 25.

370

single slit 9.3 INTERFERENCE
relative intensity
double slit
relative intensity

20 15 10 5 0 5 10 15 20 20 15 10 5 0 5 10 15 20
angle from straight through position/° angle from straight through position/°

three slits five slits

relative intensity relative intensity

20 15 10 5 0 5 10 15 20 20 15 10 5 0 5 10 15 20
angle from straight through position/° angle from straight through position/°

Figure 4 The effect of increasing the number of slits on variation of intensity. white light
source
The diffraction grating

A diffraction grating is a natural consequence of the effect on the diffraction grating
interference pattern when the number of slits is increased. Diffraction Figure 5 Diffraction grating.
gratings are used to produce optical spectra. A grating contains a large
number of parallel, equally spaced slits or “lines” (normally etched
in glass or plastic) – typically there are 600 lines per millimetre (see
figure 5). When light is incident on a grating it produces interference
maxima at angles θ given by

nλ = d sinθ

The spacing between the slits is small, which makes the angle θ large for a A θ
fixed wavelength of light and n. This means that we cannot use the small
angle approximation for relating the wavelength to the position of the dθ
maxima as we did for a double-slit. Figure 6 shows a section of a diffraction B
grating in which three consecutive slits deviate the incident waves towards
a maximum. The slits are so narrow and the screen is so far away from the θ
grating that the angles made by the diffracted waves are virtually identical. C
Providing the path difference between waves coming from the same part of
successive slits is an integral number of wavelengths, the waves will reach portion of grating light diffracted
the screen in phase and give a maximum. In the case of a diffraction grating at θ to normal
the distance d is taken as the length of both the transparent and opaque
sections of the slit (they are taken to be of equal width). From figure 6 we Figure 6 Deriving the diffraction
see that the path difference between those waves coming from A and B and grating equation.
the path between waves coming B and C will each be dsinθ. This means

371

9 WAVE PHENOMENA (AHL)

that nλ = dsinθ (where n is the “order” of the maximum and is zero for the
central maximum, 1 for the first maximum on each side of the centre, etc.).

We can see that the angular positions of the interference maxima
depend on the grating spacing, d. The shape of the diffraction envelope,
however, is determined by the width of the clear spaces (as for a
single slit). The maxima must lie within the envelope of the single slit
diffraction pattern if they are to be relatively intense; this sets an upper
limit on how wide the transparent portions of the grating can be.

Figure 7 Dispersing white light with a One of the uses of a diffraction grating is to disperse white light into
diffraction grating. its component colours: this is because different wavelengths produce
maxima at different angles. Figure 7 shows that light of greater
wavelength (for any given order) is deviated by a larger angle. This is in
line with what we would predict from using nλ = d sin θ. Each successive
visible spectrum repeats the order of the colours of the previous one but
becomes less intense and more spread out.

Grating spacing and number of lines per mm

It is usual for the “number of lines per mm” or (N) to be quoted for

a diffraction grating, rather than the spacing. N must be converted

into d in order to use the diffraction grating equation nλ = d sin θ.
_1_
This is straightforward because d = but N must first be converted to
N
the number of lines per metre by multiplying by 1000 before taking

the reciprocal.

Nature of science The wavelengths of low-energy X-rays is
around 10-10 m or 0.1 nm. The 600 lines per
Appropriate wavelengths for effective millimetre optical grating will not produce any
dispersion observable maximum ... therefore, what will
diffract X-rays? The spacing of ions in crystals is
For a diffraction grating to produce an observable of the same order of magnitude as X-rays and the
pattern, the grating spacing must be comparable regular lattice shape of a crystal can perform the
to the wavelength of the waves. The wavelength same task with X-rays as a diffraction grating does
of visible light is between approximately with visible light. The diffracted patterns are now
400–700 nm. A grating with 600 lines per mm commonly detected with charge-coupled device
has a spacing of approximately 2000 nm or four (CCD) detectors.
times the wavelength of green light. This spacing
produces very clear images.

Worked example Solution

A diffraction grating having 600 lines per N = 6.00 × 105 lines per metre.
millimetre is illuminated with a parallel beam
of monochromatic light, which is normal to the d = _1_ = ____1____ = 1.67 × 10-6 m.
grating. This produces a second-order maximum
which is observed at 42.5° to the straight- N 6.00 × 105
through direction. Calculate the wavelength of
the light. nλ = d sin θ so λ = _ds_in__θ = _1_.6_7_×__1_0_-6_×__si_n_4_2_.5_ (don’t

n 2

forget to have your calculator in degrees)

λ = 5.64 × 10-7 = 564 nm (all data is given to
three significant figures so the answer should also
be to this precision).

372

9.3 INTERFERENCE

Nature of science

Using a spectrometer

A spectrometer is a useful, if expensive, piece of used because the “lines” are the dispersed images
laboratory equipment that allows the wavelengths of the collimator slit. Before measurements are
of light emitted by sources to be analysed. The made the telescope is focused on the collimator
instrument consists of a collimator, turntable slit – this means that the dispersed light will also
and telescope. A diffraction grating or other light be in focus. It is usual for sources that are said to
disperser is placed on the turntable, which is be “monochromatic” to actually emit a variety of
carefully levelled. The collimator uses lenses to different wavelengths. By using the spectrometer,
produce a parallel beam of light from a source – the angle θ in the diffraction grating equation can
this light beam is then incident on the grating that be measured. This is usually done by measuring
disperses it. The end of the collimator furthest the angle between the two first-order images on
from the grating has a vertical adjustable slit either side of the straight-through position i.e. 2θ.
that serves as the source – when we talk about With a knowledge of the grating spacing d, the
spectral lines we infer that a spectrometer is being wavelength λ can be determined.

adjustable slit
collimator

light source diffraction grating
turntable

angular scale telescope

θ

Figure 8 Key features of a spectrometer.

Interference by division of amplitude

It was mentioned in the introduction to this sub-topic that there are two
ways of providing coherent sources that are able to interfere. Young’s double
slit and multiple slits all derive their interfering waves by taking waves from
different parts of the same wavefront. Because the interfering waves have
all come from the same wavefront they will be in phase with each other.
Wherever the waves meet they will interfere and a fringe pattern can be
obtained anywhere in front of the sources (the slits). Since this interference
can be found anywhere the fringes are said to be “non-localized”.

Division of amplitude is a method of achieving interference using two
waves that have come from the same point on a wavefront. Each wave
has a portion of the amplitude of the original wave. In order to achieve
interference by division of amplitude, the source of light must come
from a much bigger source than the slit used for division of wavefront
interference. The image produced will, however, be “localized” to one
place instead of being found anywhere in front of the sources.

Thin film interference

Figure 9 shows a wave incident at an angle θ to the normal to the
surface of a film of transparent material (such as low-density oil or
detergent) having refractive index n. This diagram is not drawn to scale

373

9 WAVE PHENOMENA (AHL)

– θ and t are both very small so that incident wave is effectively normal
to the surface. The incident wave partially reflects at the top surface
of the film and partially refracts into the film. This refracted wave, on
reaching the lower surface of the film, again partially reflects (remaining
in the film) and partially refracts into the air below the film. This process
can occur several times for the same incident wave.

incident wave ABC

Waves reflected by the film (you θ air

may be tested on this in IB Physics t transparent medium
examinations) of refractive index n

In this case A has been reflected from air
the top surface of the film and, because
the reflection is at an optically denser A′ B′ C′
medium, there is a phase change of π Figure 9 Interference at parallel-sided thin film.
radian (equivalent to half a wavelength).
The wave B travels an optical distance of Waves transmitted through the film (you will not be tested on this in
2tn before it refracts back into the air. Thus
the optical path difference between A and IB Physics examinations)
B will be 2tn. If there had been no phase
change then this optical distance would If the film is thin and θ is small the (geometrical) path difference between waves that
equal mλ for constructive interference.
However, because of A’s phase change at have passed through the film (i.e. between A' and B' or between B' and C') will be very
the top surface the overall effect will be
destructive interference. nearly 2t, in other words B' travels an extra 2t compared with A'. Because the waves

Thus for the light reflecting from the film are travelling in a material of refractive index n, the waves will slow down and the
when
wavelength becomes shorter. This means that, compared with travelling in air, they
2tn = mλ
will take longer to pass through the film. This is equivalent to them travelling through a
there will be destructive interference and
when thicker film at their normal speed. The optical path difference will, therefore, be 2tn. If

2tn = (m + _21__)λ this distance is equal to mλ where m = 0, 1, 2 etc. (we are using m to avoid confusion
there will be constructive interference.
with the refractive index n) then there will be constructive interference. If the optical
_1__
path difference is equal to an odd number of half wavelengths, (m + )λ, then there
will be destructive interference. 2

Nature of science glass (having a refractive index of 1.38). When

Coating of lenses it is used to coat a lens the waves reflecting

When light is incident on a lens some of it will from both the magnesium fluoride and the
be reflected and some transmitted. The reflected
light is effectively wasted, reducing the intensity glass will undergo a phase change of π radian
of any image formed (by the eye using corrective
lenses, in a camera or in a telescope). By coating ... so the phase changes effectively cancel each
the lens with a transparent material of quarter
of a wavelength thickness, the light reflected by other out. The optical path difference between
the coating and the light reflected by the lens
can be made to interfere destructively and thus the waves will be equal to the refractive index
eliminate the reflection altogether. Magnesium
fluoride is often used for the coating and is of magnesium fluoride multiplied by twice
optically denser than air but less dense than
the thickness of the coating. For destructive =_4λ_n_.λ2_

interference the optical path difference is 2nt
so the thickness of the coating is given by t=

With white light there will be a range of

374

9.3 INTERFERENCE

wavelengths and so it is impossible to match (= red + blue). By using multiple layers of a
the thickness of the coating to all of these material of low refractive index and one of
wavelengths. If the thickness is matched to green higher refractive index, it is possible to reduce
light, then red and blue will still be reflected the amount of light reflected to as little as 0.1%
giving the lens the appearance of being magenta for a chosen wavelength.

Worked example Solution

a) Name the wave phenomenon that is a) Although there is reflection involved, the
responsible for the formation of regions of colours come about because of interference.
different colour when white light is reflected
from a thin film of oil floating on water. b) (i) n is the refractive index of the oil. Because

b) A film of oil of refractive index 1.45 floats on the waves are travelling in oil, they move
a layer of water of refractive index 1.33 and is
illuminated by white light at normal incidence. more slowly than they would do in air and

illumination so the effective path difference between

air the waves reflected at the air–oil interface
oil
and the waves at the oil–water interface
water
is longer by a factor of 1.45. The wave
When viewed at near normal incidence a
particular region of the film looks red, with an reflected at the air–oil interface undergoes a
average wavelength of about 650 nm. _λ_
phase change equivalent to (oil is denser
(i) Explain the significance of the refractive 2
indices of oil and water with regard to than air). The waves reflected at the oil–
observing the red colour.
water interface undergo no phase change
(ii) Calculate the minimum film thickness.
on reflection at the less dense medium. So

for the bright constructive red interference
2tn = (m + _21_)λ

if m = 0,

2tn = _λ or λ = 4tn
2

(ii) t = _650 = 110 nm
(4 × 1.45)

Nature of science Figure 10 Thin film interference in a vertical soap film.

Vertical soap films

Figure 10 shows the colours of light transmitted by
a vertical soap film. Over a few seconds the film
drains and becomes thinner at the top and thicker
at the bottom. When the film is illuminated with
white light, the reflected light appears as a series
of horizontal coloured bands. The bands move
downwards as the film drains and the top becomes
thinner. The top of the film appears black just
before the film breaks. It has now become too
thin for there to be a path difference between the
waves coming from the two surfaces of the film.
The phase change that occurs for the light reflected
by the surface of the film closest to the source
means that, for all colours, there is cancellation
and so no light can be seen.

375

9 WAVE PHENOMENA (AHL)

TOK

The aesthetics of physics

The colour in the soap film makes a beautiful image. Does physics need to rely
upon the arts to be aesthetic? Herman Bondi the Anglo-Austrian mathematician
and cosmologist implied that Einstein believed otherwise when he wrote:

“What I remember most clearly was that when I put down a suggestion that
seemed to me cogent and reasonable, Einstein did not in the least contest
this, but he only said, ‘Oh, how ugly.’ As soon as an equation seemed to
him to be ugly, he really rather lost interest in it and could not understand
why somebody else was willing to spend much time on it. He was quite
convinced that beauty was a guiding principle in the search for important
results in theoretical physics.”

— H. Bondi

How does a mathematical equation convey beauty? Is the beauty in the
mathematics itself or what the mathematics represents?

9.4 Resolution

Understanding Applications and skills

➔ The size of a diffracting aperture ➔ Solving problems involving the Rayleigh
criterion for light emitted by two sources
➔ The resolution of simple monochromatic diffracted at a single slit
two-source systems
➔ Describing diffraction grating resolution

Equations
➔ Rayleigh’s criterion: θ = 1.22_λb___
➔ resolvance of a diffraction grating:
___λ____
R = = mN
△λ

Nature of science atomic and sub-atomic scale. The process of making
an observation using these waves disturbs the
How far apart are atoms? system and increases the uncertainty with which
we can locate an object. Heisenberg’s uncertainty
When we view objects, we are limited by the principle places a limit on how close we can be to
wavelength of the light used to make an observation finding the exact position of objects on the quantum
– shorter wavelengths such as X-rays, gamma scale.
rays and fast-moving electrons will improve the
resolution that is achievable. However, even using
the shortest wavelengths obtainable does not allow
us to locate the exact position of objects on the

376

9.4 RESOLUTION

Introduction

Resolution is the ability of an imaging system to be able to
produce two separate distinguishable images of two separate
objects. The imaging system could be an observer’s eye, a camera, a
radio telescope, etc. Whether objects can be resolved will depend on the
wavelength coming from the objects, how close they are to each other
and how far away they are from the observer.

Diffraction and resolution images fully resolved

We have seen that when light passes through an aperture a diffraction images just resolved
pattern is formed. For an optical system the aperture could be the pupil
of the observer’s eye or the objective lens of a telescope. When there images unresolved
are two sources of light two diffraction patterns will be formed by the ▲ Figure 1 Diffraction intensity
system.
patterns of two objects viewed
How close can these patterns be for us to still recognize that there are through a circular aperture.
two sources?

Two objects observed through an aperture will produce two diffracted
images which may or may not overlap. In the late nineteenth century
the English physicist, John William Strutt, 3rd Baron Rayleigh, proposed
what is now known as the Rayleigh criterion for resolution of images.
This states that two sources are resolved if the principal maximum
from one diffraction pattern is no closer than the first minimum
of the other pattern.

The limit to resolution is when the principal maximum of the diffraction
pattern from one source lies on the first minimum diffraction pattern
from the second source (and vice versa). Diffracted images further apart
than this limit will be resolved and those closer will be unresolved.

Figure 1 shows the diffraction intensity patterns produced by two objects
the same distance apart but viewed through a circular aperture from
different distances. The variation of intensity with angle for each of the
diffraction patterns is shown below the image. According to the Rayleigh
criterion, the uppermost pair of images are fully resolved because the
principal maximum of each diffraction pattern lies further from the other
than the first minimum. The central images are just resolved since the
principal maximum of one diffraction pattern is at the same position as
the first minimum of the second diffraction pattern. The bottom images
are unresolved as the principal maximum of one diffraction pattern lies
closer to the second pattern than its first minimum.

Nature of science at resolution about half that of Rayleigh. His
criterion is that the two diffraction patterns when
Other criteria for resolution? added together give a constant amplitude in the
regions of the two central maxima.
Rayleigh’s criterion is not the only one used
in optics. Many astronomers believe that they
can resolve better than Rayleigh predicts.
C M Sparrow developed another criterion for
telescopes that leads to an angular separation

377

9 WAVE PHENOMENA (AHL)

Resolution equation

For single slits we saw in Sub-topic 9.2 that the first minimum occurs
_λ_
when the angle with the straight-through position is given by θ =
a
where λ is the wavelength of the waves and b is the slit width. With a

circular aperture the equation is modified by a factor of 1.22, but derivation

of this factor is beyond the scope of the IB Physics course. So we have

θ =1.22 _λ_

a

in this case a is the diameter of the circular aperture (or very commonly
the diameter of the lens or mirror forming the image).

The pupil of the eye has a diameter of about 3 mm and, taking visible

light to have a wavelength in the order of 6 × 10−7 m, the minimum
t_1h._i2s_23_l×i×_m_61_i×0t_-3i_1s0_-p_7 r≈ob2ab×ly1a0l-i4ttrlaeds.mall.
angle of resolution for the eye is θ =
Optical defects in the eye mean that

The primary mirror of the Hubble Space Telescope has a diameter of 2.4 m.
_1._2_2_×__6_×__10_-_7
For this telescope, the minimum angle of resolution is θ = ≈
2.4
3 ×10-7 rad. This is a factor of about a thousand smaller than that

achieved by the unaided eye – which means that the Hubble Space

Telescope is much better at resolving images. As this telescope is in

orbit above the atmosphere, it avoids the atmospheric distortion which

degrades images achieved by Earth-based telescopes. These concepts are

covered in more detail in Option C (Imaging).

Worked example

A student observes two distant point sources of light. The wavelength
of each source is 550 nm. The angular separation between these two
sources is 2.5 × 10–4 radians subtended at the pupil of a student’s eye.

Note a) State the Rayleigh criterion for the two images on the retina to be
just resolved.
● You may wish to support
your answer by drawing b) Estimate the diameter of the circular aperture of the eye if the two
the two intensity–angle images are just to be resolved.
curves if you think your
answer may not be clear. Solution

● It is common for a) The images will be just resolved when the diffraction pattern from
students to miss out the one of the point sources has its central maximum at the same
word “diffraction” in their position as the first minimum of the diffraction pattern of the
answers – this is crucial other point source.
to score all marks.
b) θ = 1.22 _λ_ = >b = 1.22 _λ_ = _1_.2_2_×__5_50__×_1_0_-_9

b θ 2.5 × 10-4

  = 2.7 × 10−3 m ≈ 3 mm

Nature of science minimum with the straight-through is related to

Diffraction and the satellite dish diameter b by the equation

When radiation is emitted from a transmitting θ = 1.22 _λ
satellite dish the waves diffract from the b
dish. The diameter of the dish behaves as
an aperture. The angle θ made by the first

378

9.4 RESOLUTION

When the wavelength increases so does the angle ● The footprint of a satellite has social
through which the waves become diffracted. This and political implications ranging from
means that the diffracted beam now covers a unwarranted observation to sharing television
larger area. However, increasing the diameter of programmes.
a dish narrows the beam and it covers a smaller
area. In satellite communications the footprint is satellite
the portion of the Earth’s surface over which the
satellite dish delivers a specified amount of signal footprint
power. The footprint will be less than the region ▲ Figure 2 Satellite footprint.
covered by the principal maximum because the
signal will be too weak to be useful at the edges
of the principal maximum. Figure 2 shows the
footprint of a communications satellite.

● Small values of the dish diameter will give a
large footprint but the intensity may be quite
low since the energy is spread over a large area.

Resolvance of diffraction gratings double-slit angular dispersion
of principal maximum
We have seen that diffraction gratings are used to disperse light of different
colours. Such gratings are usually used with spectrometers to allow the θD
angular dispersion for each of the colours to be measured. When using a X
spectrometer, knowing the number of lines per millimetre on the diffraction
grating and measuring the angles that the wavelengths of light are deviated diffraction grating angular dispersion
through allows the wavelengths of the colours to be determined. We of principal maximum
have seen that there is a limit to how close two objects can be before their
diffraction images are indistinguishable – the same is true of the different θG
wavelengths that can be resolved using a particular diffraction grating. X

When we previously looked at interference patterns for multiple slits in ▲ Figure 3 Angular dispersion of principal
Sub-topic 9.3, we saw that increasing the number of slits improves the maximum for diffraction grating compared
sharpness of the maxima formed. Figure 3 shows that, when light of with a double-slit.
the same wavelength is viewed from the same distance x, the angular
dispersion θD for the principal maximum with the double slit is larger 379
than angular dispersion θG for the diffraction grating. A sharper principal
maximum is one with less angular dispersion. With wider maxima there
is more overlap of images from different sources and lower resolution.

Using this argument we see that, when beams of light are incident on a
diffraction grating, a wider beam covers more lines (a greater number of
slits) and will produce sharper images and better resolution.

The resolvance R for a diffraction grating (or other device used to
separate the wavelengths of light such a multiple slits) is defined as
the ratio of the wavelength λ of the light to the smallest difference in
wavelength that can be resolved by the grating λ.

The resolvance is also equal to Nm where N is the total number of slits
illuminated by the incident beam and m is the order of the diffraction.

R = _λ = Nm
λ

The larger the resolvance, the better a device can resolve.

9 WAVE PHENOMENA (AHL)

Nature of science “IB Physics” – the top uses a small number of
large pixels, while the lower one is far better
Resolution in a CCD resolved by using a large number of small pixels.
The upper image is said to be “pixellated”.
Charge-coupled devices (CCDs) were originally
developed for use in computer memory devices IB Physics
but, today, appear in all digital cameras and
smartphones. When you buy a camera or phone ▲ Figure 4 Images with a small number of large pixels and a large
you will no doubt be interested in the number of number of small pixels.
pixels that it has. The pixel is a picture element
and, for example, a 20 megapixel camera will
have 2 × 107 pixels on its CCD. The resolution of
a CCD depends on both the number of pixels and
their size when compared to the projected image.
The smaller the camera, the more convenient it
is to carry, and (at present) cameras with pixels
of dimensions as small as 2.7 µm × 2.7 µm are
mass-produced. In general CCD images are
resolved better with larger numbers of smaller
pixels. Figure 4 shows two images of the words

Worked example

Two lines in the emission spectrum of sodium R = 981.7 (no units)

have wavelengths of 589.0 nm and 589.6 nm Thus 981.7 = Nm = N × 2

respectively. Calculate the number of lines per ∴ N = 498.8 lines

millimetre needed in a diffraction grating if the This is in a beam of width 0.10 × 10−3 m so, in
1 mm, there needs to be 4988 ≈ 5000 lines.
lines are to be resolved in the second-order
Since diffraction gratings are not normally made
spectrum with a beam of width 0.10 mm. with 4988 lines mm−1, the sensible choice is to use
one with 5000 lines mm−1!
Solution

R = _λ__ = _5_8_9_.0_ (both values are in nanometres so

△λ 0.6

this factor cancels)

You would be equally justified in using 589.6 (or
589.3 – the mean value) in the numerator here.

380

9.5 THE DOPPLER EFFECT

9.5 The Doppler effect

Understanding Applications and skills

➔ The Doppler effect for sound waves and ➔ Sketching and interpreting the Doppler effect
light waves when there is relative motion between source
and observer
Nature of science
➔ Describing situations where the Doppler effect
From water waves to the expansion of can be utilized
the universe
➔ Solving problems involving the change in
In his 1842 paper, Über das farbige Licht der frequency or wavelength observed due to the
Doppelsterne (Concerning the coloured light of Doppler effect to determine the velocity of the
the double stars), Doppler used the analogy of the source/observer
measurement of the frequency of water waves to
reason that the effect that bears his name should Equations
apply to all waves. Three years later, Buys Ballot
verified Doppler’s hypothesis for sound using Doppler equation
stationary and moving groups of trumpeters. During
his lifetime Doppler’s hypothesis had no practical ➔ for a moving source:
application; one hundred years on it has far-
reaching implications for cosmology, meteorology _v_
and medicine. v ± us
( )f ′ = f

➔ for a moving observer:

_v ±_uo
v
( )f ′ = f

➔ for electromagnetic radiation:
_f _v
f = _λ ≈ c
λ

Introduction Note

When there is relative motion between a source of waves and an It is only the component
observer, the observed frequency of the waves is different to the of the wave in the source–
frequency of the source of waves. The apparent change in pitch of observer direction that
an approaching vehicle engine and a sounding siren are common is used; perpendicular
examples of this effect. The Doppler effect has wide-ranging components of the motion
implications in both atomic physics and astronomy. do not alter the apparent
frequency of the wave.
The Doppler effect with sound waves

Although we use general equations for this effect, we build up the
equations under different conditions before combining them. In the
following derivations (that need not be learned but will help you to
understand the equations and how to answer questions on this topic) the
letter s refers to the source of the waves (the object giving out the sound)
and the letter o refers to the observer; f will always be the frequency of
the source and f ’ the apparent frequency as measured by the observer.

381

9 WAVE PHENOMENA (AHL)

1. Moving source and stationary observer

vv
ff

A us B
S S'
us
f

▲ Figure 1 The Doppler effect for a moving source and stationary observer.

At time t = 0 the source is at position S and it emits a wave that travels

outwards in all directions, with a velocity v as shown in figure 1. At

time t = T (i.e. one period later) the wave will have moved a distance
_v
equivalent to one wavelength or (using v = f λ) a distance = to reach
f
positions A and B. When the source is moving to the right with a
iptowsiitlilohnaevdeattraBveitllwedilltoapSp’,eaardtihstaatntcheeupsTre(v=io_uuf_s )s.lyTo
velocity us, in time T
a stationary observer

emitted crest has reached B but the source that emitted it has moved

forwards and now is at S’. Therefore, to the observer at B, the apparent
_v _u_s
wavelength (λ’) is the distance S’B = -
f f

∴ λ’ = _v - us
f

Thus, the wavelength appears to be squashed to a smaller value. This
means that the observer at B will hear a sound of a higher frequency
than would be heard from a stationary source – the sound waves travel
at speed v (which is unchanged by the motion of the source) so

_v _v
λ’ v - us
( )f’= = f

To an observer at A the wavelength would have appeared to be stretched
to a longer value given by S’A meaning that

λ’ = _v + us
f

and the observed frequency would be lower than that of a stationary
source, meaning that

_v _v
λ’ v + us
( )f’= = f

The two equations for f ’ can be combined into a single equation

_v
v ± us
( )f’= f

in which the ± sign is changed to - for a source moving towards a
stationary observer and to + for a source moving away from a
stationary observer.

2. Moving observer and stationary source

In this case the source remains stationary but the observer at B moves
towards the source with a velocity uo. The source emits crests at a

382

9.5 THE DOPPLER EFFECT

frequency f but the observer, moving towards the source, encounters

the crests more often, in other words at a higher frequency f ’. Relative

to the observer, the waves are travelling with a velocity v + uo and the

frequency is f ’. The wavelength of the crests does not appear to have
_vf .
changed and will be λ = = The wave equation (v = f λ) applied by the
observer becomes v + uo _v
the observer will be f ’ meaning that the frequency measured by
f

_v + uo
v
( )f’= f

When the observer moves away from the source the wave speed appears

to be v - uo and so

(_)f’= f
v - uo
v

Again the two equations for f ’ can be combined into a single equation

_v ± uo
v
( )f’ = f

in which the ± sign is changed to - for an observer moving away
from a stationary source and to + for an observer moving towards a
stationary source.

Worked example

A stationary loudspeaker emits sound of frequency of 2.00 kHz.
A student attaches the loudspeaker to a string and swings the
loudspeaker in a horizontal circle at a speed of 15 m s–1. The speed of
sound in air is 330 m s–1.

An observer listens to the sound at a close, but safe, distance from the
student.

a) Explain why the sound heard by the observer changes regularly.

b) Determine the maximum frequency of the sound heard by the
observer.

Solution

a) As the loudspeaker approaches the observer, the frequency
appears higher than the stationary frequency because the
apparent wavelength is shorter – meaning that the wavefronts
are compressed. As it moves away from the observer, the
frequency appears lower than the stationary frequency
because the apparent wavelength is stretched. The overall
effect is, therefore, a continuous rise and fall of pitch heard by
the observer.

b) The maximum observed frequency occurs when the speaker is
approaching the observer so:

)___v__ ___3_30___

v - us 330 - 15
f’= f( = ( )2000 = 2100 Hz (this is 2 s.f. precision in

line with 15 m s–1)

383

9 WAVE PHENOMENA (AHL)

384 The Doppler effect with light

The Doppler effect occurs not only with sound but also with light
(and other electromagnetic waves); in which case the frequency and
colour of the light differs from that emitted by the source. There is a
significant difference in the application of Doppler effect for sound
and light waves. Sound is a mechanical wave and requires a medium
through which to travel; electromagnetic waves need no medium.
Additionally, one of the assumptions or postulates of special relativity
is that the velocity of light waves is constant in all inertial reference
frames – this means that, when measured by an observer who is
not accelerating, the observer will measure the speed of light to be
3.00 × 108 m s-1 irrespective of whether the observer moves towards
the source or away from it and, therefore, it is impossible to distinguish
between the motion of a source and an observer. This is not true for
sound waves, as we have seen.

Although the Doppler effect equations for light and sound are
derived on completely different principles, providing the speed of
the source (or observer) is much less than the speed of light, the
equations give approximately the correct results for light or sound.
The quantities us and uo in the Doppler equations for sound have no
significance for light, and so we need to deal with the relative velocity
v between the source and observer. Thus, in either of the equations

_v
v ± us
( )f’= f

or

_v ± uo
v
( )f’= f

the wave speed is that of electromagnetic waves (c) and one of the

velocities us or uo is made zero while the other is replaced by the
relative velocity v.

Note The first of these equations becomes

● Substituting values into _c _1 _v -1
the second equation will c+ v c
give the same resulting ( ) ( ) ( )f’= f =f 1 + _v =f 1 +
relationship – you
may like to try this but c
remember you do not
need to know any of This can be expanded using the binomial theorem to approximate to
these derivations.
f ’ = f (1 - _v ) ≈ f - f _v
● The equation is only c c
valid when c ≫ v and
so cannot usually be (ignoring all the terms after the second in the expansion).
used with sound (where
the wave speed is ≈ This can be written as
300 m s–1 – unless
the source or observer f- f ’≈ f _v
is moving much more c
slowly than this speed).
or

∆f ≈ f _v
c

This equation is equivalent to

λ ≈ λ _v
c

9.5 THE DOPPLER EFFECT

Worked example

As the Sun rotates, light waves received on λ = 587.5618 × _1_.9_0_×__1_0_3 = 0.003 7 nm
Earth from opposite ends of a diameter show
equal but opposite Doppler shifts. The speed 3.00 × 108
of the edge of the Sun relative to the Earth is
1.90 km s-1. What wavelength shift should be Since one edge will approach the Earth –
expected in the helium line having wavelength the shift from this edge will be a decrease
587.5618 nm? in wavelength (blue shift) and that of the
other edge (receding) will be an increase in
wavelength (red shift).

Solution λ ≈ λ _vc .

Using f = f _v this is equivalent to

c

1.90 km s−1 = 1.90 × 103 m s−1

Nature of science unshifted

Applications of the Doppler effect redshifted

1. Astronomy blueshifted
▲ Figure 2 The Doppler shifted absorption spectra.
The Doppler effect is of particular interest
in astronomy – it has been used to provide gravitational red-shift and is discussed further in
evidence about the motion of the objects Option A. The cosmological red-shift arises from
throughout the universe. the expansion of space following the Big Bang
and is what we currently detect as the cosmic
The Doppler effect was originally studied in the microwave background radiation (this is further
visible part of the electromagnetic spectrum. Today, discussed in Option D).
it is applied to the entire electromagnetic spectrum.
Astronomers use Doppler shifts to calculate the 2. Radar
speeds of stars and galaxies with respect to the
Earth. When an astronomical body emits light Radar is an acronym for “radio detection and
there is a characteristic spectrum that corresponds ranging”. Although it was developed for tracking
to emissions from the elements in the body. By aircraft during the Second World War, the
comparing the position of the spectral lines for technique has wide-ranging uses today, including:
these elements with those emitted by the same ● weather forecasting
elements on the Earth, it can be seen that the lines ● ground-penetrating radar for locating
remain in the same position relative to each other
but shifted either to longer or shorter wavelengths geological and archaeological artefacts
(corresponding to lower or higher frequencies). ● providing bearings
Figure 2 shows an unshifted absorption spectrum ● radar astronomy
imaged from an Earth-bound source together with
the same spectral lines from distant astronomical
objects. The middle image is red-shifted indicating
that the source is moving away from the Earth. The
lower image is blue-shifted showing the source to
be local to the Earth and moving towards us.

Frequency or wavelength shifts can occur
for reasons other than relative motion.
Electromagnetic waves moving close to an object
with a very strong gravitational field can be
red-shifted – this, unsurprisingly, is known as

385

9 WAVE PHENOMENA (AHL)

● use in salvaging transmitter–receiver
● collision avoidance at sea and in the air. ultrasound beam

θ blood cells
v
direction of
blood flow

blood vessel
▲ Figure 4 The Doppler effect used to measure the speed of

blood cells.

approximately 20 kHz) is transmitted towards

a blood vessel. The change in frequency of the

beam reflected by a blood cell is detected by the

receiver. The speed of sound and ultrasound is

around 1500 m s–1 in body tissue so the equation
_v
▲ Figure 3 Radar screen used in weather forecasting. f ≈ f is appropriate for blood cells moving at
c
speeds of little more than 1 m s–1. As the blood
Radar astronomy differs from radio astronomy
as it can only be used for Moon and planets close does not flow in the direction of the transmitter–
to the Earth. This depends on microwaves being
transmitted to the object which then reflects receiver, there needs to be a factor that will give
them back to the Earth for detection. In order to
use the Doppler equations, we must recognize the necessary component of the blood velocity in
that the moving object first of all behaves as a
moving observer and then, when it reflects the a direction parallel to the transmitter – receiver.
microwaves, behaves as a moving source.
As with radar, there will need to be a factor of

two included in the equation – the shift is being

caused by the echo from a moving reflector. As

This means, for microwave sensing, the equation can be seen from figure 4, the equation for the
rate of blood flow will be f ≈ 2f _v_c_oc_s _θ.

f ≈ f _v is adapted to become f ≈ 2f _vc . The Doppler effect flowmeter may be used in
preference to “in-line” flowmeters because it is
c non-invasive and its presence does not affect the
rate of flow of fluids. The fact that it will remain
3. Measuring the rate of blood flow outside the vessel that is carrying the fluid
means it will not suffer corrosion from contact
The Doppler effect can be used to measure with the fluid.
the speed of blood flow in blood vessels in the
body. In this case ultrasound (i.e. longitudinal
mechanical waves of frequency above

Worked example Using

Microwaves of wavelength 150 mm are transmitted c = fλ gives f = _c_ = _3_.0_0_×__1_0_8 = 2.00 × 109 Hz.
from a source to an aircraft approaching the source.
The shift in frequency of the reflected microwaves λ 150 × 10-3
is 5.00 kHz. Calculate the speed of the aircraft
relative to the source. Using the Doppler shift equation for radar

Solution f ≈ 2f _v =>v ≈ __f_c = _5_.0_0_×__1_03_×__3_.0_0_×__1_0_8

The data gives a wavelength and a change of c 2f 2 × 2.00 × 109
frequency so we need to find the frequency of the = 375 m s−1
microwaves.

386

QUESTIONS

Questions b) Explain why the magnitude of the tension in
the string at the midpoint of the oscillation is
1 (IB) greater than the weight of the pendulum bob.
The variation with displacement x of the
acceleration a of a vibrating object is shown c) The pendulum bob is moved to one side
below. until its centre is 25 mm above its rest
position and then released.
3000 a/m s−2
point of suspension
2000 rigid support

1000 0.80 m
x/mm

−0.6 −0.4 −0.2 0 0.2 0.4 0.6
−1000

−2000 25 mm

−3000 pendulum bob

a) State and explain two reasons why the (i) Show that the speed of the pendulum
graph indicates that the object is executing bob at the midpoint of the oscillation is
simple harmonic motion. 0.70 m s–1.

b) Use data from the graph to show that the (ii) The mass of the pendulum bob is 0.057 kg.
frequency of oscillation is 350 Hz. The centre of the pendulum bob is 0.80 m
below the support. Calculate the
c) State the amplitude of the vibrations. magnitude of the tension in the string
(9 marks) when the pendulum bob is vertically
below the point of suspension.
2 (IB) (10 marks)
a) A pendulum consists of a bob that is
suspended from a rigid support by a 3 (IB)
light inextensible string. The pendulum a) A particle of mass m attached to a light
bob is moved to one side and then spring is executing simple harmonic motion
released. The sketch graph shows how in a horizontal direction.
the displacement of the pendulum bob
undergoing simple harmonic motion State the condition (relating to the net force
varies with time over one time period. acting on the particle) that is necessary for it
to execute simple harmonic motion.
displacement b) The graph shows how the kinetic energy EK of
the particle in (a) varies with the displacement
0 0 time x of the particle from equilibrium.

Copy the sketch graph and on it clearly label 0.07 EK/J
(i) a point at which the acceleration of the 0.06

pendulum bob is a maximum. 0.05
(ii) a point at which the speed of the
0.04
pendulum bob is a maximum.
0.03

0.02

0.01

0

x/m 387
−−−−−00000.....0000051234
0.01
0.02
0.03
0.04
0.05

9 WAVE PHENOMENA (AHL)

(i) On a copy of the axes above, sketch Z Q
a graph to show how the potential ϕW Y
energy of the particle varies with the bX P
displacement x.
slit screen
(ii) The mass of the particle is 0.30 kg. Use data
from the graph to show that the frequency The angle ϕ is small.
f of oscillation of the particle is 2.0 Hz.
a) On a copy of the diagram, label the half
(8 marks) angular width θ of the central maximum of
the diffraction pattern.
4 (IB)
a) Describe what is meant by the diffraction b) State and explain an expression, in terms of
of light. λ, for the path difference ZW between the
rays ZP and XP.
b) A parallel beam of monochromatic light
from a laser is incident on a narrow slit.
The diffracted light emerging from the slit is
incident on a screen.

screen

slit

parallel light 0.40 mm C c) Deduce that the half angular width θ is
wavelength 620 nm given by the expression

θ = _λ
b

1.9 m d) In a certain demonstration of single slit
diffraction, λ = 450 nm, b = 0.15 mm and
The centre of the diffraction pattern produced the screen is a long way from the slits.
on the screen is at C. Sketch a graph to show
how the intensity I of the light on the screen Calculate the angular width of the central
varies with the distance d from C. maximum of the diffraction pattern on the
screen.
c) The slit width is 0.40 mm and it is 1.9 m
from the screen. The wavelength of the (8 marks)
light is 620 nm. Determine the width of the
central maximum on the screen. 6 (IB)
Monochromatic parallel light is incident on two
(8 marks) slits of equal width and close together. After
passing through the slits, the light is brought to a
5 (IB) focus on a screen. The diagram below shows the
Plane wavefronts of monochromatic light are intensity distribution of the light on the screen.
incident on a narrow, rectangular slit whose
width b is comparable to the wavelength λ of I
the light. After passing through the slit, the
light is brought to a focus on a screen. AB
distance along the screen
The line XY, normal to the plane of the slit, is
drawn from the centre of the slit to the screen. a) Light from the same source is incident on
The points P and Q are the first points of many slits of the same width as the widths
minimum intensity as measured from point Y. of the slits above. On a copy of the diagram,
draw a possible new intensity distribution
The diagram also shows two rays of light of the light between the points A and B on
incident on the screen at point P. Ray ZP leaves the screen.
one edge of the slit and ray XP leaves the centre
of the slit.

388

QUESTIONS

A parallel beam of light of wavelength a) State the phase change that occurs when
450 nm is incident at right angles on a light is reflected from
diffraction grating. The slit spacing of the
diffraction grating is 1.25 × 10–6 m. (i) surface A

b) Determine the angle between the central (ii) surface B.
maximum and first order principal
maximum formed by the grating. The light incident on the plastic has a
wavelength of 620 nm. The refractive index of
(4 marks) the plastic is 1.4.

7 (IB) b) Calculate the minimum thickness of the
Light of wavelength 590 nm is incident film needed for the
normally on a diffraction grating, as shown light reflected from surface A and surface B
below. to undergo destructive interference.

grating 6.0 × 105 first order (5 marks)
lines per metre zero order
first order 9 (IB)
light wavelength
590 nm The two point sources A and B emit light of
the same frequency. The light is incident on
a rectangular narrow slit and, after passing
through the slit, is brought to a focus on the
screen.

The grating has 6.0 × 105 lines per metre. A

a) Determine the total number of orders of B
diffracted light, including the zero order, point
that can be observed.
sources
b) The incident light is replaced by a beam of
light consisting of two wavelengths, 590 nm slit
and 589 nm.
screen
State two observable differences between
a first-order spectrum and a second-order a) B is covered. Sketch a graph to show how
spectrum of the diffracted light. the intensity I of the light from A varies
with distance along the screen. Label the
(6 marks) curve you have drawn A.

8 (IB) b) B is now uncovered. The images of A
Monochromatic light is incident on a thin film and B on the screen are just resolved.
of transparent plastic as shown below. Using your axes, sketch a graph to show
how the intensity I of the light from B
C varies with distance along the screen.
Label this curve B.
monochromatic AB
light film c) The bright star Sirius A is accompanied by a
much fainter star, Sirius B. The mean distance
The plastic film is in air. of the stars from Earth is 8.1 × 1016 m. Under
ideal atmospheric conditions, a telescope with
Light is partially reflected at both surface A and an objective lens of diameter 25 cm can just
surface B of the film. resolve the stars as two separate images.

Assuming that the average wavelength
emitted by the stars is 500 nm, estimate the
apparent, linear separation of the two stars.

(6 marks)

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9 WAVE PHENOMENA (AHL)

10 (IB) a) Explain, using a diagram, any difference
between f ’and f.
a) Explain what is meant by the term
resolvance with regards to a diffraction b) The frequency f is 3.00 × 102 Hz. An
grating. observer moves towards the stationary
car at a constant speed of 15.0 m s–1.
b) A grating with a resolvance of 2000 is used Calculate the observed frequency f ’ of
in an attempt to separate the red lines in the sound. The speed of sound in air is
the spectra of hydrogen and deuterium. 3.30 × 102 m s-1.

(i) The incident beam has a width of 0.2 mm. (5 marks)
For the first order spectrum, how many
lines per mm must the grating have? 13 (IB)
The wavelength diagram shown below
(ii) Explain whether or not the grating is represents three lines in the emission spectrum
capable of resolving the hydrogen lines sample of calcium in a laboratory.
which have wavelengths 656.3 nm and
656.1 nm. AB C

(6 marks)

11 A source of sound approaches a stationary

observer. The speed of the emitted sound

and its wavelength, measured at the source, wavelength

are v and λ respectively. Compare the wave

speed and wavelength, as measured by the A distant star is known to be moving directly

observer, with v and λ . Explain your away from the Earth at a speed of 0.1c. The

answers. (4 marks) light emitted from the star contains the

12 The sound emitted by a car’s horn has emission spectra of calcium. Copy the diagram
frequency f, as measured by the driver. An
observer moves towards the stationary car at and sketch the emission spectrum of the
constant speed and measures the frequency of
the sound to be f ’. star as observed in the laboratory. Label the

lines that correspond to A, B, and C with the

letters A*, B*, and C*. Numerical values of the

wavelengths are not required. (3 marks)

390