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A-LEVEL FOUNDATION PHYSICS
CAMBRIDGE MODULAR SCIENCES AND OCR
Wave motion and progressive
waves
• A wave motion is a means of transferring
energy from one point to another without
there being any transfer of matter between
the points.
• Waves that move through a material or a
vacuum and distribute energy from the
source to the surroundings are called
progressive waves.
Classification of waves
• Mechanical
waves
•
•
•
•
Water waves
Sound waves
Shock waves
Waves in a stretched
string and spring.
• Electromagnetic
waves
• Light
• Radio
• X-rays
• Microwaves
• Gamma rays
• Infra-red
• Ultraviolet
Describing waves (definitions)
• Frequency (f) in Hz.
• Wavelength () in m or
nm.
• Displacement(x) in m.
• Amplitude (a) in m.
• Speed (v) in m/s.
• Period (T) in s.
• Phase difference in
degrees.
• Wave equation
• V=fx
• This equation applies
to any wave.
• The wave speed is
measured in m, if
frequency is measured
in Hz and wavelength
in m.
Frequency calculations
• The calculation of the frequency of sound waves can
be done by using an oscilloscope and a signal
generator with or without a speaker and microphone.
• On CRO, you set time to 0.5 ms and count the
number of divisions on screen of one complete wave
(distance between two crests)
• Then find the time period T = divisions x time
• You must change time from milliseconds to seconds
before calculating T.
• Then calculate f = 1/T in Hz.
• You can also calculate the wavelength of sound
waves using the same method.
More on polarisation
• Electromagnetic waves are transverse waves and
can therefore be polarised.
• If a radio wave is transmitted from a vertical aerial
then the wave will have vertical polarisation and
any receiving aerial must also be positioned
vertically.
• The main transmitters in the U.K. send out signals
which are horizontally polarised.
• Many areas of the country are now served by
mixed polarisation transmitters.
• Examples: Sunglasses reduce the glare of light,
LCD displays on calculators are usually polarised.
The laws of reflection
• The angle of incidence is equal to the angle
of reflection, this means sin i = sin r .
• The incident ray, the reflected ray and the
normal all lie in the same plane.
Normal
• Rules for image position- when an object is
placed in front of a plane mirror:
• The image formed is the same size as the
object.
• The image is as far behind the mirror as the
object is in front.
• The line joining the image to the object is at
right angles to the mirror.
• The image is virtual and laterally inverted.
If the glancing
angle is g, then the A
angle of deviation
of a ray by a plane
mirror is 2g.
P
2
B
Deviation of a
reflected ray
by a rotated
mirror.
Angle COB = 2g
M2
g
X


O
The mirror is rotated through an
angle  to a position M2 from M1.
The ray is now reflected in an OP
direction.
M1
The glancing angle with M2 is
g + . Hence the new angle of
deviation COP = 2 (g +)
C
Therefore the deviation of a reflected ray through an
angle BOP = COP - COB = 2 (g + ) - 2g = 2
In a number of instruments a beam of light is used as a
pointer; this has a negligible weight and so is sensitive
to deflections of the moving system.
A mirror galvanometer is used for measuring very small electric
currents by using a rotating mirror.
The formula to calculate the deflection of the ray of light from a
rotating mirror when a small current flows through it, is as below:
tan  = x / d
where x = deflection of the spot of light on scale
and d = distance between the beam of light and the mirror.
Angle of
incidence
i
Optically less dense
medium, e.g. air
Optically more
dense medium,
e.g. glass
Angle of
refraction
r
Sine of angle of incidence (sin i)
Refractive index n = ----------------------------------------Sine of angle of refraction (sin r)
Speed of light in less dense medium (C1)
Refractive index n = ------------------------------------------------------
Speed of light in more dense medium (C2)
Wavelength of light in less dense medium (1)
Refractive index n = ------------------------------------------------------------Wavelength of light in more dense medium (2)
Material
Refractive index
Diamond
2.42
Ruby
1.76
Glass
1.50
Glycerol
1.47
Water
1.33
Ice
1.31
Air
1.00STP
STD = standard
temperature and pressure
AIR
GLASS
90
LESS DENSE
1
MEDIUM
Sin (C) = ------MORE DENSE
MEDIUM
n
Where n = refractive
index of the interface
Less dense medium
C
r
More dense medium
Uses: View finder
systems of many
cameras, prismatic
binoculars and
periscopes.
The critical angle of the glass/air boundary is about 42°.
Thus, whenever light which is travelling in glass is
incident on such a boundary at an angle of more than
42, it undergoes a total internal reflection.
CLADDING n2 less dense
CORE n1 more dense
Refractive index of core n1
Refractive index of fibre optics = ---------------------------------------Refractive index of cladding n2
The transmission of light through a glass fibre is one of the most
important applications of total internal reflection.
There is no energy lost at the reflecting surface, therefore it is
possible to send a light wave along a glass fibre and for the light to
be reflected millions of times without any loss of brightness.
Glass fibres have a core a few micrometers (.000001m = 1 micron)
in diameter and are covered with a layer of glass cladding.
The refractive indices of the core and the cladding are different.
When the light beam travels down the fibre, total internal
reflection occurs at the interface between the core and the
cladding, as long as the angle of incidence is large enough.
USES OF FIBRE OPTICS
• To see inside the human body, for
endoscopy inspection and keyhole
surgery.
• For lighting road signs.
• In security fences.
• For telephone and cable TV.
100 Fibre optics
1 cable
Polyurethane
sheath
Q1.
What are the critical angles for the interfaces between the
following materials and the air? a) ruby (n= 1.76)
b) diamond (n= 2.42). (air n = 1.00 at STP)
Q2.
Light is travelling along an optical fibre of refractive index
1.52. What must be the R.I. Of the cladding if the critical
angle is to be 85º?
Q3.
Q4.
What is meant by polarisation of a wave? Explain whether or
not a) sound waves b) electromagnetic waves can be
polarised.
Describe an experiment to determine the frequency and the
wavelength of a sound wave in air.
Q5.
Q6.
A)
Distinguish between longitudinal waves and transverse
waves.
The figure shows a cross-section through a bicycle reflector in
the form of a transparent red plastic prism of refractive index
(R.I.) 1.6.
90
45
B) Ray X is incident normally on one face of the
prism, as shown in the figure. Use the result of
your calculation in A to complete the path of the
ray X through the plastic prism.
SOLVE OLD QUESTION PAPERS TO GET SUCCESS
X