9749 H2 PHYSICS
Lecture Notes
Chapter 11
WAVE MOTION
Content




Progressive waves
Transverse and longitudinal waves
Polarisation
Determination of frequency and wavelength of sound waves
Learning Outcomes
Candidates should be able to:
(a)
show an understanding and use the terms displacement, amplitude, period, frequency,
phase difference, wavelength and speed
(b)
deduce, from the definitions of speed, frequency and wavelength, the equation v  f
(c)
recall and use the equation v  f
(d)
show an understanding that energy is transferred due to a progressive wave
(e)
recall and use the relationship, intensity  (amplitude)2
(f)
show an understanding of and apply the concept that a wave from a point source and
travelling without loss of energy obeys an inverse square law to solve problems
(g)
analyse and interpret graphical representations of transverse and longitudinal waves
(h)
show an understanding that polarisation is a phenomenon associated with transverse waves
(i)
recall and use Malus’ law (intensity  cos2θ) to calculate the amplitude and intensity of a
plane polarised electromagnetic wave after transmission through a polarising filter
(j)
determine the frequency of sound using a calibrated oscilloscope
(k)
determine the wavelength of sound using stationary waves. **
**
to be taught in the topic “Superposition”
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Lecture Notes
Waves
A wave is a propagation of a disturbance which transfers energy from one point to another without
the physical transfer of matter. The source of any wave is a vibration or an oscillation.
A wave which results in a net transfer of energy from one place to another is known as a
progressive (travelling) wave. Energy is transferred from the source outwards, along the direction
of propagation of the wave.
The contrast of a progressive wave is a stationary (standing) wave, which will be covered in the
topic “Superposition”.
Types of waves
Mechanical and Electromagnetic Waves
Waves may be classified as either mechanical or electromagnetic.
Mechanical waves require a material medium for their propagation. Examples of mechanical waves
are water waves and sound waves.
Electromagnetic (EM) waves consist of mutually perpendicular time-varying electric and magnetic
fields, and travel through vacuum at a speed of 3.0  108 m s-1. The energy of an EM wave is
transmitted as a result of the oscillations of its electric and magnetic fields. Examples of EM waves
include light, radio waves and X-rays.
Transverse and Longitudinal Waves
We can also distinguish waves by considering how the motions of the particles of matter are
related to the direction of propagation of the waves themselves. Waves can be classified as being
either transverse or longitudinal.
Transverse Wave
A transverse wave is a wave in which the oscillations of the particles are perpendicular to the
direction of propagation of the wave.
In a transverse progressive mechanical wave, the particles in the medium oscillate in a direction
perpendicular to the direction of wave propagation. An example of transverse mechanical wave is
water waves.
In a transverse progressive electromagnetic wave, the electric and magnetic fields oscillate in
mutually perpendicular directions, and both of these oscillations are in turn perpendicular to the
direction of wave propagation. An example of transverse electromagnetic wave is light waves.
Longitudinal Wave
A longitudinal wave is a wave in which the oscillations of the particles are parallel to the direction
of propagation of the wave. An example of longitudinal wave is sound waves.
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Lecture Notes
Terminology
displacement from
equilibrium position
amplitude, A
0
distance from
source
wavelength,
Fig. 1 Displacement-distance graph of a sinusoidal wave
displacement, x/y
amplitude, A
Position of an oscillating particle from its equilibrium position. (Unit: m)
The maximum distance (magnitude of displacement) of an oscillating
particle from its equilibrium position. (Unit: m)
For a progressive wave, it is the distance between any two successive
particles that are in phase. (e.g. the distance between 2 adjacent
maximum displacements) (Unit: m)
wavelength, 
displacement from
equilibrium position
amplitude, A
0
time
period,T
Fig. 2 Displacement-time graph of a sinusoidal wave
period, T
frequency, f
wave speed, v
wavefront
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Time taken for a particle to undergo one complete cycle of oscillation. It is
also the time taken for the wave to travel one wavelength. (Unit: s)
Number of oscillations performed by a particle per unit time. It is also the
number of wavelengths that pass a fixed point in one second. (Unit: Hz)
If the frequency of a wave is f, then f wavelengths pass a fixed point in
one second. The time taken for the wave to travel one wavelength, T, is
1
1
therefore seconds. Hence T  .
f
f
The distance travelled by the wave profile per unit time. (Unit: m s1)
A wavefront is a locus or imaginary line joining all the points of the wave
that have the same phase. It is useful to draw the wavefront by joining all
the crests of a wave, and then seeing it from a bird’s eye view. See
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Lecture Notes
figures below.
The distance between successive wavefronts is therefore equivalent to
one wavelength.
direction of wave motion
Fig. 3a Wavefronts

Fig. 3b Plane waves from a bird’s eye view
ray
A line drawn in the direction of the wave motion which is used to indicate
the path taken by the wave. Rays are always at right angles to the wave
fronts i.e. wave fronts are always perpendicular to the direction of
propagation.
wavefronts
ray
ray
wavefront
Fig. 4a Circular or spherical waves
Fig. 4b Plane waves
Derivation of v  f
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Lecture Notes
Since the wave travels a distance of  in time T, from definition of speed,
distance
speed 
time taken
d
v
t


T
 f
1

 f  
T

v  f
Note: The v here refers to the wave speed and not the speed of the oscillating particle.
velocity of rope particle
velocity of wave
Fig. 5 Velocity of particle and of wave of a transverse wave
Example 1
A sound wave of frequency 512 Hz travels from a rod into the air. The speed of sound is
4800 m s-1 in the rod, and 330 m s-1 in the air. Find the wavelength of sound in the rod and in air.
v
wavelength of sound in rod, rod  rod
f
4800

512
 9.375
 9.38 m
Since frequency of the sound wave remains unchanged as it leaves the rod and enters air,
v
wavelength of sound in air, air  air
f
330

512
 0.645 m
Example 2
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Lecture Notes
A wave is represented by its displacement-time graph and displacement-distance graph below.
Determine the speed of propagation of the wave.
y/cm
y/cm
10.0
10.0
0
t/s
0.1
0.2
0
x/m
0.1
0.2
0.3
0.4
0.5
From displacement-time graph, period of wave, T  0.20 s.
1
f 
T
1

0.20
 5.0 Hz
From displacement-distance graph, wavelength of wave,   0.40 m.
v  f
 5.0  0.40 
 2.0 m s1
Phase and Phase Difference
The phase () of a particle, in degrees () or radians (rad) gives a measure of the fraction of a
cycle that has been completed by an oscillating particle. One cycle corresponds to 360 or 2 rad.
Phase difference () is a measure of how much one wave is out of step with another or one
particle in a wave is out of step with another particle in the same wave. It is expressed in terms of
angle from 0 to 360 or 0 to 2 radians.
Two particles are in phase when they are in the same stage of oscillation at the same time (in step
with one another), i.e. zero phase difference. In Fig. 6, particles A and B are in phase; so are
particles C and D.
displacement
C
A
D
B
distance
0
Fig. 6 Particles that are in phase
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Lecture Notes
On the other hand, two particles are said to be out of phase when they are in different stages of
oscillation at a particular instant. In Fig. 7, particles E and F are out of phase.
displacement
E
distance
0
F
G
Fig. 7 Particles that are out of phase
Two particles are in anti-phase (exactly out of phase) if they are out of phase by half a cycle, i.e.
180 or  rad. In Fig.7, E and G are in anti-phase.
Phase difference given a displacement-distance graph
displacement

c
distance
0
x
Fig. 8 Displacement-distance graph of wave with wavelength 
In Fig. 8, the phase difference between two particles in a wave separated by a distance ∆x in the
direction of the wave having wavelength  is given by
 
displacement
x

 2 
x
distance
0

Fig. 9 Displacement-distance graphs of two waves with wavelength 
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Lecture Notes
Similarly the phase difference between two waves of the same frequency and wavelength 
travelling in the same direction, with their maximum/minimum displacements at a distance ∆x apart
(as shown in Fig. 9), is also given by
 
x

 2 
Phase difference given a displacement-time graph
displacement
∆t
time
0
T
Fig. 10 Displacement-time graphs of two waves with period T
In Fig. 10, the phase difference between two waves of the same frequency and period T travelling
in the same direction, with their maximum/minimum displacements at a time ∆t apart, is given by
 
t
 2 
T
Note:
1.
In order to compare phase or finding phase difference, amplitudes of oscillating particles
need not be the same but they must have the same frequency and wavelength.
2.
Two particles or two waves are said to be in phase when their phase difference is zero. This
implies that particles that are one wavelength apart on the same wave are always in phase
and in general particles or waves that are n apart (where n is a positive integer) are also in
phase, e.g. , 2, 3, etc.
3.
Two particles or two waves are said to be in anti-phase (exactly out of phase) when their
phase difference is  radian. Particles that are half a wavelength apart on the same wave are
1

2

positive integer) are also exactly out of phase, e.g. /2, 3/2, 5/2, etc.
always exactly out of phase. This implies that particles that are  n   apart (where n is a
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Lecture Notes
Example 3
A sound wave of frequency 400 Hz is travelling in air at a speed of 320 m s1. Calculate the
difference in phase between two points on the wave 0.20 m apart in the direction of travel.
x
v  f
 
 2 

v

 0.20 
f

  2 
320
 0.80 

400

 rad
 0.80 m
2
Example 4
The displacement-time graphs of two waves of the same frequency received by a detector are
shown below. Find the phase difference between the two waves.
t
 2 
T
1 
 4T 
 2

 
T
 


2
rad
Example 5
A water wave travels with a speed of 5.0 m s1 and has a frequency of 0.50 Hz as shown.
Calculate
(a) the distance between points A and B
D
(b) the phase difference between points B and C
A
B
(c) the phase difference between points B and D
(a)
(b)
v  f
v

f
5.0

0.50
 10 m
distance between A and B  2
C
direction of wave motion
 2 10 
 20 m
Since B and C are 2 apart, they are in phase. Hence their phase difference is 0 rad.
Alternatively,
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 
x

Lecture Notes
 2 
 2 

  2 
  
 4 rad
 0 rad
(c)
Since B and D are
1
 apart, they are exactly out of phase. Hence their phase difference is
2
 rad. Alternatively,
x
 
 2 

1 
2
 2

 

  rad
Graphical Representations of Transverse and Longitudinal Waves
Both the displacement-time and displacement-distance graphs apply to both transverse and
longitudinal progressive waves. It is necessary to adopt some convention to present the direction
of the vibration of the wave particle;
 For transverse waves, the positive axis usually refers to the upward displacement of a particle.
 For longitudinal waves, the positive axis usually refers to the displacement of a particle to the
right.
A
0
A
Fig. 11 Representation of transverse wave
Fig. 12 Representation of longitudinal wave
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Lecture Notes
Displacement-Time Graph
The displacement-time graph shows how displacement of a single particle varies with time.
Period and amplitude can be obtained from the displacement-time graph.
displacement from the
equilibrium position/m
For a transverse
wave particle
t2
T
A
t1, t3
t0
t1
t2
t3
t4
t5
t6
t7
t8
time/s
t0, t4, t8
t5, t7
A
t6
For a longitudinal wave particle
t6
t5, t7
t0, t4, t8
t1, t3
t2
Fig. 13 Displacement-time graph
All the particles move in a similar manner with the same amplitude and frequency as the wave i.e.
 frequency of particle  frequency of the wave
 amplitude of particle  amplitude of the wave
From the graph, t0 to t8 signifies the completion of one oscillation, and is the period of the wave.
Displacement-distance Graph
The displacement-distance graph shows how the displacements of the particles (from their
individual equilibrium position) vary with the distance from the source at a particular instant in
time.
For transverse waves, this is similar to a snapshot of the actual wave travelling through the
medium.
For longitudinal waves, however, unlike transverse waves, the displacement-distance graph is not
a snapshot of the actual wave travelling through the medium and has to be found by finding the
displacement of individual particles.
Wavelength and amplitude can be determined from the displacement-distance graph.
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Lecture Notes
displacement from the
equilibrium position/m
A
0
C
R
C
distance from
source/m

A
Fig. 14 Displacement-distance graph (right is taken as positive)
For longitudinal waves, a compression occurs where the air molecules are closest together,
while a rarefaction occurs where the air molecules are furthest apart from each other.
Compressions and rarefaction are labelled as C and R in Fig. 14. The wavelength of a longitudinal
wave is therefore equal to the distance between successive compressions or successive
rarefactions. When the particle is at the centre of a compression or rarefaction, it has zero
displacement. Pressure is highest at the centre of a compression, and is lowest at the centre
of a rarefaction.
rarefaction
compression
compression
displacement
distance
along wave
pressure
distance
along wave
Fig. 15 Displacement-distance and pressure-distance graph of a longitudinal wave
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Lecture Notes
Example 6
The figure below shows a particular instant of a transverse wave travelling from left to right on a
rope at time t  0 s.
B
C
A
(a)
(b)
(c)
(a)
(b)
(c)
What is the direction of motion of particles A and B at this instant?
At the next instant, what is the direction of motion of particles A, B and C?
Sketch a displacement-time graph for particle A.
Direction of motion of A is downward, while B is momentarily at rest.
Direction of motion of both A and B is downward, while that of C is upward.
displacement
0
time
Energy and Intensity of Waves
Wave motion involves the transportation of energy from one place to another.
The intensity of a wave is the energy delivered per unit area per unit time (normal to area).
The unit of intensity is W m2.
E
... Eqn. 1
At
P

 where A is area 
A
intensity, I 
power, P
area, A
Fig. 16 Intensity of a wave
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Lecture Notes
Energy of sinusoidal waves
If the source of the wave vibrates in simple harmonic motion, i.e. sinusoidal, each particle of the
wave will also oscillate with simple harmonic motion. For simple harmonic motion, the energy of
the oscillating particles is proportional to the square of its amplitude, A, i.e. E  A2, recall
1
E  m 2 A2 . Hence, From Eqn. 1, it can be deduced that the intensity of the wave will therefore
2
be proportional to the square of its amplitude, i.e.
I  A2
Intensity of waves with spherical wavefronts
The wave coming from a point source travels out uniformly in all directions is known as a
spherical wave e.g. sound travelling in open air and light wave. As the wave moves outward, the
energy that it carries is spread over a larger and larger area. Since the surface area of a sphere
with radius r is 4 r 2 , by the definition of intensity of a wave, the intensity of a spherical wave with
a fixed (constant) source of power, P, is therefore,
P
A
P

4 r 2
intensity, I 
Hence,
I
1
r2
i.e. the further from the source, the smaller the intensity as power of the wave has been distributed
over a larger surface area.
As the intensity is proportional to the square of its amplitude, I  A2, the amplitude of a wave also
decreases with increasing distance from the source i.e.
A
1
r
Example 7
A surface of area S is placed perpendicular to the direction of travel of a plane wave. The energy
per unit time intercepted by the surface is E when the amplitude of the wave is A. The area of the
1
surface is reduced to S and the amplitude of the wave is increased to 2A. What is the energy
2
per unit time intercepted by this smaller surface?
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Lecture Notes
 where S is area 
P  IS


E  kA2 S
2
E '  A'   S ' 

E  A   S 
 1  
S
2 
 2 A    2  
E'
E

 A   S 




 2E
The energy per unit time intercepted by this smaller surface is 2E.
Example 8
A person is initially 8.0 m from a point source which emits energy uniformly in all directions at a
constant rate. If the power of the source is to be halved but the sound is to be as loud as before, at
what distance should the person be from the source?
P
Using I 
, since intensity is the same,
4 r 2
1 
2P
r
r' 
P
1

r
2
1

 8.0 
2
 5.7 m
P' P

r '2 r 2
2
P'
r '
r   P
 
r'
P'
r
P
Example 9
A 20 W loudspeaker is emitting sound at its full power in all directions. Find
(a) the intensity of sound at a distance of 10 m away
P
I
A
P

4 r 2
20

2
4 10 
 0.016 W m2
(b)
the power received by a square microphone of length 2.0 cm placed at a distance 10 m away
from the loudspeaker
P
I
A
P  IA

 0.016 2.0  10 2

2
 6.4  10 6 W
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(c)
Lecture Notes
the amplitude of the vibrations at 10 m, given that at 2.0 m, the amplitude of the vibrations is
4.0 cm
1
r
A
A'  A
r
r'
A' r
 2.0 
2


 4.0  10
A r'
 10 
 0.0080 m


Determination of the frequency of sound
The frequency of sound can be determined using a cathode ray oscilloscope (C.R.O.) by placing a
microphone in the path of the sound wave. The microphone converts the sound energy into
electrical energy, such that the amplitude of microphone’s voltage output is proportional to the
amplitude of the sound wave.
Since the x-axis of the oscilloscope is the time base, the oscilloscope displays the voltage-time
graph of the microphone output, which will give the displacement time graph of the sound wave.
The time-base can be adjusted to appropriate value until a stationary trace is obtained. The period
of the sound wave is given by the product of the length of one wave cycle, as determined from the
1
screen, and the time-base setting. Then the frequency of the signal can be calculated using f  .
T
The frequency of sound from a loudspeaker can be measured by using the setup illustrated in
Fig. 17.
audio frequency
generator
loudspeaker
oscilloscope
y-input
microphone
metre rule
time base set to
50 ms/div
Y-gain set to 0.5 V/div
Fig. 17 Set-up to determine frequency of sound using a C.R.O.
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Lecture Notes
From Fig. 17, the horizontal distance from peak to peak (1 wave) is 8 divisions. As time-base is set
1
to 50 ms/div, the period, T, is therefore, T  8  50  400 ms. Using frequency, f  ,
T
1
f 
400  103
 2.5 Hz
Example 10
A sinusoidal sound wave of unknown frequency is fed into a C.R.O. and the waveform on C.R.O is
shown below. The length of each division for the time-base is 1 cm. Find the frequency of the
sound.
5
time base
2
1
0.5
10
0.2
50
0.1
OFF
ms cm1
From C.R.O, 8 divisions correspond to 2.5 wavelengths,

2.5T  8 2  10 3

T  0.0064 s
1
f 
T
1

0.0064
 156.25
 160 Hz
Polarisation
In a transverse wave, vibrations can occur in many different planes that are perpendicular to the
direction of wave propagation, at the same time. This is known as an unpolarised wave.
Polarisation is a process by which a wave’s oscillations are made to occur in a single plane only. It
is a property exhibited only by transverse waves, and does not occur for longitudinal waves. In a
longitudinal wave the direction of vibration is the same as the direction of wave propagation, hence
longitudinal waves cannot be polarised.
If the oscillations in a transverse wave are confined to only one direction, (which is at right angle to
the propagation of the wave), the wave is said to be linearly polarised in that direction.
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Lecture Notes
Consider a rope that is set into vibration in a vertical plane such that it forms a transverse wave. If
we place a vertical slit in the path of the wave, it will pass through unaffected. However, if we place
a horizontal slit in its path, it will not pass through.
direction of rope
vibrations
direction of wave
travel
(a)
(b)
A
unpolarised
light
B
polariser
A
C
(no light)
Fig. 18 Polarisation of transverse waves
Plane-polarised light can be obtained by passing unpolarised light through the polariser (Polaroid).
In Fig. 18, the transmitted wave through polariser A is said to be plane-polarised, or more
specifically, it is polarised in the vertical plane. This vertically polarised wave is able to pass
through polariser B as polariser B has the same transmission (polarisaton) axis as polariser A.
However the polarised wave is completely blocked by polariser C which has a transmission axis
perpendicular to A and no light is able to pass through.
unpolarised light
polariser
A0
transmission axis
polarised light
Fig. 19 Illustration of polarisation of light wave using a polariser
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Lecture Notes
Did you know?
1.
Polaroid is used in sunglasses to reduce the glare from reflected light (e.g. from windows and
water).
2.
In 3-D movies, a pair of pictures is taken a short distance apart. The pictures are then
projected onto a screen through a pair of projectors fitted with polarised filters. The
polarisation axes of the filters are at right angles to each other – one horizontal and the other
vertical. To the naked eye, the pictures on the screen look blurred.
To see the pictures in 3-D, the viewer needs to wear glasses made of Polaroid, as shown in
Fig. 20. Each eye sees a separate picture, just as in real life. The brain then interprets the
two pictures as a single picture with depth.
Polaroid with a
vertical
polarisation axis
Polaroid with a horizontal
polarisation axis
Fig. 20 Glasses used to watch 3-D movies
Example 11
Explain why it would not be possible to polarise sound waves.
Sound waves in a gas or liquid do not have polarization because the medium vibrates only along
the direction in which the waves are travelling.
Polarisation of Electromagnetic Waves
Electromagnetic waves may be polarised by passing the unpolarised wave through a polariser, e.g.
a Polaroid.
Polaroid with vertical
polarisation axis
unpolarised
light
vibration in
vertical
plane only
vertical
plane
polarised
light
Fig. 21 Plane-polarised light
Fig. 21 shows unpolarised light being passed through a Polaroid with a vertical polarisation axis.
The result is light that is polarised in the vertical plane, i.e. the direction of vibration of the polarised
light is along the polarisation axis of the polariser.
Unpolarised light consists of light with random directions of oscillation. Each of these oscillations
can be resolved into components along two mutually perpendicular directions. On average an
unpolarised beam of light comprises two plane-polarised beams of equal magnitude perpendicular
to one another. When this beam passes through a polariser, one component is eliminated. Hence
the intensity of light that passes through the polariser is reduced by half, since half the light is
eliminated.
Nanyang Junior College
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9749 H2 PHYSICS
Lecture Notes
If I0 is the intensity of unpolarised light incident on a Polaroid, the intensity of light transmitted by
the Polaroid is
1
I0 .
2
Malus’ Law
polarisation axis
polarisation axis


A
Acos
unpolarised
light
Polaroid P
Polaroid Q
Fig. 22 Variation of intensity of polarised light
Fig. 22 shows unpolarised light incident on Polaroid P which has a vertical polarisation axis. Light
transmitted by Polaroid P is plane-polarised with amplitude of vibration A. The polarised light then
falls on Polaroid Q whose polarisation axis is inclined at an angle  to the plane of polarisation of
the incident light. The light passing through Polaroid Q has an amplitude of vibration given by
A cos .
Since the intensity of light I is proportional to the square of the amplitude of vibration, I   A cos  
2
i.e. I  cos2  .
Therefore the intensity of light emerging from Polaroid Q is
I  I0 cos2 

I  A2

where
I0  intensity of plane-polarised light incident on Polaroid Q, and
  angle between the polarisation axis of Polaroid Q and the plane of polarisation of the incident
light.
If Polaroid Q is slowly rotated about the axis of wave propagation from   0 to   90, the
intensity of the transmitted light decreases from maximum to zero. When   90 or 270o, no light
passes through Polaroid Q.
Nanyang Junior College
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9749 H2 PHYSICS
Lecture Notes
Example 12
Two pieces of Polaroid are arranged such that their polarisation axes are parallel and vertical, as
shown below. The intensity of the emergent light is I0. Through what angle must Polaroid Q be
rotated so that the intensity of the emergent light decreases to
1
I0 ?
2
incident
light
I0
Q
P
I  A2
I '  A' 

I  A 
2
 A cos  


 A 
 cos2 
cos2  

1
2

4
I  I0 cos2 
2
1 

 I '  2 l0 


I0
 I0 cos2 
2
1
cos2  
2


4
rad
rad
Nanyang Junior College
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9749 H2 PHYSICS
Lecture Notes
Appendix
Properties of Electromagnetic Waves
B
E
electric field
oscillation
(horizontal)
magnetic field
oscillation
(vertical)







distance
Consist of mutually perpendicular electric and magnetic fields, with directions of both fields
perpendicular to the direction of propagation of the wave (see above). The fields vary
sinusoidally but are always in phase with each other.
Travel at a speed of 3.0  108 m s1 in free space (vacuum)
Transverse waves and therefore can be polarised
Unaffected by electric and magnetic fields
Obey laws of reflection and refraction
Exhibit interference and diffraction
Made up of photons, where the energy of each photon is E  hf (quantum physics)
type of EM
radiation
range of
wavelength/
m
Order of magnitude of wavelength in electromagnetic spectrum
radio
infra-red
visible
ultra-violet
microwave
X-ray
wave
ray
light
ray
> 1  101
1  103
to 1  101
7  107
to 1  103
4  107
to 7  107
1  109
to 4  107
1  1011
to 1  109
indigo
violet
gamma
ray
< 1  1011
Visible light spectrum:
increasing
frequency
red
orange
yellow
green
blue
increasing
wavelength
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