Waves (AQA Unit 2)

Report
Unit 2 Chapters 12 and 13
Waves and Optics
Dr K. Newson TGSG Physics Dept.
Chapter 12
AQA PHY 2
Section 1
Wave Types and Polarisation
Objectives:
 Know the differences between transverse and
longitudinal waves
 Be able to describe a plane-polarised wave
 Know the physical test that can distinguish between
transverse and longitudinal waves.
Types of wave
 There are two types:


Longitudinal waves
Transverse waves
 Waves can also be categorized in terms of their origins,
i.e. either:
 Electromagnetic e.g. light, UV
 Mechanical e.g. sound or seismic
Longitudinal Waves
1) Longitudinal Wave: The vibration of particles is
parallel to the direction of the wave.
 E.g. sound waves, some seismic waves.
Longitudinal waves in springs
Transverse Wave: The vibration (red arrow) is parallel to
the wave’s direction (black arrow).
 E.g. sound waves.
Transverse Waves
Transverse Wave: The vibration (red arrow) is
perpendicular to the wave’s direction (black arrow).
 E.g. light waves (or any electromagnetic waves).
Polarisation
 In reality, unpolarised transverse waves have displacements in
every plane, but all the directions are at right angles to the
direction of the wave.
Direction of
wave
Wave vibrations
are in more than
one plane
End-on view
Polarisation contd.
 Transverse waves that are plane polarised have vibrations that
are in one plane only.
Wave vibrations are
in just one plane
Note: Longitudinal waves cannot be
polarised as the vibration is parallel to
the wave’s direction of motion.
End-on view
Polarising filters
 A polarising filter is a sheet of plastic material. The sheet
allows vibrations to pass through in one plane only.
 The molecules that make up polarising filters are aligned so
that they act like billions of slits. The slits will only allow the
component of light that is in the same direction as the slit
to pass through.
Wave passes through
Wave doesn’t pass through
Polarising filters
Filter will polarise the light in the
vertical plane
If rotated by 90o the light will be
polarised in the horizontal plane
Polarisation of light
Unpolarised
light
Polarizing
filter
Plane polarised light
Polarisation of light
A second filter will stop the light if it is at right angles to the
first filter.
Light is stopped
by two filters at
right-angles
Polarisation of light
If the second filter is in the same orientation as the first then
the light will still only be polarised in the same plane.
Section 2
Measuring Waves
Objectives:
 Know what amplitude of a wave is
 Know between which two points a wavelength can be
measured
 Know how to calculate frequency from time period
Waves – Key terms
All waves can be characterised in terms of their:
 Displacement: The distance and direction of a particle





from its equilibrium position.
Amplitude: The maximum displacement of a vibrating
particle.
Cycle: this refers to one full wave passing.
Frequency f : The number of complete cycles (vibrations)
of a particle per unit time. This measured in Hertz (Hz).
Wavelength λ: The distance between two neighbouring
points on a wave that are vibrating in phase e.g. two
successive peaks or troughs.
Period T: The time for one complete cycle of a wave.
Amplitude
Representingλ waves
λ
Length/m
Amplitude
RepresentingT waves
T
Time
Wave speed (c)
 The time taken for one cycle = 1/f
 Since frequency = number of waves per second,
and the wavelength = length of each wave, then:
 Frequency × wavelength
= total length per second = speed
 In symbols:
c = fλ
Phase angle and phase difference
 The Phase angle is a measure of how far through
a cycle an oscillation is.
 Note: 1 complete cycle = 360o in degrees, this is
equal to 2π radians.
 We can use degrees, radians or wavelength as a
measure of phase difference.
 Remember:
Angle in degrees
Angle in radians
0
0
90
π/2
180
π
270
3π/2
360
2π
In terms of λ
0
λ/4
λ/2
3λ/4
λ
Phase difference
These oscillations are in step, they are said to be in
phase. Their phase difference is zero.
0
180
360
Phase difference
 These oscillations are a ¼ of a cycle out of phase. Their
phase difference is π/2 radians (90o), or λ/4
0
180
360
Phase difference
These oscillations are ½ of a cycle out of phase. They are
said to be in anti-phase. Their phase difference is π
radians (180o) = λ/2.
0
180
360
Section 3
Reflection, Refraction and Diffraction
Objectives:
 Know what causes waves to refract when they pass
across a boundary
 Know which way light waves refract when passing from
glass into air
 Be able to define diffraction
Wavefronts
 Wavefronts are lines joining points of a wave that are in phase
 In the diagrams below line PP, QQ, and RR represent wavefronts
 The distance between wavefronts is equal to the wavelength
Circular wavefronts
Plain wavefronts
R
P Q
Q
R
P
λ
λ
λ
λ
λ
P
Q
R
P Q
R
λ
λ
Wavefronts contd.
 The expanding circles
in the wave train are
called wave fronts.
Reflection
 Straight waves that hit a hard flat surface reflect off at the
same angle.
 That means that:
the angle of incidence = the angle of reflection
 This effect occurs when light strikes a plane mirror.
 More later in the Optics work.
Refraction
 When waves pass a boundary where their wave speed
changes, the wavelength also changes. For EM waves this
occurs where there is a change in density of the medium.
 If the wave fronts cross the boundary at an angle that is
not aligned with the normal, then they change direction
as well as speed. This is called refraction.
 More later in the Optics work.
Diffraction
 Diffraction is the spreading of waves after passing a
gap or around the edge of an obstacle.
 Diffraction is at a maximum when:
 The wavelength is large.
 The gap is the same size as the wavelength of the wave.
 The wavelength is approximately the same size as the
obstacle.
Diffraction
Strong diffraction as gap width ≈ λ
Diffraction
If the wavelength does not match the size of the gap,
then only a little diffraction will occur (at the edge of
the wave).
Section 4
Superposition
Objectives:
 Know what causes reinforcement
 Be able to explain phase difference
 Know the phase difference that produces maximum
cancellation
 Be able to explain why maximum cancellation is
difficult to achieve
Travelling waves
 Water, sound and electromagnetic waves convey
energy from the source that produces them. For this
reason they are known as travelling or progressive
waves.
Superposition
 When two or more waves meet, their displacements superpose.
 The Principle of superposition states that when two waves overlap,
the resultant displacement at a given instant is equal to the vector
sum of the individual displacements.
Constructive superposition
+
=
Two waves of the same wavelength that are in phase add
constructively. This is also known as constructive interference.
The end result is an increase in amplitude, with no change in λ
Destructive superposition
+
=
 If the two waves are in antiphase, they cancel each other.
The resulting amplitude will be zero. This is also known as
destructive interference.
Coherence
Coherence is an essential condition for the interference of
waves.
Two sources of waves are described as coherent if:
 The waves have the same wavelength.
 The waves are in phase or have a constant phase difference.
Lasers are an example of a coherent source of light. Light
from a normal filament bulb is not coherent since the light
is emitted randomly from the individual atoms in the
filament
Sections 5&6
Stationary and Progressive Waves
Objectives:
 Describe what’s needed to form a stationary wave
 Know if stationary waves are formed by superposition
 Explain why nodes are in fixed positions
 Know how frequencies of overtones compare to the
fundamental
Experiments

1.
Three demonstrations of stationary waves
Melde’s apparatus (string) using sheet wv6
Stationary waves (a.k.a. standing waves)
If two waves of equal frequency and amplitude, travelling
along the same line with the same speed, but in opposite
directions meet, they undergo superposition to produce a
wave pattern in which the positions of the crests and troughs
do not move. This pattern is called a stationary wave.
Nodes and Antinodes
N
N
A
N
A
A = antinode
N = node
N
A
N
λ/2
N
Nodes and Antinodes
 The places labelled N are called nodes, the
amplitude at nodes is zero. Nodes are caused by
destructive superposition.
 The places labelled A are called antinodes, the
amplitude is at a maximum at the antinodes.
Antinodes are caused by constructive superposition.
 The distance between adjacent nodes (or
adjacent antinodes) is half a wavelength.
Stationary wave patterns
 Different stationary wave patterns are obtained at certain resonant
frequencies.
 The simplest standing wave pattern consists of a single loop
 If the string has a length L, then L= λ/2
N
X
A
N
Y
•This simplest mode of oscillation is called the fundamental
and occurs at the fundamental frequency.
Harmonics
 If the frequency is doubled, the
next stationary wave pattern is
formed. This is called the first
harmonic.
N
A
A
N
If the string has a length L, then λ= L
N
Harmonics
 The trend continues for the other resonant frequencies,
remember each single loop has a length of λ/2. Each
subsequent harmonic adds one single loop
 The frequencies only occur at certain values because the
length of string (or air column or other) must equal a
whole number of half wavelengths i.e. L = nλ/2.
 This effect is often used in music written for violins.
Stationary Waves on a Vibrating String
 Carry out the experiment described on p.184
 Setting for the fundamental, 1st and 2nd overtones.
 Fundamentals have nodes at either end and an antinode
in the middle, so λ0=2L. Hence f0=c/λ0=c/2L
 The next stationary wave pattern is the first overtone this
has a node in the middle, so the string has two loops, so
λ1=L. Hence f1=c/λ1=c/L=2f0
 The next stationary wave pattern is the first overtone this
has a node in the middle, so the string has two loops, so
λ2=2/3L. Hence f2=c/λ2=3c/2L=3f0
 Stationary waves occur at f0, 2f0, 3f0, 4f0, etc.
Energy and stationary waves
 A travelling wave carries energy in its direction of
travel.
 A stationary wave consists of two identical travelling
waves acting in opposite directions. Therefore, the
two waves convey equal energies in opposite
directions.
 The net effect of this is that NO energy is transferred
by a stationary wave.
Demonstration of stationary waves
1) On a string/cord
X
Signal
generator
Thick rubber cord
Y
Vibration
generator
The cord is fixed at positions X and Y.
Simulation of Stationary waves
 http://www.walter-fendt.de/ph14e/stwaverefl.htm
Chapter 13
AQA PHY 2
Section 1
Refraction of Light
Objectives:
 Know what rays are
 Know Snell’s Law
Rays
Light rays represent the direction of travel of wavefronts.
wavefronts
Ray
Refraction
 Refraction is the change in the direction of a wave
when it is directed towards a boundary between
different media.
 For example when light moves from air into glass.
Refraction through a glass block:
The wave slows down and
bends towards the normal due
to entering a more dense
medium e.g. air into glass
i1
i2
r1
Wave speeds up and bends
away from the normal due to
entering a less dense medium
e.g. glass into air
r2
The wave slows down but
is not bent, due to
entering along the normal
Prism Refraction
Copy fig. 4 p.189
 Notice that the refraction rule still applies at each
surface, only one special angle will cause the spectrum
to be formed.
Investigating Refraction
 By using a ray box and a glass block, it should be
possible to find the glass block’s refractive index, n,
by using Snell’s Law:-
n=
sin i
sin r
 Carry out experiment described on p188
 What else is observed?
Comparing Refraction
If you use a rectangular glass block, then the two sides where
refraction occur are parallel, and so i1 = r2 and i2 = r1
This means that if the air to glass boundary has a known
refractive index of n, then the glass to air boundary has a
refractive index of 1/n.
Section 2
Explaining Refraction
Objectives:
 Know what happens to the speed of the light waves
when changing medium
 Know how refractive index relates to this speed change
 Be to explain why a prism can split white light into a
spectrum
Refraction Explained
 To explain refraction we have to think of a series of
wavefronts arriving at a boundary.
 The part of the wavefront striking the boundary
first, will slow down, whilst the rest of the
wavefront continues at the higher velocity, until it
meets the boundary.
 To maintain a single wavefront the wave has to
change direction.
Refraction explained
Air
λ
Wavelength in air >
Wavelength in glass
Glass
Y
Air
i
X
X’
Glass
Y’
r
XX’ = Cst
YY’ = Ct
Refraction Explained
See Handout/Fig.1 p190
 The wave front lines (XY and X’Y’) are the same
wave after a time t.
 Hence the wavefront moves a distance ct when at
speed c between Y and Y’.
 Hence the wavefront moves a distance cst when at
speed cs in between X and X’.
Refraction explained
 YY’ is perpendicular to the XY wavefront, so using triangle
XX’Y’ and the angle (i) we can say ct = XY’sini
 Similarly for triangle XX’Y’, XY’ is perpendicular to XX’, and
the angle r we can say cst =XY’sinr
 Combining these gives:
sini
sinr
c
=
cs
 Therefore the refractive index of a substance is given by:
ns
sini
c
=
=
sinr
cs
Refraction explained
 Speed and wavelength change but frequency remains
constant
 Therefore as c= λf it also follows that:
c
λ
ns =
=
cs
λs
Refraction between two transparent
substances
 (See P191) Using similar theory to previously, but
considering the speed of light through the two substances
as c1 and c2.
sini
c1
 We can then state that:
n1
sinr
n2
r
i
=
c2
Refraction between two substances
 This can be rearranged:
1/c1sini = 1/c2sinr
 Multiply both sides by c gives:
c /c1sini = c /c2sinr
 Substitute n1 for c/c1, and similarly for n2 gives:
n1sini = n2sinr
Refraction between vacuum and
transparent medium
sini / sinr = n
and:
n = c / cs
The speed of light in air is 99.97% of its speed in a
vacuum, thus giving air a refractive index of 1.0003 –
thus for most purposes air can be considered a
vacuum, with a refractive index of 1.
HINT: This means you know c for light travelling in air.
Dispersion & the light spectrum
Why does a prism cause dispersion of light?
 Remember the sequence ROYGBIV
 Red light has a wavelength of about 650nm, whilst
violet is about 400nm.
 The dispersion occurs because the refractive index
changes for different wavelengths. i.e. the speed of
light in glass depends on its wavelength.
 The shorter a wavelength, the greater its refraction.
Hence each colour has a different wavelength and so is
refracted by a different amount.
Section 3
Total Internal Reflection
Objectives:
 Know the conditions required for TIR
 Know how the critical angle and refractive index are
related
Total internal reflection (TIR)
At angles θ < θc the light is
refracted at the boundary
Air
θ
Glass
Total internal reflection (TIR)
At the critical angle θc
the light is refracted
through 90o
θc
Total internal reflection (TIR)
At angles greater than the critical angle
all the light is reflected, none is
refracted. Hence we have TIR.
Critical angle & Refractive Index



At the critical angle, ic, the angle of refraction is 90°
because the light ray emerges along the boundary.
Thus:
n1sin ic = n2sin90 (where n1 is the refractive index of
the incident substance and n2 is the refractive index
of the other.
But sin90 = 1, so…
sin ic = n2/n1
TIR Uses

1.
2.

1.
2.
Optical Fibres for:
Communications (both light and Infra-Red can be
used)
Medical Endoscope
Exercise:
Use p193/194 to define multipath dispersion,
spectral dispersion.
Answer Q4 p195
Sections 4&5
Double Slit Interference (including Young’s Double Slits
Experiment) & Coherence
Aims and objectives
 Understand the concepts of path difference and coherence.
 Understand the requirements of two source and single source
double slit systems for the production of maxima and minima.
Path difference and interference
First subsidiary maximum
Slit 1 (s1)
O
Central Maximum
Microwave
source
Slit 2
A
Minimum
(s2)
B
First subsidiary maximum
Path difference
 Slits S1 and S2 act as two sources of microwaves, the waves
leave the slits in phase.
 The point O is the same distance from both slits (zero path
difference), the waves are in phase: giving constructive
superposition, this point is known as the central
maximum.
X
Path difference (d1 – d2)
X
S1
d1
S2
d2
maximum
O
0
X A
λ/2
B
λ
X C
3λ/2
X minimum
Finding wavelengths using path difference method
 Set up the microwave apparatus with a double slit and the
probe receiver.
 Mark the central positions of the slits on a piece of paper.
 Move the detector along a line parallel to the slits to find the
points of central maximum, the first minimum (at A), the
first subsidiary maximum (at B), then successive minima and
maxima.
 Measure the distance from the centre of each slit to the
points of maximum and minimum signal strength.
Finding wavelengths using path difference
method
 Path differences between maxima = 0, λ, 2λ, 3 λ, ……
 Path differences between minima = λ/2, 3λ/2, 5λ/2, ……
In general:
 For constructive superposition, path difference = nλ
 For destructive superposition, path difference = (n + ½)λ
 Where n = 0, 1, 2, 3,…………
Sample question 15.4
 The diagram shows the position P of the third maximum of the
superposition pattern of two coherent microwave sources s1
and s2. Calculate the wavelength of the sources.
P
66 cm
s1
110cm
28 cm
s2
Answer
P
66 cm
s1
110cm
28 cm
s2
Since P is the position of the third maximum, the path difference (s2P – s1P) = 3λ
Using Pythagoras (s2P)2 = 1102 + (66 + 14)2
And
therefore s2P = 136.0cm
(s1P)2 = 1102 + (66 - 14)2
therefore s1P = 121.7cm
Therefore (s2P – s1P) = (136.01 - 121.67) = 14.34 = 3λ
so = λ = 14.3/3 = 4.8cm
Interference of waves
Think light or laser not
microwaves
Young’s double slit experiment
 Young’s double slit experiment demonstrates the
interference of light and shows that light has wave-
like properties.
 The experiment involves a coherent light source,
usually a laser, being directed onto a twin slit.
Light is diffracted from the two slits, these slits act
as separate light sources.
Young’s double slit experiment
 Waves diffract as they pass through the slits.
 The two sets of waves overlap and superposition
occurs. This causes the waves to produce an
interference pattern on a screen. The pattern
consists of light and dark fringes.
 The light fringes are caused by constructive
superposition (amplitude increased).
 The dark fringes (minima) are caused by
destructive interference (waves cancel).
Apparatus
Screen
w
D
Double
slit
HeNe Laser
Reminder about interference
Young’s double slit experiment
1st Subsidiary max
= P
P
w
S2
w
S1
O (central max)
Q
Double slits
D
Screen
 Path difference (slits to 1st sub max) = S1B
S2
s
Q
S1
For 1st Subsidiary maximum
path difference = λ
Young’s double slit experiment
S2
θ
s
Therefore distance S1B = λ
Q
S1
Also, S1B = sSinθ = λ
Young’s double slit experiment
If D>>w (or s) then triangles COQ and S1S2Q are similar
1st Subsidiary max
P =P
w
S2
O (central max)
w
S1
B
Double slits
D
Screen
Superposition formula
 Sinθ = S1Q/s = λ/s.
 However, as triangles COP and S1S2B are similar
 it therefore follows that: sinθ = w/D = λ/s which in turn gives:
λ =ws/D
Double slit equation
Key equation:
λD
w =
s
Where:
 D = distance from the slits to the screen (1 - 10m for laser),
 w = separation between adjacent fringes,
 s = separation of the twin slits (typically ~ 0.2 - 0.3mm)
 λ = wavelength of laser light
 Note: this equation assumes that D >> w or s
 So if D is not much greater than w or s, use the path difference
method (i.e. with microwaves)
White Light Fringes
Each component of light show its own fringe pattern, with
each pattern coming together at the centre, hence it
remains white.
Each colour has its own fringe pattern, with different width
fringes, due to the differing wavelengths of the light.
Double slit diffraction
Single slit diffraction
Single slit diffraction
Single slit diffraction
 Main points (see page 206)
 The central fringe is much more intense (bright) than the





others.
The central fringe is twice as wide as the outer fringes
The intensity of each fringe decreases with distance from the
central fringe.
The greater the wavelength, the wider the fringes.
The narrower the slit width (a) the wider the fringes (W).
The width W of the fringe observed on a screen a distance D
from the slit is given by:
W =
λ
a
× 2D
Diffraction gratings
 Diffraction gratings consist of an opaque glass slide with many
parallel slits (300mm-1) etched on them.
 Each slit acts as a separate wave source and we get interference
between the waves that emerge from each slit.
 The reinforcement of waves occurs in certain directions only.
 The central beam is called the zero order, the other orders are
directed at an angle θ to the zero order beam.
Diffraction grating
 The diffraction grating equation is given by:
dsinθ = nλ
 Where d = grating spacing, n is the number of the order and λ is
the wavelength.
 The maximum order number n is given by d/λ rounded down to
the nearest whole number.
 Sometimes N the number of slits per metre is given instead of d.
In this case N = 1/d
Example (question 3 page 207)
 Light of wavelength 430nm is directed normally at
a diffraction grating. The first order transmitted
beams are at 28o to the zero order beam.
 Calculate
 A) the number of slits per mm
 B) the angle of diffraction for each order of
diffracted beam.
Answer to example question
 A) Rearranging the equation dsinθ = nλ for the 1st order
(28o) we get:
 d = (nλ ÷ sinθ) = (1× 430 x 10-9) ÷ sin28 = 9.16 x 10-7
 N = 1/d = 1 ÷ 9.16 x 10-7 = 1.09 x 106 m-1
 So there are 1.09 x 103 slits per mm





B) The highest order n = d/λ (rounded down)
So max n = 9.16 x 10-7 ÷ 430 x 10-9 = 2
Again using dsinθ = nλ rearranged we get sinθ = nλ/d
sinθ = (2 × 430 x 10-9) ÷ 9.16 x 10-7 = 0.9388
Therefore θ = 69.9o
Types of spectra
C: constructive interference
D: destructive interference
All travelling waves transfer energy from one place to
another
•Waves may be classed as being: mechanical, e.g. Sound
waves, or electromagnetic, e.g. Radio or UV, in origin
•

similar documents