3Section 1 Wave Types and Polarisation Objectives: Know the differences between transverse and longitudinal wavesBe able to describe a plane-polarised waveKnow the physical test that can distinguish between transverse and longitudinal waves.
4Types of wave Longitudinal waves Transverse waves There are two types:Longitudinal wavesTransverse wavesWaves can also be categorized in terms of their origins, i.e. either:Electromagnetic e.g. light, UVMechanical e.g. sound or seismic
5Longitudinal Waves1) Longitudinal Wave: The vibration of particles is parallel to the direction of the wave.E.g. sound waves, some seismic waves.
6Longitudinal waves in springs Transverse Wave: The vibration (red arrow) is parallel to the wave’s direction (black arrow).E.g. sound waves.
7Transverse WavesTransverse Wave: The vibration (red arrow) is perpendicular to the wave’s direction (black arrow).E.g. light waves (or any electromagnetic waves).
8PolarisationIn 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 waveEnd-on viewWave vibrations are in more than one plane
9Polarisation contd.Transverse waves that are plane polarised have vibrations that are in one plane only.Wave vibrations are in just one planeEnd-on viewNote: Longitudinal waves cannot be polarised as the vibration is parallel to the wave’s direction of motion.
10Polarising filtersA 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 throughWave doesn’t pass through
11Polarising filtersFilter will polarise the light in the vertical planeIf rotated by 90o the light will be polarised in the horizontal plane
12Polarisation of light Unpolarised light Polarizing filter Plane polarised light
13Polarisation of lightA 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
14Polarisation of lightIf the second filter is in the same orientation as the first then the light will still only be polarised in the same plane.
15Section 2 Measuring Waves Objectives: Know what amplitude of a wave is Know between which two points a wavelength can be measuredKnow how to calculate frequency from time period
16Waves – 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.
19Wave 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 = speedIn symbols:c = fλ
20Phase 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 degrees90180270360Angle in radiansπ/2π3π/22πIn terms of λλ/4λ/23λ/4λ
21Phase differenceThese oscillations are in step, they are said to be in phase. Their phase difference is zero.
22Phase differenceThese oscillations are a ¼ of a cycle out of phase. Their phase difference is π/2 radians (90o), or λ/4
23Phase differenceThese 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.
24Section 3 Reflection, Refraction and Diffraction Objectives: Know what causes waves to refract when they pass across a boundaryKnow which way light waves refract when passing from glass into airBe able to define diffraction
25Wavefronts Circular wavefronts Plain wavefronts P Q R λ λ λ λ λ P Q R Wavefronts are lines joining points of a wave that are in phaseIn the diagrams below line PP, QQ, and RR represent wavefrontsThe distance between wavefronts is equal to the wavelengthCircular wavefronts Plain wavefrontsRP Q RQPλλλ λ λ λ λPQRP Q R
26Wavefronts contd.The expanding circles in the wave train are called wave fronts.
27ReflectionStraight waves that hit a hard flat surface reflect off at the same angle.That means that:the angle of incidence = the angle of reflectionThis effect occurs when light strikes a plane mirror.More later in the Optics work.
28RefractionWhen 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.
29DiffractionDiffraction 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.
31DiffractionIf the wavelength does not match the size of the gap, then only a little diffraction will occur (at the edge of the wave).
32Section 4 Superposition Objectives: Know what causes reinforcement Be able to explain phase differenceKnow the phase difference that produces maximum cancellationBe able to explain why maximum cancellation is difficult to achieve
33Travelling wavesWater, sound and electromagnetic waves convey energy from the source that produces them. For this reason they are known as travelling or progressive waves.
35SuperpositionWhen 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.
36Constructive 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
37Destructive 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.
38CoherenceCoherence 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
39Sections 5&6 Stationary and Progressive Waves Objectives: Describe what’s needed to form a stationary waveKnow if stationary waves are formed by superpositionExplain why nodes are in fixed positionsKnow how frequencies of overtones compare to the fundamental
40Experiments Three demonstrations of stationary waves Melde’s apparatus (string) using sheet wv6
41Stationary 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.
42Nodes and AntinodesN N N N N NA A A λ/2A = antinodeN = node
43Nodes and AntinodesThe 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.
44Stationary wave patterns Different stationary wave patterns are obtained at certain resonant frequencies.The simplest standing wave pattern consists of a single loopIf the string has a length L, then L= λ/2N A NYXThis simplest mode of oscillation is called the fundamental and occurs at the fundamental frequency.
45Harmonics If the string has a length L, then λ= L If the frequency is doubled, the next stationary wave pattern is formed. This is called the first harmonic.A AN N NIf the string has a length L, then λ= L
46HarmonicsThe trend continues for the other resonant frequencies, remember each single loop has a length of λ/2. Each subsequent harmonic adds one single loopThe 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.
47Stationary Waves on a Vibrating String Carry out the experiment described on p.184Setting 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/2LThe 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=2f0The 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=3f0Stationary waves occur at f0, 2f0, 3f0, 4f0, etc.
48Energy 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.
49Demonstration of stationary waves 1) On a string/cordSignal generatorVibration generatorThick rubber cordXYThe cord is fixed at positions X and Y.
52Section 1 Refraction of Light Objectives: Know what rays are Know Snell’s Law
53Rays Light rays represent the direction of travel of wavefronts.
54RefractionRefraction 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.
55Refraction through a glass block: The wave slows down and bends towards the normal due to entering a more dense medium e.g. air into glassi1i2r1r2Wave speeds up and bends away from the normal due to entering a less dense medium e.g. glass into airThe wave slows down but is not bent, due to entering along the normal
56Prism 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.
57Investigating 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:-Carry out experiment described on p188What else is observed?sin isin rn =
58Comparing RefractionIf you use a rectangular glass block, then the two sides where refraction occur are parallel, and so i1 = r2 and i2 = r1This 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.
59Section 2 Explaining Refraction Objectives: Know what happens to the speed of the light waves when changing mediumKnow how refractive index relates to this speed changeBe to explain why a prism can split white light into a spectrum
60Refraction ExplainedTo 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.
61Refraction explained λ Air Wavelength in air > Wavelength in glass
63Refraction 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’.
64Refraction explained sini = c sinr cs ns = sini c sinr cs YY’ is perpendicular to the XY wavefront, so using triangle XX’Y’ and the angle (i) we can say ct = XY’siniSimilarly for triangle XX’Y’, XY’ is perpendicular to XX’, and the angle r we can say cst =XY’sinrCombining these gives:Therefore the refractive index of a substance is given by:sini=csinrcsns=sinicsinrcs
65Refraction explained ns = c = λ cs λs Speed and wavelength change but frequency remains constantTherefore as c= λf it also follows that:ns =c=λcsλs
66Refraction 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.We can then state that:sini=c1sinrc2n1n2ri
67Refraction between two substances This can be rearranged:1/c1sini = 1/c2sinrMultiply both sides by c gives:c /c1sini = c /c2sinrSubstitute n1 for c/c1, and similarly for n2 gives: n1sini = n2sinr
68Refraction between vacuum and transparent medium sini / sinr = n and:n = c / csThe speed of light in air is 99.97% of its speed in a vacuum, thus giving air a refractive index of – 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.
69Dispersion & the light spectrum Why does a prism cause dispersion of light?Remember the sequence ROYGBIVRed 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.
70Section 3 Total Internal Reflection Objectives: Know the conditions required for TIRKnow how the critical angle and refractive index are related
71Total internal reflection (TIR) At angles θ < θc the light is refracted at the boundaryAirθGlass
72Total internal reflection (TIR) At the critical angle θc the light is refracted through 90oθc
73Total internal reflection (TIR) At angles greater than the critical angle all the light is reflected, none is refracted. Hence we have TIR.
74Critical 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
75TIR Uses Optical Fibres for: Communications (both light and Infra-Red can be used)Medical EndoscopeExercise:Use p193/194 to define multipath dispersion, spectral dispersion.Answer Q4 p195
77Aims and objectivesUnderstand 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.
78Path difference and interference First subsidiary maximumSlit 1 (s1)O Central MaximumMicrowave sourceA MinimumSlit 2(s2)B First subsidiary maximum
79Path differenceSlits 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.
80X Path difference (d1 – d2) X O d1 X A λ/2 d2 B λ X C 3λ/2 maximum S1XOd1S2XAλ/2d2BλXC3λ/2maximumX minimum
81Finding 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.
82Finding 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 = nλFor destructive superposition, path difference = (n + ½)λWhere n = 0, 1, 2, 3,…………
83Sample question 15.4The 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.28 cm110cmP66 cms1s2
84Answer28 cm110cmP66 cms1s2Since P is the position of the third maximum, the path difference (s2P – s1P) = 3λUsing Pythagoras (s2P)2 = ( )2 therefore s2P = 136.0cmAnd (s1P)2 = ( )2 therefore s1P = 121.7cmTherefore (s2P – s1P) = ( ) = = 3λso = λ = 14.3/3 = 4.8cm
86Young’s Double slit experiment Think light or laser not microwaves
87Young’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.
88Young’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).
94Young’s double slit experiment If D>>w (or s) then triangles COQ and S1S2Q are similar1st Subsidiary max = PPwS2O (central max)wS1BDouble slitsDScreen
95Superposition formula Sinθ = S1Q/s = λ/s.However, as triangles COP and S1S2B are similarit therefore follows that: sinθ = w/D = λ/s which in turn gives:λ =ws/D
96Double slit equation w = λD s Key equation: 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 ~ mm)λ = wavelength of laser lightNote: this equation assumes that D >> w or sSo if D is not much greater than w or s, use the path difference method (i.e. with microwaves)
99White Light FringesEach 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.
103Single 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 fringesThe 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×2Da
104Diffraction gratingsDiffraction 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.
105Diffraction grating The diffraction grating equation is given by: dsinθ = nλWhere 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
107Example (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.CalculateA) the number of slits per mmB) the angle of diffraction for each order of diffracted beam.
108Answer 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 10-7N = 1/d = 1 ÷ 9.16 x 10-7 = 1.09 x 106 m-1So there are 1.09 x 103 slits per mmB) The highest order n = d/λ (rounded down)So max n = 9.16 x 10-7 ÷ 430 x 10-9 = 2Again using dsinθ = nλ rearranged we get sinθ = nλ/dsinθ = (2 × 430 x 10-9) ÷ 9.16 x 10-7 =Therefore θ = 69.9o
111Waves and Vibrations: Types of wave All travelling waves transfer energy from one place to anotherWaves may be classed as being: mechanical, e.g. Sound waves, or electromagnetic, e.g. Radio or UV, in origin