Presentation is loading. Please wait.

Presentation is loading. Please wait.

Dr K. Newson TGSG Physics Dept.

Similar presentations

Presentation on theme: "Dr K. Newson TGSG Physics Dept."— Presentation transcript:

1 Dr K. Newson TGSG Physics Dept.
AQA Physics A Unit 2 Chapters 12 and 13 Waves and Optics Dr K. Newson TGSG Physics Dept.

2 Waves and their properties
Chapter 12 AQA PHY 2

3 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.

4 Types of wave Longitudinal waves Transverse waves
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

5 Longitudinal Waves 1) Longitudinal Wave: The vibration of particles is parallel to the direction of the wave. E.g. sound waves, some seismic waves.

6 Longitudinal waves in springs
Transverse Wave: The vibration (red arrow) is parallel to the wave’s direction (black arrow). E.g. sound waves.

7 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).

8 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 End-on view Wave vibrations are in more than one plane

9 Polarisation contd. Transverse waves that are plane polarised have vibrations that are in one plane only. Wave vibrations are in just one plane End-on view Note: Longitudinal waves cannot be polarised as the vibration is parallel to the wave’s direction of motion.

10 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

11 Polarising filters Filter will polarise the light in the vertical plane If rotated by 90o the light will be polarised in the horizontal plane

12 Polarisation of light Unpolarised light Polarizing filter
Plane polarised light

13 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

14 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.

15 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

16 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.

17 Representing waves λ Amplitude λ Length/m

18 Representing waves T Amplitude T Time

19 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

20 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 90 180 270 360 Angle in radians π/2 π 3π/2 In terms of λ λ/4 λ/2 3λ/4 λ

21 Phase difference These oscillations are in step, they are said to be in phase. Their phase difference is zero.

22 Phase difference These oscillations are a ¼ of a cycle out of phase. Their phase difference is π/2 radians (90o), or λ/4

23 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.

24 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

25 Wavefronts 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 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 R Q P λ λ λ λ λ λ λ P Q R P Q R

26 Wavefronts contd. The expanding circles in the wave train are called wave fronts.

27 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.

28 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.

29 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.

30 Diffraction Strong diffraction as gap width ≈ λ

31 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).

32 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

33 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.


35 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.

36 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

37 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.

38 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

39 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

40 Experiments Three demonstrations of stationary waves
Melde’s apparatus (string) using sheet wv6

41 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.

42 Nodes and Antinodes N N N N N N A A A λ/2 A = antinode N = node

43 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.

44 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 A N Y X This simplest mode of oscillation is called the fundamental and occurs at the fundamental frequency.

45 Harmonics 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 A N N N If the string has a length L, then λ= L

46 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.

47 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.

48 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.

49 Demonstration of stationary waves
1) On a string/cord Signal generator Vibration generator Thick rubber cord X Y The cord is fixed at positions X and Y.

50 Simulation of Stationary waves

51 Optics Metre rule r Chapter 13 AQA PHY 2

52 Section 1 Refraction of Light Objectives: Know what rays are
Know Snell’s Law

53 Rays Light rays represent the direction of travel of wavefronts.

54 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.

55 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 r2 Wave speeds up and bends away from the normal due to entering a less dense medium e.g. glass into air The wave slows down but is not bent, due to entering along the normal

56 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.

57 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:- Carry out experiment described on p188 What else is observed? sin i sin r n =

58 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.

59 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

60 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.

61 Refraction explained λ Air Wavelength in air > Wavelength in glass

62 Y YY’ = Ct Air i Y’ X r XX’ = Cst X’ Glass

63 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’.

64 Refraction 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’sini Similarly for triangle XX’Y’, XY’ is perpendicular to XX’, and the angle r we can say cst =XY’sinr Combining these gives: Therefore the refractive index of a substance is given by: sini = c sinr cs ns = sini c sinr cs

65 Refraction explained ns = c = λ cs λs
Speed and wavelength change but frequency remains constant Therefore as c= λf it also follows that: ns = c = λ cs λs

66 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. We can then state that: sini = c1 sinr c2 n1 n2 r i

67 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

68 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 – 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.

69 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.

70 Section 3 Total Internal Reflection Objectives:
Know the conditions required for TIR Know how the critical angle and refractive index are related

71 Total internal reflection (TIR)
At angles θ < θc the light is refracted at the boundary Air θ Glass

72 Total internal reflection (TIR)
At the critical angle θc the light is refracted through 90o θc

73 Total internal reflection (TIR)
At angles greater than the critical angle all the light is reflected, none is refracted. Hence we have TIR.

74 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

75 TIR Uses 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

76 Sections 4&5 Double Slit Interference (including Young’s Double Slits Experiment) & Coherence

77 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.

78 Path difference and interference
First subsidiary maximum Slit 1 (s1) O Central Maximum Microwave source A Minimum Slit 2 (s2) B First subsidiary maximum

79 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.

80 X Path difference (d1 – d2) X O d1 X A λ/2 d2 B λ X C 3λ/2 maximum
S1 X O d1 S2 X A λ/2 d2 B λ X C 3λ/2 maximum X minimum

81 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.

82 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,…………

83 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. 28 cm 110cm P 66 cm s1 s2

84 Answer 28 cm 110cm P 66 cm s1 s2 Since P is the position of the third maximum, the path difference (s2P – s1P) = 3λ Using Pythagoras (s2P)2 = ( )2 therefore s2P = 136.0cm And (s1P)2 = ( )2 therefore s1P = 121.7cm Therefore (s2P – s1P) = ( ) = = 3λ so = λ = 14.3/3 = 4.8cm

85 Interference of waves

86 Young’s Double slit experiment
Think light or laser not microwaves

87 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.

88 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).

89 Apparatus Screen w D Double slit HeNe Laser

90 Reminder about interference

91 Young’s double slit experiment
Screen Double slits D O (central max) 1st Subsidiary max = P P w S1 S2 w Q

92 Path difference (slits to 1st sub max) = S1B
For 1st Subsidiary maximum path difference = λ Q S1

93 Young’s double slit experiment
θ s Therefore distance S1B = λ Also, S1B = sSinθ = λ Q S1

94 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

95 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

96 Double 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 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)



99 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.

100 Double slit diffraction

101 Single slit diffraction
                                                                                                           Single slit diffraction

102 Single slit diffraction

103 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 × 2D a

104 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.

105 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


107 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.

108 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 = Therefore θ = 69.9o

109 Types of spectra

110 C: constructive interference
D: destructive interference

111 Waves and Vibrations: Types of wave
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

Download ppt "Dr K. Newson TGSG Physics Dept."

Similar presentations

Ads by Google