2What are electromagnetic waves? EM waves are energy emitted resulting from acceleration of electric charges.-
3EM Waves They can travel through vacuum. In vacuum, their speed = 3 × 108 ms-1c = f.λAn EM wave consists of electric and magnetic fields, oscillating in phase and at right angles to each other.
4Electromagnetic spectrum The range of the wavelength of EM waves is enormous.10-14 m – 1 kmThe electromagnetic spectrum is named according to the range of the wavelength and the method of production.
5Radio Waves Production: Apply an a.c. voltage of high frequency to a pair of metal rods (dipole).If the rods are vertical, the radio wave is also said to be vertically polarized.transmitterdirection of propagation ofradio waveoscillating a.c.oscillating electric field
6Radio WavesReceiver:The receiver dipole is parallel to the direction of polarization. In this case, it is vertical.transmitterreceiveroscillating electric fielddirection of propagation ofradio waveoscillating a.c.
7Spectrum of Radio Waves WavelengthLong waves1 km – 10 kmMedium waves100 m – 1 kmShort waves10 m – 100 mVery high frequency (VHF)1 m – 10 mUltra high frequency(UHF)0.1 m – 1 m
8Microwaves Microwave is polarized along the length of the dipole. transmitterreceiveroscillating electric fielddirection of propagation ofmicrowaveoscillating a.c.
9Microwaves Vertical metal rod can absorb the energy of the microwave. transmitterreceiveroscillating electric fielddirection of propagation ofmicrowaveno responseoscillating a.c.
10MicrowavesHorizontal metal rod cannot absorb the energy of the microwave.metal rodtransmitterreceiveroscillating electric fielddirection of propagation ofmicrowaveoscillating a.c.
11Interference of Microwaves At P, the wave from the transmitter meets the reflected wave. Interference occurs.Ptransmitterimageof transmittermetal plate
12Interference of Microwaves We may consider it as an interference from two coherent sources, the transmitter and its image.Ptransmitterimageof transmittermetal plate
13Interference of Microwaves The two sources are in anti-phase because there is a phase change of on reflection.Ptransmitterimageof transmittermetal plate
14Interference of Microwaves If the path difference at P = n., there is destructive interference.Ptransmitterimageof transmittermetal plate
15Interference of Microwaves If the path difference at P = , there is constructive interference.Ptransmitterimageof transmittermetal plate
16Microwave CookingOne possible frequency of microwave is 2.45 GHz which is equal to the natural frequency of water molecules.Microwave can set water molecules into oscillation. The water molecules absorb the energy from microwave.
17Microwave in satellite communications Reading the following
22Scattering of Light Light energy is absorbed by an atom or molecule. The atom (molecule) re-emits the light energy in all direction.The intensity of light in initial direction is reduced.incident lightatomoscillating E-field
23Scattering of Light Light energy is absorbed by an atom or molecule. The atom (molecule) re-emits the light energy in all directions.The intensity of light in initial direction is reduced.axis along which the atom oscillatingatomscattered light
24Scattering of LightNote that there is not any scattered light along the direction of oscillation of the atom.The scattered light is maximum at right angle to the axis.axis along which the atom oscillatingatomscattered lightstrongeststrongest
25Why is the sky blue at noon and red at sunrise and sunset? Why is the sky blue in daytime?Why is the sky red in sunset/sunrise?
26Why is the sky blue at noon and red at sunrise and sunset? At noon, we see the most scattered light.Note that the natural frequency of air molecules is in the ultraviolet region. Blue light is easily scattered by air molecules.white light fromthe sunBlue lightis most scatteredRed light is least scattered
27Why is the sky blue at noon and red at sunrise and sunset? At sunset, we see the least scattered light.Red light is least scattered.white light fromthe sunRed lightis least scatteredBlue light is most scattered
28Polarization of lightLight is transverse wave so it exhibits polarization.Unpolarized light: the electric field is not confined to oscillate in a plane.Plane-polarized light: the electric field at every point oscillates in the same fixed plane.Plane of polarization: the plane in which the electric field of a plane polarized light oscillates.
29Polarization of Light Plane polarized light: Unpolarised light: electric vectorelectric vector
30Polarization by Absorption An array of parallel conducting wires.It can absorb electric field of microwave oscillating in a plane parallel to its conducting wires.Plane-polarizedmicrowaveConducting wires are verticalNo microwaveE-field is vertial.
31Polarization by Absorption An array of parallel conducting wires.It cannot absorb electric field of microwave oscillating in a plane perdpndicular to its conducting wires.Plane-polarizedmicrowaveConducting wires are verticalE-field is horizontalPlane-polarizedmicrowave
32Polarization by Absorption An array of parallel conducting wires.It can be a polarizer of microwavesUnpolarizedmicrowaveConducting wires are verticalPlane-polarizedmicrowave
33Polarization by Absorption Polaroid is a plastic sheet consisting of long chains of molecules parallel to one another.It can absorb electric field of light oscillating in a plane parallel to its chains of molecules.Chains of molecules are verticalPlane-polarized lightNo lightE-field is vertical
34Polarization by Absorption Polaroid is a plastic sheet consisting of long chains of molecules parallel to one another.It cannot absorb electric field of light oscillating in a plane perpendicular to its chains of molecules.Chains of molecules are verticalPlane-polarized lightPlane-polarized lightE-field is horizontal
35Polarization by Absorption Polaroid is a plastic sheet consisting of long chains of molecules parallel to one another.It can be a polarizer of light.Chains of molecules are verticalUnpolarized lightPlane-polarized light
36Polarization by Reflection Assume that the direction of the reflected light and that of the refractedlight are perpendicular.plane-polarizedincident lightrefracted lightNo reflected lightairglass
37Polarization by Reflection The electric field sets theelectrons in the glass tooscillate at right angles tothe refracted ray.The intensity perpendicularto the axis of oscillation isstrongest The refracted ray is bright.The intensity parallel tothe axis of oscillation is zero no reflected ray.plane-polarizedincident lightrefracted lightNo reflected lightairglass
38Polarization by Reflection Assume that the direction of the reflected light and that of the refractedlight are perpendicular.The plane of polarizationis parallel to the surfaceof medium.Unpolarizedincident lightPolarized reflected lightairglassUnpolarizedrefracted light
39The Brewster’s Angle p = Brewster’s angle r = Angle of refraction. incidentrayreflectedray is completelypolarizedrefractedairmedium
40The Brewster’s Anglen = tan p where n is the refractive index of the medium.princidentrayreflectedrefractedairmediumProve it!
41Example 1The Brewster’s angle for glass is about 56.3o.
42Polarization by Scattering When light energy is absorbed by an atom, the atom re-radiates the light.incident rayatomThe atom absorbsthe wave energy.The atom re-radiatesthe wave energy.
43Polarization by Scattering vertically polarizedlightno scattered lightvertically polarizedlightwater mixed with milkvertically polarizedlightvertically polarizedlightno scattered light
44Polarization by Scattering horizontally polarizedlightno scattered lighthorizontally polarizedlightwater mixed with milkhorizontally polarizedlightno scattered lighthorizontally polarizedlight
45Polarization by Scattering horizontally polarizedlightvertically polarizedlightunpolarizedlightwater mixed with milkunpolarizedlightvertically polarizedlighthorizontally polarizedlight
46Polaroid SunglassesWhy are the polaroid sunglasses designed to absorb horizontally polarized light?Study p.233 of the textbook.
47Interference of Light Light is a kind of wave. Interference is a wave property.
48Interference of LightConditions for an observable interference pattern of light:Coherent sources : two sources emit light of the same frequency and maintain a constant phase difference.The light waves are of same frequency and almost equal amplitude.The separation of the two sources is of the same order as the wavelength.The path difference must be not too large.
49Interference of Light Young’s double-slit experiment The incident ray is split into two coherent sourcesS1 and S2 by the double-slit.S1 and S2 are in phase.The screen is far away from the slit. D>> a.The angles are very small.
50Young’s double-slit experiment Suppose that there is a maximum at point P.A constructive interference occurs at P.PS1acentral lineS2Dscreen
51Young’s double-slit experiment As point P is far away from the double slit,the light rays of the same fringe are parallel.parallel rays meet at point PS1acentral lineS2The path difference = a.sin
52Young’s double-slit experiment For points with constructive interference,= a.sin = m. where m = 0, 1, 2,…m is the order of the fringes.parallel rays meet at point P withmaximum intensity.S1acentral lineS2The path difference = a.sin
53Young’s double-slit experiment For points with destructive interference,= a.sin = (m ). where m = 0, 1, 2,…parallel rays meet at a point withminimumintensity.S1acentral lineS2The path difference = a.sin
54separation between the fringes Suppose the the order of the fringe at P is m.The distance from P to the central line is ym.The distance between the double slitand the screen is D.PymS1aMcentral lineS2D
55separation between the fringes The line from mid-point M to P makes the same angle with the central line.Pym = D.tan D.sin =ymS1aMcentral lineS2D
56separation between the fringes ym =(1)By similar consideration, for the m+1 bright fringeym+1=(2)The separation between the two fringes iss = ym+1 – ym =
57separation between the fringes s = ym+1 – ym =The fringes are evenly separated.For well separated fringes.s D Place the screen far away from the slit.s Different separation for waves of different wavelength.s The slits should be close.
58Variation of intensity If the slits are sufficiently narrow, light spreads out evenly from each slit and the bright fringes are equally bright.
59Variation of intensity If the intensity on the screen using one slit is Io,the intensity is 4.Io at the position of bright fringes andthe intensity is 0 at the position of dark fringes.Energy is re-distributed on the screen.
60Variation of intensity In practice, light waves do not diffract evenly out from each slit. There is an angle of spread.
61Variation of intensity The intensity of the fringes is enclosed in an envelope as shown.
62Example 2Separation of fringes in Young’s double-slit experiment.
63White light fringes Separation of violet fringes is shortest. Separation of red fringes is longest.
64Submerging in a LiquidIf the Young’s double-slit experiment is done in a liquid, what would happen to the separation of fringes?ymLiquid with refractive index n
65Submerging in a Liquid The wavelength changes! Let be the wavelength in vacuum/airand n the wavelength in liquid.Let n be the refractive index of the liquid.The fringe separation in liquidsn =
66Submerging in a LiquidThe fringe separation is reduced by a factor of n.The fringe separation in liquidsn =
67Optical pathOptical path of light in a medium is the equivalent distance travelled by light in vacuum.thickness = tmedium ofrefractive indexnLight requires the sametime to travel throughthe two paths.Incidentlightthickness = optical pathvacuum
68Optical path Show that the optical path = n.t thickness = t medium of refractive indexnLight requires the sametime to travel throughthe two paths.Incidentlightthickness = optical pathvacuum
69Optical pathThe number of waves in the medium = the number of waves in the optical paththickness = tmedium ofrefractive indexnLight requires the sametime to travel throughthe two paths.Incidentlightthickness = optical pathvacuum
70Example 3Note that light rays pass through different media. We need to consider their path difference in terms of the optical paths.
71Shifting a System of Fringes Note that ray A has passes through a medium of refractiveindex n and thickness t.We need to find the path difference in terms of the opticalpath.AB
72Shifting a System of Fringes Without the medium, the central maximum is atthe central line. (Path difference = 0)Now the central maximum shifts to another position.Find the central maximum.The central maximum shifts through a distance
73Shifting a System of Fringes If the central line now has the mth bright fringe,the central maximum has shifted through m fringes.
74Multiple-slit More slits than two. 3 slits 4 slits
75Multiple-slit N = number of slits. Compare N = 2 with N = 3
76Multiple-slit N = 2 N = 3 N = number of slits. Compare N = 2 with N = 3N = 2N = 3Maximum occurs at positions with a.sin = m.The intensity at the maximum is 4.IoThe intensity at the maximum is 9.IoBetween two maxima, it is a minimum.Between two maxima, it is a peak.The width of bright fringes is large.The width of bright fringes is less
77Multiple-slit N = number of slits. Compare N = 2 with large N. N = 2 Maximum occurs at positions with a.sin = m.The intensity at the maximum is 4.IoThe intensity at the maximum is N2.IoBetween two maxima, it is a minimum.Between two maxima, there are (N-2) peaks. The intensity of the peak is almost zeroThe width of bright fringes is very narrow
78Multiple-slit N = 3 Maximum occurs at a.sin = m. (same as N = 2). Central maximum1st order maximumphase difference= 0aaphase difference= 0aam= 0 = 0All three rays are in phase.m= 1 a.sin = All three rays are in phase.
79Multiple-slit Textbook, p.238. Fig. 30. Position of maximum when N = 2 is still a maximum.There are peak(s) between two successive maxima.The number of peaks = N – 2.The intensity of peaks drops with N.
80Multiple-slit (N = 3)m = 0m = 1Why is there a peak between two maxima when N = 3?
81Multiple-slit (N = 3) There is a peak at position with Δ1 =phase difference= πaθθΔ1Δ2 =λphase difference= 0aΔ2Add three rotating vectors for the resultant wave.
82Multiple-slit (N = 3)m = 1m = 0m = 1Why are there 2 minima between two maxima when N = 3?
83Multiple-slit (N = 3) There is a minimum at position with Δ1 =phase difference=aθθΔ2 =phase difference=Δ1aΔ2Add three rotating vectors for the resultant wave.
84Multiple-slit (N = 3) There is a minimum at position with Δ1 =phase difference=Δ2 =aθθΔ1aΔ2Add three rotating vectors for the resultant wave.
85Diffraction gratingA diffraction grating is a piece of glass with many equidistant parallel lines.
86Diffraction grating The mth order maximum is given by a.sin = m.
87Diffraction grating The maximum order is given by m = 3 m = 2 m = 1
89coarse grating and fine grating The separation between lines on a coarse grating is longer than that of a fine grating.Example of a coarse grating: 300 lines/cm.Example of a fine grating: 3000 lines/cm.Find the maximum order of the above two gratings.
105Air Wedge A normal incident ray is reflected at the boundary between the slide and the air wedge.reflected ray Anormal incidentrayslideair wedgeglass block
106Air Wedge The normal incident ray goes into the wedge, passing through a distance t and reflected at the boundary betweenthe air wedge and the glass block.reflected ray Areflected ray Bnormal incidentrayair wedgetslideglass block
107Air Wedge The ray into the glass block is ignored. reflected ray A reflected ray Bnormal incidentrayair wedgetslideglass block
108Air Wedge The ray reflected at the top of the slide is ignored. normal incidentrayslideair wedgeglass block
109Air WedgeDepending on the the distance t and the wavelength of the incident wave, the two reflected rays may have interference.The pattern is a series of bright and dark fringes when we view through the travelling microscope.
110Air Wedge To produce bright fringes, 2.t = (m - )., m = 1, 2, 3,… reflected ray Areflected ray Bnormal incidentrayair wedgetconstructiveinterferenceslideglass block
111Air Wedge Ray B has a phase change on reflection. The path difference must be m.reflected ray Areflected ray Bnormal incidentrayair wedgetconstructiveinterferenceslideglass block
112Air Wedge To produce dark fringes, 2.t = m ., m = 0, 1, 2,… reflected ray Areflected ray Bnormal incidentrayair wedgetdestructiveinterferenceslideNote that ray B has a phase change on reflection.glass block
113Air Wedge Ray B has a phase change on reflection. The path difference must be (m + ).reflected ray Areflected ray Bnormal incidentrayair wedgetdestructiveinterferenceslideNote that ray B has a phase change on reflection.glass block
114Air Wedge At the vertex, t = 0 m = 0 dark fringe. reflected ray A reflected ray Bnormal incidentrayair wedgetdestructiveinterferenceslideNote that ray B has a phase change on reflection.glass block
115Air Wedge To find the separation s between two successive bright fringesLet D be the height of the high end of the slide.Let L be the length of the slide.(N+1)th fringeNth fringeslideLDtN+1air wedgetNsglass block
116Air Wedge To find the separation s between two successive bright fringesAngle of inclination of the slide can be found from(N+1)th fringeNth fringeslideLDtN+1air wedgetNsglass block
117Air Wedge Separation s between two successive bright fringes tN+1 – tN =tNtN+1s
118Air Wedge Separation s between two successive bright fringes For small angle , sin tan andtNtN+1s
119Air Wedge For flat surfaces of glass block and slide the fringes are parallel and evenly spaced.
120Air WedgeFor flat surface of glass block and slide with surface curved upwardsthe fringes are parallel and become more closely packed at higher orders.
121Air WedgeFor flat surface of glass block and slide with surface curved downwardsthe fringes are parallel and become more widely separated at higher orders.
122Air Wedge We can use this method to check a flat glass surface. The surface is flat.The surface is not flat.
123Measuring the Diameter D of a Wire Measure the quantities on the righthand side and calculate D.air wedgeDL
124Example 8 There are 20 dark fringes m = 19. L 19.s s = D =
126Soap film The two transmitted rays A and B may have interference depending on the thickness t of the film and the wavelength.ray Atincident rayray Bsoapwater
127Soap film Note that ray B has two reflections. No phase change is due to reflection.ray Atincident rayray B
128Soap film To observe bright fringes, the path difference = m. 2.t = m. ray Atconstructiveinterferenceincident rayray B
129Soap film To observe dark fringes, the path difference = (m+ ). 2.t = (m+ ). ray Atconstructiveinterferenceincident rayray B
130Soap film The two reflected rays A and B may have interference depending on the thickness t of the film and the wavelength.ray Aray Bincident raytsoapwater
131Soap filmNote that this time there is a phase change due to reflection.ray Aray Bincident raytsoapwater
132Soap film Find out how the interference depends on t and ray Aray Bincident raytsoapwaterinterference
133Soap filmAs the interference depends on , there will be a colour band for white incident light.In each colour band, violet is at the top and red is at the bottom.
134Soap film Soap water tends to move downwards due to gravity. The soap film has a thin vertex and a thick base. The fringes are not evenly spaced.The fringes are dense near the bottom and less dense near the vertex.
135Soap filmThe pattern of the reflected rays and that of the transmitted rays are complementary.incident rayCDC: constructiveinterferenceD: destructivereflected raystransmitted rays
136Example 9 The reflected rays have constructive interference. Note that there is a phase change on reflection.
140Newton’s rings The two reflected rays have interference depending on the thickness t of the air gap and the wavelength .reflected ray Ainterferenceincident rayreflected ray Blensairt = thickness of air gaptglass block
141Newton’s rings For bright fringes, reflected ray A interference incident rayreflected ray Blensairt = thickness of air gaptglass block
142Newton’s rings For dark fringes, reflected ray A interference incident rayreflected ray Blensairt = thickness of air gaptglass block
143Newton’s rings At the center of the lens, there is a dark spot. reflected ray Ainterferenceincident rayreflected ray Blensairt = thickness of air gaptglass block
144Newton’s rings The spacing of the rings are not even. Near the center, the rings are widely separated.Near the edge, the rings are close together.
145Newton’s ringsFind the radius of the mth dark ring.Rm
146Newton’s rings C R = radius of curvature of the lens lens B A Rm air t t = thickness of air gapOglass block
147Newton’s rings for the mth dark fringe C R = radius of curvature of the lenslensBARmairttt = thickness of air gapOglass block
148Newton’s rings C R = radius of curvature of the lens lens B A Rm air t t = thickness of air gapOglass block
149Newton’s rings Separation between two successive rings s = Rm+1 – Rm = The separation approaches zero for high orders.
150Example 10To find the radius of curvature of a lens by Newton’s rings.If there is distortion of the Newton’s rings, the lens is not a good one.
151Thin films Light is incident obliquely onto a thin film of refractive index n and thickness t.thin filmtnincidentray
152Thin films The reflected rays have interference depending on the angle of view and wavelength .thin filmtnincidentray
153Thin film coloured spectrum incident white light
154Diffraction of Light laser tube single slit screen
156Diffraction of LightConsider a light source which is far away from the single slit and the light is normal to the slit.In general, the position of the mth dark fringe due to a slit of width d is given byd.sinm = m.
158Theory of diffraction Formation of 1st order dark fringe. Divide the slit into two equal sections A1 and A2.The light from section A1 cancels the light from section A2dA1A21Prove that
159Theory of diffraction Formation of 2nd order dark fringe. Divide the slit into 4 equal sections A1 , A2, A3 and A4.The light from section A1 cancels the light from section A2.The light from section A3 cancels the light from section A4Prove thatA12A2dA3A4
160Theory of diffraction Formation of 2nd order dark fringe. In general for the mth dark fringe,A12A2dA3A4
161Diffraction of LightConsider a light source which is far away from the single slit and the light is normal to the slit.For a circular hole with diameter d, the center is a bright spot and the 1st dark ring is given by