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Chapter 11 Electromagnetic Waves

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1 Chapter 11 Electromagnetic Waves
Physics Beyond 2000 Chapter 11 Electromagnetic Waves

2 What are electromagnetic waves?
EM waves are energy emitted resulting from acceleration of electric charges. -

3 EM Waves They can travel through vacuum.
In vacuum, their speed = 3 × 108 ms-1 c = f.λ An EM wave consists of electric and magnetic fields, oscillating in phase and at right angles to each other.

4 Electromagnetic spectrum
The range of the wavelength of EM waves is enormous. 10-14 m – 1 km The electromagnetic spectrum is named according to the range of the wavelength and the method of production.

5 Radio 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. transmitter direction of propagation of radio wave oscillating a.c. oscillating electric field

6 Radio Waves Receiver: The receiver dipole is parallel to the direction of polarization. In this case, it is vertical. transmitter receiver oscillating electric field direction of propagation of radio wave oscillating a.c.

7 Spectrum of Radio Waves
Wavelength Long waves 1 km – 10 km Medium waves 100 m – 1 km Short waves 10 m – 100 m Very high frequency (VHF) 1 m – 10 m Ultra high frequency (UHF) 0.1 m – 1 m

8 Microwaves Microwave is polarized along the length of the dipole.
transmitter receiver oscillating electric field direction of propagation of microwave oscillating a.c.

9 Microwaves Vertical metal rod can absorb the energy of the microwave.
transmitter receiver oscillating electric field direction of propagation of microwave no response oscillating a.c.

10 Microwaves Horizontal metal rod cannot absorb the energy of the microwave. metal rod transmitter receiver oscillating electric field direction of propagation of microwave oscillating a.c.

11 Interference of Microwaves
At P, the wave from the transmitter meets the reflected wave. Interference occurs. P transmitter image of transmitter metal plate

12 Interference of Microwaves
We may consider it as an interference from two coherent sources, the transmitter and its image. P transmitter image of transmitter metal plate

13 Interference of Microwaves
The two sources are in anti-phase because there is a phase change of  on reflection. P transmitter image of transmitter metal plate

14 Interference of Microwaves
If the path difference at P = n., there is destructive interference. P transmitter image of transmitter metal plate

15 Interference of Microwaves
If the path difference at P = , there is constructive interference. P transmitter image of transmitter metal plate

16 Microwave Cooking One 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.

17 Microwave in satellite communications
Reading the following

18 Infrared Radiation Self-reading.

19 Ultraviolet Radiation
Self-reading.

20 Visible light Self-reading.

21 Colored Video Pictures
Self-reading.

22 Scattering 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 light atom oscillating E-field

23 Scattering 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 oscillating atom scattered light

24 Scattering of Light Note 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 oscillating atom scattered light strongest strongest

25 Why 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?

26 Why 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 from the sun Blue light is most scattered Red light is least scattered

27 Why 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 from the sun Red light is least scattered Blue light is most scattered

28 Polarization of light Light 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.

29 Polarization of Light Plane polarized light: Unpolarised light:
electric vector electric vector

30 Polarization 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-polarized microwave Conducting wires are vertical No microwave E-field is vertial.

31 Polarization 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-polarized microwave Conducting wires are vertical E-field is horizontal Plane-polarized microwave

32 Polarization by Absorption
An array of parallel conducting wires. It can be a polarizer of microwaves Unpolarized microwave Conducting wires are vertical Plane-polarized microwave

33 Polarization 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 vertical Plane-polarized light No light E-field is vertical

34 Polarization 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 vertical Plane-polarized light Plane-polarized light E-field is horizontal

35 Polarization 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 vertical Unpolarized light Plane-polarized light

36 Polarization by Reflection
Assume that the direction of the reflected light and that of the refracted light are perpendicular. plane-polarized incident light refracted light No reflected light air glass

37 Polarization by Reflection
The electric field sets the electrons in the glass to oscillate at right angles to the refracted ray. The intensity perpendicular to the axis of oscillation is strongest  The refracted ray is bright. The intensity parallel to the axis of oscillation is zero  no reflected ray. plane-polarized incident light refracted light No reflected light air glass

38 Polarization by Reflection
Assume that the direction of the reflected light and that of the refracted light are perpendicular. The plane of polarization is parallel to the surface of medium. Unpolarized incident light Polarized reflected light air glass Unpolarized refracted light

39 The Brewster’s Angle p = Brewster’s angle r = Angle of refraction.
incident ray reflected ray is completely polarized refracted air medium

40 The Brewster’s Angle n = tan p where n is the refractive index of the medium. p r incident ray reflected refracted air medium Prove it!

41 Example 1 The Brewster’s angle for glass is about 56.3o.

42 Polarization by Scattering
When light energy is absorbed by an atom, the atom re-radiates the light. incident ray atom The atom absorbs the wave energy. The atom re-radiates the wave energy.

43 Polarization by Scattering
vertically polarized light no scattered light vertically polarized light water mixed with milk vertically polarized light vertically polarized light no scattered light

44 Polarization by Scattering
horizontally polarized light no scattered light horizontally polarized light water mixed with milk horizontally polarized light no scattered light horizontally polarized light

45 Polarization by Scattering
horizontally polarized light vertically polarized light unpolarized light water mixed with milk unpolarized light vertically polarized light horizontally polarized light

46 Polaroid Sunglasses Why are the polaroid sunglasses designed to absorb horizontally polarized light? Study p.233 of the textbook.

47 Interference of Light Light is a kind of wave.
Interference is a wave property.

48 Interference of Light Conditions 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.

49 Interference of Light Young’s double-slit experiment
The incident ray is split into two coherent sources S1 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.

50 Young’s double-slit experiment
Suppose that there is a maximum at point P. A constructive interference occurs at P. P S1 a central line S2 D screen

51 Young’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 P S1 a central line S2 The path difference  = a.sin

52 Young’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 with maximum intensity. S1 a central line S2 The path difference  = a.sin

53 Young’s double-slit experiment
For points with destructive interference, = a.sin = (m ).  where m = 0, 1, 2,… parallel rays meet at a point with minimumintensity. S1 a central line S2 The path difference  = a.sin

54 separation 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 slit and the screen is D. P ym S1 a M central line S2 D

55 separation between the fringes
The line from mid-point M to P makes the same angle  with the central line. P ym = D.tan  D.sin  = ym S1 a M central line S2 D

56 separation between the fringes
ym = (1) By similar consideration, for the m+1 bright fringe ym+1= (2) The separation between the two fringes is s = ym+1 – ym =

57 separation 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.

58 Variation of intensity
If the slits are sufficiently narrow, light spreads out evenly from each slit and the bright fringes are equally bright.

59 Variation of intensity
If the intensity on the screen using one slit is Io, the intensity is 4.Io at the position of bright fringes and the intensity is 0 at the position of dark fringes. Energy is re-distributed on the screen.

60 Variation of intensity
In practice, light waves do not diffract evenly out from each slit. There is an angle of spread.

61 Variation of intensity
The intensity of the fringes is enclosed in an envelope as shown.

62 Example 2 Separation of fringes in Young’s double-slit experiment.

63 White light fringes Separation of violet fringes is shortest.
Separation of red fringes is longest.

64 Submerging in a Liquid If the Young’s double-slit experiment is done in a liquid, what would happen to the separation of fringes? ym Liquid with refractive index n

65 Submerging in a Liquid The wavelength changes!
Let  be the wavelength in vacuum/air and n the wavelength in liquid. Let n be the refractive index of the liquid. The fringe separation in liquid sn =

66 Submerging in a Liquid The fringe separation is reduced by a factor of n. The fringe separation in liquid sn =

67 Optical path Optical path of light in a medium is the equivalent distance travelled by light in vacuum. thickness = t medium of refractive index n Light requires the same time to travel through the two paths. Incident light thickness = optical path vacuum

68 Optical path Show that the optical path = n.t thickness = t medium of
refractive index n Light requires the same time to travel through the two paths. Incident light thickness = optical path vacuum

69 Optical path The number of waves in the medium = the number of waves in the optical path thickness = t medium of refractive index n Light requires the same time to travel through the two paths. Incident light thickness = optical path vacuum

70 Example 3 Note that light rays pass through different media. We need to consider their path difference in terms of the optical paths.

71 Shifting a System of Fringes
Note that ray A has passes through a medium of refractive index n and thickness t. We need to find the path difference in terms of the optical path. A B

72 Shifting a System of Fringes
Without the medium, the central maximum is at the central line. (Path difference = 0) Now the central maximum shifts to another position. Find the central maximum. The central maximum shifts through a distance

73 Shifting a System of Fringes
If the central line now has the mth bright fringe, the central maximum has shifted through m fringes.

74 Multiple-slit More slits than two. 3 slits 4 slits

75 Multiple-slit N = number of slits. Compare N = 2 with N = 3

76 Multiple-slit N = 2 N = 3 N = number of slits.
Compare N = 2 with N = 3 N = 2 N = 3 Maximum occurs at positions with a.sin = m. The intensity at the maximum is 4.Io The intensity at the maximum is 9.Io Between 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

77 Multiple-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.Io The intensity at the maximum is N2.Io Between two maxima, it is a minimum. Between two maxima, there are (N-2) peaks. The intensity of the peak is almost zero The width of bright fringes is very narrow

78 Multiple-slit N = 3 Maximum occurs at a.sin = m. (same as N = 2).
Central maximum 1st order maximum phase difference = 0 a a phase difference = 0 a a m= 0   = 0 All three rays are in phase. m= 1  a.sin =  All three rays are in phase.

79 Multiple-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.

80 Multiple-slit (N = 3) m = 0 m = 1 Why is there a peak between two maxima when N = 3?

81 Multiple-slit (N = 3) There is a peak at position with
Δ1 = phase difference = π a θ θ Δ1 Δ2 =λ phase difference = 0 a Δ2 Add three rotating vectors for the resultant wave.

82 Multiple-slit (N = 3) m = 1 m = 0 m = 1 Why are there 2 minima between two maxima when N = 3?

83 Multiple-slit (N = 3) There is a minimum at position with
Δ1 = phase difference = a θ θ Δ2 = phase difference = Δ1 a Δ2 Add three rotating vectors for the resultant wave.

84 Multiple-slit (N = 3) There is a minimum at position with
Δ1 = phase difference = Δ2 = a θ θ Δ1 a Δ2 Add three rotating vectors for the resultant wave.

85 Diffraction grating A diffraction grating is a piece of glass with many equidistant parallel lines.

86 Diffraction grating The mth order maximum is given by a.sin = m.

87 Diffraction grating The maximum order is given by m = 3 m = 2 m = 1

88 Intensity of diffraction grating

89 coarse 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.

90 Colour Spectrum from a Diffraction Grating
a..sinm = m.  The spatial angle m depends on  m = 2 m = 3 m = 2 m = 1 m = 1 white light m = 0 m = 1 m = 1 m = 2 m = 3 m = 3 m = 2

91 Example 4 Monochromatic light = light with only one colour (frequency)

92 Example 5 Overlapping of colour spectrum

93 Blooming of lenses Coat a thin film on a lens to reduce the reflection of light. reflected ray with coating incident ray no reflected ray without coating incident ray film glass glass

94 Blooming of lenses Ray A is reflected at the boundary between air and the film. Reflected ray A incident ray with coating

95 Blooming of lenses Ray B is reflected at the boundary between the thin film and the glass. Reflected ray A incident ray Reflected ray B with coating

96 Blooming of lenses It is designed to have destructive interference for the reflected light rays.  No reflected light ray. Reflected ray A incident ray Reflected ray B with coating

97 Blooming of lenses Suppose that the incident ray is normal to the lens. The reflected light rays are also along the normal. reflected rays incident ray film film glass glass

98 Blooming of lenses Let n’ be the refractive index and t be the thickness of the thin film. Let  be the wavelength of the incident light. incident ray reflected rays film film glass glass

99 Blooming of lenses To have destructive interference for the reflected rays, incident ray reflected rays film film glass glass

100 Blooming of lenses Energy is conserved. As there is not any reflected light rays, the energy goes to the transmitted light ray. incident ray No reflected rays film film glass glass transmitted ray

101 Blooming of lenses The reflected ray from the bottom of the glass is so dim that it can be ignored. This reflected ray is ignored. incident ray film film glass glass

102 Blooming of lenses Limitation: For normal incident ray only.
For wave of one particular wavelength only. This reflected ray is ignored. incident ray film film glass glass

103 Examples Example 6 Find the minimum thickness of the thin film.
Find the wavelength.

104 Air Wedge: Experimental setup

105 Air Wedge A normal incident ray is reflected at the boundary
between the slide and the air wedge. reflected ray A normal incident ray slide air wedge glass block

106 Air Wedge The normal incident ray goes into the wedge, passing
through a distance t and reflected at the boundary between the air wedge and the glass block. reflected ray A reflected ray B normal incident ray air wedge t slide glass block

107 Air Wedge The ray into the glass block is ignored. reflected ray A
reflected ray B normal incident ray air wedge t slide glass block

108 Air Wedge The ray reflected at the top of the slide is ignored.
normal incident ray slide air wedge glass block

109 Air Wedge Depending 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.

110 Air Wedge To produce bright fringes, 2.t = (m - )., m = 1, 2, 3,…
reflected ray A reflected ray B normal incident ray air wedge t constructive interference slide glass block

111 Air Wedge Ray B has a phase change  on reflection.
The path difference must be m. reflected ray A reflected ray B normal incident ray air wedge t constructive interference slide glass block

112 Air Wedge To produce dark fringes, 2.t = m ., m = 0, 1, 2,…
reflected ray A reflected ray B normal incident ray air wedge t destructive interference slide Note that ray B has a phase change on reflection. glass block

113 Air Wedge Ray B has a phase change  on reflection.
The path difference must be (m + ). reflected ray A reflected ray B normal incident ray air wedge t destructive interference slide Note that ray B has a phase change on reflection. glass block

114 Air Wedge At the vertex, t = 0  m = 0  dark fringe. reflected ray A
reflected ray B normal incident ray air wedge t destructive interference slide Note that ray B has a phase change on reflection. glass block

115 Air Wedge To find the separation s between two successive
bright fringes Let D be the height of the high end of the slide. Let L be the length of the slide. (N+1)th fringe Nth fringe slide L D tN+1 air wedge tN s glass block

116 Air Wedge To find the separation s between two successive
bright fringes Angle of inclination of the slide can be found from (N+1)th fringe Nth fringe slide L D tN+1 air wedge tN s glass block

117 Air Wedge Separation s between two successive bright fringes
tN+1 – tN = tN tN+1 s

118 Air Wedge Separation s between two successive bright fringes
For small angle , sin   tan  and tN tN+1 s

119 Air Wedge For flat surfaces of glass block and slide
the fringes are parallel and evenly spaced.

120 Air Wedge For flat surface of glass block and slide with surface curved upwards the fringes are parallel and become more closely packed at higher orders.

121 Air Wedge For flat surface of glass block and slide with surface curved downwards the fringes are parallel and become more widely separated at higher orders.

122 Air Wedge We can use this method to check a flat glass surface.
The surface is flat. The surface is not flat.

123 Measuring the Diameter D of a Wire
Measure the quantities on the right hand side and calculate D. air wedge D L

124 Example 8 There are 20 dark fringes  m = 19. L  19.s  s =  D =

125 Soap film Why soap film is coloured?

126 Soap film The two transmitted rays A and B may have interference
depending on the thickness t of the film and the wavelength . ray A t incident ray ray B soap water

127 Soap film Note that ray B has two reflections. No phase change is due
to reflection. ray A t incident ray ray B

128 Soap film To observe bright fringes, the path difference = m.
 2.t = m.  ray A t constructive interference incident ray ray B

129 Soap film To observe dark fringes, the path difference = (m+ ).
 2.t = (m+ ).  ray A t constructive interference incident ray ray B

130 Soap film The two reflected rays A and B may have interference
depending on the thickness t of the film and the wavelength . ray A ray B incident ray t soap water

131 Soap film Note that this time there is a phase change due to reflection. ray A ray B incident ray t soap water

132 Soap film Find out how the interference depends on t and 
ray A ray B incident ray t soap water interference

133 Soap film As 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.

134 Soap 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.

135 Soap film The pattern of the reflected rays and that of the transmitted rays are complementary. incident ray C D C: constructive interference D: destructive reflected rays transmitted rays

136 Example 9 The reflected rays have constructive interference.
Note that there is a phase change on reflection.

137 Newton’s rings: Experimental setup

138 Newton’s rings What do we see through the travelling microscope with white incident light?

139 Newton’s rings If we use red incident light,

140 Newton’s rings The two reflected rays have interference depending on
the thickness t of the air gap and the wavelength . reflected ray A interference incident ray reflected ray B lens air t = thickness of air gap t glass block

141 Newton’s rings For bright fringes, reflected ray A interference
incident ray reflected ray B lens air t = thickness of air gap t glass block

142 Newton’s rings For dark fringes, reflected ray A interference
incident ray reflected ray B lens air t = thickness of air gap t glass block

143 Newton’s rings At the center of the lens, there is a dark spot.
reflected ray A interference incident ray reflected ray B lens air t = thickness of air gap t glass block

144 Newton’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.

145 Newton’s rings Find the radius of the mth dark ring. Rm

146 Newton’s rings C R = radius of curvature of the lens lens B A Rm air t
t = thickness of air gap O glass block

147 Newton’s rings for the mth dark fringe C
R = radius of curvature of the lens lens B A Rm air t t t = thickness of air gap O glass block

148 Newton’s rings C R = radius of curvature of the lens lens B A Rm air t
t = thickness of air gap O glass block

149 Newton’s rings Separation between two successive rings s = Rm+1 – Rm =
The separation approaches zero for high orders.

150 Example 10 To 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.

151 Thin films Light is incident obliquely onto a thin film of refractive
index n and thickness t. thin film t n incident ray

152 Thin films The reflected rays have interference depending on the angle
of view  and wavelength . thin film t n incident ray

153 Thin film coloured spectrum incident white light

154 Diffraction of Light laser tube single slit screen

155 Diffraction of Light Intensity variation

156 Diffraction of Light Consider 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 by d.sinm = m.

157 Diffraction of Light Variation of intensity

158 Theory 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 A2 d A1 A2 1 Prove that

159 Theory 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 A4 Prove that A1 2 A2 d A3 A4

160 Theory of diffraction Formation of 2nd order dark fringe.
In general for the mth dark fringe, A1 2 A2 d A3 A4

161 Diffraction of Light Consider 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


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