Chapter 37 HW Q: 2, 4, 7, 10, P: 4, 8, 13, 19, 21, 28, 31, 36, 42, 59, 62

Transcription

1 Chapter 37 HW Q: 2, 4, 7, 10, P: 4, 8, 13, 19, 21, 28, 31, 36, 42, 59, 62

2 Wave Interference y = y 1 +y 2 = 2A cos (φ/2) sin (kx - ωt + φ/2) Δφ Resultant Amplitude: 2Acos 2 Constructive Interference: Δ φ = 2 nπ, n= 0,1,2,3... Destructive Interference: Δ φ = (2n+ 1) π, n= 0,1, 2,3...

3 1-D Sound Wave Interference

4 Spherically Symmetric Waves

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8 Intensity

9 Quiet Min Loud Max Quiet Min Loud Max

10 Constructive or Destructive? (Indentical in phase sources) 2π Phase Difference at P: Δ φ = Δr λ P 2 π Δ φ = (1 λ ) = 2 π λ Constructive! Δφ Resultant Amplitude: 2Acos 2 Constructive Interference: Δ r = nλ, Δ φ = 2 nπ, n= 0,1,2,3... λ Destructive Interference: Δ r = (2n+ 1), Δ φ = (2n+ 1) π, n= 0,1, 2,3... 2

11 Where does the Phase Difference at P Come from? v 2π 2π Δ φ = ωδ t = 2π fδ t = 2 π Δ t = ( vδ t) = Δr λ λ λ Phase Difference at P: Δφ is different from the phase difference φ between the two source waves! 2π Phase Difference at P: Δ φ = Δr λ λ Path Difference at P: Δ r = Δφ 2π

12 Constructive or Destructive? (Source out of Phase by 180 degrees) 2π Phase Difference at P: Δ φ = Δ r +Δφ0 λ P 2 π Δ φ = (1 λ ) + π = 3 π λ Destructive! Δφ Resultant Amplitude: 2Acos 2 Constructive Interference: Δ r = nλ, Δ φ = 2 nπ, n= 0,1,2,3... λ Destructive Interference: Δ r = (2n+ 1), Δ φ = (2n+ 1) π, n= 0,1, 2,3... 2

13 Constructive or Destructive? (Indentical in phase sources) 2π Phase Difference at P: Δ φ = Δr λ P

14 Constructive or Destructive? (Indentical out of phase sources) 2π Phase Difference at P: Δ φ = Δr λ P

15 Contour Map of Interference Pattern of Two Sources

16 In Phase or Out of Phase? A B

17 Constructive or Destructive at P, R & Q?

18 Intensity

19 Quiet Min Loud Max Quiet Min Loud Max

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21 Interference of 2 Light Sources

22 Interference Patterns Constructive interference occurs at point P The two waves travel the same distance Therefore, they arrive in phase As a result, constructive interference occurs at this point and a bright fringe is observed

23 Interference Patterns The upper wave has to travel farther than the lower wave to reach point Q The upper wave travels one wavelength farther Therefore, the waves arrive in phase A second bright fringe occurs at this position

24 Interference Patterns The upper wave travels one-half of a wavelength farther than the lower wave to reach point R The trough of the bottom wave overlaps the crest of the upper wave This is destructive interference A dark fringe occurs

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26 Diffraction depends on SLIT WIDTH: the smaller the width, relative to wavelength, the more bending and diffraction.

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28 Double Slit is VERY IMPORTANT because it is evidence of waves. Only waves interfere like this.

29 If electron were hard bullets, there would be no interference pattern.

30 Particle Picture

31 In reality, electrons do show an interference pattern, like light waves.

32 Double Slit for Electrons shows Wave Interference

33 Electrons act like waves going through the slits but arrive at the detector like a particle.

34 Interference pattern builds one electron at a time. Electrons act like waves going through the slits but arrive at the detector like a particle.

35 Electron Diffraction

36 Particle Wave Duality

37 Conditions for Interference To observe interference in light waves, the following two conditions must be met: 1) The sources must be coherent They must maintain a constant phase with respect to each other 2) The sources should be monochromatic Monochromatic means they have a single wavelength

38 Double-Slit Assumptions Assumptions L >> d d >> λ a<< λ Approximation: θ is small and therefore the small angle approximation tan θ ~ sin θ can be used δ = r 2 r 1 = d sin θ y = L tan θ L sin θ Draw this! 2π 2π φ = δ = d sin θ λ λ

39 Young s Double-Slit Experiment: Geometry The path difference, δ, is found from the tan triangle δ = r 2 r 1 = d sin θ This assumes the paths are parallel Not exactly true, but a very good approximation if L is much greater than d 2π 2π φ = δ = d sin θ λ λ

40 Constructive Interference The path difference must be either zero or some integral multiple of the wavelength δ = d sin θ bright = mλ m = 0, ±1, ±2, m is called the order number When m = 0, it is the zeroth-order maximum When m = ±1, it is called the first-order maximum

41 Destructive Interference When destructive interference occurs, a dark fringe is observed This needs a path difference of an odd half wavelength δ = d sin θ dark = (m + ½)λ m = 0, ±1, ±2,

42 Interference Conditions Constructive: l l = mλ, m= 0,1,2,3, Destructive: l2 l1 = ( m+ ) λ, m= 0,1,2,3,... 2

43 Interference Fringes λ Bright fringe: sin θ = m m= 0,1,2,3,... d 1 λ Dark fringe: sin θ = ( m+ ) m= 0,1, 2,3,... 2 d Δl

44 Problem Red light (λ=664nm) is used in Young s double slit as shown. Find the distance y on the screen between the central bright fringe and the third order bright fringe. λ Bright fringe: sin θ = m m= 0,1,2,3,... d

45 Last Time: Fringe Equations For bright fringes λl ybright = m ( m = 0, ± 1, ± 2 K ) d For dark fringes λl 1 ydark = m+ ( m= 0, ± 1, ± 2 K ) d 2

46 Reality

47 Double Slit Interference Hyperphysics

48 Double Slit Interference Dependence on Slit Separation

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50 Iridescence Diffraction & Interference

51 Thin Film Interference

52 Where does light actually come from? Light comes from the acceleration of charges.

53 Light is emitted when an electron in an atom jumps between energy levels either by excitation or collisions.

54 Atoms are EM Tuning Forks They are tuned to particular frequencies of light energy.

55 Atomic Emission of Light Each chemical element produces its own unique set of spectral lines when it burns

56 Hydrogen Spectra

57 Light Emission

58 Incandescent Light Bulb Full Spectrum of Light All frequencies excited!

59 Light Waves E = Emax cos (kx ωt) B = Bmax cos (kx ωt) Emax ω E = = = c B k B max I E B E cb = Sav = = = 2μ 2μ c 2μ 2 2 max max max max o o o I E 2 max

60 The Electromagnetic Spectrum

61 Visible Light Different wavelengths correspond to different colors The range is from red (λ ~ 7 x 10-7 m) to violet (λ ~4 x 10-7 m)

62 I am Watching YOU!!

63 Human Retina Sharp Spot: Fovea Blind Spot: Optic Nerve

64 Human Vision An optical Tuning Fork

65 Optical Antennae: Rods & Cones Rods: Intensity Cones: Color

66 If you pass white light through a prism, it separates into its component colors. long wavelengths R.O.Y. G. B.I.V spectrum short wavelengths

67 Radiation of Visible Sunlight

68 Additive Primary Colors Red, Green, Blue

69 RGB Color Theory

70 Additive Complementary Colors Yellow, Cyan, Magenta The color you have to add to get white light. Red + Green = Yellow Blue + Green = Cyan Red + Blue = Magenta Red + Blue + Green = White White light red light =?? White light yellow light =??

71 FYI: Mixing Colored Pigments Subtractive Colors Pigments subtract colors from white light. Yellow + Cyan = Green Cyan + Magenta = Purple Yellow + Magenta = Red Yellow + Cyan + Magenta = Black

72 Why are some materials colored? Colored materials absorb certain colors that resonate with their electron energy levels and reject & reflect those that do not.

73 Why is the Ocean Cyan? White light minus cyan is red. Ocean water absorbs red.

74 Thin Film Interference

75 Thin Films When reflecting off a medium of greater refractive index, a light wave undergoes a phase shift of ½ a wavelength. Wave 1 undergoes a phase shift of 180 degrees.

76 The Index of Refraction Refraction: Light Bends in Transmission The speed of light in any material is less than its speed in vacuum The index of refraction, n, of a medium can be defined as speed of light in a vacuum c n = speed of light in a medium v n λ λin vacuum = λn λin a medium For a vacuum, n = 1 We assume n = 1 for air also For other media, n > 1 n is a dimensionless number greater than unity, not necessarily an integer

77 Some Indices of Refraction

78 Following the Reflected and Refracted Rays Ray 1 is the incident ray Ray 2 is the reflected ray Ray 3 is refracted into the lucite Ray 4 is internally reflected in the lucite Ray 5 is refracted as it enters the air from the lucite

79 Index of Refraction Light may refract into a material where its speed is lower The angle of refraction is less than the angle of incidence The ray bends toward the normal n < n 1 2

80 Index of Refraction Light may refract into a material where its speed is higher The angle of refraction is greater than the angle of incidence The ray bends away from the normal n > n 1 2

81 Index of Refraction The path of the light through the refracting surface is reversible For example, a ray that travels from A to B If the ray originated at B, it would follow the line AB to reach point A

82 Interference in Thin Films Assume the light rays are traveling in air nearly normal to the two surfaces of the film Ray 1 undergoes a phase change of 180 with respect to the incident ray Ray 2, which is reflected from the lower surface, undergoes no phase change with respect to the incident wave

83 Phase Changes Due To Reflection An electromagnetic wave undergoes a phase change of 180 upon reflection from a medium of higher index of refraction than the one in which it was traveling Analogous to a pulse on a string reflected from a rigid support

84 Phase Changes Due To Reflection, cont. There is no phase change when the wave is reflected from a boundary leading to a medium of lower index of refraction Analogous to a pulse on a string reflecting from a free support

85 Interference in Thin Films An electromagnetic wave traveling from a medium of index of refraction n 1 toward a medium of index of refraction n 2 undergoes a 180 phase change on reflection when n 2 > n 1 There is no phase change in thereflected wave if n 2 < n 1

86 Interference in Thin Films Ray 2 also travels an additional distance of 2t before the waves recombine For constructive interference 2nt = (m + ½)λ (m = 0, 1, 2 ) This takes into account both the difference in optical path length for the two rays and the 180 phase change For destructive interference 2nt = mλ (m = 0, 1, 2 )

87 Interference in Thin Films Two factors influence interference Possible phase reversals on reflection Differences in travel distance The conditions are valid if the medium above the top surface is the same as the medium below the bottom surface If there are different media, these conditions are valid as long as the index of refraction for both is less than n

88 Interference in Thin Films If the thin film is between two different media, one of lower index than the film and one of higher index, the conditions for constructive and destructive interference are reversed With different materials on either side of the film, you may have a situation in which there is a 180 o phase change at both surfaces or at neither surface Be sure to check both the path length and the phase change

89 Problem Solving Strategy with Identify the thin film causing the interference The type of interference constructive or destructive that occurs is determined by the phase relationship between the upper and lower surfaces Thin Films

90 Problem Solving with Thin Films Phase differences have two causes differences in the distances traveled phase changes occurring on reflection Both causes must be considered when determining constructive or destructive interference The interference is constructive if the path difference is an integral multiple of λ and destructive if the path difference is an odd half multiple of λ

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92 A thin film of gasoline floats on a puddle of water. Sunlight falls almost perpendicularly on the film and reflects into your eyes a yellow hue. Interference in the the thin gasoline film has eliminated blue (469nm in vacuum) from the reflected light. The refractive indices of the blue light in gasoline and water are 1.40 and 1.33 respectively. Determine the minimum nonzero thickness of the film.

93 Newton s Rings Interference Pattern

94 Newton s Rings Another method for viewing interference is to place a plano-convex lens on top of a flat glass surface The air film between the glass surfaces varies in thickness from zero at the point of contact to some thickness t A pattern of light and dark rings is observed These rings are called Newton s rings The particle model of light could not explain the origin of the rings Newton s rings can be used to test optical lenses

95 Light Waves E = Emax cos (kx ωt) B = Bmax cos (kx ωt) Emax ω E = = = c B k B max I E B E cb = Sav = = = 2μ 2μ c 2μ 2 2 max max max max o o o I E 2 max

96 Intensity Distribution, The total magnitude of the electric field at any point on the screen is the superposition of the two waves The magnitude of each wave at point P can be found E 1 = E o sin ωt E 2 = E o sin (ωt + φ) Both waves have the same amplitude, E o Electric Fields

97 Intensity Distribution, Phase Relationships The phase difference between the two waves at P depends on their path difference δ = r 2 r 1 = d sin θ A path difference of λ corresponds to a phase difference of 2π rad A path difference of δ is the same fraction of λ as the phase difference φ is of 2π This gives 2π 2π φ = δ = d sin θ λ λ

98 Intensity Distribution Resultant Field The magnitude of the resultant electric field comes from the superposition principle E P = E 1 + E 2 = E o [sin ωt + sin (ωt + φ)] This can also be expressed as φ φ EP = 2Eocos sin ωt E P has the same frequency as the light at the slits The magnitude of the field is multiplied by the factor 2 cos (φ / 2)

99 Intensity Distribution, Equation The expression for the intensity comes from the fact that the intensity of a wave is proportional to the square of the resultant electric field magnitude at that point The intensity therefore is 2 sin 2 I Imax cos πd θ Imax cos πd = y λ λl

100 Light Intensity, Graph The interference pattern consists of equally spaced fringes of equal intensity This result is valid only if L >> d and for small values of θ 2 sin 2 I Imax cos πd θ Imax cos πd = y λ λl

101 Phasor Addition of Waves, E 1 The sinusoidal wave can be represented graphically by a phasor of magnitude E o rotating about the origin counterclockwise with an angular frequency ω E 1 = E o sin ωt It makes an angle of ωt with the horizontal axis E 1 is the projection on the vertical axis

102 Phasor Addition of Waves, E 2 The second sinusoidal wave is E 2 = E o sin (ωt + φ) It has the same amplitude and frequency as E 1 Its phase is φ with respect to E 1

103 Phasor Addition of Waves, E R The resultant is the sum of E 1 and E 2 E R rotates with the same angular frequency ω The projection of E R along the vertical axis equals the sum of the projections of the other two vectors

104 E R at a Given Time From geometry at t = 0, E R = 2E 0 cos α = 2E o cos (φ / 2) The projection of E R along the vertical axis at any time t is φ EP = ERsin ωt + 2 φ φ = 2 Eo cos sin ωt + 2 2

105 Finding the Resultant of Several Waves Represent the waves by phasors Remember to maintain the proper phase relationship between one phasor and the next The resultant phasor E R is the vector sum of the individual phasors

106 Finding the Resultant of Several Waves, cont. At each instant, the projection of E R along the vertical axis represents the time variation of the resultant wave The phase angle α is between E R and the first phasor The resultant is given by the expression E P = E R sin (ωt + φ)

107 Phasor Diagrams for Two Coherent Sources, Comments E R is a maximum at φ = 0, 2π, 4π, The intensity is also a maximum at these points E R is zero at φ = π, 3π, The intensity is also zero at these points These results agree with the results obtained from other procedures

108 Phasor Diagrams for Two Coherent Sources

109 Three-Slit Interference Pattern Assume three equally spaced slits The fields are: E 1 = E o sin ωt E 2 = E o sin (ωt + φ) E 3 = E o sin (ωt + 2φ) Phasor diagrams can be used to find the resultant magnitude of the electric field

110 Three Slits Phasor Diagram The phasor diagram shows the electric field components and the resultant field The field at P has a maximum value of 3E o at φ = 0, ±2π, ± 4π... These points are called primary maxima

111 Three Slits, Additional Maxima The primary maxima occur when the phasors are in the same direction Secondary maxima occur when the wave from one slit exactly cancels the wave from another slit The field at P has a value of E o These points occur at φ = 0, ±π, ±3π...

112 Three Slits, Minima Total destructive interference occurs when the wave from all the slits form a closed triangle The field at P has a value of 0 These points occur at φ = 0, ±2π/3, ±4π/3...

113 Three Slits, Phasor Diagrams

114 Michelson Interferometer The interferometer was invented by an American physicist, A. A. Michelson The interferometer splits light into two parts and then recombines the parts to form an interference pattern The device can be used to measure wavelengths or other lengths with great precision

115 Michelson Interferometer, A ray of light is split into two rays by the mirror M o The mirror is at 45 o to the incident beam The mirror is called a beam splitter It transmits half the light and reflects the rest Schematic

116 Michelson Interferometer, Schematic Explanation, cont. The reflected ray goes toward mirror M 1 The transmitted ray goes toward mirror M 2 The two rays travel separate paths L 1 and L 2 After reflecting from M 1 and M 2, the rays eventually recombine at M o and form an interference pattern

117 Michelson Interferometer Operation The interference condition for the two rays is determined by their path length difference M 1 is moveable As it moves, the fringe pattern collapses or expands, depending on the direction M 1 is moved

118 Michelson Interferometer Operation, cont. The fringe pattern shifts by one-half fringe each time M 1 is moved a distance λ/4 The wavelength of the light is then measured by counting the number of fringe shifts for a given displacement of M 1

119 Michelson Interferometer Applications The Michelson interferometer was used to disprove the idea that the Earth moves through an ether Modern applications include Fourier Transform Infrared Spectroscopy (FTIR) Laser Interferometer Gravitational-Wave Observatory (LIGO)

120 Fourier Transform Infrared Spectroscopy This is used to create a high-resolution spectrum in a very short time interval The result is a complex set of data relating light intensity as a function of mirror position This is called an interferogram The interferogram can be analyzed by a computer to provide all of the wavelength components This process is called a Fourier transform

121 Laser Interferometer Gravitational-Wave Observatory General relativity predicts the existence of gravitational waves In Einstein s theory, gravity is equivalent to a distortion of space These distortions can then propagate through space The LIGO apparatus is designed to detect the distortion produced by a disturbance that passes near the Earth

122 LIGO, cont. The interferometer uses laser beams with an effective path length of several kilometers At the end of an arm of the interferometer, a mirror is mounted on a massive pendulum When a gravitational wave passes, the pendulum moves, and the interference pattern due to the laser beams from the two arms changes

123 LIGO in Richland, Washington