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

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

2 Superposition of Sinusoidal Waves Assume two waves are traveling in the same direction, with the same frequency, wavelength and amplitude The waves differ in phase y 1 = A sin (kx - ωt) y 2 = A sin (kx - ωt + φ) y = y 1 +y 2 = 2A cos (φ/2) sin (kx - ωt + φ/2) Resultant Amplitude Depends on phase: Spatial Interference Term

3 1-D Wave Interference y = y 1 +y 2 = 2A cos (φ/2) sin (kx - ωt + φ/2) The resultant wave function, y, is also sinusoidal The resultant wave has the same frequency and wavelength as the original waves The amplitude of the resultant wave is 2A cos (φ/2) The phase of the resultant wave is φ/2 Constructive Destructive Interference φ = 0 φ = π φ

4 1-D Wave Interference y = y 1 +y 2 = 2A cos (φ/2) sin (kx - ωt + φ/2) φ Resultant Amplitude: 2Acos 2 Constructive Interference (Even π): Δ φ = 2 nπ, n= 0,1, 2,3... ( ) Destructive Interference Odd π : Δ φ = (2n+ 1) π, n= 0,1, 2,3... Constructive Destructive Interference φ = 0 φ = π φ

5 1-D Sound Wave Interference

6 2-D Wave Interference? 2-D Phase Difference at P: Δφ is different from the phase difference φ between the two source waves! It depends on the path difference traveled by the two waves! 2π Phase Difference at P: Δ φ = Δ r + φ0 λ Δφ Amplitude at P: 2Acos( ) 2 P

7 Constructive or Destructive? (Indentical in phase sources) 2π Phase Difference at P: Δ φ = Δ r + φ0 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

8 Constructive or Destructive? (Source out of Phase by 180 degrees) 2π Phase Difference at P: Δ φ = Δ r + φ0 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

9 Constructive or Destructive? (Indentical in phase sources) 2π Phase Difference at P: Δ φ = Δ r + φ0 λ P

10 Constructive or Destructive? (Sources out of phase) Phase Difference at P: 2π Δ φ = Δ r + φ0 λ P

11 Exam Extra Credit Phase Difference at P: 2π Δ φ = Δ r + φ0 λ

12 Exam Extra Credit 2π Phase Difference at P: Δ φ = Δ r + φ0 λ y = y + y = 2Acos( Δφ/ 2)sin( kx ωt+δφ/ 2) 1 2 Δφ Amplitude at P: 2Acos( ) 2 I A 2 I = I Δφ 2 max cos ( / 2)

13 In Phase or Out of Phase? Constructive? Destructive? A B

14 Quiet Min Loud Max Quiet Min Loud Max

16 Light Waves: Coherent Sources Intensity 2π Phase Difference at P: Δ φ = Δr, Δ φ0 = 0 λ

17 Interference of 2 Light Sources

18 Conditions for Interference Use a Double Slit! 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

19 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

20 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

21 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

23 Interference of 2 Light Sources

24 Double Slit Interference Hyperphysics Website

25 Light Intensity 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

26 Reality

27 Double-Slit

28 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, L>>d Not exactly true, but a very good approximation if L is much greater than d 2π 2π φ = δ = d sin θ λ λ

29 Interference Assumptions If L>>d, then the rays are approximated parrallel, and the pink triangle is a right triangle and the angles are equal as shown and the path difference is: Δr If L >> d, then Δ r = d sinθ

30 Interference Conditions Constructive: Δ r = mλ, m= 0,1,2,3,... 1 Destructive: Δ r = ( m+ ) λ, m= 0,1, 2,3,... 2

31 Interference Fringes Bright fringe: Δ r = dsin θ = mλ, m= 0,1, 2,3,... 1 Dark fringe : Δ r = dsin θ = ( m+ ) λ m= 0,1, 2,3,... 2 Δr

32 Derive 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

33 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. Find the width of the central bright maxima. λl ybright = m ( m = 0, ± 1, ± 2 K ) d

34 You Try 2 Slit The image shows the light intensity on a screen behind a double slit. The slit spacing is 0.20 mm and the wavelength of light is 600 nm. What is the distance from the slits to the screen? λl ybright = m ( m = 0, ± 1, ± 2 K ) d

35 Double Slit Interference Dependence on Slit Separation λl ybright = m ( m = 0, ± 1, ± 2 K ) d

37 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

38 Double Slit Interference Hyperphysics Website

39 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 amplitude at P is given by 2E o cos (φ / 2) Intensity is proportional to the square of the amplitude: I = I 2 cos ( Δφ / 2) I A 2 max

40 Light Intensity 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

41 HO 2 Slit

42 You Try In a double-slit experiment, the distance between the slits is 0.2 mm, and the distance to the screen is 150 cm. What wavelength (in nm) is needed to have the intensity at a point 1 mm from the central maximum on the screen be 80% of the maximum intensity? a. 900 b.700 c. 500 d.300 e. 600

43 Thin Film Interference

44 RGB Color Theory

45 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 =??

$The Index of Refraction Refraction: Light Bends in Transmission The speed of light in any material is less than its speed in vacuum The index of refraction, n, of a medium can be defined as speed of$

46 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

47 Some Indices of Refraction

48 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

$Thin Films When reflecting off a medium of greater refractive index, a light wave$

49 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.

50 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

$medium of lower index of refraction Analogous to a pulse on a string reflecting from a free support$

51 Phase Changes Due To 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 Reflection

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

53 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

$Interference in the the thin gasoline film has eliminated blue (469nm in vacuum) from the reflected light. The refractive indices of the blue light in gasoline and water are 1.40 and 1.$

54 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.

55 Newton s Rings Interference Pattern

56 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

57 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 After reflecting from M 1 and M 2, the rays eventually recombine at M o and form an interference pattern 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

58 Michelson Interferometer The fringe pattern shifts by one-half fringe each time M1 is moved a distance λ/4

59 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)

60 James Clerk Maxwell 1860s Light is wave. The medium is the Ether. 1 8 c= = 3.0x10 m/ s με 0 o

61 Measure the Speed of the Ether Wind The Luminiferous Aether was imagined by physicists since Isaac Newton as the invisible "vapor" or "gas aether" filling the universe and hence as the carrier of heat and light.

62 Rotate arms to produce interference fringes and find different speeds of light.

63 Michelson-Morely Experiment 1887 The speed of light is independent of the motion and is always c. The speed of the Ether wind is zero. OR. Lorentz Contraction The apparatus shrinks by a factor : 1 v / c 2 2

64 On the Electrodynamics of Moving Bodies 1905

65 Clocks slow down and rulers shrink in order to keep the speed of light the same for all observers!

66 Time is Relative! Space is Relative! Only the SPEED OF LIGHT is Absolute!

67 As you approach c, lengths contract.

68 LIGO in Richland, Washington

69 LISA

70 Double Slit for Electrons shows Wave Interference

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

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

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

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

75 Particle Wave Duality

76 Particle Picture: Trying to detect which slit the electron went through destroys the wave behavior

77 Feynman version of the Uncertainty Principle It is impossible to design an apparatus to determine which hole the electron passes through, that will not at the same time disturb the electrons enough to destroy the interference pattern. -Richard Feynman

PH 222-3A Spring 2010

PH 222-3A Spring 2010 PH -3A Spring 010 Interference Lecture 6-7 Chapter 35 (Halliday/Resnick/Walker, Fundamentals of Physics 8 th edition) 1 Chapter 35 Interference The concept of optical interference is critical to understanding