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Chapter 25

Chapter 25. Waves and Particles. Wave Phenomena. Interference Diffraction Reflection. Wave Description. – wavelength: distance between crests (meters) T – period: the time between crests passing fixed location (seconds) v – speed: the distance one crest moves in a second (m/s)

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Chapter 25

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  1. Chapter 25 Waves and Particles

  2. Wave Phenomena • Interference • Diffraction • Reflection

  3. Wave Description • – wavelength: distance between crests (meters) T – period: the time between crests passing fixed location (seconds) v – speed: the distance one crest moves in a second (m/s) f – frequency: the number of crests passing fixed location in one second (1/s or Hz)  – angular frequency: 2f: (rad/s)

  4. Wave: Variation in Time

  5. Wave: Variation in Space

  6. Wave: Variation in Time and Space ‘-’ sign: the point on wave moves to the right

  7. Wave: Phase Shift phase shift, =0…2 Two waves are ‘out of phase’ But E @ t=0 and x =0, may not equal E0 (Shown for x=0)

  8. Wave: Amplitude and Intensity Intensity I (W/m2): E0 is a parameter called amplitude (positive). Time dependence is in cosine function Often we detect ‘intensity’, or energy flux ~ E2. For example: Vision – we don’t see individual oscillations Works also for other waves, such as sound or water waves.

  9. Interference Superposition principle: The net electric field at any location is vector sum of the electric fields contributed by all sources. Can particle model explain the pattern? Laser: source of radiation which has the same frequency (monochromatic) and phase (coherent) across the beam. Two slits are sources of two waves with the same phase and frequency.

  10. Interference: Constructive E1 E2 Two emitters: Fields in crossing point Superposition: Amplitude increases twice: constructive interference

  11. Interference: Energy E1 E2 Two emitters: What about the intensity (energy flux)? Energy flux increases 4 times while two emitters produce only twice more energy There must be an area in space where intensity is smaller than that produced by one emitter

  12. Interference: Destructive E1 E2 Two waves are 1800 out of phase: destructive interference

  13. Interference Superposition principle: The net electric field at any location is the vector sum of the electric fields contributed by all sources. Constructive: Destructive: Amplitude increases twice Two waves are 1800 out of phase Constructive: Energy flux increases 4 times while two emitters produce only twice more energy

  14. Interference Intensity at each location depends on phase shift between two waves, energy flux is redistributed. Maxima with twice the amplitude occur when phase shift between two waves is 0, 2, 4, 6… (Or path difference is 0, , 2 …) Minima with zero amplitude occur when phase shift between two waves is , 3, 5… (Or path difference is 0, /2, 3/2…) Can we observe complete destructive interference if 1  2 ?

  15. Predicting Pattern For Two Sources C normal Path difference: Point C on screen is very far from sources Need to know phase difference Very far: angle ACB is very small Path AC and BC are equal If l = 0, , 2, 3, 4 … - maximum If l = /2, 3/2, 5/2 … - minimum

  16. Predicting Pattern For Two Sources C normal Path difference: If l = 0, , 2, 3, 4 … - maximum If l = /2, 3/2, 5/2 … - minimum What if d <  ? complete constructive interference only at =00, 1800 What if d < /2 ? no complete destructive interference anywhere Note: largest Dl for q=p/2

  17. Intensity versus Angle Path difference: If l = 0, , 2, 3, 4 … - maximum If l = /2, 3/2, 5/2 … - minimum d = 4.5  Why is intensity maximum at =0 and 1800 ? Why is intensity zero at =90 and -900 ? What is the phase difference at Max3?

  18. Intensity versus Angle Path difference: d = /3.5 If l = 0, , 2, 3, 4 … - maximum If l = /2, 3/2, 5/2 … - minimum Two sources are /3.5 apart. What will be the intensity pattern?

  19. Two-Slit Interference Path difference: If l = 0, , 2, 3, 4 … - maximum If l = /2, 3/2, 5/2 … - minimum L=2 m, d=0.5 mm, x=2.4 mm What is the wavelength of this laser? Small angle limit: sin() tan() 

  20. Application: Interferometry Detector Using interference effect we can measure distances with submicron precision laser

  21. Multi-Source Interference: X-ray Diffraction Coherent beam of X-rays can be used to reveal the structure of a crystal. Why X-rays? - they can penetrate deep into matter - the wavelength is comparable to interatomic distance Diffraction = multi-source interference

  22. Multi-Source Interference X-ray lattice Diffraction = multi-source interference Electrons in atoms will oscillate causing secondary radiation. Secondary radiation from atoms will interfere. Picture is complex: we have 3-D grid of sources We will consider only simple cases

  23. Generating X-Rays Accelerated electrons X-rays Copper Electrons knock out inner electrons in Cu. When these electrons fall back X-ray is emitted. (Medical equipment) Synchrotron radiation: Electrons circle around accelerator. Constant acceleration leads to radiation

  24. X-Ray: Constructive Interference Simple crystal: 3D cubic grid first layer Simple case: ‘reflection’ incident angle = reflected angle phase shift = 0

  25. X-Ray: Constructive Interference Condition for intense X-ray reflection: where n is an integer Reflection from the second layer will not necessarily be in phase Path difference: Each layer re-radiates. The total intensity of reflected beam depends on phase difference between waves ‘reflected’ from different layers

  26. Simple X-Ray Experiment crystal x-ray diffracted turn crystal May need to observe several maxima to find n and deduce d

  27. X-ray of Tungsten

  28. Using Crystal as Monochromator incident broadband X-ray reflected single-wavelength X-ray crystal Suppose you have a source of X-rays which has a continuum spectrum of wavelengths. How can one make it monochromatic?

  29. X-Ray of Powdered Crystals Powder contains crystals in all possible orientations Note: Incident angle doesn't have to be equal to scattering angle. Crystal may have more than one kind of atoms. Crystal may have many ‘lattices’ with different d polycrystalline LiF

  30. X-Ray of Complex Crystals (Myoglobin) 1960, Perutz & Kendrew

  31. Reflection of Visible Light Condition for intense X-ray reflection: where n is an integer Why do we see reflection of light from any smooth surface? Visible light:  ~ 6000 Å >> interatomic spacing Reflection from many layers is almost in-phase

  32. Reflection of Visible Light Constructive interference: The only possible difference in path length is zero. There will be maxima only when incident angle is equal to scattering angle.

  33. Thin-Film Interference Thin films such as soap bubbles are often colored: interference Consider thin /2-thick film There are ~3000 atomic layers Layer 1 and (N/2+1): destructive interference For each layer i=1…N/2 there is a layer i+N/2 which re-radiates with 1800 phase shift resulting in zero intensity – there will be no reflection of light for this particular wavelength

  34. Thin-Film Interference Destructive interference: for film thickness n/2. Constructive interference: for film thickness /4, 3/4, 5/4… Why are soap bubbles so colorful? Why after a while soap bubbles lose their color? Why there is no such effect for thick glass plates? Other examples or thin-film interference: oil or gasoline on water butterfly wings (in some cases) bird feathers (in some cases)

  35. Index of Refraction Wavelength of light in dense materials is shorter than in vacuum. Atoms get polarized due to the E induced by EM wave and due to the field created by other polarized atoms. The crest-to-crest distance in the net electric field is reduced since  is reduced, the speed v is slower v=f Index of refraction: n=c/v, or v=c/n

  36. Index of Refraction Index of refraction: n=c/v, or v=c/n Water: n=1.33 Glass: n~1.5 Frequency of light: not affected v=f Wavelength: ’ =/n X-rays: very high frequency, barely polarize atoms, speed almost not affected 1 =/n1 2 =/n2 1 n1= 2 n2

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