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Part 2 Diffraction of light

Part 2 Diffraction of light. Diffraction phenomena of light. §17-6 Diffraction of light Huygens-Fresnel’s principle. I. Diffraction phenomena of light. Condition :. The width of the diffracting obstacle is not very largely compared to the wavelength.

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Part 2 Diffraction of light

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  1. Part 2 Diffraction of light

  2. Diffraction phenomena of light

  3. §17-6 Diffraction of light Huygens-Fresnel’s principle I. Diffraction phenomena of light Condition : The width of the diffracting obstacle is not very largely compared to the wavelength.

  4. a >> ，Diffraction negligible a >103，Diffraction is not obvious a ~102 —10，Diffraction fringes appear a ~时， Diffraction is obvious a < ，Scatterance (散射)

  5. obstacle source screen II. Classification Fresnel diffraction --the source or the screen or both are at finite distance from the diffracting obstacle. Fraunhofer diffraction --the source and the screen are at infinite distance from the diffracting obstacle.

  6. dA  n  dA  r P III. Huygens-Fresnel’s principle Every elementdS of wave frontSis the source of a secondary spherical wavelet. The amplitude dA of the secondary spherical wavelet emitted by dS is proportional to the size of dS, dA  dS k()—inclination factor

  7. n  dA  r Interference appears. P The intensity of light changes in space. The light oscillation at P produce by dS is The resultant oscillation of light at P is the coherent superposition of all spherical wavelets emitted by all elements on the S

  8. Coherent superposition screen P1 Wave front 1 1 a Optical axis 2 Wave ray P2 IV. Parallel beams interference—a simplified discussion 1 , 2– diffraction angle

  9. §17-7Single slit Fraunhofer diffraction I. Diffraction device

  10. C λ λ λ λ 2 2 2 2 II. Distribution law of diffraction fringes ---- Fresnel half wave zone method (半波带法)

  11. BC=asin=2(l /2) --The wave frontABis divided into 2 half wave zones(2个半波带) The optical path difference between two corresponding points on A0 and 0B is /2， --Destructive interference. --point P is dark.

  12.  BC=asin=3(l /2) --The wave frontABis divided into 3 half wave zones(3个半波带) The optical path difference between two corresponding points on AA1 and A1A2is /2， Theyproduce destructive interference. The light oscillations coming from wave front A2B produce constructive interference. --point Pis bright.

  13. --dark --bright BC=asin =n(l /2) --dark fringes • If nis even number(偶数)： • If nis odd number(奇数)： --bright fringes  Ifasin integral times of/2, the intensity of light is between maximum and minimum.

  14. Discussion Central diffraction maximum fringe：the region between the first positive and negative dark fringes Half-angle width： Half width： --inverse proportion

  15.  The distance between other adjacent fringes (bright or dark)

  16. The intensity distribution of diffraction fringes: Most of the light intensity is concentrated in the broad central diffraction maximum.

  17. [Example]In experiment of Fraunhofer diffraction from a single slit, f = 0.5m，=5000Å，the width of the slit a=0.1mm. Find the width of central maximum, the width of the secondary maximum. Solution  the width of central maximum

  18. The width l of the secondary maximum equals to the space of the first minimum and the second minimum.

  19. I. Fraunhofer diffraction by circular aperture §17-9Resolving power of optical instrument Airy disk The diffraction angle of first dark ring,

  20. II. Resolution of optical instrument distinguish

  21. Airy disk Can’t be distinguished Minimum resolving angle Just distinguished

  22. Rayleigh criterion：Two images are just resolved when the center of central maximum of one pattern coincides with the first dark ring of another. 爱里斑 Minimum resolving angle Resolving power of an optical instrument: ImproveR：  increasing D—astronomical telescope with large radius  decreasing --electronic microscope

  23. I. Grating An optical device which consists of a large number of equally and parallel slits with same distance. §17-8 Diffraction grating classification： Reflecting grating Transmitting grating

  24. ---optical path difference between rays from adjacent slits. --grating constant

  25. 1. The interference of multi-slits (多缝干涉) II. The formation of grating diffracting fringes diffraction + interference The phase difference between rays from adjacent slits is When  =2k , The rays coming from all slits are in phase. ----constructive interference

  26. According to We get --grating equation The principle maximum appears at the direction with the diffraction angle .

  27. N=6 N=4 N=2 Principle maximum secondary maximum Principle maximum and secondary maximum

  28. The intensity N2 N=3 N=2 N=1 2.The influence of diffraction by each slit to the interference fringes The diffraction patterns of all slits coincide.

  29. Slit diffraction Interference of multi-slits

  30. Missing order -2 0 1 -3 3 2 5 6 -6 -5 -4 4 -1 differactin + interference  Grating differactin

  31. The missing order phenomenon of grating: On some direction, if diffraction angle  satisfies, --constructive interference and --minimum of slit diffraction then the k-th principle maximum will disappear. -- The k-th fringe ismissing order.

  32. III. The incident ray inclination The optical path difference of two adjacent rays is The grating function for inclination incidence is

  33. Third order spectrum Second order spectrum First order spectrum IV. Grating spectrum When a polychromatic light (复色光) is incident a grating, except for the central fringe, all others principle maximum with different  have different  for each same order.

  34. ? d a f [Example]Two slits with d=0.40mm, the width is a=0.08mm. A parallel light with =4800Å is emitted on the two slits. A lens with f =2.0m is put on the slits. Calculate: The distance x of the interference fringes on the focal plane of the lens.  The numbers of interference fringes located in the width of the central maximum producing by single diffraction.

  35. Solution For two beams interference, the bright fringes (principle maxima) satisfies: the position of k-th order bright fringe is the distance between two adjacent bright fringes is

  36. 5 x 4 3 2 1 x0 0 -1 -2 -3 -4 -5  For a single diffraction, the width of central maximum is bright fringes appears in the width of x0 .

  37. [Example]A diffraction grating has 500 slits per millimeter. It is irradiated by Sodium(钠) light with =0.59×10-3mm. Find  The maximum order of spectrum can be observed when the beams of Sodium light are incident normally? How many orders of spectrum can be seen whenthe incident angle is300 ? Solution  The grating constant is

  38. When ，k gets maximum. Take integral=3 i.e., when the beams are incident normally, the maximum order which can be observed is 3-th.  When the incident rays and diffraction rays are in the same side of the optical axis,

  39. Take integral= -1 Take integral=5 the order numbers which can be observed in this case is 5-th. When the incident rays and diffraction rays are in the two side of the optical axis, the order numbers which can be observed in this case is 1-th.

  40. 1 order 0-th order 5 orders • The spectrum with order -1, 0,1,2,3,4,5can be observed

  41. Take integral= 4 For , [Example] A monochromatic light with =7000Å is incident normally on a grating. The grating has d= 3×10-4 cm, a=10-4cm. Find The maximum order of the spectrum can be observed ?  Which orders are missing? Solution The maximum order which can be observed is 4-th.

  42. for same  , When , corresponding to k=3,6…  for interference bright fringes. for diffraction dark fringes. As The order k=3 is missing.

  43. Missing order 0 1 2 3 4 -4 -3 -2 -1 i.e., the orders of the spectrum that can be observed are 4-1=3. They correspond with k= -4,-2,-1,0,1,2,4 ( seven principle maxima)

  44. §17-10 x –Ray diffraction by Crystal • It was discovered by W.K Roentgen ( a German physicist) in 1895.11. I. x -Ray • The first x–ray photo: his wife’s hand. • He got the first Noble Prize of Physics in 1901 as the discovery of x-ray.

  45. Anode Cathode x-ray tube • x-ray: produced by bombarding a target element (Anode) with a high energy beam of electrons in a x-ray tube. • It’s a type of electromagnetic waves with wavelength ranges about 0.1--100Å, between Ultraviolet and -ray.

  46. x-ray is a wave Lead plate crystal film In 1912, a collimated beam of x-ray which contain a continuous distribution of wavelengths strikes a single crystal, the diffraction pattern of x-ray was observed by German physicist M. Von Laue. Laue (劳厄) spots

  47. x-ray can be used widely to study the internal structure of crystals. • Laue got Noble Prize of Physics in 1914 because verified that x-ray is a wave.

  48. II.Bragg equation 掠射角 • W.H.Bragg, and W.L.Bragg, two British physicists ( son and father) took another method to study x-ray diffraction. 晶面间距d 晶面 The optical path difference of the two x-ray beams scattering by the atoms that they locate in two parallel planes is

  49. They found that when ，the x-ray beams produceconstructive interference. ---- Bragg equation

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