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 >10 3 ， Diffraction is not obvious a ~10 2 —10 ， Diffraction fringes appear a ~ 时， Diffraction is obvious a < ， Scatterance ( 散射 )

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

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

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

8. IV. Parallel beams interference—a simplified discussion  1,  2 – diffraction angle Optical axis Wave front Wave ray a screen P1P1P1P1 1111 1111 P2P2P2P2 2222 Coherent superposition

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

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

11. asin   BC=asin  =2(  /2) --The wave front AB is 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 ， --point P is dark. --Destructive interference.

12.  BC=asin  =3(  /2) The wave front AB is divided into 3 half wave zones (3 个半波带 ) -- The wave front AB is divided into 3 half wave zones (3 个半波带 ) The optical path difference between two corresponding points on AA 1 and A 1 A 2 is /2 ， --point P is bright. They produce destructive interference. The light oscillations coming from wave front A 2 B produce constructive interference.

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

14.  Central diffraction maximum fringe ： the region between the first positive and negative dark fringes Half-angle width ： Discussion 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. §17-9 Resolving power of optical instrument I. Fraunhofer diffraction by circular aperture Airy disk The diffraction angle of first dark ring,

20. distinguish II. Resolution of optical instrument

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

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 Resolving power of an optical instrument: Improve R ：  increasing D—astronomical telescope with large radius  decreasing --electronic microscope Minimum resolving angle

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

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

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

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

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

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

29. Interference of multi-slitsSlit diffraction

30. Missing order Grating differactin differactin + interference  0 1 2 34 5 6 -6 -5 -4 -3 -2

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

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

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

34. [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. f d a ?

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

36.  For a single diffraction, the width of central maximum is bright fringes appears in the width of x 0. 0 1 2 3 4 5 -2 -3 -4 -5 x0x0x0x0 xx

37. [Example]A diffraction grating has 500 slits per millimeter. It is irradiated by Sodium ( 钠 ) light with =0.59×10 -3 mm. 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 when the incident angle is 30 0 ? Solution  The grating constant is

38. 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, When gets maximum. ， k gets maximum. Take integral=3

39. 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 5-th. Take integral=5 Take integral= -1 the order numbers which can be observed in this case is 1-th.

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

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

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

43. 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) Missing order 0 1 2 34 -4 -3 -2

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

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

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

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

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

50. They found that when ---- ---- Bragg equation ， the x-ray beams produceconstructive interference. produce constructive interference.

51. got Noble Prize of Physics in 1915 because they found a new method to study the properties of x-ray. F W.H.Bragg, W.L.Bragg got Noble Prize of Physics in 1915 because they found a new method to study the properties of x-ray.