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Part 2 Diffraction of light Diffraction phenomena of light §17-6 Diffraction of light Huygens-Fresnel’s principle I. Diffraction phenomena of light E E S S Condition : The width of the diffracting obstacle is not very largely compared to the wavelength. a >> ,Diffraction negligible a >103,Diffraction is not obvious a ~102 —10,Diffraction fringes appear a ~时, Diffraction is obvious a < , Scatterance (散射) II. Classification obstacle Fresnel diffraction S --the source or the screen or both are at finite distance source from the diffracting obstacle. Fraunhofer diffraction --the source and the screen are at infinite distance from the diffracting obstacle. S E screen E III. Huygens-Fresnel’s principle Every element dS of wave front S is 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, 1 dA dS dS S r n dA P dA r dS dA k ( ) r k()—inclination factor dS S The light oscillation at P produce by dS is n dA 2r dy dA cos( t ) P r The resultant oscillation of light at P is the coherent superposition of all spherical wavelets emitted by all elements on the S y( P ) dA cos( t S 2r ) Interference appears. The intensity of light changes in space. IV. Parallel beams interference—a simplified discussion screen Coherent superposition P1 Wave front Wave ray 1 a 2 1 Optical axis 1 , 2 – diffraction angle P2 §17-7 Single slit Fraunhofer diffraction I. Diffraction device E S II. Distribution law of diffraction fringes ---- Fresnel half wave zone method (半波带法) BC a sin a A A1 A2 A3 B P C f E BC=asin =2( /2) --The wave front AB is divided into 2 half wave zones(2个半波带) A a 0 The optical path difference between two corresponding points on A0 and 0B is /2, --Destructive interference. --point P is dark. B C BC=asin=3( /2) --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 AA1 and A1A2 is /2, A A1 A2 a B C They produce destructive interference. The light oscillations coming from wave front A2B produce constructive interference. --point P is bright. n 1,2 BC=asin =n( /2) If n is even number (偶数): --dark fringes If n is odd number (奇数): --bright fringes --dark a sin 2 k 2 k 1,2 a sin ( 2k 1) --bright 2 If asin integral times of /2, the intensity of light is between maximum and minimum. Discussion Central diffraction maximum fringe:the region between the first positive and negative dark fringes Half-angle width: 1 E 0 1 sin1 a 2 f Half width:x 2 f tg 2 f 0 a 1 --inverse proportion x0 a The distance between other adjacent fringes (bright or dark) x xk 1 xk f tg k 1 f tg k ( k 1) k f[ ] a a f 1 x0 2 a The intensity distribution of diffraction fringes: Most of the light intensity is concentrated in the broad central diffraction maximum. I I0 5 2a 3 2a 3 2a a a 5 2a sin [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 x0 2 f tg 1 2 f sin1 P1 x 0 1 f 2f a 2 0.5 5000 10 0.1 103 10 5 mm The width l of the secondary maximum equals to the space of the first minimum and the second minimum. 2 l f sin 2 f sin1 f ( ) a a f 2.5 mm a §17-9 Resolving power of optical instrument I. Fraunhofer diffraction by circular aperture Airy disk E S The diffraction angle of first dark ring, sin 1 0.61 r 1.22 D I 0.610 r I0 0.610 sin r II. Resolution of optical instrument E S1 S2 A2 A1 distinguish E S1 S2 S1 A2 A1 Can’t be distinguished Minimum resolving angle S2 Just distinguished E Airy disk A2 A1 Rayleigh criterion:Two images are just resolved when the center of central maximum of one pattern coincides with the first dark ring of another. sin 1.22 1 1 爱里斑 D Minimum resolving angle Resolving power of an optical 1 D instrument: R 1 1.22 Improve R: increasing D—astronomical telescope with large radius decreasing --electronic microscope §17-8 Diffraction grating I. Grating An optical device which consists of a large number of equally and parallel slits with same distance. classification: Reflecting Transmitting grating grating d sin b da E 0 x P d a b --grating constant d sin ---optical path difference between rays from adjacent slits. II. The formation of grating diffracting fringes diffraction + interference 1. The interference of multi-slits (多缝干涉) The phase difference between rays from adjacent slits is 2 (a b) sin When =2k , A B (a b) sin C The rays coming from all slits are in phase. ----constructive interference According to We get 2 (a b) sin k d sin k k 0,1,2 --grating equation The principle maximum appears at the direction with the diffraction angle . Principle maximum and secondary maximum N=2 N=4 N=6 Principle maximum I I secondary maximum I 2.The influence of diffraction by each slit to the interference fringes The diffraction patterns of all slits coincide. The intensity N2 N=3 N=2 N=1 Slit diffraction Interference of multi-slits differactin + interference Missing order Grating differactin -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 The missing order phenomenon of grating: On some direction, if diffraction angle satisfies, (a b) sin k --constructive interference and a sin 2k ' ab k k' a 2 --minimum of slit diffraction k' 1,2, then the k-th principle maximum will disappear. -- The k-th fringe is missing order. III. The incident ray inclination The optical path difference of two adjacent rays is A B C AB AC d sin d sin The grating function for inclination incidence is d (sin sin ) k k 0,1,2 IV. Grating spectrum Third order spectrum Second order spectrum First order spectrum When a polychromatic light (复色光) is incident a grating, d sin k except for the central fringe, all others principle maximum with different have different for each same order. [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. ? d a f Solution For two beams interference, the bright fringes (principle maxima) satisfies: d sin k the position of k-th order bright fringe is k xk f tg f sin f d the distance between two adjacent bright fringes is f 3 x xk 1 xk 2.4 10 m d For a single diffraction, the width of central maximum is x0 2 f tg 1 2 f sin1 2 f x0 10 x 0,1,2,3,4 bright fringes appears in the width of x0 . N 9 x0 5 4 3 2 1 0 -1 -2 -3 -4 -5 a 2 2.4 10 m x [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 when the incident angle is 300 ? Solution The grating constant is 1 3 ab 2 10 mm 500 (a b) sin k k 0,1,2, When 2 ,k gets maximum. ab kmax 3.39 Take integral=3 i.e., when the beams are incident normally, the maximum order which can be observed is 3-th. (a b)(sin sin ) k When the incident rays and diffraction rays are in the same side of the optical axis, ( a b)(sin 90 sin 30 ) kmax 0 0 5.08 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, ( a b)[sin 90 sin( 30 )] kmax 0 1.69 0 Take integral= -1 the order numbers which can be observed in this case is 1-th. The spectrum with order -1, 0,1,2,3,4,5can be observed 1 order 0-th order A B C 5 orders [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? d sin k Solution k 0,1,2 For 90 , 0 kmax ab 4.28 Take integral= 4 The maximum order which can be observed is 4-th. d sin k for interference bright fringes. a sin 2k 2 for diffraction dark fringes. d for same , k k ' a When k' 1, 2, , corresponding to k=3,6… As kmax 4 The order k=3 is missing. 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 -4 -3 -2 -1 0 1 2 3 4 §17-10 x –Ray diffraction by Crystal I. x -Ray It was discovered by W.K Roentgen ( a German physicist) in 1895.11. first x–ray photo: his wife’s hand. The He got the first Noble Prize of Physics in 1901 as the discovery of x-ray. x-ray: produced by bombarding a target element (Anode) with a high energy beam of electrons in a x-ray tube. Anode Cathode K A x-ray tube It’s a type of electromagnetic waves with wavelength ranges about 0.1--100Å, between Ultraviolet and -ray. In 1912, a collimated beam of x-ray which contain a continuous distribution of wavelengths strikes a single crystal, x-ray is a wave the diffraction pattern of x-ray was observed by German physicist M. Von Laue. Lead plate E crystal film Laue (劳厄) spots 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. II.Bragg equation and W.L.Bragg, two British physicists ( son and father) took another method to study x-ray diffraction. 掠射角 W.H.Bragg, O A 晶面间距d B C 晶面 The optical path difference of the two x-ray beams scattering by the atoms that they locate in two parallel planes is AC BC 2d sin They found that when 2d sin k ,the x-ray beams produce constructive interference. k 1,2, ---- Bragg equation 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.