Chapter17 Diffraction of light 17.1 The diffraction of light wave 17.2 Huygens---Fresnel principle 17.3 Fraunhofer diffraction by a slit 17.4 Fraunhofer diffraction by a circular aperture The resolving power of optical instruments 17.5 Diffraction by a grating 17.6 The diffraction of X- rays Bragg condition 17.7 Application

17.1 Diffraction of light wave 光的衍射 1 The phenomena of diffraction (1) Diffraction by a single-slit 单缝衍射 (2) Diffraction by a circular aperture 圆孔衍射 (3) Diffraction by a grating 光栅衍射

17.1 Diffraction of light wave 光的衍射 2 The classification of diffractions 衍射分类 (1) Frensnel diffraction 菲涅耳衍射 The distance between obstacle and light source or screen are finite  Light source screen obstacle 法国

17.1 Diffraction of light wave (2) Fraunhofer diffraction 夫琅和费衍射 The light source and screen are infinite from obstacle  light source screen obstacle 德国

17.1 Diffraction of light wave 3 The purpose to research diffraction (1) In theory: Research wave property of light (2) In practice Optical spectrometer 光谱仪 (3) Resolving power 分辨率 分辨本领  Eyes 眼睛： The limit of resolution of eye: 1'；  Visible light microscope可见光显微镜： The limit of resolution of visible light microscope: 0.03" The resolving power is 2000 times of eyes；  Electron microscope电子显微镜： The limit of resolution of electron microscope: 0.00003" The resolving power is 1000 times of visible light microscope ； 可以分辨约为30Å 的两物点

17.2 Huygens-Frensnel principle 1 Huygens principle 惠更斯原理 Every point on a wave front can be regarded as the center of a secondary disturbance which gives rise to spherical wavelets; The position of the wave –front at any later time is the envelope of all such wavelets. 2 Huygens-Frensnel principle “the secondary wavelets mutually interfere” 子波相干叠加 The wavelets generated by all points in the same wave front is coherent wave 同一波面上各点发出的子波是相干波 (2) All the wavelets mutually interfere 各子波相遇进行相干叠加

17.2 Huygens-Frensnel principle The amplitude at P： Proportional to area and inverted proportional to distance 正比于面元，反比于距离 Related to initial phase 与初位有关 Related to direction 与方位有关 Coherence superposition 相干叠加 子波不发射倒退波 An inclination factor

17.3 Fraunhofer diffraction by a slit 单缝夫琅和费衍射 1 The feature of diffraction fringes衍射条纹特点 (1) The alternate bright and dark fringes are located symmetrically in the both side of the central bright fringe; (2) The central bright fringe is nearly two times widths of fringes lying on both sides; (3) The farther the distance from the central fringe is, the fainter the bright fringes is. (4) Reducing slit width, the bright fringes become wider; increasing slit width, the bright fringes become narrower

17.3 单缝夫琅和费衍射 2 To explain phenomena 17.3 单缝夫琅和费衍射 2 To explain phenomena The central bright fringe: The optical difference is zero 光程差为零：中央明纹 With the optical path difference increasing, the bright and dark fringes symmetrically appear 随着光程差增加，两边对称出现明暗条纹 其它条纹特点在讨论完半波损失后解释。

17.3 Fraunhofer diffraction by a slit 3 Frensnel’s half-wave zone 菲涅耳半波带法 a=AB Two half-wave zones: dark fringe Four half-wave zones: dark fringe Condition for dark fringes:

17.3 Fraunhofer diffraction by a slit For the bright fringes: Three half-wave zones 亮纹 Five half-wave zones 亮纹 Conditions for bright fringes C The central places of bright and dark fringes can be found with the conditions for bright and dark fringes. So the diffraction pattern intensity also can be drawn.

17.3 Fraunhofer diffraction by a slit 4. To explain the features of diffraction pattern (2), (3) and (4) The diffraction pattern of intensity 强度分布曲线 I k 为级次，k 增大半波带越多，每个半波带光强越弱。解释了(3)  定量计算中央明纹和其它明纹的角宽度： 解释了（2） （4）现象

17.3 Fraunhofer diffraction by a slit Questions: 1 The physical meaning of Consider it from half-wave zone 2 当缝向下平移时，此时衍射条纹向何处移动？ Answer： It does not move 3 当光源向上平移时，此时衍射条纹向何处移动？ Answer： It moves downwards

1 Fraunhofer diffraction by a circular aperture 17.4 圆孔的夫琅和费衍射 光学仪器的分辨率 1 Fraunhofer diffraction by a circular aperture 圆孔的夫琅和费衍射 Airy disk 爱里斑 Airy disk： A very intense central bright fringe 第一个暗环所围的中央亮斑 D d Half angular width of airy disk

17.4 圆孔的夫琅和费衍射 光学仪器的分辨率

2 The resolving power of optical instruments 17.4 圆孔的夫琅和费衍射 光学仪器的分辨率 2 The resolving power of optical instruments The Rayleigh’s criterion 瑞利判据 The center of one Airy disk must come no closer than the first dark fringe of the other Airy disk. The smallest angular resolution 最小分辨角 The resolving power 分辨率(本领) 例 对于可见光，平均波长为=550nm，试比较物镜直径为D1=5.0cm的普通望远镜和直径为D2=6.0m的反射式天文望远镜的分辨本领。 解

17.4 圆孔的夫琅和费衍射 光学仪器的分辨率

17.5 Diffraction by a grating 光栅衍射 1 The apparatus and the diffraction pattern (1) Diffraction grating A set of many narrow slits that are side by side平行、等宽而又等间距的多条狭缝 (2) The grating constants 光栅常数 d = (a+b) The slit width 缝宽 a The slit separation 间距 b

17.5 Diffraction by a grating 光栅衍射 (3) The feature of the fringes  a+b 愈小，条纹分得愈开  Bright fringes are very sharp亮纹尖锐 missing order 存在缺级现象

17.5 Diffraction by a grating 光栅衍射 插入实验装置示意图和衍射图样

17.5 Diffraction by a grating 光栅衍射 2 Conditions for bright and dark fringes The diffraction grating: a compound effect of single slit diffraction and slits interference。 双缝干涉 (1) Condition for dark fringes (partially) (2) Condition for bright fringes — grating equation constructive interference grating equation (3) The reason that the bright fringes are sharp Condition for bright fringes: the grating equation Condition for dark fringes: dark fringes of single slit diffraction dark fringes of slits interference

17.5 Diffraction by a grating 光栅衍射 (4) To explain missing order 解释缺级 If a  satisfy both equations in the same time Missing order 出现缺级 例：某一光栅，光栅常数为(a+b), 且a =b，a+b=2a。试观察此光栅的衍射亮纹位置。 解： k=0 k=1 k =2 k =3 k =4 单缝衍射暗纹条件 k ' =1 k '=2 缺级级数 除0级外，偶数级亮纹缺级。

17.6 The diffraction of X- rays Bragg condition 德国 1 The discovery and application of X- ray X- 射线的发现及应用 (1) The discovery of X-ray (Rontgen rays) X-射线的发现 It is discovered by Rontgen in 1895. X-rays are produced when high-velocity electrons strike a solid target. It could penetrate untransparent medium for visible light—x-ray 当高速电子撞击某些固体时, 会产生一种看不见的射线, 能透过许多对可见光不透明的物质 (2) The determination of X-ray 伦琴射线波长的测定 In 1912，the wavelength was tested by Laue and Bragg.

17.6 The diffraction of X- rays Bragg condition (3) The reason it can not to be determined for a long time 长期未被测定的原因 When  is very small: X 射线是波长很短的电磁波，波长范围在10-11m----10-8m。它是原子内壳层电子跃迁产生的一种辐射。 X射线特点： 在电磁场中不发生偏转，使某些物质发荧光，使气体电离，使底片感光，具有极强的穿透力。

17.6 The diffraction of X- rays Bragg condition (4) The application of X-ray 伦琴射线的应用 1912年，劳厄利用一块晶体作为衍射光栅，直接在屏幕上看到了x射线的衍射图样，证实了其波动性。 By sending a beam of X-rays, through a crystal diffraction pattern are produced, and from their analysis information about the structure of crystal may be deduce. X rays reveal the inner structure of crystals. X射线衍射实验已发展成为晶体结构研究的重要手段 伦琴夫人手的X 片 戒指 X 射线的应用不仅开创了研究晶体结构的新领域，而且用它可以作光谱分析，在科学研究和工程技术上有着广泛的应用。在医学和分子生物学领域也不断有新的突破。

17.6 The diffraction of X- rays Bragg condition 因原子间距约为10-10m，与X 射线的波长同数量级，故天然晶体可以看作是光栅常数很小的空间三维衍射光栅。1912年德国物理学家劳尼，设想将晶体做为三维光栅，他设计了如下实验：X 射线经晶体片衍射后使底片感光，得到一些规则分布的斑点（劳尼斑）。劳尼斑的出现是X 射线通过晶体点阵发生衍射的结果。 红宝石晶体 硅单晶体

17.6 The diffraction of X- rays Bragg condition 劳尼解释了劳尼斑的形成，但他的方法比较复杂。 不久，英国物理学家布拉格父子提出一种比较简单的方法来说明X 射线的衍射。 Bragg condition expresses the condition for constructive interference of rays reflected from successive lattices so as to produce intensity maximums The condition for constructive interference 光相遇互相加强 — 亮

17.6 The diffraction of X- rays Bragg condition Using Bragg condition： (1) To determine the wavelength of unknown waves 求未知光波的波长 (2) To determine the crystal constant of unknown crystals 求出未知晶体的晶格常数 When you receive a chest X-ray at a hospital, the X-rays pass through a series of parallel ribs（肋骨）in your chest. Do the ribs act as a diffraction grating for X-rays? Solution：Strictly speaking, the ribs do act as a diffraction grating, but the separation distance of the ribs is so much larger than the wavelength of the X-rays that there are no observable effects.

17.7 Application Ex1 Light of wavelength 589nm passes through a slit of width a=0.50mm，the diffraction pattern is seen on a screen that is f =0.80m from the slit. What is the width of the central bright fringe? Other bright fringes? The first dark fringe f x1 第一级暗纹到中央亮纹中心的距离： 中央亮纹宽度： 其它亮纹宽度：

17.7 Application Ex 2 Monochromatic light of =600nm falls normally on a diffraction grating, and the second-order maximum is deviated 30o. The third-order bright fringe is missing order. Calculate the grating constant and the slit width, then give the order of the bright fringes observed on the screen. (1) (2) (3) 考虑到3级缺级， 4级在/2处（实际上看不到） 在屏幕上可能出现的主极大的级次： k =0, 1, 2。

17.7 Application 例3 试估算人眼瞳孔在视网膜上所形成的爱里斑的大小，以及人眼所能分辨的、20m远处的最小线距离。 解： 人的瞳孔基本上是圆的，直径D=2~8mm 取=550nm， D=2mm 人眼的最小分辨角 人眼基本上是球形，婴儿眼球直径约为16mm， 成人眼球直径约为24mm，取 f =20mm， =550nm 估算视网膜上爱里斑的直径 人眼所能分辨的、20m远处的最小线距离：

17.7 Application 例4 以波长为589.3nm的钠黄光垂直入射到光栅上，测得第二级谱线的偏角0为28o8'，用另一未知波长的单色光入射时，它的第一级谱线的偏角为13o30'。（1）试求未知波长；（2）试问未知波长的谱线最多能观测到第几级？ 解： （1） 设 0=589.3nm 为未知波长 （2） 最多能观测到第四级谱线

17.7 Application Ex5 A diffraction grating is illuminated by white light (400nm ~760nm) at normal incidence. The third-order spectrum and the second-order spectrum overlaps. What are the wavelength region of the second-order of the overlap region？ 解： 开始重叠条件： 第二级谱线从600nm开始与第三级谱线的400nm重叠 第二级光谱被重叠的波长范围 例6 试分析单色光斜入射时光栅衍射的极大条件。 主极大位置已 不在透镜焦点上