Chapter 4 Diffraction.

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1 Chapter 4 Diffraction

2 4.1 Diffraction Phenomena 光的衍射现象
Huygens-Fresnel’s Assumptions 惠更斯—菲涅耳原理 Fresnel Zones 菲涅耳波带(菲涅耳带） Fresnel Diffraction(Circular Aperture and Opaque Circular Disc) 菲涅耳衍射（圆孔和圆屏） Fresnel Zone Plate 菲涅耳波带片 Fraunhofer Diffraction at Single Slit 夫琅禾费单缝衍射 Fraunhofer Diffraction at Double Slit 夫琅禾费双缝衍射 Plane Diffraction Grating 平面衍射光栅 Diffraction of X rays through Crystal 晶体对X射线的衍射 Fraunhofer Diffraction at Circular Aperture 夫琅禾费圆孔衍射 4.11 Resolving power of vision-aided instruments 助视仪器的分辨本领 4.12 Resolving power of spectrographic devices 分光仪器 的分辨本领

3 4.1 Diffraction Phenomena 光的衍射现象
obstacle obstacle screen 衍射屏 观察屏 a 衍射屏 观察屏 L L * S  S   ~ a Definition:the bending of light round the edges of an obstacle or the encroachment of light within the geometrical shadow of an obstacle is call diffraction 定义：光在传播过程中能绕过障碍物的边缘而偏离直线传播的现象叫光的衍射

4 4.2 Huygens ─ Fresnel Assumptions
惠更斯—菲涅耳原理 4.2.1 Huygens principle a wavefront can be divided into a large number of small area , each area is called as a point source of secondary wave , the new wavefront is the envelop of all the secondary wave 波前（波阵面）上的每一点都可作为次波的波源，各自发出球面 次波；在后一时刻这些次波的包络面就是新的波前。 wavefront: locus of all the point of the medium which are vibrating in same phase at same time confinement:no explanation to interference, quantitative diffraction and absence of backward wavefront *局限：不能解释干涉现象、不能定量解释衍射 现象，也不能解释无倒退波现象发生。

5 · r n ~~~~~~~~~~~~~~~~~~  dE(p) dS p Q S
4.2.2 Huygens ─ Fresnel Assumptions a wavefront S can be divided into a large number of small area dS , each area is as a new point source,which emanate secondary wave . the vibration at any point is the combination of all secondary wave. 波面S上每个面积元dS都可看成新的波源，它们 均发出次波。波传播方向上某一点P 的振动可由S面上 所有面积元发出的次波在该点叠加后的合振动来表示。  p dE(p) r Q dS S n ~~~~~~~~~~~~~~~~~~ all the secondary waves have: 1. Same initial phase 2. the vibration magnitude is inverse proportion to the distance of the point P from dS , wavefront 3. the phase at P depends on the optical path

6  K( )  = 0， K=Kmax   90o，K = 0 a. Vibration at P
A(Q)is determined by the energy at Q  = 0， K=Kmax K()：direction factor  K( )   90o，K = 0 ——Fresnel integrals 菲涅耳衍射积分

7 * S P D L B source L or D is finite distance b. sort screen obstacle
（1） Fresnel diffraction 菲涅耳衍射 L or D is finite distance （2） Fraunhofer diffraction夫琅禾费衍射 L and D infinite distance ,or by lens parallel beams

8 —Fresnel half-wave zone
4.3 Fresnel zone 菲涅耳半波带 4.3.1 Fresnel zone Or half wave zone o p B0 B1 R r0 rk r1 dividing the wavefront into a number of circular zones. secondary waves from neighbouring zones have a phase difference of  a phase difference of  ~ optical path difference —Fresnel half-wave zone 菲涅耳半波带（简称半波带）

9 a1 ,a2 ,… ,ak 4.3.2 calculation of resultant amplitude at P
amplitudes produced by each Fresnel’s zone at P resultant amplitude: according to Huygens ─ Fresnel Assumptions:

10 → ∴ due to trigonometric relation not relate to k
Ak only depends on K(θ) K(θ) ↑， ak ↓ ∴ resultant amplitude if k is odd if k is even

11 ｝ → 4.4 Fresnel Diffraction(Circular Aperture and Opaque
Circular Disc) 菲涅耳衍射（圆孔和圆屏） 4.4.1 diffraction at a circular aperture 圆孔衍射 o p h B1 R r0 rk  according to Fresnel’s zones: ~~ ρ2=R2-(R-h)2=2Rh-h2 ρ2 =r2k-(r0+h)2=r2k-r20-2r0h-h2 ｝ →

12 (4) ∴ from formula (1) combine (2)and(3) trait of the intensity at P：
1.continuously altering ρ or moving screen(altering r0)， alternately bright and dark 2. if ρ→∞(自由传播),k→∞,ak→∞, 3. if at a very small pin hole, k=1, 4.R→∞parallel beam

13 diffraction pattern due to a circular aperture
Fresnel diffraction Fraunhofer diffraction

14 4.4.2 diffraction at a opaque circular disc
圆屏衍射 resultant amplitude at P： Poisson point if the circular disc is small enough , only a few of half period zone is covered over , the center of screen(at P) will be bright

15 4.4.3 Fresnel zone plate 菲涅耳波带片
Zone plate is a specially constructed screen. According the Fresnel’s method, the wavefront is divided half period zones, the light is obstructed from every alternate zone. The light is cut off due to the even numbered zones, or the odd numbered zones 设计成只让奇数或偶数半波带透光， resultant amplitude at P: or bright at P. the function of a zone plate is similar to that of a convex lens. 类似于透镜成像 can be rewrited to:

16 focus： 1. 2. ——similar to imaging formula of lens 与薄透镜物象公式相似
Trait for the focus of zone plate: 1. 2. 3. have a number of foci, which depend on the number of zones and wavelength 存在次焦距，如f´/3, f´/5

17 · * 4.6 Fraunhofer Diffraction at Single Slit 夫琅禾费单缝衍射 x P B  S b δ A
4.6.1 set up and optical path 装置和光路 Slit plane screen lens L x S：line source P lens L B  S b δ slit width *  ： diffraction angle A f  f intensity distribution according to Huygens -Fresnel Assumptions:

18 α=ωt-kr0， β =ksinθ=(2π / λ)sinθ
Suppose the slit is divided to a number of rectangle zones of dx width Secondary waves from each zone are plane wave the resultant amplitude A0 due to whole slit the amplitude of the secondary wave from the zone： dx x r0 θ resultant amplitude at P： α=ωt-kr0， β =ksinθ=(2π / λ)sinθ let

19 or 令 ∴ intensity at P：

20 4.6.3 diffraction pattern due to single slit 单缝衍射花样
1 intensity curve sin 0.047 0.017 I / I0 from 1.central maximum 主最大（中央明纹中心）位置： 即为几何光学像点位置

21 · · · · · y1 = tgu y y2 = u u  2.minima (dark)极小(暗纹)位置: and
3.secondary maxima 次极大位置： y1 = tgu u  2 - -2 y y2 = u +2.46 +1.43 -1.43 -2.46

22 Trait of diffraction pattern：
alternate bright and dark bands paralleling to the slit,central bright maximum, secondary maxima are of much less intensity,which falls off rapidly with increase of the order 平行于光源的亮暗直条纹，中央主最大光强最大， 次最大光强远小于主最大的值，且随着级数的增大 而很快减小； width of central maximun is unequal to width of secondary maxima 条纹不等间隔 angular width of central maximun 中央主最大条纹角宽度为 dark bands with with equal width等间距的暗条纹 secondary maxima with unequal width 次最大不等间隔 3. if with white light , white bright at the center,color bands for secondary maxima 白光作为光源，中央仍为白色,次最大形成彩色条纹。

23 diffraction pattern at a single slit

24 discussion讨论题： change the set up, how about the pattern 当单缝衍射装置有如下变动时，衍射图样的变化 1、increase the focus增大观察屏前透镜的焦距(观察屏仍在焦平面上) 与焦距无关，但位置x=f’tgθ,条纹增宽 2、move forward or back the single slit前后移动衍射屏(单缝) 不变 3、move up or dawn single slit 上下移动衍射屏（单缝） 不变，同样衍射角的光仍会聚于同一地方 4、 move up or dawn light source 上下移动缝光源 衍射花样下上移动（反方向平移，衍射主最大位于 光源的几何光学成像位置）

25 I / I0 Where, 4.7 Fraunhofer Diffraction at Circular Aperture夫琅禾费圆孔衍射
intensity 1.22(/D) sin 1 I / I0 screen circular aperture L  1 中央亮斑 (爱里斑) Airy disc （Airy disk） f diameter D intensity distribution： (具体推导参见附录2.2) 84% energy intensity curve is similar to the curve due to single slit, but the circular ring Where,

26 trait of diffraction pattern：
1. alternate bright and dark rings 同心的明暗相间的圆环 2. central bright disc is called Airy Disc 中心亮斑称为爱里斑 half angular width半角宽度: D: diameter of the circular aperture

27 4.8 Plane Diffraction Grating 平面衍射光栅
a. definition any screen with spatial periodicity can be called grating 任何具有空间周期性的衍射屏都可叫做衍射光栅 usually ,grating consists of a very large number of narrow slits (or reflecting bands) side by side 光栅是由大量的等宽等间距的平行狭缝或(反射面)组成 transmission grating reflecttion grating b. sort d d according to the making manner： score grating and holographic grating 刻划光栅和全息光栅

28 d = a+b  grating constant 光栅常数
c. grating constant 光栅常数 a+b is called grating element retating to the periodary of the grating 光栅常数是光栅空间周期性的表示 b a b is the width of slit 透光（或反光）部分的宽度 a is the width of opaque portion不透光(或不反光）部分的宽度 d = a+b  grating constant 光栅常数 decades of strips/mm thousands of strips/mm， even ten thousands of strips /mm(d 10-1m)。

29 coherent superposition相干叠加
4.8.2 Fraunhofer Diffraction at double Slit 夫琅和费双缝衍射 b d f lens P superposition of diffraction waves from each slit θ θ θ 1  I 2 coherent superposition相干叠加 amplitude add d at P： interference of double slit：

30 interference due to double slit Single slit d=3b
强度受单缝衍射因子调制 的双缝干涉花样。 diffraction due to double slit I/4A02 1. principal maximum干涉主最大位置 2. maximum intensity 3. missing orders 缺级 4. 中央主最大中条纹数

31 ｝ ｛ 4.8.3 Fraunhofer Diffraction at N Slits—grating 多缝衍射——光栅
a. intensity distribution ｛ diffraction due a slit ｝ perform together interference between N slits diffraction pattern： alternate bright and dark bands paralleling to the slit ， which intensity is modulated by diffraction factor 平行于缝的明暗相间的条纹，其强度受单缝衍射因子调制 (2) maximum intensity is N2 times the intensity due to single slit 最大光强为单缝衍射的N2倍。

32 (3) missing orders 缺级： (4)between interference principal maxima there are N-1 secondary minima and N-2 secondary maxima 干涉主最大之间存在N-1个极小，N-2个次最大 N is a large number, widths of principal maxima are very narrow and the position is related to . The principal maxima appear sharp bright lines called spectral lines N大,条纹为暗的背景下锐细的亮线，这种条纹称为光谱线。 Is sin I0s -2 -1 1 2 ( /b) I N2I0单 sin 4 8 -4 -8 ( /d ) N = 4 d = 4b missing orders ±4,±8

33 19 bright fringes missing order missing order comparison between diffractions due to single slit and multiple slits （d =10 a）

34 o L p d sinθ0 λ d sin b. grating equation 光栅方程
angular position of spectral line while parallel beams are incident Normally: order grating equation grating equation for oblique incidence斜入射光栅方程： screen  L λ grating d sinθ0 p K=0,±1, ±2,… o n  > 0 θ0 < 0 incident diffractive (normal) grating （+） （-） f d sin

35 ——grating spectra 光栅光谱
Sign of θ0 and  : n  > 0 θ0 < 0 incident diffractive (normal) grating （+） （-） “+” :θ0 and  are at same side of the normal “-” :θ0 and  are at opposite sides of the normal c. grating spectra 光栅光谱 while white light is incident , the color fringes disperse on screen ——grating spectra 光栅光谱 (1) angular dispersion光栅的角色散 definition: difference in angle of deviation between two spectral lines with unit wavelegth difference 定义：单位波长间隔光谱线 所散开的角度。

36 ▲overlapping of spectral lines 光谱的重叠
trait of angular dispersion： ▲ ▲ k=0 no dispersion ,when k1 dispersion exists 零级条纹无色散，一级以后有色散，紫在内侧，红在外侧； ▲overlapping of spectral lines 光谱的重叠 ▲ ▲ linear dispersion on screen 线色散： angular and linear dispersion are not related to the number N 角色散和线色散都与光栅缝数N无关

37 ｝ → ∴ definition: half the angular width of the principal maximum
(2)half angular width 半角宽度 definition: half the angular width of the principal maximum 从主最大的中心到其一侧相邻最小值之间的角距离 principal maximum ： ｝ neighboring minima： subtraction： → ∴

38 例题２：一光栅刻划面宽５cm，光栅常数d=10-3cm, 缝宽b=5×10-4cm, 光垂直入射，求： １）第一级可见光的光谱角宽度．
expresses: sharper； ▲ sharper； ▲ wider ▲ 例题２：一光栅刻划面宽５cm，光栅常数d=10-3cm, 缝宽b=5×10-4cm,　光垂直入射，求： １）第一级可见光的光谱角宽度． ２）何处缺级？ ３）第一级极大与第三级极大的强度比． ４）波长500nm光的第一级谱线的半角宽度．

39 N N’ 4.8.4 brazed grating 闪耀光栅
disadvantage of transmission grating：central principal maximum(no dispersion) have the majority of total energy b d grating plane N N’ satisfy： diffraction maximum at B advantage of grating: energy is transferred to the spectral lines from the central maximum having no dispersion 闪耀光栅的优点：将单缝的中央最大值的位置从零级光谱转移 到其他有色散的光谱级上。

40 - + *4.9 Diffraction of X rays through Crystal 晶体对X射线的衍射
in 1895 Röntgen 伦琴( ） found x ray RÖntgen got the first Nobel Physical Prize in 1901 1901年伦琴获首届诺贝尔物理奖 - K A X ray X ray tube + A K ten thousands high voltage

41 · X ray  : 10 -2  101nm （10 -1  10 2 Å） 600 strips/mm，
Laue experiment劳厄实验（1912）： Laue disc Collimation slit crystal crystal as a three- dimension grating 晶体相当于三维光栅 劳厄斑 X ray verify the wave behavior of x ray 证实了X射线的波动性。 X ray  :  101nm （10 -1  10 2 Å） using general grating： 600 strips/mm，  very short  small θ,can not be detected using crystal , d is very very small larger θ,can not be detected

42 d :crystal lattice constant
晶格常数 1 2 α α d  d A B dsinα C NaCl d =0. 28nm α: glancing angle 掠射角 d Bragg’s equation 布喇格公式

43 ▲ α、 →determining d ▲ α、d →determining Application:
— analysis of crystal structure ▲ α、d →determining — spectral analysis of x ray W.H.Bragg， W.L.Bragg (father and son) got the Nobel Prize in 1915 for the outstanding work on analysis of crystal structure 布喇格父子共同获得了1915年的诺贝尔物理学奖。 end

44 { • • • 4.11 Resolving power of vision-aided instruments preface •
助视仪器的分辨本领 The diffraction can not be neglected. Airy disc Circular aperture diffraction In an optical system, the lens can be looked as circular aperture 圆孔衍射 preface For ideal image formation, we can obtain point image from point object Considering diffraction, we can obtain Airy disc from point object 点物 爱里斑 • • • • • Two images cannot be distinguished

45 一、Resolving power 1. Definition 2. Rayleigh criterion
The ability of an optical instrument to resolve the image of two nearby points. 光学系统分辨细微结构的能力。 2. Rayleigh criterion Two images are said to be just resolved if the radius of the central disc of either pattern is equal to the distance between the centers of the two patterns. 观察屏上甲亮斑（衍射图样）的主极大正好落在 乙亮斑（衍射图样）的第一极小处，两个亮斑刚 好能被分辨。

46 the two images can be distinguished can be just distinguished can not be distinguished

47 二、resolving power of human eye人眼的分辨本领
The visual angle of two light points for the entrance pupil of lenses: 两发光点对光具组入射光瞳中心所张视角 R：radius of lens D：diameter of lens（aperture） • u u 二、resolving power of human eye人眼的分辨本领 The radius of pupil wavelength =555 nm Resolving limiting angle of pupil： The minimum distance of two points at the distance of distinct vision

48 三、Resolving power of the telescope objective
On the retina： The distance between retina and pupil is about to 2.2 cm 三、Resolving power of the telescope objective 望远镜物镜的分辨本领 objective——effective stop, entrance pupil D/f '—relative aperture The resolving limit distance of two image points The resolving limit distance of two object points:

49 四、 Resolving power of the microscope objective
显微镜物镜的分辨本领 objective——effective stop, entrance pupil The distance between the object and objective is very small， 物离物镜很近 R u l The limit resolving distance of two object points: The object in the medium （an oil immersion objective） nsinu ，y  Numerical aperture

50 用紫外光照射，分辨本领增至2倍，即增大1倍。
例 （1）显微镜用波长为250nm的紫外光照射比用波长为500nm的 可见光照射时，其分辨本领增大多少倍？ （2）它的物镜在空气中的数值孔径约为0.75，用紫外光所能 分辨的两线之间的距离是多少？ （3）用折射率为1.56的油浸系统时，这个最小距离为多少？ （4）若照相底片上的感光微粒的大小约为0.45mm，问油浸 系统紫外光显微镜的物镜横向放大率为多少时，在底片上刚 好能分辨出这个最小距离。 解： （1） 用紫外光照射，分辨本领增至2倍，即增大1倍。

51 （2） 用紫外光照射时分辨的极限距离： （3） （4） 物镜的横向放大率：

52 4.12 Resolving power of spectrographic devices
分光仪器 的分辨本领 Definition： The ability of the instrument to form separate spectral images of two neighboring wavelength, λ and λ+d λ in the wavelength region λ. 衡量分开光谱中两波长很接近的谱线的能力 For example：prism, grating, Fabry-perot interferometer 棱镜、光栅、法布里—珀罗干涉仪 一、prism 0 ） （ A i1 i2 b ） ） 

53 0—the angle of minimum deviation
The face of the prism limits the incident beam to a rectangular section of width b, therefore , there is diffraction. 通过棱镜的光束是限制在一定宽度b以内，因此要发生单缝衍射 The first minimum of diffraction by 第一衍射极小 A—the angle of prism by 0—the angle of minimum deviation Differentiating two sides of equation and， 且 i2=A/2 then，

54 Spectral lines splits farther
The dispersion angle between two spectral lines of wavelength difference  波长相差的两谱线间的色散角： 1. Angular dispersion 角色散 just resolve 2. The resolving power of prism 色分辨本领 3. Linear dispersion 线色散 discussion A， Spectral lines splits farther 光谱展得越开 P  ，the resolving power gets higher 色分辨本领越高

55 * 二、 Grating spectrograph 光栅光谱仪 or Through grating equation
Differentiating equation Angular dispersion or Linear dispersion

56 Half angular width of spectral line
谱线半角宽度 when = ，the two spectral lines can just be resolved 两谱线刚好分辨 Resolving power 色分辨本领

57 例 解： 已知： （1） 一个棱角为50°的棱镜由某种玻璃制成，它的色散特性 由
决定，其中a= ，b=4.6528×103nm3. 当其对550nm的光处于最小偏向角时，试求： （1）这个棱镜的角色散率为多少？ （2）若该棱镜的底面宽度为2.7cm时，对该波长的光的 色分辨本领为多少？ （3）若会聚透镜的焦距为50cm，这个系统的线色散率为多少？ 解： （1） 已知：

58 角色散率： （2） 已知： 色分辨本领： （3） 线色散率：

59 例 解： （1） 由光栅的色分辨本领 可得： 用一宽度为5cm的平面透射光栅分析钠光谱，钠光垂直
投射在光栅上，若需在第一级分辨波长分别为589nm和 589.6nm的钠双线，试求： （1）平面光栅所需的最小缝数应为多少？ （2）钠双线第一级最大之间的角距离为多少？ （3）若会聚透镜的焦距为1m，其第一级线色散率为多少？ 解： 由光栅的色分辨本领 （1） 可得：

60 由光栅方程 （2） k=1 角距离： （3） 线色散率为： chap. 4

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