Laser Principle

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激光技术介绍(1)

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激光武器：激光武器：激光武器有它的独特性，令它被广泛应用于防空，反坦克，轰炸机自卫等军事用途.激光之所以能成为威力强大的武器，是因为它有三个层次的破坏能力: 1.烧蚀效应跟激光热加工原理一样，当高能激光束射到目标时，激光的能量会被目标的材料吸收，转化为热能.这些热能足以令目标部分或完全穿孔，断裂，熔化，蒸发，甚至产生爆炸. 2.激波效应如目标材料被气化，目标材料会在极短时间内产生反冲作用，形成压缩波使材料表面层裂碎开，碎片向外飞时造成进一步破坏. 3.辐射效应目标材料气化的同时会形成等离子体云，能产生辐射紫外线及X光线，使目标内部的电子零件被破坏。
7、激光蚀刻、
Laser lithography
• 激光蚀刻技术比传统的化学蚀刻技术工艺简单、可大幅度降低生产成本，可加工 0.125～1微米宽的线，非常适合于超大规模集成电路的制造。
八、激光的发展激光的发展
• 激光手术：激光手术： • 激光能产生高能量﹑聚焦精确的单色光﹐具有一定的穿透力﹐作用于人体组织时能在局部产生高热量。激光手术就是利用激光的这一特点 ﹐去除或破坏目标组织﹐达到治疗的目的。主要包括激光切割和激光换肤。
四、激光的特性激光的特性
• 激光被广泛应用是因为它的特性。激光几乎是一种单色光波，频率范围极窄，又可在一个狭小的方向内集中高能量，因此利用聚焦后的激光束可以对各种材料进行打孔。以红宝石激光器为例，它输出脉冲的总能量不够煮熟一个鸡蛋，但却能在3毫米的钢板上鉆出一个小孔。激光拥有上述特性，并不是因为它有与别不同的光能，而是它的功率密度十分高，这就是激光被广泛应用的原因。 • 激光有以下三大特性： • · 单色波长 • · 同调性 • · 平行光束

激光原理及应用课件

1.1 激光发展的历史
• 突破
– 1958年Schawlow和Townes在Phy. Rev. 上发表论文 “Infrared and Optical Maser”，标志着激光作为一种新事物登上了历史舞台。
– 1960年5月，休斯实验室的Maiman研制的红宝石激光器发出了694.3nm的红色激光，这是公认的世界上第一台激光器。
1.1 激光发展的历史
– 1960年年中，IBM实验室利用CaF2中的三价铀制成了第一台四能级固体激光器；
– 1960年12月，BELL实验室的Javan，Bennett和Herriott制成了第一台氦氖气体激光器；
– 1962年，GaAs半导体激光器； – 1963年，液体激光器； – 1964年，CO2激光器； – 1964年，离子激光器； – 1964年，Nd：YAG固体激光器； – 1965年，HCl化学激光器； – 1966年，生物染料激光器； – 从1917年爱因斯坦提出受激辐射的概念到1960年第一台激光器诞
– 17世纪—对光的本性的探求：
• 波动说：以一定方式沿空间传输的波动过程，惠更斯、虎克；
• 微粒说：以经典方式运动着的微小粒子，牛顿；
– 19世纪：
• 光的波动本性有了进一步发展，杨氏双缝干涉，菲涅耳波动理论等
• 电磁场理论、麦克斯韦方程组（法拉第、麦克斯韦、坡印廷、赫兹等）
1.1 激光发展的历史
激光概述
辐射跃迁：受激吸收；自发辐射；受激辐射；
激光概述
粒子数反转
激光原理就是要研究光的受激辐射是如何在激光器内产生并占主导地位而抑制自发辐射！
He-Ne激光器工作原理图
Байду номын сангаас

principles of lasers激光原理第5章

The resonance frequency
The eigenmode and eigenvalue solutions are
The diffraction loss is
Elm ( x, y,0)
lm 1 lm
2
vlmn
c lm n 2L 2
dN P N P dt P t N P (t ) N P (0) exp P t I (t ) I (0) exp P
127
Due to the mirror loss, internal loss, diffraction loss, the photon in the cavity can
2kL 2m
Longitudinal modes
Mode spacing/ FSR
Concentric, confocal cavities
Concentric cavity (共心腔): L=R1+R2; L=2R (Spherical waves)
121
Confocal cavity (共焦腔): L=R1=R2=R
Eigenmodes in a cavity
propagating each round trip, i.e.
E ( x, y, 2 L) E ( x, y,0) exp( j 2kL) The constant
125
Inside the cavity, the electric field of the cavity mode must reproduce the its shape after
A B
131
Assume the transfer matrix of the optical cavity is C D After one-round trip propagation

laser(激光器)——外文原著(大量彩图)第一章

Introduction in Optics I:I-1)Electro-magnetic WavesThe discussion of electric and magnetic fields can be classified in two general categories.The first includes fields that do not vary with time.The electrostatic field of a distribution of charges at rest and the magnetic field of a steady current in a conductor are examples of fields,which,while they may vary from point to point in space,do not vary with time at any individual point.For such situations it is possible to treat the electric field and magnetic fields independently,without worrying about interactions between the fields.The second category includes situation in which the fields do vary with time,and in all such cases it is not possible to treat the fields independently.Faraday’s law tells us that a time-varying magnetic field acts as source of electric field.This field is manifested in the induced electromotive forces(emf’s)in inductances and transformers.Similarly,in developing the general formulation of Ampere’s law,which is valid for charging capacitors and similar situations as well as for ordinary conductors, we found it necessary to regard a changing electric field as a source of magnetic field.Therefore,when either field is changing with time,a field of the other kind is induced in adjacent regions of space.We are led to consider the possibility of an electromagnetic disturbance,consisting of time-varying electric and magnetic fields,which can propagate through space from one region to another,even when there is no matter in the intervening region. Such a disturbance,if it exists will have the properties of a wave,and the appropriate descriptive term is electromagnetic wave.Such waves do exist; radio and television transmission,x-ray and in the current context most important:light.Figure I-1-1visualizes such electromagnetic wave propagation.yxFigure I-1-1:Electromagnetic monochromatic wave.E and B correspond to the electric and magnetic field.Note the transverse character of the wave.The magnitudes of the field vectors E and B are in phase and are related by E =cB ,with01εµ=c ,(I-1-1)where,c (=2.9979246×108m s -1)is the speed of light in the vacuum,where µ0(=1.2566×10-6Vs V -1m -1)and ε0(=8.8542×10-12As V -1m -1)are the permeability and permittivity (i.e.,the dielectric constant)of the vacuum.The space (x )and time (t )dependence of the electric and magnetic field is described byE y =E y0sin(kx-ωt )(I-1-2)andB z =B z0sin(kx -ωt ),(I-1-3)where E y0and B z0is the amplitude of the E and B field,respectively;k=2π/λis the wave number and ω=2πν is the angular frequency which depend on the wavelength λand frequency ν,respectively.As shown in figure I-1-1,in vacuum (and air)the E and B fields at any point are in phase.In a dissipative medium,however,a phase shift between the fields takes place.In good conductors,the magnetic field is much larger than the electric field and exhibits a phase delay of approximately 45o .In non-dissipative media,as e.g.glass for visible light,the E and B fields behave similar as in vacuum and are in phase.The energy per photon of a monochromatic (=one color)wave is given byλνchh E ==,(I-1-4)where h is Planck’s constant (=6.626×10-34J s),νis the frequency and λis the wavelength.The electromagnetic spectrum is shown in figure I-1-2.Specifically in semiconductor optics the energy is expressed in eV rather than in J.Hence,it is convenient to apply the following relation to convert nm (10-9m)into eV,)nm (1240)eV (λ=E .(I-1-5)Example:One of the possible emissions of an Argon laser is at514.5nm. What is the energy of the emission in nm?Solution:E=1240/514.5=2.41eV.The electromagnetic waves cover an extremely broad spectrum of wavelengths,as shown in figure I-1-2.We can detect only a very small segment of this spectrum directly through our sense of sight from approximately750to430nm.Figure I-1-2:The electromagnetic spectrum.I-2Refraction and ReflectionWe shall begin our introduction in optical phenomena with reflection and refraction at a boundary surface that has been formed by the meeting of two different media.The velocity of light in a medium is below the velocity of light in the vacuum and is given by v=c/n,where n is the refractive index of the medium.We will see that the refractive index does not only determine the light velocity in a medium but it is also an essential parameter for the reflection.Let’s consider that we investigate the directions of the incident, reflected and refracted rays of monochromatic light.We fill find the following results illustrated by Fig.I-2-1:1.The incident,reflected and refracted beams and the normal to thesurface,all lie in the same plane.2.The angle of reflectionφr is equal to the angle of incidenceφa(φr=φa).3.For a given pair of substances,a and b,on opposite sides of the surfaceof separation,the ratio of the sine of the angleφa(between the beam in substance a and the normal)and the sine of angleφb(between the beam in substance b and the normal)is a constant(sinφa/sinφb=constant).If the a beam of monochromatic light travels in vacuum (or air),making an angle of incident φ0with the normal to the surface of a substance a ,we writea aon =φsin sin ,(I-2-1)where n a is the refractive index of substance a .The refractive index is always greater than unity and depends not only on the substance but on the wavelength of the light.I-3Snell’s law of refractionApplying equation (I-2-1)to the to substances a and b in figure I-3-1,we have sin φ0/sin φa =n a and sin φ0/sin φb =n b .Dividing the second equation by the first,we obtain sin φa /sin φb =n b /n a and from this the best known form of Snell’s law of refraction ,n a sin φa =n b sin φb .(I-3-1)The angles in figure I-3-1are independent of the thickness and space between the two plates and are the same when the space shrinks to nothing,as in figureI-3-2.Figure I-3-1:The transmission of light through parallel plates of different substances.The incident and emerging rays areparallel.Figure I-3-2:The figure shows the light rays at the interface of substances a and b without space between the plates.The angles are the same as these in figureI-3-1.Example:In figure I-3-3material a is water and b is glass with index of refraction of 1.52.If the incident ray makes an angle of 60o with the normal,find the directions of the reflected and refractedrays.Solution:Using equation (I-3-1),we find (1.33)(sin600)=(1.52)(sin θb )and θb =arcsin[(1.33)(sin60o )/1.52]=49.3o .I-4Total Internal ReflectionFigure I-4-1shows a number of rays diverging from a point source P in a medium a of index n a and striking the surface of a second medium b of index n b ,where n a >n b.Figure I-4-1:Total internal reflection.The angle of incidence φa ,for which the angle of refraction is 90o ,is called the criticalangle.The angle of incidence for which the refracted ray emerges tangent to the surface is called critical angle φcrit .At this angle φb =90o and Snell’s law becomes n a sin φa =n b ,since sin90o =1.We then have with φa =φcritabcrit sin n n =φ.(I-4-1)For a glass/air interface with n =1.52for the glass,sin φcrit =1/1.52and it follows φcrit =41.1o .The fact that φcrit is less than 45o makes it possible to use a triangular prism with angles 45o ,45o ,and 90o as a totally reflecting surface Such a prism is called Porro prism and is shown in figure I-4-2(a).An application of total internal reflection is shown in figure I-4-2(b).Example:A persiscope uses two totally reflecting 45o -45o -90o prisms.It springs a leak,and the bottom prisms is covered with water.Explain why the periscope no longer works.Solution:The critical angle for water (n b =1.33)on glass (n a =1.52)is φcrit =arcsin(1.33/1.52)=61.0o .The 45o angle of incidence is less than the 61o critical angle for a totally reflecting prism,so total internal reflection does not occur at the glass/water interface.Most of the light is transmitted into the water,and very little is reflected back into the prism.A very important application of total internal reflection is the fiber-optic cable shown in figure I-4-3(a).Figure I-4-3(b)shows the working principle of the cable.When a beam of light enters at one end of the transparent fiber,the light is totally reflected internally and is trapped within therod.Figure I-4-3:(a)Fiber-optic cable,used to transmit a modulated laser beam for communication purposes.(b)The so-called light pipe.The light is trapped by internal reflection,provided that the angles shown exceed the criticalangle.(b)I-5DispersionOrdinarily,white light is a superposition of waves with wavelengths extending through-out the visible spectrum.The speed of light in vacuum is the same for all wavelengths,but the speed in a material substance is different for different wavelengths.Therefore the index of refraction of a material depends on the wavelength.The dependence of the index of refraction on the wavelength is called dispersion .Figure I-5-1shows the variation of the refractive index with the wavelength for different optical materials.The value of n usually decreases with increasing wavelength and thus increases with increasing frequency.Light of longer wavelength usually has greater speed in a material than light of shorter wavelength.The brilliance of diamond is due in part to its large dispersion and in part to its unusually large refractive index (2.417).When you experience the beauty of a rainbow,you are seeing the combined effects of dispersion and total internalreflection.Figure I-5-2shows the ray of white light incident on a prism.The deviation (change of direction)produced by the prism increases with increasing the refractive index and frequency (i.e.,the energy,see equationI-1-4).Refractive IndexbyAlphabetical Listing of MaterialRefractive IndexbyIncreasing RI Value Material RI MaterialRI Air (STP)1.00029Vacuum 1.00000Amethyst (Quartz) 1.54(+1.55)Air (STP) 1.00029Beryl (Emerald) 1.57(+1.60)Water 1.333Citrine1.55Glass 1.517Corundum (Ruby,Sapphire) 1.76(+1.77)Quartz1.54(+1.55)Emerald (Beryl) 1.57(+1.60)Amethyst (Quartz) 1.54(+1.55)Diamond2.417Rock Crystal (Quartz) 1.54(+1.55)Garnet (Pyropes) 1.73-1.75Citrine1.55Garnet (Almandine) 1.76-1.83Beryl (Emerald) 1.57(+1.60)Garnet (Rhodolite) 1.76Emerald (Beryl) 1.57(+1.60)Glass1.517Topaz1.61(+1.62)Peridot (Olivine) 1.65(+1.69)Tourmaline1.62(+1.64)Quartz1.54(+1.55)Peridot (Olivine) 1.65(+1.69)Rock Crystal (Quartz) 1.54(+1.55)Garnet (Pyropes) 1.73-1.75Ruby (Corundum) 1.76(+1.77)Garnet (Rhodolite) 1.76Sapphire (Corundum) 1.76(+1.77)Garnet (Almandine) 1.76-1.83Topaz1.61(+1.62)Ruby (Corundum) 1.76(+1.77)Tourmaline 1.62(+1.64)Sapphire (Corundum)1.76(+1.77)Vacuum 1.00000Corundum (Ruby,Sapphire) 1.76(+1.77)Water1.333'High'Zircon 1.96(+2.01)High'Zircon1.96(+2.01)Diamond2.417Table I-5-1:Refractive index of various materials (from /HTML/Materials3.htm).Note:refractive index listings which have two numbers [ex.1.54(+1.55)]denote materials with double refractionproperties.I-6PolarizationPolarization occurs to all transverse waves.Figure I-6-1illustrates the idea of polarization by showing a transverse wave as it travels long a rope toward a slit.The wave is said to be linearly polarized,which means that its vibration always occur along one direction.Figure I-6-1:The principle of polarization:A transverse wave is linearly polarized when its vibrations always occur along one direction.(a)The rope passes a slit parallel to the vibrations,but(b)does not pass trough a slit that is perpendicular to the vibrations.Linearly polarized light can be produced from unpolarized light with the aid of certain materials.One commercially available material goes under the name of Polaroid.As shown in figure I-6-2,such materials allow only the component of the electric field along one direction to pass through,while absorbing the field component perpendicular to this direction.Light from ordinary sources is not polarized.The“antennas”that radiate light waves are the molecules that makes up the sources.The waves emitted by any one molecule may be linearly polarized.However,any actual light source contains a tremendous number of molecules with random orientations,so the light emitted is a random mixture of waves that are linearly polarized in all-possible directions.In figure I-6-3unpolarized light is incident on a polarizer.The blue line represents the polarizing axis.The E vectors of the incident wave exhibit random directions.The polarizer transmits only the components of E parallel to the polarizing axis.The intensity of the transmitted light is exactlyFigure I-6-3:Unpolarized light is incident on the polarizer.The intensity of the transmitted linearly polarized light,measured by the photocell,is the same for all orientations of the polarizer.half of the incident unpolarized light,no matter how the polarizing axis is oriented.Here’s why:We can resolve the E field of the incident wave into a component parallel to the polarizing axis and a component perpendicular to it.Because the incident light is a random mixture of all states of polarization, these two components are,on average,equal.The(ideal)polarizer transmits only the component that is parallel to the polarizing axis,so half of the incident intensity(I0/2)is transmitted.What happens when the linearly polarized light emerging from a polarizer passes through a second polarizer,as shown in figure I-6-4?To find the transmitted intensity at intermediate values of the angleφ,we bear in mind that the intensity of an electromagnetic wave is proportional to the square of the amplitude of the wave.The ratio of the transmitted to incidentamplitude is cosφ,so the ratio of transmitted to incident intensity is cos2φ. Thus,the intensity of the light transmitted through the analyzer isI=(I0/2)cos2φ,(I-6-1)Where,I0is the maximum light intensity atφ=0.Equation(I-6-1)is called Malus’s law.Figure I-6-4:The analyzer transmits only the component that is parallel to its polarization axis.Example:In figure I-6-4the incident unpolarized light has the intensity I0. Find the intensity transmitted by the first polarizer and the second if the angle between the axes of the two filters is30o.Solution:As explained above,the intensity after the first filter is I0/2. According to equation(I-6-1)with30o,the second polarizer reduces the intensity by a factor cos230o=3/4.Thus the intensity transmitted by the second polarizer is I0/2×(3/4)=(3/8)×I0.Example:What value ofφshould be used in figure I-6-4,so that the average intensity of the polarized light reaching the photocell is one-tenth the average intensity of the unpolarized light?Solution:Using equation(I-6-1),we find I0/10=(I0/2)cos2φ.Solving this relation forφyieldsφ=arccos(1/5)(1/2)=63.4o.A further possibility to create either partially or totally polarized light is by reflection.In figure I-6-5,unpolarized light is incident on a reflectingsurface between two transparent optical materials.The plane containing the incident and reflected rays and the normal to the surface is called the plane of incidence .Figure I-6-5:When light is incident at the polarizing angle,the reflected light is linearly polarized.At one particular angle of incidence,called the polarizing angle θp ,only the light for which the E vector is perpendicular to the plane of incidence is reflected.The reflected light is therefore linearly polarized perpendicular to the plane of incidence (i.e.,parallel to the reflecting surface).In 1812,Sir David Brewster noticed that when the angle of incidence is equal to the polarizing angle θp ,the reflected and refracted ray are perpendicular to each other.The situation is shown in figure I-6-6.In this case θb =90o -θp .Using equation (I-3-1),we find sin θp /sin(90o -θp )=sin θp /cos θp =n b /n a and finally.tan a b n n p =θ(I-6-2)This relation is known as Brewster’s law .Light and other electromagnetic radiation can also have circular or elliptical polarization,i.e.,the E describes a circular or elliptical rotation.In this context polarization by birefringence is important.Birefringence occurs in calcite and other noncubic materials (hence also in various semiconductors)and some stressed plastics and cellophane.Most materials are isotropic ,that is,the speed of light passing through the material is the same in all directions.Because of their atomic structure,birefringent materials are anisotropic .The speed of light depends on its direction of propagation through the material.When a light ray is incident on such materials it may be separated into two rays called the ordinary and extraordinary ray.There is one particular direction in a birefringent material in which both rays propagate with the same speed.This direction is called the optic axis of the material.However,when light is incident at an angle to the optic axis,as shown in figure I-6-7,the rays travel in different directions and emerge separated inspace.If light is incident on a birefringent plate perpendicular to its crystal face and perpendicular to the optic axis,the two rays travel in the same direction but different speeds.The rays emerge with a phase difference that depends on the thickness of the plate and on the wavelength of the incident light.In a quarter-wave plate,the thickness is such that there is a90o phase difference between the waves of a particular wavelength when they emerge. In a half-wave plate,the rays emerge with a phase difference of180o.Suppose that the incident light is linearly polarized such that E is45o to the optic axis,as illustrated in figure I-6-8.The ordinary and extraordinary rays start out in phase and have equal amplitudes.With a quarter-wave plate,they emerge with a phase difference of90o,so the resultant components of E are E x=E0sinωt and E y=E0sin(ωt+90o)=E0cosωt (ω=2πνis the angular frequency and t represents the time).The electric field vector thus rotates in a circle and the wave is circularly polarized.Figure I-6-8:Polarized light is incident on a birefringent crystal such that E makes45o angle with the optic axis,which is perpendicular to the light beam.If the crystal is a quarter-wave plate the light behind the crystal is circularly polarized.Figure I-6-9shows the propagation of circular polarized light.If the advancing wave revolves clockwise(looking toward the source),then it’s said to be right-circularly polarized;if counterclockwise,it’s left-circularly polarized.The magnitude of E remains constant while revolving once around with every advance of one wavelength.With a half-wave plate,the wave emerge with a phase difference of 180o ,so the resultant electric field is linearly polarized with components E x =E 0sin ωt and E y =E 0sin(ωt +180o )=-E 0sin ωt .Hence,the direction of the wave polarization is rotated by 90o relative to that of the incident wave,as shown in figureI-6-10.If the phase difference between the two components of E is something other than a quarter wavelength,or if the two component wave have different amplitudes,the resulting wave is elliptically polarized.I-7Huygens’PrincipleThe principles governing reflection and refraction of light rays,discussed in I-2and I-3,were discovered experimentally long before the wave nature of the light was firmly established.These principles however can be derived from wave considerations and thus shown to be consistent with the wave nature of light.To establish this connection we use a principle called Huygens’principle(Christian Huygens in1678).According to this principle,each point on a given wavefront can be considered to be a point source of secondary wavelets.Figure I-7-1shows the plane wavefront AA’striking a mirror at point A.As can be seen from the figure,the angleφ1between the wavefront and the mirror is the same as the angle of incidenceθ1.From figure I-7-1it isFigure I-7-1:Plane wave reflected at a plane mirror.readily shown that the angle of reflection equals the angle of incidence. Figure I-7-2shows an enlargement of a portion of figure I-7-1showing AP, which is part of the original wavefront.The reflected BB’’makes anangle φ1’with the mirror that is equal to the angle of reflection θ1’between reflected ray and the normal to the mirror.The triangles ABP and BAB’’are both right triangles with a common side AB and an equal side AB’’=BP =ct .Hence,these triangles are congruent and the angles φ1and φ1’are equal,implying that θ1’=θ1.Figure I-7-3shows a plane wave incident on an air/glass interface.We apply Huygen’s construction to find the wavefront of the transmitted wave.The new wavefront BB’is not parallel to the original wavefront AP because the speeds v 1and v 2are different.From the triangle APB ,sin φ1=v 1t /AB or AB =v 1t /sin φ1=v 1t/sin θ1using the fact that φ1=θ1.Similarly,from triangle AB’B ,sin φ2=v 2t/AB or AB =v 2t /sin φ2=v 2t /sin θ2,where θ2=φ2is the angle of refraction.Equating the two values for AB ,we obtain2211sin sin v v θθ=.(I-7-1)Substituting v 1=c /n 1and v 2=c /n 2in this equation delivers Snell’s law,n 1sin θ1=n 2sin θ2.I-8Thin filmsYou have probably noticed the colored bands in a soap bubble or in the film on the surface of oily water.The bands are due to the interference of light reflected from the top to the bottom surfaces of the film.The different colors arise because of the variation in the thickness of the film,causing interference for different wavelength at different points.Such an interference effect is shown in figureI-8-1.We consider now a thin film of uniform thickness d and index of refraction n shown in figure I-8-2.To determine whether the reflected light rays interfere constructively or destructively,we must note the following fact:A wave traveling in a medium of low refractive index (air)undergoes a 180o phase change upon reflection from a medium of higher refractive index.There is no phase change in the reflected wave if it reflects from a medium of lower refractiveindex.dRay 1is reflected from the upper surface A undergoes a phase change of 180o with respect to the incident wave.On the other hand,ray 2,which is reflected from the lower surface B undergoes no phase change with respect to the incident wave.Therefore,ray 1is 180o out of phase with ray 2corresponding to path difference of λn /2.However,we must consider that ray 2travels an extra distance equal to 2d before the waves recombine.Hence,if 2d =λn /2=λ/(2n )the phase difference between both rays is 360o and the waves recombine in phase and constructive interference takes palace.In general,the condition for constructive interference is expressed as2nd =(m +1/2)λ(m =0,1,2,…)(I-8-1)and for destructive interference we have2nd =m λ.(m =0,1,2,…)(I-8-2)Thin films are of considerable importance for the formation semiconductor devices.Almost all optoelectronic devices are composed of the combination of various thin films.Concerning research and development,by means of optical spectroscopy not only the optical or optoelectronic features of semiconductors are investigated but also other features as the film thickness.Hence,in many cases optical characterization methods accompany the manufacturing steps of electronic and optoelectronic devices.For optical thickness measurements,equations (I-8-1)and (I-8-2)can be used to determine the film thickness.According to equation (I-8-2),the fringe of order m lies at λ1and that of order (m +1)at λ2.Hence,we have m λ1=(m +1)λ2so that m =λ2/(λ1-λ2).With (I-8-2)we find,2nd =λ1λ2/(λ1-λ2)and the thickness of the film is)(22121λλλλ−=n d ,(I-8-3)where λ1and λ2is the wavelength of two adjacent maxima or minima in the spectrum.Example:Figure I-8-3shows the transmission spectrum of a thin CdS film on glass.The transmittance starts at the band-gap of the material (≈500nm)and pronounced fringes at 582and 631nm are observed.More details concerning semiconductor will follow in Semiconductor Optics I &II.3003504004505005506006507007500.00.10.20.30.40.50.6582nm631nmT r a n s m i t t a n c e W avelength (nm)Figure I-8-3:Transmittance of thin film CdS on glass at room temperature.Solution:The thin film CdS exclusively causes the fringes in figure I-8-3.The glass substrate does not influence the interference effect.Hence,we insert the wavelengths of the two indicated maxima and n CdS =2.5in equation (I-8-3)and get the thickness of the film,µm 1.5m)10(495m)10m)(63110(582999=××××=−−−d .The calculation of the transmitted and reflected intensities of thin films requires the consideration of the internal reflections.Figure I-8-4shows the concept.I 0is the intensity of the incident beam,R is the reflection coefficient of the surface and backface,αis the absorption coefficient and x the thickness of the film.The transmitted and reflected intensities are summed up with a geometrical series delivering the following results for the transmittance and reflectance,xxe R e R Tr αα2221)1(−−−−=(I-8-4)and}{2α22α2e 1e )(11x x R R R Re −−−−+=.(I-8-5)Figure I-8-4:The transmitted and reflected intensities through a thin film.For many applications,the formulas)}2exp()1(1{2x R R Re α−−+=(I-8-6)and)exp()1(2x R Tr α−−=(I-8-7)are accurate enough.Example:Calculate the transmittance of the film with a thickness of 1µm,an absorption coefficient of 100cm -1and an reflection coefficient of 0.2.Solution:Tr =(1-0.2)2exp(-100cm -1×10-4cm)≈(1-0.2)2=0.64.We see,in case of effective absorption the reflection of the surface determines the transmission features of the film.。

Laser Principle

kink in output power 扭结、弯曲
Laser history
Javan invents He-Ne laser Maiman builds first Townes invents and Schawlow and Townes (ruby) LASER builds first MASER propose LASER Spectra introduces first fiber optic communication IBM builds first system Hall buildsAlferov builds first laser printer (Chicago) player Einstein predicts CD Ti:Sapphire laser heterostructure laser semiconductor stimulated emission laser Nakamura builds builds quantum built Faist nanowire laser at UCB blue laser diode cascade laser
Optical Electronics, Fall 2010
Optical Electronics, Fall 2010
Optical Electronics, Fall 2010
Optical Electronics, Fall 2010
Optical Electronics, Fall 2010
laser as amplifier
Input Monitor tap
隔离器
Isolator
隔离器
Isolator Monitor tap Output

激光原理与技术课程实验及实践教学改革

2021年3月总第357期ISSN1672-1438CN11-4994/T 激光原理与技术课程实验及实践教学改革王萌何铁锋王宁项炳锡张良静深圳技术大学中德智能制造学院广东深圳 518118摘要：根据对激光原理与技术课程教学现状的分析，从教学内容、教学方法和实验教学3个方面对激光原理与技术课程进行改革。

激发学生的学习兴趣和学习的主动性，加强基础实验教学并增加开设设计研究性实验，弥补当前实验内容不足、不新、不全的缺陷，培养学生的探索精神和创新能力。

关键词：激光原理与技术；实践教学；教学效果；探究式教学作者简介：王萌，工学博士，助理教授；何铁锋，工学硕士，实验师；王宁，工学博士，助理教授；项炳锡，工学博士，助理教授；张良静，理学博士，副研究员。

激光原理与技术课程是激光智能制造专业的专业必修课，是一门理论性很强的专业基础课。

本课程主要介绍激光产生的基本原理和特性、激光器的基本结构和类型、激光器的主要应用以及光通信和信息光电子领域常用的激光技术和典型激光器件，培养学生对激光、激光器件及各种激光技术的感性认识，使学生掌握激光原理、技术和器件的基本知识，并引导学生了解激光在当今信息光电子领域的主要应用和前沿动态，通过本课程的学习可以为学生今后从事激光技术、光通信、信息处理、红外探测、环境检测、激光医疗诊断和材料加工等方面的相关光学工程研究打下基础。

但是这门课程内容具有抽象性、理论性强等特点，对于本科生而言，学习难度较大，容易使学员产生畏难情绪。

实验教学是这门专业基础课的一种重要组成部分，是培养学生实践能力、创造能力的主要手段，是学员真正认识激光器、理解激光理论与技术的必要步骤。

但是目前的实验器材不能涵盖高斯光束的特性、光斑质量测定等关键教学内容。

基于以上因素，笔者结合课堂讲授激光原理课程的体会，对本门课程进行了教学改革。

1 目前针对激光原理与技术课程教学改革的必要性激光智能制造具有低成本、环境要求低、无须掩模等优点，目前，德国的工业激光加工技术处于世界领先地位，欧美主要国家在大型制造产业，如机械、汽车、航空、钢铁、造船、电子等行业中，基本完成了激光加工工艺的更新换代，进入“光加工”时代。

第一章激光基本原理--Part1

• 在物质与辐射场的相互作用中，构成物质的原子或分子可以在光子的激励下产生光子的受激发射或吸收。 • 粒子数反转：能利用受激发射实现光放大 • 受激辐射光子与激励光子具有相同的频率、方向、相位、偏振态，是相干光。
Einstein
1947年，Lamb和Reherford在氢原子光谱中发现了明显的受激辐射，这是受激辐射第一次被实验验证。Lamb由于在氢原子光谱研究方面的成绩获得1955年诺贝尔物理学奖； "for his discoveries concerning the fine structure of the hydrogen spectrum" 1950年，Kastler提出了光学泵浦的方法，两年后该方法被实现。他因为提出了这种利用光学手段研究微波谐振的方法而获得诺贝尔奖。 "for the discovery and development of optical methods for studying Hertzian resonances in atoms"
1966年研制成了固体锁模激光器获得了超短脉冲。 1970年研制成了准分子激光器。 1977年研制成了红外波段的自由电子激光器（FEL） 1984年研制出光孤子激光器（SL）美国电话电报公司贝尔实验室的研究人员于1992年研制出当时世界上最小的固体激光器，它在扫描电子显微镜下看起来就像一个个微型图钉，其直径只有 2 至 10 微米。在一个大头针的针头上，可以装下1万个这样的新型半导体激光器。
DARPA built the megawatt-class Alpha HF chemical laser during the 1980s
An electron-beam pumped ArF laser experiment at Sandia National Laboratories (1975, Courtesy Sandia National Labs)

激光原理十二、激光器的输出功率

（a1 t1）I（ 2L） (3-54)
(6) 平均行波光强
对腔内任何一处z都有两束传播方向相反的行波I+(z)和I-(2L-z)引
起粒子数反转分布值发生饱和，增益系数也发生饱和，近似用
平均光强2I代替腔内光强 I+(z)+ I -(2L-z)，用
G G0 1 2I
Is
G阈
由上式可知，作为腔内的平均增益系数，则腔内的平均行波光
二、腔内光强达到稳态的过程
外界激发很弱时 G0 G阈激光器无输出
增强外界激发
G 0 G阈
I
G
当 G G阈时，稳态建立
I 恒定
3.5 激光器的输出功率
3.5.1 均匀增宽型介质激光器的输出功率
1.稳定出光时激光器内诸参数的表达式 2.激光器的输出功率 3.输出功率与诸参量之间的关系
3.5.2 非均匀增宽型介质激光器的输出功率
此时，输出镜具有最佳透射率时激光器的输出功率为：
2LG0
P
1 2
Is
A
a1 (
2LG0
a1 ) (
1 2
Is
A(
2LG0
a1 )2
2LG0 1) 2 LG 0 a1
(3-61)
3.5.2 非均匀增宽型介质激光器的输出功率
在非均匀增宽型介质中，频率为υ的光波只能使速度为 ±uz的粒子数密度反转分布值饱和，对其它速度的粒子数密
(3-60)
图中画出往返损耗率a取不同数值 t1
时的输出镜的最佳透射率t1与2LG0 0.4
的关系曲线。
0.3
a 0.1 a 0.05
0.2
将 t1 a1 ( 2LG代0 入 a1 )

激光原理十四、气体激光器

① 选用对3.39um的光具有低反射率的谐振腔反射镜,使339um达不到阈值条件，如图，在腔内加色散棱镜，将两谱线分开，通过调整谐振腔反射镜的位置，只允许0.6328um的辐射起振，而使3.39um的辐射偏离出谐振腔外。
②腔内放置甲烷吸收盒，因为甲烷对3.39um的光具有强吸收而对0.6328um的光透明，因此可用甲烷抑制3.39um振荡；
He-Ne激光器的放电电流对输出功率影响很大。
① 图中所示为He-Ne激光器的的输出功率与放电电流的关系曲线。表明:在气压比为定值时,每个总气压都存在一个输出最大的放电电流 ,其大小随着总气压的升高而降低, 这是因为气压升高,只需要较小的放电电流就能得到相同的电子密度。在最佳充气条件下，使输出功率最大的放电电流叫最佳放电电流。
5.2.2 二氧化碳激光器
5.2.2 二氧化碳激光器
C02激光器的主要特点是输出功率大,能量转换效率高,输出波长(10.6um) ，广泛用于激光加工、医疗、大气通信及其它军事应用。
② He-Ne激光器存在着最佳混合比和最佳充气总压强，即存在最佳充气条件。实验发现，氦气与氖气的分压比为7／1时是最佳分压比。而总压强在100Pa～400Pa。选用He气作辅助气体的原因:
Ne原子不能直接被电子碰撞激发到激光上能级；
He*与Ne*能级极相近, 易发生能量共振转移。 ③ 若放电毛细管的直径为d，充气压强为p，则存在一个使输出功率最大的最佳p和d 值。 ④ 最佳放电条件下，工作物质增益系数和毛细管直径成反比。
五、He-Ne激光器的寿命 He-Ne激光器使用一段时间或存放一段时间后，它的输出
功率会逐渐降低，以致最后没有激光输出。现在一般规定输出功率下降到最高功率的1/e的工作时间为器件的寿命。影响器件寿命的因素大致有以下几方面：

国外激光原理经典教材述评

一、《激光(Lasers)》《激光(Lasers)》是一本物理学领域的名著。其作者是美国斯坦福大学的安东尼·E，西格曼(AnthonyE，Siegman)教授。该书1986年由英国牛津大学出版社出版，1283页，为美国的大学科学丛书。
Siegman教授于1957年获得斯坦福大学的电子工程学博士学位起，从事微波电子学、微波激射以及激光器领域的教学和科研工作。其研究领域包括激光物理、激光仪器和激光应用等方面，最主要集中在激光振荡器及其模式和激光光束的传播上。
《激光》一书，是他全面阐述激光原理的著作，它系统深入地论述了激光的产生、激光原理以及激光应用等基本内容。
本书分为三大部分，
对所选教材进行了客观地评价与介绍。
关键词：国外教材：激光原理；激光技术；激光器件；评介
教育部南开大学外国教材中心，多年来一直承担全国物理类外国教材的引进、研究评介及交流，积累了一定的外国教材研究工作经验。曾于1989年出版过《全国高等学校物理类外国教材研究文集》。2007年1月，该中心正式承担了教育部2006年外国教材研究项目“国外激光原理教材研究评介”。本文为该课题研究的成果之一。
本文选择国外知名大学三本激光原理教材作为研究对象，并着重对其进行评介分析。这三本激光原理经典教材分别是：美国斯坦福大学Anthony E，Siegman教授1986年出版的Lasers~美国佛罗里达中央大学WilliamtSilfvast教授2004年出版的LaserFundamentals(SecondEdition)和意大利米兰理工学院Orazio Svelto教授1998年出版的Principles of Lasers(Fourth Edition)。

光学专业英语词汇总结

Vocabulary 2
Ultraviolet 紫外的 visible 可见的 infrared 红外的 scalar function 标量函数 vector function 矢量函数 wavelength 波长 frequency 频率 Angular frequency 角频率 Radian 弧度
Vocabulary 9
frequency conversion 频率转换 Down conversion 下转换 Parametric process 参量过程 Nonparametric process 非参量过程 Spontaneous Parametric Down conversion 自发参量下转换 quasi-phase match 准相位匹配 Phase mismatch 相位失配
Hologram 全息图 holography 全息术 holographic reconstruction 全息再现 holographic recording 全息记录 volume holography 体全息术 reference wave 参考波 object wave 物波 coherent light 相干光
Hologram vocabulary
Emulsion
感光乳剂 slit 缝
Orthogonal
正交的 monochromatic 单色的
Exposure
曝光
bragg condition
布拉格（布喇格）条件 conjugate 共扼 rainbow hologram 彩虹全息图
Vibrate 振动 Apparatus 器械，仪器 Minimal 最小的 Fluctuation 波动，起伏 illuminate 照明 Transparency 透明物 Planar 平面的 Three-dimensinal 三维的

镭射工作原理

镭射工作原理
镭射（Laser）是一种通过激光来产生和放大一束高度聚焦、
具有高单色性、相干性和方向性的光束的技术。

它的工作原理基于三个主要的过程：受激辐射、自发辐射和光学放大。

在工作原理的首个过程中，光子的受激辐射是通过将外界的能量输入到激活介质中来实现的。

这个过程被称为受激吸收，其中一个高能态的原子通过吸收一个外界光子的能量激发到一个更高的能级。

这个高能态的原子然后处于一个不稳定的状态，而当另一个光子被引入到激活介质中时，它会通过受激辐射的过程释放出两个光子。

这两个光子与入射光子具有相同的波长、相位和方向，并且它们会进一步激发另外的原子，形成一个连锁反应。

第二个过程是自发辐射，其中一个被激发的原子从高能态跃迁到一个较低的能级，并释放出一个光子。

这个过程是随机的，并且光子的波长、相位和方向是无法控制的。

相比之下，受激辐射产生的光子具有特定的波长、相位和方向。

最后一个过程是光学放大，其中的光子会不断通过被放置在两个反射镜之间的激活介质，而且在每次通过时会受到受激辐射的影响而被增强。

其中一个反射镜是部分透明的，允许一部分光子通过形成输出光束，而另一个镜子是完全反射的，使得光子可以在激活介质中来回传播。

这种来回传播的光子会与越来越多的原子相互作用，从而促使更多的受激辐射发生。

因此，输出的光束将会是一束高度相干、聚焦和具有高能量密度的激光光束。

总的来说，镭射的工作原理是通过受激辐射、自发辐射和光学放大这三个过程实现的。

这种技术可以在医学、通信、制造业、科学研究等领域发挥重要作用。

激光原理(英文)-文档资料

Special cases
Types of resonators
10
Classical mechanics analog
Fabry-Perot boundary conditions
Multi-mode laser
Multiple resonant frequencies
4
Single longitudinal mode lasers
• Insert etalon into cavity • Use low reflectivity etalon
Laser basics
Optics, Eugene Hecht, Chpt. 13; Optical resonator tutorial:
/tutorials/lasers/leot/
1
Laser oscillation
Laser is oscillator • Like servo with positive feedback • Greater than unity gain
Multi-longitudinal Multi-transverse&long. Single mode
7
Gaussian beams
• Zero order mode is Gaussian
• •
Intensity profile:
I
I e2r2/w2 0
beam waist: w0
2
w w0
9
Resonator options
• Best known -- planar, concentric, confocal • Confocal unique
– mirror alignment not critical – position is critical – transverse mode frequencies identical

2019华中科技大学839激光原理考试大纲

2019华中科技大学硕士研究生入学考试《激光原理》考试复习大纲一、课程名称：激光原理与技术Laser Principle and Technology二、课程编码：三、学分与学时：64/4四、先修课程：量子力学、几何光学、物理光学五、课程教学目标：《激光原理与技术》课程是光电子专业本科生的专业基础课，其教学目标是使学生能够掌握本课程的基本理论、基本分析方法和基本技能。

初步具备应用所学到的基本理论和方法分析和解决本专业的一般性问题。

六、适用学科专业：高等院校光电子技术、光通讯、光电器件应用物理等本科专业。

七、基本教学内容与学时安排：第一章绪论（4学时）一、激光的诞生及发展二、激光产生的机理三、激光的特性四、激光器实例第二章光线矩阵及高斯光束（10学时）一、光线的传播１．光线矩阵２．双周期性透镜波导３．相同周期性透镜波导４．光线在反射镜之间的传播二、光束在均匀介质中传输１．均匀介质中的基本高斯光束２．ＡＢＣＤ法则３．高斯光束在透镜波导中的传输４．均匀介质中的高阶高斯光束三、高斯光束的变换１．高斯光束通过薄透镜的传输２．高斯光束的聚焦、准直和匹配３．高斯光束的自再现变换与稳定球面腔第三章激光谐振腔（10学时）一、光学谐振腔的稳定性条件１．光学谐振腔的稳定性２．光学谐振腔的构成与分类３．光学谐振腔的作用二、光学谐振腔的模式１．光学谐振腔中光波模的谐振频率２．光学谐振腔内的多纵模振荡和单纵模的选取３．纵模的频率漂移三、平行平面腔的迭代法１．开腔衍射理论的分析方法四、平稳定球面腔１．对称共焦腔的模式２．一般稳定球面腔与对称共焦腔的等价性３．一般稳定球面腔的模式４．非稳定球面腔第四章光场与物质的相互作用（8学时）一、光场与物质相互作用的理论１．光场与物质相互作用的理论体系２．电介质的极化３．原子自发辐射的经典模型二、谱线加宽与线型函数１．光谱线的频率分布２．爱因斯坦辐射系数在谱线加宽时的修正３．原子与有谱线线宽辐射场的相互作用三、均匀加宽和非均匀加宽四、激光器的速率方程理论１．三能级速率方程组２．四能级速率方程组第五章连续和脉冲激光器的运行特性（8学时）一、小信号增益系数１．增益系数正比于反转粒子数２．增益系数与入射光场频率的关系二、均匀加宽时的增益饱和１．增益饱和现象及其物理机制２．均匀加宽条件下反转粒子数的饱和３．均匀加宽条件下的大信号增益三、均匀加宽时的增益饱和１．非均匀加宽条件下反转粒子数的饱和２．非均匀加宽条件下的大信号增益四、连续激光器的稳态工作特性１．激光器的阈值条件２．稳态工作时的腔内光强３．连续激光的输出功率和最佳透过率。

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第一章激光基本原理--Part1

激光原理十二、激光器的输出功率

激光原理十四、气体激光器