光电子学与光子学的原理及应用S.O.Kasa课后答案详解
SM Ch1
The beam width at a distance of 10 m is 2w = 2wo[1 + (z/zo)2]1/2 = (0.810-3 m){1 + [(10 m)/(0.79 m)]2}1/2 = 0.01016 m or 10.16 mm. 1.5 Gaussian beam in a cavity with spherical mirrors Consider an optical cavity formed by two aligned spherical mirrors facing each other as shown in Figure 1.54. Such an optical cavity is called a spherical mirror resonator, and is most commonly used in gas lasers. Sometimes, one of the reflectors is a plane mirror. The two spherical mirrors and the space between them form an optical resonator because only certain light waves with certain frequencies can exist in this optical cavity. The radiation inside a spherical mirror cavity is a Gaussian beam. The actual or particular Gaussian beam that fits into the cavity is that beam whose wavefronts at the mirrors match the curvature of the mirrors. Consider the symmetric resonator shown in Figure 1.54 in which the mirrors have the same radius of curvature R. When a wave starts at A, its wavefront is the same as the curvature of A. In the middle of the cavity it has the minimum width and at B the wave again has the same curvature as B. Such a wave in the cavity can replicate itself (and hence exist in the cavity) as it travels between the mirrors provided that it has right beam characteristics, that is the right curvature at the mirrors. The radius of curvature R of a Gaussian beam wavefront at a distance z along its axis is given by R(z) = z[1 + (zo/z)2] ; zo = wo2/ is the Rayleigh range Consider a confocal symmetric optical cavity in which the mirrors are separated by L = R. (a) Show that the cavity length L is 2zo, that is, it is the same as the Rayleigh range, which is the reason the latter is called the confocal length. (b) Show that the waist of the beam 2wo is fully determined only by the radius of curvature R of the mirrors, and given by 2wo = (2R/)1/2 (c) If the cavity length L = R = 50 cm, and = 633 nm, what is the waist of the beam at the center and also at the mirrors?
光电子技术课后答案
习 题11. 设在半径为R c 的圆盘中心法线上，距盘圆中心为l 0处有一个辐射强度为I e 的点源S ，如图所示。
试计算该点源发射到盘圆的辐射功率。
ΩΦd d e e I =， 202πd l R c =Ω 202e πd d l R I I c e e ==ΩΦ2. 如图所示，设小面源的面积为∆A s ，辐射亮度为L e ，面源法线与l 0的夹角为θs ；被照面的面积为∆A c ，到面源∆A s 的距离为l 0。
若θc 为辐射在被照面∆A c 的入射角，试计算小面源在∆A c 上产生的辐射照度。
用定义r r ee A dI L θ∆cos =和A E e e d d Φ=求解。
4. 霓虹灯发的光是热辐射吗？不是热辐射。
6. 从黑体辐射曲线图可以看出，不同温度下的黑体辐射曲线的极大值处的波长λm 随温度T 的升高而减小。
试由普朗克热辐射公式导出常数=T m λ。
这一关系式称为维恩位移定律，其中常数为2.898⨯10-3m ∙K 。
普朗克热辐射公式求一阶导数，令其等于0，即可求的。
9. 常用的彩色胶卷一般分为日光型和灯光型。
你知道这是按什么区分的吗？按色温区分。
习 题21. 何为大气窗口，试分析光谱位于大气窗口内的光辐射的大气衰减因素。
对某些特定的波长，大气呈现出极为强烈的吸收。
光波几乎无法通过。
根据大气的这种选择吸收特性，一般把近红外区分成八个区段，将透过率较高的波段称为大气窗口。
2. 何为大气湍流效应，大气湍流对光束的传播产生哪些影响？ l 0 SR c 第1题图L e∆A s ∆A c l 0 θs θc第2题图是一种无规则的漩涡流动，流体质点的运动轨迹十分复杂，既有横向运动，又有纵向运动，空间每一点的运动速度围绕某一平均值随机起伏。
这种湍流状态将使激光辐射在传播过程中随机地改变其光波参量，使光束质量受到严重影响，出现所谓光束截面内的强度闪烁、光束的弯曲和漂移（亦称方向抖动）、光束弥散畸变以及空间相干性退化等现象，统称为大气湍流效应。
光电子学与光子学讲义-作业答案(第1、2章)13版.doc
第一章1.10 Refractive index(a) Consider light of free-space wavelength 1300 nm traveling in pure silica medium. Calculate the phase velocity and group velocity of light in this medium. Is the group velocity ever greater than the phase velocity?(b) What is the Brewster angle(the polarization angle qp) and the critical angle(qc) for total internal reflection when the light wave traveling in this silica medium is incident on a silica/air interface. What happens at the polarization angle?(c) What is the reflection coefficient and reflectance at normal incidence when the light beam traveling in the silica medium is incident on a silica/air interface?(d) What is the reflection coefficient and reflectance at normal incidence when a light beam traveling in air is incident on an air/silica interface? How do these compare with part (c) and what is your conclusion?1.18 Reflection at glass-glass and air-glass interfaceA ray of light that is traveling in a glass medium of refractive index n1=1.460 becomes incident on a less dense glassmedium of refractive index n2=1.430. Suppose that the free space wavelength of the light ray is 850 nm.(a) What should the minimum incidence angle for TIR be?(b) What is the phase change in the reflected wave when the angle of incidence qi =85 ° and when qi =90° ?(c) What is the penetration depth of the evanescent wave into medium 2 when qi =85 ° and when qi =90° ?(d) What is the reflection coefficient and reflection at normal incidence (qi =0 ° )when thelight beam traveling in the glass medium (n=1.460) is incident on a glass-air interface?(e) What is the reflection coefficient and reflectance at normal incidence when a light beam traveling in air is incident on an air/-glass interface (n=1.460)? How do these compare with part (d) and what is your conclusion?1.20 TIR and polarization at water-air interface(1) Given that the refractive index of water is about 1.33, what is the polarization angle for light traveling in air and reflected from the surface of the water?(2) consider a diver in sea pointing a flashlight towards the surface of the water. What is the critical angle for the light beam to be reflected from the water surface?1.22 phase changes on TIRConsider a lightwave of wavelength 870nm traveling in a semiconductor medium (GaAs) of refractive index 3.6. It is incident on a different semiconductor medium (AIGaAs) of°. Will this result in total internal refractive index 3.4, and the angle of incidence is 80reflection? Calculate the phase change in the parallel and perpendicular components of the reflected electric field?1.25 Goos-Haenchen phase shiftAray of light that is traveling in a glass medium(1) of refractive index n1 =1.460 becomes incident on a less dense glass medium(2) of refractive index n2=1.430. Suppose that the free space wavelength of the light ray is 850nm.the angle of incidence(9=85 . Estimate the lateral Goos-Haenchen shift in the reflected wave for the perpendicular field component. Recalculate the Goos-Haenchen shift in the secondmedium has n2=1 (air). What is your conclusion?Assume that the virtual reflection occurs from a virtual plane in medium B at a distance d that is roughly the same as the penetration depth.Note that d actually depends on the polarization ,the direction of the field,but we will ignore this dependence.第二章作业习题：2.7Dielectric slab waveguide Consider a dielectric slab waveguide that has a thin GaAs layer of thickness 0.25 /w between two AIGaAs layers. The refractive index of GaAs is3.6 and that of the AIGaAs layers is 3.40. What is the cut-off wavelength beyond which only a single mode can propagate in the waveguide, assuming that the refractive index does not vary greatly with thewavelength? If a radiation of wavelength 860 nm (corresponding to bandgap radiation) is propagating in the GaAs layer, what is the penetration of the evanescent wave into the AIGaAs layer? What is the mode field width (MFW) of this radiation? Point out the effect of change of radiation wavelength (为on the MFW.2.9 Dielectric slab waveguide Consider a planar dielectric waveguide with a core thickness 10/m,ni=1.4446, n2=1.4440. Calculate the V -number, the mode angle 劣for m=0 (use a graphical solution, if necessary), penetration depth, and mode field distance (MFW=2 c(^2 9, for light wavelengths of 1.0 /xn and 5 /jn. What is your conclusion? Compare your MFW calculationwith 2»o =2a (V +1 )/V .The model angle 3is given as &=88.85?for ^=1 /m and ^=88.72 ?for ^=1.5 仰for the fundamental mode m=0.2.10 A multimode fibe r Consider a multimode fiber with a core diameter of 60 /m, core refractive index of 1.47, and a cladding refractive index of 1.45 both at 870 nm. Consider operating this fiber at ^=870nm.(e) Calculate the numerical aperture.(f) Find out the normalized core-cladding index difference.(g) Calculate the V-number for the fiber and estimate the number of guided modes.(h) Calculate the wavelength beyond which the fiber becomes single-mode.(i) Calculate the modal dispersion △ T and hence the bit rate x distance product.2.12 Single mode fiber Consider a fiber with a S1O2-13.5% GeOz core of diameter of 6 Wi and refractive index of 1.47 and a cladding refractive index of 1.46 both refractive indices at 1300 nm where the fiber is to be operated using a laser source with a half maximum width (FWHM) of 2 nm.. (j) Calculate the V -number for the fiver.(k) what is the maximum allowed diameter of the core that maintains oprations in single-mode?(l) Calculate the wavelength below which the fiber becomes multimode.(m) C alculate the numerical aperture.(n) Calculate the maximum acceptance angle.(o) Obtain the material dispersion and wavelength dispersion and hence estimate the bit rate x distance product ( B x L) of the fiber.。
光电子课后习题答案讲解
第一章1. 光电子器件按功能分为哪几类？每类大致包括哪些器件？光电子器件按功能分为光源器件、光传输器件、光控制器件、光探测器件、光存储器件、光显示器件。
光源器件分为相干光源和非相干光源。
相干光源主要包括激光器和非线性光学器件等。
非相干光源包括照明光源、显示光源和信息处理用光源等。
光传输器件分为光学元件(如棱镜、透镜、光栅、分束器等等)、光波导和光纤等。
光控制器件包括调制器、偏转器、光开关、光双稳器件、光路由器等。
光探测器件分为光电导型探测器、光伏型探测器、热伏型探测器等。
光存储器件分为光盘(包括CD、VCD、DVD、LD等)、光驱、光盘塔等。
光显示器件包括CRT、液晶显示器、等离子显示器、LED显示。
2．谈谈你对光电子技术的理解。
光电子技术主要研究物质中的电子相互作用及能量相互转换的相关技术，以光源激光化，传输波导（光纤）化，手段电子化，现代电子学中的理论模式和电子学处理方法光学化为特征，是一门新兴的综合性交叉学科。
⒌据你了解，继阴极射线管显示（CRT）之后，哪几类光电显示器件代表的技术有可能发展成为未来显示技术的主体？等离子体显示（PDP），液晶显示（LCD），场致发射显示（EL），LED显示。
第二章：光学基础知识与光场传播规律⒈ 填空题⑴ 光的基本属性是光具有波粒二象性，光粒子性的典型现象有光的吸收、发射以及光电效应等；光波动性的典型体现有光的干涉、衍射、偏振等。
⑵ 两束光相干的条件是频率相同、振动方向相同、相位差恒定；最典型的干涉装置有杨氏双缝干涉、迈克耳孙干涉仪；两束光相长干涉的条件是(0,1,2,)m m δλ==±±，δ为光程差。
⑶两列同频平面简谐波振幅分别为01E 、02E ，位相差为φ，则其干涉光强为22010201022cos E E E E φ++，两列波干涉相长的条件为2(0,1,2,)m m φπ==±±⑷波长λ的光经过孔径D 的小孔在焦距f 处的衍射爱里斑半径为1.22f D λ。
激光原理及应用(第二版)课后习题答案(全)
思考练习题4
2．He-Ne激光器辐射632.8nm光波，其方形镜对称共焦腔，腔长L＝
0.2m。腔内同时存在，，横模。若在腔内接近镜面处加小孔光阑选取
横模，试问：
(1)如只使模振荡，光阑孔径应多大？ (2)如同时使，模振荡而抑制振荡，光阑孔径应多大？ 答：(1)TEM00模在镜面处的光斑半径为 所以光阑孔径应该为0.2mm (2)TEM11模在镜面处的光斑半径为 所以光阑孔径为0.35mm
思考练习题1
1． 试计算连续功率均为1W的两光源，分别发射＝0.5000m， =3000MHz的光，每秒从上能级跃迁到下能级的粒子数各为多 少？
答：粒子数分别为：
11．静止氖原子的3S22P4谱线的中心波长为0.6328m，设氖原子分别以
0.1c,0.5c的速度向着接收器运动，问接收到的频率各为多少？ 答： 同理可求：；
答：（a）；
（b）
4.
稳定谐振腔的两块反射镜，其曲率半径分别为R1＝40cm，2＝
100cm，求腔长L的取值范围。
答：
10. 实验测得He-Ne激光器以波长＝0.6328工作时的小讯号增益系数 为G0＝310－4/d(cm-1)，d为腔内毛细管内径(cm)。以非均匀增宽 计算腔内光强I＝50W／cm2的增益系数G(设饱和光强Is＝30W／ cm2时，d＝1mm)，并问这时为保持振荡稳定，两反射镜的反射
3．一高斯光束束腰半径＝0.2mm，＝0.6328，今用一焦距f为3cm的短 焦距透镜聚焦，已知腰粗离透镜的距离为60cm，在几何光学近似下求 聚焦后光束腰粗。 答：
5．用如图（4－33）所示的倒置望远镜系统改善由对称共焦腔输出的光
光电子学与光子学的原理及应用s.o.kasa chapter4课后答案
(le3vpe1)lsarnedsuthlteinrgemfraoimninvgareioleucstrdoinffse(r2enp5t)vwaliutheisnoqfumanl atunmd mmsetchhaat nciacnalarsusilgesn.edNtoot
e all
sixth excited transitions to
spins parallel has a lower energy than that with opposite electron spins as indicated in Figure 4Q1. These
two He states can excite Ne atoms to either the 2p54s1 or 2p55s1. There is then a population inversion between these levels and the 2p53p1 and between 2p55s1 and 2p54p1 which leads to the above lasing
electron these
energy levels are allowed as photon emission requires quantum number selection rules to be obeyed.
a Table 4Q1, shows some typical commercial He-Ne laser characteristics (within 30-50%) for various wavelengths. Calculate the overall efficiencies of these lasers.
光电子学与光子学的原理及应用第二章-课后答案
光电子学与光子学的原理及应用第二章-课后答案1. 选择题1.1 题目一答案：C解析：光电效应是指物质受到光的照射后，吸收光能，将光能转化为电能的一种现象。
光电效应首先是由爱因斯坦在1905年提出的，他在描述光电效应时，引入了光子概念，假设光是由一组个别粒子组成的（即光量子），这些粒子就是后来被称为光子的电磁辐射量子。
1.2 题目二答案：A解析：光电倍增管是指通过光电效应，在光电面上光电发射物质外壳的钨丝和灯管之间加一个高达2000-3000伏的电压使其产生光电流，再对光电流进行电子倍增，最后输出检测的一种光电探测器。
光电倍增管的结构与普通的电子管相似，但是在各个电极和玻璃壳之间加入了紧密和高度真空的保护，同时在阳极和阳极网之间还添加了一个用直流电压加电的光电体。
当阳极对外加正电压使阳极电流开始增大时，就成为光电倍增管。
1.3 题目三答案：D解析：光电二极管是一种能够将光信号转化为电信号的器件。
光电二极管的基本原理是利用半导体材料的PN结在光照射下产生光电效应，使得PN结两端产生电荷，从而产生电压信号。
光电二极管的结构和普通二极管类似，主要由P型和N型的半导体材料组成，当光照射到光电二极管上时，光子能量被半导体材料所吸收，产生的热力激发电子，从而引起半导体PN结的载流子的复合和流动，产生感光电流。
光电二极管应用广泛，如光通信、光电测量、光谱分析等领域。
1.4 题目四答案：B解析：光导纤维是一种能够传输光信号的特殊纤维材料。
光导纤维的核心部分是由高折射率的材料构成，而外部由低折射率的材料构成。
当光线传输到光导纤维中时，会发生全反射现象，使得光线能够沿着光导纤维进行传输，最终到达目标地点。
光导纤维具有传输距离远、损耗小、带宽大、抗电磁干扰等优点，在通信、医疗、传感等领域得到广泛应用。
2. 填空题2.1 题目一答案：钠解析：钠具有低电离电势，激发电子的能量比较低，是光电电子极容易脱离的材料之一。
2.2 题目二答案：光电效应解析：光电效应是指物质受到光的照射后，吸收光能，将光能转化为电能的一种现象。
光电子学与光子学讲义-作业答案（第1、2章）13版.doc
光电⼦学与光⼦学讲义-作业答案（第1、2章）13版.doc 第⼀章1.10 Refractive index(a) Consider light of free-space wavelength 1300 nm traveling in pure silica medium. Calculate the phase velocity and group velocity of light in this medium. Is the group velocity ever greater than the phase velocity?(b) What is the Brewster angle(the polarization angle qp) and the critical angle(qc) for total internal reflection when the light wave traveling in this silica medium is incident on a silica/air interface. What happens at the polarization angle?(c) What is the reflection coefficient and reflectance at normal incidence when the light beam traveling in the silica medium is incident on a silica/air interface?(d) What is the reflection coefficient and reflectance at normal incidence when a light beam traveling in air is incident on an air/silica interface? How do these compare with part (c) and what is your conclusion?1.18 Reflection at glass-glass and air-glass interfaceA ray of light that is traveling in a glass medium of refractive index n1=1.460 becomes incident on a less dense glassmedium of refractive index n2=1.430. Suppose that the free space wavelength of the light ray is 850 nm.(a) What should the minimum incidence angle for TIR be?(b) What is the phase change in the reflected wave when the angle of incidence qi =85 ° and when qi =90° ?(c) What is the penetration depth of the evanescent wave into medium 2 when qi =85 ° and when qi =90° ?(d) What is the reflection coefficient and reflection at normal incidence (qi =0 ° )when thelight beam traveling in the glass medium (n=1.460) is incident on a glass-air interface?(e) What is the reflection coefficient and reflectance at normal incidence when a light beam traveling in air is incident on an air/-glass interface (n=1.460)? How do these compare with part (d) and what is your conclusion?1.20 TIR and polarization at water-air interface(1) Given that the refractive index of water is about 1.33, what is the polarization angle for light traveling in air and reflected from the surface of the water?(2) consider a diver in sea pointing a flashlight towards the surface of the water. What is the critical angle for the light beam to be reflected from the water surface?1.22 phase changes on TIRConsider a lightwave of wavelength 870nm traveling in a semiconductor medium (GaAs) of refractive index 3.6. It is incident on a different semiconductor medium (AIGaAs) of°. Will this result in total internal refractive index 3.4, and the angle of incidence is 80reflection? Calculate the phase change in the parallel and perpendicular components of the reflected electric field?1.25 Goos-Haenchen phase shiftAray of light that is traveling in a glass medium(1) of refractive index n1 =1.460 becomes incident on a less dense glass medium(2) of refractive index n2=1.430. Suppose that the free space wavelength of the light ray is 850nm.the angle of incidence(9=85 . Estimate the lateral Goos-Haenchen shift in the reflected wave for the perpendicular field component. Recalculate the Goos-Haenchen shift in the secondmedium has n2=1 (air). What is your conclusion?Assume that the virtual reflection occurs from a virtual plane in medium B at a distance d that is roughly the same as the penetration depth.Note that d actually depends on the polarization ,the direction of the field,but we will ignore this dependence.第⼆章作业习题：2.7Dielectric slab waveguide Consider a dielectric slab waveguide that has a thin GaAs layer of thickness 0.25 /w between two AIGaAs layers. The refractive index of GaAs is3.6 and that of the AIGaAs layers is 3.40. What is the cut-off wavelength beyond which only a single mode can propagate in the waveguide, assuming that the refractive index does not vary greatly with thewavelength? If a radiation of wavelength 860 nm (corresponding to bandgap radiation) is propagating in the GaAs layer, what is the penetration of the evanescent wave into the AIGaAs layer? What is the mode field width (MFW) of this radiation? Point out the effect of change of radiation wavelength (为on the MFW.2.9 Dielectric slab waveguide Consider a planar dielectric waveguide with a core thickness 10/m,ni=1.4446, n2=1.4440. Calculate the V -number, the mode angle 劣for m=0 (use a graphical solution, if necessary), penetration depth, and mode field distance (MFW=2 c(^2 9, for light wavelengths of 1.0 /xn and 5 /jn. What is your conclusion? Compare your MFW calculationwith 2?o =2a (V +1 )/V .The model angle 3is given as &=88.85?for ^=1 /m and ^=88.72 ?for ^=1.5 仰for the fundamental mode m=0.2.10 A multimode fibe r Consider a multimode fiber with a core diameter of 60 /m, core refractive index of 1.47, and a cladding refractive index of 1.45 both at 870 nm. Consider operating this fiber at ^=870nm.(e) Calculate the numerical aperture.(f) Find out the normalized core-cladding index difference.(g) Calculate the V-number for the fiber and estimate the number of guided modes.(h) Calculate the wavelength beyond which the fiber becomes single-mode.(i) Calculate the modal dispersion △ T and hence the bit rate x distance product.2.12 Single mode fiber Consider a fiber with a S1O2-13.5% GeOz core of diameter of 6 Wi and refractive index of 1.47 and a cladding refractive index of 1.46 both refractive indices at 1300 nm where the fiber is to be operated using a laser source with a half maximum width (FWHM) of 2 nm.. (j) Calculate the V -number for the fiver.(k) what is the maximum allowed diameter of the core that maintains oprations in single-mode?(l) Calculate the wavelength below which the fiber becomes multimode.(m) C alculate the numerical aperture.(n) Calculate the maximum acceptance angle.(o) Obtain the material dispersion and wavelength dispersion and hence estimate the bit rate x distance product ( B x L) of the fiber.。