工程光学英语补充内容和习题
应用光学习题”、“物理光学习题”、“工程光学-练习题.doc
本习题供复习参考。
更多的内容请参考“应用光学习题”、“物理光学习题”、“工程光学+练习题”等。
所有资料均可在网络课件资源处下载。
一.选择题1、几何光学有三大基本定律，它们是是：(D )A、折射与反射定律，费马原理，马吕斯定律；B、直线传播定律，折射与反射定律，费马原理；C、独立传播定律，折射与反射定律，马吕斯定律；D、直线传播定律，独立传播定律，折射与反射定律。
2、对理想光学系统，下列表述正确的是：(C )A、位于光轴上的物点的共轨像点不在光轴上；B、物方焦点与像方焦点共觇；C、基点与基面为：焦点、主点、节点，焦平面、主平面、节平面；D、牛顿物像位置关系，它是以主点为坐标原点。
3、关于光阑，下列表述正确的是：(B )A、孔径光阑经其前面的光学系统所成的像称为入窗；B、若孔径光阑在光学系统的最前面，则孔径光阑本身就是入瞳；C、孔径光阑、入窗、出窗三者是物像关系；D、视场光阑是限制轴上物点孔径角的大小，或者说限制轴上物点成像光束宽度、并有选择轴外物点成像光束位置作用的光阑。
4、关于人眼，下列描述正确的是：(A )A、眼睛自动改变焦距的过程称为眼睛的视度调节；B、近视眼是将其近点矫正到明视距离，可以用负透镜进行校正；C、眼睛可视为由水晶体、视网膜和视神经构成的照相系统。
；D、人眼分辨率与极限分辨角成正比关系。
5、关于典型光学系统，下列表述正确的是：(B )A、增大波长可以提高光学系统的分辨率；B、显微镜的有效放大率,放大率高于1000NA时，称作无效放大率，不能使被观察的物体细节更清晰；C、目视光学仪器，其放大作用可以由横向放大率来表示；D、减小孔径可以提高光学系统的分辨率。
6、关于光的电磁理论，下列表述正确的是：(D )A、两列光波相遇后又分开，每列光波不再保持原有的特性；B、两列光波叠加后其光强为两列光波的强度之和；C、等振幅面传播的速度称为相速度；D、两个振幅相同、振动方向相同、传播方向相同，但频率接近的单色光波叠加形成拍现象。
材料科学与工程专业英语第三版-翻译以及答案
UNIT 1一、材料根深蒂固于我们生活的程度可能远远的超过了我们的想象，交通、装修、制衣、通信、娱乐（recreation）和食品生产，事实上（virtually），我们生活中的方方面面或多或少受到了材料的影响。
历史上，社会的发展和进步和生产材料的能力以及操纵材料来实现他们的需求密切（intimately）相关，事实上，早期的文明就是通过材料发展的能力来命名的（石器时代、青铜时代、铁器时代）。
二、早期的人类仅仅使用（access）了非常有限数量的材料，比如自然的石头、木头、粘土（clay）、兽皮等等。
随着时间的发展，通过使用技术来生产获得的材料比自然的材料具有更加优秀的性能。
这些性材料包括了陶瓷（pottery）以及各种各样的金属，而且他们还发现通过添加其他物质和改变加热温度可以改变材料的性能。
此时，材料的应用（utilization）完全就是一个选择的过程，也就是说，在一系列有限的材料中，根据材料的优点来选择最合适的材料，直到最近的时间内，科学家才理解了材料的基本结构以及它们的性能的关系。
在过去的100年间对这些知识的获得，使对材料性质的研究变得非常时髦起来。
因此，为了满足我们现代而且复杂的社会，成千上万具有不同性质的材料被研发出来，包括了金属、塑料、玻璃和纤维。
三、由于很多新的技术的发展，使我们获得了合适的材料并且使得我们的存在变得更为舒适。
对一种材料性质的理解的进步往往是技术的发展的先兆，例如：如果没有合适并且没有不昂贵的钢材，或者没有其他可以替代（substitute）的东西，汽车就不可能被生产，在现代、复杂的（sophisticated）电子设备依赖于半导体（semiconducting）材料四、有时，将材料科学与工程划分为材料科学和材料工程这两个副学科（subdiscipline）是非常有用的，严格的来说，材料科学是研究材料的性能以及结构的关系，与此相反，材料工程则是基于材料结构和性能的关系，来设计和生产具有预定性能的材料，基于预期的性能。
工程光学英语补充内容和习题
Refracted ray Rarer medium
6. Fermat’ principle
Light takes the path that requires the least time.
5
7. Key words and concept
Image space extends from infinity on one side to infinity on the other. The same holds for the object space; both completely overlap. Whether a given point is in the object space or the image space depends on whether it is part of a ray before or after refraction. If a given point is in the object space, it is a point object, otherwise it is a point image.
Angular magnification (magnifying power) The total magnification of a combination of lenses is the product of
the magnification of the individual lenses.
17
Keywords and concept
2. Nodal points are where no refraction occurs. Whenever the refractive indices on either side of the lens are the same, the nodal points coincide with the principal points. If the refractive indices on the two sides of the lens are different, the N points would move away from the principal planes, toward the side of higher index.
工程英语测试题及答案
工程英语测试题及答案一、选择题（每题2分，共20分）1. What is the term used to describe the process of turning raw materials into finished products?A. FabricationB. AssemblyC. MachiningD. Casting答案：A2. The primary function of a ________ is to convert electrical energy into mechanical energy.A. MotorB. GeneratorC. TransformerD. Inverter答案：A3. In engineering, the term "stress" refers to:A. The internal resistance of a material to deformationB. The force applied to a materialC. The change in shape of a materialD. The rate of change of force答案：A4. Which of the following is not a type of welding process?A. Arc weldingB. Gas weldingC. Ultrasonic weldingD. Friction welding答案：C5. The process of designing and building a structure is known as:A. EngineeringB. ArchitectureC. ConstructionD. All of the above答案：D6. What does the abbreviation "CAD" stand for in the field of engineering?A. Computer-Aided DesignB. Computer-Aided DraftingC. Computer-Aided DevelopmentD. Computer-Aided Documentation答案：A7. The SI unit for pressure is:A. PascalB. NewtonC. JouleD. Watt答案：A8. A ________ is a type of joint that allows for relative movement between connected parts.A. Rigid jointB. Revolute jointC. Fixed jointD. Pin joint答案：B9. The process of removing material from an object to achieve the desired shape is known as:A. MachiningB. CastingC. ForgingD. Extrusion答案：A10. In engineering, the term "specification" refers to:A. A detailed description of the requirements of aprojectB. A list of materials to be used in a projectC. The estimated cost of a projectD. The timeline for a project答案：A二、填空题（每题1分，共10分）11. The ________ is the process of cutting a flat surface ona material.答案：sawing12. A ________ is a type of bearing that allows for rotation.答案：ball bearing13. The term "gearing" refers to the use of gears to transmit ________.答案：motion14. The ________ is the study of the properties of materials.答案：material science15. In a hydraulic system, a ________ is used to control the flow of fluid.答案：valve16. The ________ is the process of heating and cooling a material to alter its physical properties.答案：heat treatment17. The ________ is a tool used to measure the hardness of a material.答案：hardness tester18. A ________ is a type of joint that connects two parts ata fixed angle.答案: hinge joint19. The ________ is the process of joining two pieces ofmetal by heating them to a molten state.答案：fusion welding20. The ________ is the study of the behavior of structures under load.答案：structural analysis三、简答题（每题5分，共30分）21. Define the term "mechanical advantage" in engineering.答案：Mechanical advantage is the ratio of output force to input force in a simple machine, indicating how much the machine amplifies the force applied to it.22. Explain the concept of "factor of safety" in engineering design.答案：The factor of safety is a ratio used in engineering to ensure that a structure or component can withstand loads beyond the maximum expected in service, providing a margin of safety against failure.23. What is the purpose of a "stress-strain curve" in material testing?答案：A stress-strain curve is a graphical representation of the relationship between the stress applied to a material and the resulting strain, used to determine the material's mechanical properties such as elasticity, yield strength, and ultimate strength.24. Describe the difference between "static" and "dynamic" loads in engineering.答案：Static loads are constant forces that do not changeover time, while dynamic loads are forces that vary in magnitude or direction over time, often due to movement or vibrations.25. What is "creep" in the context of material behavior under load?答案：Creep。
光电英语期末考试题及答案
光电英语期末考试题及答案一、选择题（每题2分，共20分）1. What is the fundamental principle of the photoelectric effect?A. Electromagnetic inductionB. Quantum theoryC. Newton's third lawD. Ohm's lawAnswer: B. Quantum theory2. Which of the following is not a type of optical fiber?A. Single-mode fiberB. Multi-mode fiberC. Polarization fiberD. All of the above are types of optical fibersAnswer: C. Polarization fiber3. What does LED stand for?A. Light Emitting DiodeB. Large Emission DiodeC. Limited Energy DeviceD. Linear Energy DistributionAnswer: A. Light Emitting Diode4. In the context of optical communication, what does the term 'attenuation' refer to?A. The increase in signal strength over distanceB. The decrease in signal strength over distanceC. The frequency of the signalD. The phase of the signalAnswer: B. The decrease in signal strength over distance5. What is the primary function of a photodiode?A. To convert electrical energy into lightB. To convert light into electrical energyC. To amplify electrical signalsD. To filter out noise in electrical signalsAnswer: B. To convert light into electrical energy二、填空题（每空2分，共20分）6. The ________ is the study of the behavior and effects of light, including its interactions with matter.Answer: Optics7. A ________ is a device that can change the direction of light by reflection or refraction.Answer: Lens8. The process of converting electrical signals into light pulses for transmission is known as ________.Answer: Electro-Optical Conversion9. The ________ effect is the phenomenon where electrons are emitted from a material when it is exposed to light of sufficient energy.Answer: Photoelectric10. In fiber optics, the ________ is the point at which the light is fully confined within the core of the fiber.Answer: Critical Angle三、简答题（每题15分，共30分）11. Explain the difference between a laser and a LED in terms of their applications and operational principles.Answer:A laser, or Light Amplification by Stimulated Emission of Radiation, produces a highly focused and coherent beam of light. It is used in various applications such as communication, surgery, and material processing due to its precision and power. On the other hand, a LED, or Light Emitting Diode, emits incoherent light and is used for general illumination, indicator lights, and displays due to its efficiency and longevity.12. Describe the process of total internal reflection and its significance in fiber optics.Answer:Total internal reflection occurs when light traveling from a medium with a higher refractive index to a medium with a lower refractive index hits the boundary at an anglegreater than the critical angle. Instead of refracting, the light is completely reflected back into the original medium. In fiber optics, this phenomenon is crucial as it allowslight signals to travel long distances with minimal loss, making it ideal for communication networks.四、论述题（每题15分，共30分）13. Discuss the role of photodetectors in modern communication systems.Answer:Photodetectors play a pivotal role in modern communication systems, particularly in optical communication. They convert the incoming optical signals back intoelectrical signals, which can then be processed and interpreted by electronic devices. The high-speed and high-sensitivity characteristics of photodetectors make them suitable for handling the vast amounts of data transmitted in fiber optic networks. Moreover, their ability to operate over a wide range of wavelengths allows for the efficient use of the optical spectrum.14. Explain the concept of optical amplification and its importance in long-haul communication.Answer:Optical amplification is the process of increasing the power of an optical signal without converting it to an electrical signal. This is achieved using devices such as erbium-doped fiber amplifiers (EDFAs), which can amplifysignals over a broad range of wavelengths. Optical amplification is crucial for long-haul communication as it compensates for signal loss due to attenuation, thereby extending the reach of optical signals without the need for frequent signal regeneration, thus improving the efficiency and reliability of the communication system.光电英语期末考试题答案。
八年级物理光学现象英语阅读理解20题
八年级物理光学现象英语阅读理解20题1. We can see our reflection in a mirror. This is an example of _____.A. light absorptionB. light reflectionC. light refractionD. light dispersion答案解析：B。
选项A 光的吸收，镜子不是吸收光才让我们看到自己的影像；选项 C 光的折射，在镜子中看到自己不是折射现象；选项D 光的色散，与看到自己在镜子中的影像无关。
镜子中看到自己是光的反射现象。
2. When light hits a smooth surface like water, it can be _____.A. completely absorbedB. partly reflectedC. totally refractedD. randomly dispersed答案解析：B。
选项 A 完全被吸收不符合水面的情况；选项 C 完全折射不准确；选项D 随机色散也不对。
水面等光滑表面会使光部分反射。
3. The reflection of light from a mirror follows a certain law. Which law is it?A. Newton's first lawB. Newton's second lawC. Law of reflectionD. Law of universal gravitation答案解析：C。
选项 A 和 B 是牛顿运动定律与光的反射无关；选项D 是万有引力定律也与光的反射无关。
光从镜子反射遵循反射定律。
4. What is the angle of incidence equal to in the case of reflection?A. Angle of refractionB. Angle of dispersionC. Angle of reflectionD. Angle of absorption答案解析：C。
工程光学练习答案(带样题).doc
工程光学练习答案(带样题)期末，东北石油大学审查了09级工程光学的测量和控制材料。
第一章练习1，假设真空中的光速为3米/秒，则计算水中(n=1.333)、皇冠玻璃(n=1.51)、燧石玻璃(n=1.65)、加拿大树胶(n=1.526)、钻石(n=2.417)和其他介质中的光速。
解决方案:当灯在水中时，n=1.333，v=2.25m米/秒，当灯在皇冠玻璃中时，n=1.51，v=1.99m米/秒，当灯在燧石玻璃中时，n=1.65，v=1.82m米/秒，当灯在加拿大树胶中时，n=1.526，v=1.97m米/秒，当灯在钻石中时，n=2.417，v=1.24米/秒。
2.一个物体穿过针孔照相机，在屏幕上形成一个60毫米大小的图像。
如果屏幕被拉开50毫米，图像的尺寸变成70毫米，计算出从屏幕到针孔的初始距离。
解决方案:在同一个均匀的介质空间中，光直线传播。
如果选择通过节点的光，方向不会改变，从屏幕到针孔的初始距离为x，则可以根据三角形的相似性得到:因此，x=300mm毫米意味着从屏幕到针孔的初始距离是300毫米。
3、一块厚度为200毫米的平行平板玻璃(n=1.5)，下面放一块直径为1毫米的金属板。
如果玻璃板上覆盖有圆形纸片，则要求玻璃板上方的任何方向都不能看到纸片。
这张纸的最小直径是多少？解决方案:如果纸片的最小半径是x，那么根据全反射原理，当光束从玻璃发射到空气中的入射角大于或等于全反射临界角时，就会发生全反射，正是由于这个原因，在玻璃板上方看不到金属片。
全反射的临界角由下式确定:(1)其中N2=1，n1=1.5，根据几何关系，利用平板的厚度和纸张与金属片的半径计算全反射临界角的方法如下:(2)纸张的最小直径x=179.385mm毫米可以通过组合等式(1)和(2)来获得，因此纸张的最小直径为358.77毫米4.光纤芯的折射率是n1.包层的折射率为n2，光纤所在介质的折射率为n0。
计算光纤的数值孔径(即n0sinI1，其中I1是光在光纤中以全反射模式传播时，光在入射端面的最大入射角)。
工程光学第二章练习参考答案
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工程光学英文题加中文题含问题详解
English Homework for Chapter 11.In ancient times the rectilinear propagation of light was used to measure the height of objects by comparing the length of their shadows with the length of the shadow of an object of known length. A staff 2m long when held erect casts a shadow 3.4m long, while a building’s shadow is 170m long. How tall is the building?Solution. According to the law of rectilinear propagation, we get, 4.32170 x x=100 (m)So the building is 100m tall.2.Light from a water medium with n=1.33 is incident upon a water-glass interface at an angle of 45o. The glass index is 1.50. What angle does the light make with the normal in the glass?Solution. According to the law ofrefraction, We get,''sin sin I n I n =626968.05.145sin 33.1sin =⨯='οIο8.38='I So the light makenormal in the glass.3. A goldfish swims 10cm from the side of a spherical bowl of water of radius 20cm. Where does the fish appear to be? Does it appear larger or smaller?Solution. According to the equation. rn n l n l n -'=-'' and n ’=1 , n=1.33, r=-20we can get11416.110133.15836.8)(5836.81165.02033.01033.11>-=⨯⨯-=''=-='∴-=--+-=-'+='l n l n cm l r n n l n l βΘ So the fish appears larger.A4.An object is located 2cm to the left of convex end of a glass rod which has a radius of curvature of 1cm. The index of refraction of the glass is n=1.5. Find the image distance.Solution. Refer to the figure. According to the equationrn n l n l n -'=-'' and n=1, n ’=1.5, l 1=-2cm,r 1=1cm , we getcm l l d l l l 2021115.15.121211='∴-∞='-=∞='∴=-+-='English Homework for Chapter 21.An object 1cm high is 30cm in front of a thin lens with a focal length of 10cm. Where is the image? Verify your answer by graphical construction of the image. Solution. According toequation, f l l '=-'11 and l=-30cm f ’we get)(15)30(10)30(10cm l f l f l =-+-⨯=+''='Others are omitted.′2.A lens is known to have a focal length of 30cm in air. An object is placed 50cm to the left of the lens. Locate the image and characterize it.Solution.f l l '=-'11 and f′=30cm l we get (75)50(30)50(30l f l f l =-+-⨯=+''='5.15075-=-='=l l βThe image is a real, larger one.3.The object is transparent cube, 4mm across, placed 60cm in front of 20cm focal length. Calculate the transverse and axial magnification and describe what the image looks like?Solution. From Gauss’s equation, we find for the rear surface of the cube (the face closer to the lens)that,)(3020)60()20()60(111cm f l f l l +=+-⨯-='+'='′For the front surface (the face farther away from the lens),)(9.29204.6020)4.60(2cm l +=+-⨯-='The transverse magnification for therear surface is ⨯-=-+=5.06030t MBut the axial magnification is⨯+=----=∆'∆=25.0)4.60(609.2930l l M aSince atM M ≠,the cube doesn’t look likea cube.4.A biconvex lens is made out of glass of n=1.52. If one surface has twice the radius of curvature of the other, and if the focal length is 5cm, what are the two radii?Solution. Supposing r 1= -2r 2 (ρ2=-2ρ1),according to the lens equation))(1(21ρρϕ--=n we get,)(152.1(51ρ-=1282.01=∴ρ2564.02-=ρ∴r 1=7.8(cm) r 2=-3.9(cm)返回English Homework for Chapter 4 1. A stop 8mm in diameter is placed halfway between an extended object and a large-diameter lens of 9cm focal length. The lens projects an image of the object onto a screen 14cm away. What is the diameter of the exit pupil?Solution. Refer to the figure. First, from the known focal length and the image distance,we find the object distance. f l l '=-'111Θand l ’=14 f ’=9l =-25.2(cm)The stop is one-half that distance is front of the lens, so l s =12.6(cm) ∴l s ’=31.5(cm)22.255.31-='==ss stop ex l l D D βΘ∴)(28.05.2cm D ex=⨯=2. Two lenses, a lens of 12.5cm focal length and a minus lens of unknown power, are mounted coaxially and 8cm apart. The system is a focal, that is light entering the system parallel at one side emerges parallel at the other. If a stop 15mm in diameter is placed halfway between the lenses:1) Where is the entrance pupil?2) Where is the exit pupil?3) What are their diameters?Solution.Refer to the figure. For thesystem to be a focal, the focal points of the two lenses must coincide. Since f 1’=12.5cm, and the two lenses are 8cm apart, so f 2’=-4.5cm. The entrance pupil is the image of stop formed by the first lens.According to Gauss’s equation, 111111f l l '=-'and l 1’=4cm, f 1’=12.5cm. We get())(88.55.845.1211111cm l f l f l =⨯='-'''=)(05.22488.5151mm D D stopentrance =⨯==β The exit pupil’slocation is)(95.715412.2)(12.25.818)4()5.4()4()5.4(222222mm D D cm f l l f l stop exit =⨯=•=-=-=-+--⨯-='+'='β返回English Homework for Chapter 71. A person wants to look at the image of his or her own eyes, without accommodation, using a concave mirror of 60cm radius of curvature. How far must the mirror be from the eye if the person has1) Normal vision?2) 4diopter myopia, without correction?3) 4diopter hyperopia, without correction?Solution.1) When the person has normal vision, according to the following scheme 1, we get ∞='lso, cm r l 302== 141-=m l r cm l l r 25-==' Scheme 1r l l 211=+'Θ and )(25cm l l +'= cm r 60=265852253048585025308522±=⨯⨯-±==⨯+-l l l Θ ⎩⎨⎧==∴)(50')(7511cm l cm l , or ⎩⎨⎧-==)(15')(1022cm l cm lSo the mirror must be 75cm or 10cm from the eye.rl l 211=+'Θ and )(25'cm l l += cm r 60= 265352253043535025303522±=⨯⨯+±==⨯--l l l Θ ⎩⎨⎧==∴)(75')(5011cm l cm l , or ⎩⎨⎧=-=)(10')(1522cm l cm l (Since the object isreal, so we can give up this answer)So the mirror must be 50cm from the eye.Scheme 2 Scheme 32. Discussion: What differences between the following situations:1) a microscope is used for projection;2) the microscope is used for visual observation.返回工程光学（上）期末考试试卷一．问答题：（共12分，每题3分）1．摄影物镜的三个重要参数是什么？它们分别决定系统的什么性质？2．为了保证测量精度，测量仪器一般采用什么光路？为什么？3．显微物镜、望远物镜、照相物镜各应校正什么像差？为什么？4．评价像质的方法主要有哪几种？各有什么优缺点？二．图解法求像或判断成像方向：（共18分，每题3分）1．求像A'B'2．求像A'B'3．求物AB经理想光学系统后所成的像，并注明系统像方的基点位置和焦距4．判断光学系统的成像方向5．求入瞳及对无穷远成像时50%渐晕的视场6．判断棱镜的成像方向三．填空：（共10分，每题2分）1．照明系统与成像系统之间的衔接关系为：①________________________________________________②________________________________________________2．转像系统分____________________和___________________两大类，其作用是：_______________________________ __________3．一学生带500度近视镜，则该近视镜的焦距为_________________,该学生裸眼所能看清的最远距离为_________________。
材料科学与工程专业英语第三版-翻译以及答案
UNIT 1一、材料根深蒂固于我们生活的程度可能远远的超过了我们的想象，交通、装修、制衣、通信、娱乐（recreation）和食品生产，事实上（virtually），我们生活中的方方面面或多或少受到了材料的影响。
历史上，社会的发展和进步和生产材料的能力以及操纵材料来实现他们的需求密切（intimately）相关，事实上，早期的文明就是通过材料发展的能力来命名的（石器时代、青铜时代、铁器时代）。
二、早期的人类仅仅使用（access）了非常有限数量的材料，比如自然的石头、木头、粘土（clay）、兽皮等等。
随着时间的发展，通过使用技术来生产获得的材料比自然的材料具有更加优秀的性能。
这些性材料包括了陶瓷（pottery）以及各种各样的金属，而且他们还发现通过添加其他物质和改变加热温度可以改变材料的性能。
此时，材料的应用（utilization）完全就是一个选择的过程，也就是说，在一系列有限的材料中，根据材料的优点来选择最合适的材料，直到最近的时间内，科学家才理解了材料的基本结构以及它们的性能的关系。
在过去的100年间对这些知识的获得，使对材料性质的研究变得非常时髦起来。
因此，为了满足我们现代而且复杂的社会，成千上万具有不同性质的材料被研发出来，包括了金属、塑料、玻璃和纤维。
三、由于很多新的技术的发展，使我们获得了合适的材料并且使得我们的存在变得更为舒适。
对一种材料性质的理解的进步往往是技术的发展的先兆，例如：如果没有合适并且没有不昂贵的钢材，或者没有其他可以替代（substitute）的东西，汽车就不可能被生产，在现代、复杂的（sophisticated）电子设备依赖于半导体（semiconducting）材料四、有时，将材料科学与工程划分为材料科学和材料工程这两个副学科（subdiscipline）是非常有用的，严格的来说，材料科学是研究材料的性能以及结构的关系，与此相反，材料工程则是基于材料结构和性能的关系，来设计和生产具有预定性能的材料，基于预期的性能。
工程光学+练习题
一、简答题1、 什么是共轴光学系统、光学系统物空间、像空间？2、 几何光学的基本定律及其内容是什么？3、 什么是理想像和理想光学系统。
4、 理想光学系统的基点和基面有哪些？5、 如何确定光学系统的视场光阑？6、 什么是光学系统的孔径光阑和视场光阑?7、 成像光学系统有哪两种色差？试说明它们的成因？8、 常见非正常眼有哪两种？如何校正常见非正常眼？ 9、 瑞利判据和道威判据。
10、 光学系统极限分辨角为多大？采取什么途径可以提高极限分辨角？ 11、 相速度、群速度。
12、 等厚干涉、等倾干涉。
13、 惠更斯—菲涅耳原理14、 请分别写出多缝干涉亮纹和单缝衍射暗纹条件，并由此简述光栅缺级条件。
15、 常见非正常眼有哪两种？如何校正常见非正常眼？二、填空题1、在空气和折射率为2的介质界面上发生全反射的临界角是30。
2、空气中，光焦度可表示为 ，它表示光学系统对光束汇聚（或发散）的本领。
或越小，越大。
若()00>'>Φf ，光组是汇聚（汇聚或发散）光组，愈大，汇聚（汇聚或发散）本领愈大，反之亦然。
3、凡1次镜面反射或奇数次镜面反射的像称为镜像，物的坐标关系若为右手，则像的坐标关系为左手。
4、入瞳孔径越大，则景深越小；对准平面距离越远，则景深越大。
5、照相机中的可变光阑起的是孔径光阑的作用，它限制的是进入系统的光的能量。
6、单色像差有（写出其中的三个）球差, 慧差, 像散, 场曲。
7、望远镜分辨率以能分辨的两物点对望远镜的张角表示；照相系统的分辨率以11f fΦ==-'像平面上每毫米能分辨的伴对数N 表示。
8、光波的空间周期是指 光波波长，空间频率是指波长的倒数。
9、干涉条纹的可见度定义为K = (IM--Im) / (IM+Im)，当两束相干光振幅相同时，条纹可见度K = 1 。
10、如图所示，用波长为的单色光垂直照射折射率为的劈尖薄膜，图中各部分折射率的关系为321n n n <<。
工程光学考试题及答案高中
工程光学考试题及答案高中一、选择题1. 光学成像的基本条件是什么？A. 物体必须在焦点以内B. 物体必须在焦点以外C. 物体必须在焦点上D. 物体必须在焦距以内答案：B2. 凸透镜成像的规律是什么？A. 物远像近像变小B. 物近像远像变大C. 物远像远像变大D. 物近像近像变小答案：B3. 以下哪种情况不属于光的折射现象？A. 光从空气进入水中B. 光从水中进入空气中C. 光从玻璃进入空气中D. 光从空气直接传播答案：D4. 光的三原色是什么？A. 红、绿、蓝B. 红、黄、蓝C. 红、橙、绿D. 蓝、绿、紫答案：A5. 以下哪种光学仪器是利用光的反射原理制成的？A. 望远镜B. 放大镜C. 显微镜D. 潜望镜答案：D二、填空题6. 凸透镜的焦距越短，其成像能力越________。
答案：强7. 当物体位于凸透镜的焦点上时，成像情况是________。
答案：不成像8. 光的折射定律中，入射角和折射角的关系是________。
答案：入射角越大，折射角越大9. 光的干涉现象是指两个或两个以上的________相互叠加的现象。
答案：光波10. 光的衍射现象是指光绕过障碍物继续传播的现象，这种现象说明了光具有________。
答案：波动性三、简答题11. 请简述光的干涉条件。
答案：光的干涉条件包括：光波的频率相同、光波的相位差恒定、光波的振动方向相同。
12. 什么是全反射现象？请简述其产生条件。
答案：全反射现象是指当光从光密介质射向光疏介质，且入射角大于临界角时，光将完全反射回光密介质中。
产生全反射的条件包括：光从光密介质进入光疏介质，入射角大于临界角。
四、计算题13. 已知凸透镜的焦距为10cm，物体距离透镜15cm，求像的性质和位置。
答案：根据凸透镜成像公式1/f = 1/u + 1/v，其中f为焦距，u为物距，v为像距。
代入数据得1/10 = 1/15 + 1/v，解得v = 30cm。
由于物距大于焦距，像距大于物距，所以成像为倒立、放大的实像。
韩军《工程光学》课后习题答案之第4章
注意：
切记不要思维固化，不要死记硬背X、Y、Z轴，拿到一个新的题目时要 明确各个坐标轴的属性，明确到底哪个坐标轴是入射光轴，哪个坐标轴是垂 直于主截面的轴，哪个坐标轴是平行于主截面的轴。
已知FF'＝1mm，所以
f22 = 1
h 2 =
f
'
- = 2
5 rad 二 0 . 0 0 0
中 书中52⻚公式（4-17） ,
n=
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判断步骤：
1、O'X'：与光轴出射方向一致。 2、O'Z': 看屋脊面个数，垂直纸面向里、向外奇变偶不变。 3、O'Y': 先看物方坐标系是左手还是右手，再看反射次数（其中
韩军《工程光学》课后习题答案 ——几何光学部分
赛先生卢仕强 ⻄安工业大学 光电工程学院 2018届毕业生
2020.03.11
第4章
根据平面镜成像性质可画出光路简图如下：
头顶
i眼睛
镜子
脚
由图可知镜子的高度为人身高的一半，和人离镜子前 后距离无关。
中 , 根据题意可画出光路示意图如下：
f -
== 由图示有： f!22
应用光学第七章 习题-英文习题
Chapter SevenClassical Optical System补充：NA is invariant across plane refractive surface, whether measured in air or in the cover glass. Why? (Think about the characters of PPP)1. A man has a diopter of -2D, The accommodation range is 8D.1) Where is the far point? (Try to find r l ) 2) What about near point? (Try to findp l )3) If he wears a pair of spectacles with -1D, how large is 'f ? 4) What ’s the far point distance from the eye? 5) What ’s the near point distance from the eye? Solution:1) m l r 2.0-= (reciprocal value of r l ) 2) m l p 1.0-= 3)m f 1'-=4)、5)⎪⎩⎪⎨⎧-=-=-==m m l m l l p r 11.09111 2. A magnifier withmmf 25'=, entrance pupil diametermm D 18=. The eye separates from the magnifier by mm 50, Theimage has a distinct distance of mm 250, vignetting coefficientis%50=K .Try to find visual magnifying power , y 2andposition of the object. Solution:1) ⨯=9r2) mm y 102=3) mml 2.22-3. The objective of a microscope has the magnifying power⨯-=3β, numerical aperture1.0=NA , conjugate distancemm L 180=.The objective is served as the aperture stop. Focallength of eyepiece is mm f e 25'=. Find:(1) Visual magnifying power of the microscope. (2) Diameter of exit pupil. (3) Eye relief.(4) What ’s the resolution of microscope when illuminatedobliquely withm μλ55.0=.(5) Entrance pupil diameter of objective.(6) If mm y 62=, %50=K . Find the diameter of eyepiece Solution: (1)⨯=Γ30(2) mm D Xp 67.1=(3) eye relief is mm l 6.29'=(4) mm 00275.0=σ (5) mm D Ep 9= (6)mm D e 33.21=4. To resolve a small object with the size ofmm 000725.0, andilluminate it obliquely with mm 00055.0=λ. Try to find : (1) The minimum visual magnifying power of microscope. (2) What ’s the NA? Solution:38.0=NA Γ≤NA 500 ∴ ⨯==Γ190500min NA5．A biological microscope has NA=0.5, 2y=0.4mm, the area of illumination filament is 1.2×1.2mm 2,The distance from filament to the object plane is 100mm.If critical illumination is adopted. What’s the focal length of condens er and its diameter? Solution:1) F’= 18.75mm2) D=25mm6.To observe two points separated by 150mm which are 4km away (set 1ˊ=0.0003rad), if Kaplerian telescope is adopted , try to answer the following questions. 1) Effective magnifying power of Kaplerian telescope. 2) If tube length L=100mm,what’s the0f 'and e f '?3) If objective is served as the aperture stop, where is the eye relief?4) To meet the requirement of working magnifying power, findEPD .5) The accommodation range of diopter is 5D ±,whatis theshifting quantity of the eyepiece?6) If FOV in object space is 28ω=︒, what’s the FOV in image space?7) If vignetting coefficient K=50％,What’s the diameter of eyepiece? Solution: 1)0.000380.0000375T ⨯==2) 011.1,88.9e L f mm f mm ''⇒===. 3) eye relief: 12.5mm 4) D=18.4mm5) + 0.62mm , -0.62mm6) 58.4 7) D e =147.A kaplerian telescope has0200f mm '=,25e f mm '=.FOV inobject space is 028ω=, vignetting coefficient K=50％, Byusing a field lens in the back focal plane of objective, the diameter of eye piece is shortened to 23.7mm. 1) What’s the focal length of field lens?2) If i t’s a thin plano -convex lens (with plane in front), refraction index n=1.5. Try to find the radius of surface. Solution: 1)89.7f mm '=or 90mm2) 1211.5,,4590n r r mm ϕ==∞=⇒=-8.A film projector with the film size 22×16mm 2 ，adopt a carbon lamp as light source. The illuminance of screen is required to be 100lx, the distance from projector to screen is 50m, Width of screen is 7m,(Lightness of carbon lamp is L=1.5×108cd/m2). What the focal length of projector, relative aperture, and FOV? Solution: 1) f ’=157mm2) 1: 3.5 3) 169. An illuminating apparatus is composed of a bulb and a condenser. Focal length of condenser is400f mm '=, D=200mm,To illuminate a circle diameter 3m away from the illuminator 5m, where should the bulb be placed? Solution:188.7mm10) A liquid crystal television has the diagonal size 3 inch （76.2mm ）,The maximum diagonal size of large screen is 100 inch(2540mm), the projection distance is 3100mm,If the size of projected screen can be adjusted continuously from 45~90 inch(1143~2286mm).Try to find scope of the focal length and FOV . Solution: y=76.2mm1111111110011111433100206.772.6111193.81/211430.1844,10.45,220.93100l l mml l f mm l l f tg βωωω'=-==⇒=-'-=⇒=''⨯'''====2)2222222222863100103.376.21111001228620.36873100l mm l f mm l l f tg βω=-=⇒=-'-=⇒=''⨯'==0220.24ω'=02240.48ω'=11.A TV camera is used to monitor targets in sky, set lightness of target is 2500cd/m 2, the transmittance of optical system is 0.6, If we need to obtain illuminance of 20lx on video tube surface, how large aperture/iris should be adjusted ? Solution:21()4/#7.6720,0.6,2500D E L f f F DE lx L τπτ⎫'='⎪'⇒==⎬⎪'===⎭Take F/#=812. Design a laser beam expander with beam expanding ratio 10⨯, The tube length is 220mm. 1) What’s the focal length1f 'and 2f '?2) What kind of aberration of the laser beam expander should be corrected?3) If adopt two thin lens to compose a beam expander, what’sthe radius (set n=1.6, 23r r ==∞).Solution:1)1220200f mmf mm '='=2) Without off-axis FOV and chromatic aberration ⇒Spherical aberration should be corrected well. 3) r 1=12mm r 2=infinity r 3=infinityr 4=-120mm13. A stereo rangefinder has the base line of 1m. Relative error is less than 1℅ when object is 4km away . how large visual magnifying power of the rangefinder should be? Solution:2014．A kaplerian telescope has the tube length 225mm,8,26,'5D mm ω⨯Γ=-==o ,without vignetting, Find:(1)''0?,?e f f ==(2) The aperture of eyepiece and eye relief.(3) A field lens is put at the back focal plane of objective withfocal length '75f mm =.what’s the new aperture of eyepiece andeye relief?(4) The accommodation range of eyepiece is ±4D. what’s the shifting quantity? Solution: (1)''0'''00'225200,258e e e f f f mm f mmf f ⎫+=⎪⇒==⎬Γ=-=-⎪⎭(2)''111,225,25'28.13'e el mm f mm l mm l l f -==-=⇒= eye relief8,405EP EPEP XP D D D mm D ⨯Γ=-=--='0'20082252.5,225tan 311.792()2(11.79 2.5)28.58EP eeye f D a f a mm b mm D a b mm=======⨯+=⨯+=o(3)'200tan 310.48y mm ==oFor field lens,'111'l l f -=200,'75'120l f l mm =-=⇒=So the distance from the image of field lens to the eyepiece is'2595l mm -=. For eyepiece'111,95,'25'19.8'l mm f mm l mm l l f -===⇒=(eye relief)For L1, 20381.62tan 30l mm ==oApplying Gaussian equation, find200'131.23200ll mml ==+200'68.77x l mm =-=For field lens ,68.77,'75l x mm f mm =-=-='111'827.9'l mm l l f -=⇒=-Similar triangle :10.48827.921./2852.9e e D mm D ==⇒= (4) '2±4 2.51000e f x mm mm ==±15. A rangfinder of model 59-1 has the base line 3m ,the EP diameter of telescope' 1.6,32D mm ⨯=Γ=.Try to find:（1）the diffraction resolution and visual resolution.（2）To resolve a target of 0.1mm,what ’s the maximumdistance from the target to the rangfinder?(3) Could we increase visual magnifying power Γto resolve smaller detail? Why?Solution:(1)Diffraction resolution:120''120''120'' 2.34'''51.2D D ϕ====Γ Visual resolution:60''60'' 1.875''32ϕϕ⨯⋅Γ=⇒== (2)0.1 2.34''0.0000113448.815L L m ϕ==== (3) NO.To see more detail, we need to increase the diameter of objective D. 120''D ϕ=16. An 80mm focal length thin lens is used to image an object with magnification of -1/2, The lens diameter is 25mm and a stop of diameter 20mm is located 40mm in front of the lens. How big is the field of view without uvignetting ? How big is the field of view with fully vignetting ?Solution:'1/2,2'1111''80l l l l l l f β⎫==-=-⎪⎪⇒⎬⎪-==⎪⎭240'120l mm l mm =-=For unvignetting condition:/2/24020,25160/2201605200/240st st st D x D x D D x mmD x y mm x y y =+==⇒====⇒=-For fully vignetted condition:/24/2405160/9/240245/2200st D x D x x mmD x y mmy x ==-=-=⇒=+补充：1．what functions is F number? write down the equation of F number.答：控制像面的照度、系统的分辨率 和焦深（景深）。
《光学工程补充》.(DOC)
2、一物体经针孔相机在屏上成一60mm 大小的像，若将屏拉远50mm，则像的大小变为70mm,求屏到针孔的初始距离。
解：在同种均匀介质空间中光线直线传播，如果选定经过节点的光线则方向不变，令屏到针孔的初始距离为x，则可以根据三角形相似得出：所以x=300mm即屏到针孔的初始距离为300mm。
3、一厚度为200mm 的平行平板玻璃（设n=1.5），下面放一直径为1mm 的金属片。
若在玻璃板上盖一圆形纸片，要求在玻璃板上方任何方向上都看不到该金属片，问纸片最小直径应为多少？解：令纸片最小半径为x,则根据全反射原理，光束由玻璃射向空气中时满足入射角度大于或等于全反射临界角时均会发生全反射，而这里正是由于这个原因导致在玻璃板上方看不到金属片。
而全反射临界角求取方法为：(1)其中n2=1, n1=1.5,同时根据几何关系，利用平板厚度和纸片以及金属片的半径得到全反射临界角的计算方法为：(2)联立（1）式和（2）式可以求出纸片最小直径x=179.385mm，所以纸片最小直径为358.77mm。
4、光纤芯的折射率为n1、包层的折射率为n2,光纤所在介质的折射率为n0，求光纤的数值孔径（即n0sinI1,其中I1 为光在光纤内能以全反射方式传播时在入射端面的最大入射角）。
解：位于光纤入射端面，满足由空气入射到光纤芯中，应用折射定律则有：n0sinI1=n2 sinI2 (1)而当光束由光纤芯入射到包层的时候满足全反射，使得光束可以在光纤内传播，则有：(2)由（1）式和（2）式联立得到n0 .2、已知一个透镜把物体放大-3x 投影在屏幕上，当透镜向物体移近18mm 时，物体将被放大-4x 试求透镜的焦距，并用图解法校核之。
解：3．一个薄透镜对某一物体成实像，放大率为-1x,今以另一个薄透镜紧贴在第一个透镜上，则见像向透镜方向移动20mm，放大率为原先的3/4 倍，求两块透镜的焦距为多少？解：4．有一正薄透镜对某一物成倒立的实像，像高为物高的一半，今将物面向透镜移近100mm，则所得像与物同大小，求该正透镜组的焦距。
工程光学英文版课后练习题含答案
工程光学英文版课后练习题含答案IntroductionEngineering Optics is a branch of optics that studies the application of optical principles and devices to solve engineering problems, including optical design, imaging systems, and measurement techniques. As an important part of Engineering Optics, the homework exercises help students understand the theoretical knowledge and familiarize themselves with practical problems. In this document, we provide a set of homework exercises with answers for Engineering Optics, which are designed to help students review the knowledge they learned in class and prepare for exams.Chapter 1: Introduction1.What is the definition of light?–Light is an electromagnetic wave that travels through space and has both electric and magneticcomponents perpendicular to each other and to thedirection of propagation.2.What are the primary properties of light?–The primary properties of light include reflection, refraction, diffraction, interference,and polarization.3.What is the difference between coherent andincoherent light?–Coherent light is light that has a constant phase relationship between two or more waves, whileincoherent light is light that has a random phaserelationship between two or more waves.4.What is the difference between monochromatic andpolychromatic light?–Monochromatic light consists of a single wavelength, while polychromatic light consists ofmultiple wavelengths.5.Define dispersion.–Dispersion is the phenomenon of different wavelengths of light traveling at different speedsthrough a medium, leading to a separation of thecolors of light.Chapter 2: Geometrical Optics1.Define ray and expln how rays are used ingeometrical optics.–A ray is an idealized model of the path that light travels through space, represented as a linewith an arrow indicating the direction ofpropagation. Rays are used in geometrical optics to determine the behavior of light as it passesthrough lenses, mirrors, and other optical devices.2.Define optical axis and principal plane.–The optical axis is the imaginary line passing through the center of curvature of a sphericallysymmetric optical system. The principal plane isthe plane perpendicular to the optical axis thatpasses through the focal point of the system.3.Define focal length and expln how it relates to the curvature of a lens.–The focal length is the distance between the center of curvature of a lens and the point whereparallel rays of light converge after passingthrough the lens. The curvature of a lensdetermines its focal length.4.Define the focal plane and expln how it relates to the focal length.–The focal plane is the plane perpendicular to the optical axis that passes through the focalpoint of a lens or mirror. The distance from thelens or mirror to the focal plane is equal to thefocal length.5.Expln the concept of conjugate planes.–Conjugate planes are prs of object and image planes that are related by an optical system suchthat an object in one plane is imaged onto theother plane. The distance between the two planes isequal to the sum of the object distance and imagedistance.Chapter 3: Optical Instruments1.Define the resolving power of an optical system.–The resolving power of an optical system is its ability to distinguish two closely spacedobjects as separate entities. It is determined bythe numerical aperture and wavelength of the lightused in the system.2.Define the magnification of an optical system.–The magnification of an optical system is the ratio of the size of the image produced by thesystem to the size of the object being imaged.3.What is a camera and how does it work?–A camera is an optical instrument that uses a lens to focus an image onto a light-sensitivesurface, such as film or a digital sensor. Theimage is formed by the interaction of light withthe surface, creating a chemical or electronicpattern that can be developed into a visible image.4.What is a microscope and how does it work?–A microscope is an optical instrument that uses a lens or a series of lenses to magnify small objects that cannot be seen with the naked eye. The specimen is placed on a stage and illuminated witha light source, and the image is formed by lensesthat focus the light onto the observer’s eye or a camera sensor.5.What is a telescope and how does it work?–A telescope is an optical instrument that usesa lens or a mirror or a combination of both tocollect and focus light from distant objects, such as stars, galaxies, or planets. The image is formed by lenses that magnify the light and focus it onto the observer’s eye or a camera sensor.ConclusionIn conclusion, the homework exercises and answers provided in this document are intended to help students review key concepts and prepare for exams in Engineering Optics. By solving these problems, students can deepen their understanding of optical principles and devices and develop their problem-solving skills. We hope that this resource will be useful for students and instructors alike in the study of Engineering Optics.。
工程学科英语复习
Unit 1Passage 11. 科学知识scientific knowledge2. 实践知识practical knowledge3. 应用科学applied science4. 科学原理scientific principle5. 操作条件operating condition6. 预期功能intended function7. 机械原理mechanical principle8. 民用建筑civilian structure9. 技术学科technical discipline10. 分支学科sub-discipline11. 基础设施infrastructure12. 制造工程manufacturing engineeringTask 11. Tissue engineering has been a newly developed _________ which represents the new direction of biological medicine engineering.disciplinea branch of knowledge组织工程学是一门新兴的学科，它代表了生物医学工程领域发展的新方向。
2. What sort of preferential policies can foreign investors in software and integrated_________ industry enjoy?circuitan electrical device that provides a path for electrical current to flow外商投资软件产业和集成电路产业能享受何种优惠政策？3. NC machine tools are difficult to _________ and often result in heavy economic losses when they go wrong due to the high prices.maintainkeep in a certain state数控机床由于价格昂贵，一旦出现故障，维修困难，常带来较大的经济损失。
应用光学第六章-英文习题
第六章1.Design an aplanatic lens with r 1=-95mm, object is at the center of first surface, second spherical surface satisfies aplanatic condition. If the lens thickness d=5mm,index n=1.5,and lens is in air, find: 1) the second radius of the lens, 2) the lens magnification solution:(1)first surface: 5.11'111==n n β；second surface : 22222/)'(n r n n L +=；'/)'('22222n r n n L +=1',5.1,100'2212==-=-=n n mm d L Lmm r 602-= （2 21βββ=222222222)15.1()'(''===n n L n L n β5.1=β2.what does mean by aplanatic imaging? what does mean by isoplanatic imaging? If an object is in the vertical plane cross the each of tree aplanatic points, can the refractive surface forms a perfect image of the object?3.if the primary tangential coma of an optical system is equalto width of focus )''/(u n λ,what is ∑II SSolution:II T S u n K ∑-=''23' ∑II S =32λ-4. if the primary spherical aberration of an optical systemis equal to depth of focus)''/(2u n λ,what is ∑I S ? Solution:∑-=ISu n L 2''21'（初）δ ∑I S =λ2-5.if an object is at the vertex of the lens, second surface satisfies aplanatic condition, set lens index n=1.5，thickness d=4,find the angular magnification and r 2?Solution: 11=β；22222/)'(n r n n L +=；'/)'('22222n r n n L +=1',5.1,4'2212==-=-=n n mm d L Lmm r 4.22-=21βββ=222222222)15.1()'(''===n n L n L n β 25.2=β；44.01==βγ6.lHow many spherical aberration free points are there in a spherical mirror?7.design a achromatic telescope with cemented doublet,'f =100mm ， K9(D n =1.5163，D ν=64.1)and F2(D n =1.6128，D ν=36.9)，I if positive lens 21r r =，find(1)the focal length of positive and negative lenses,(2)tree radius. Solution:(1) 0//221121=Φ+ΦΦ+Φ=Φνν mm f mm f 7.73';4.42'21-==（2）))(1())(1(32222111ρρρρ--=Φ--=Φn n 21r r =，6128.1,5163.121==n nmm r mm r r 3.1429,78.43321-==-=8. in fig.6-22, point out(1)marginal spherical aberration'mL δ =?0 (2) 0.707 zone spherical aberration '707.0L δ=?-0.03 (3) marginal chromatic aberration 'FCL ∆ =?0.06 (4) paraxial chromatic aberration'FCl ∆=?-0.06 (5) 0.707 zone chromatic aberration '707.0FC L ∆=?(6)chromatic aberration 'FCL δ=?0.12 (7)secondary spectrum 'FCDL ∆=?0.079.fesign a cemented doublet lens with 'f =100mm ，the glasses are K9(D n =1.5163, D ν=64.1) and F2(D n =1.6725, D ν=32.2)，in order to compensate chromatic aberration of other element, the lens should reserve primary longitudinal chromatic aberration'FCl ∆=-0.26mm, find(1)the power of positive and negative lenses(2)How large is the secondary spectrum of the lens.solution ：（1）∑Φ-=∆kkk FCh u n l 122''1'ν)('4'26.0212122νννΦ+Φ-=Φ-=-∑f u h kk21Φ+Φ=Φmmf 83.50'/065.011222211=⇒-Φ-Φ=Φνννννmmf 38.103'/065.021212211-=⇒-Φ-Φ-=Φννννν（2）mm f L FCD 052.0'00052.0'==∆10.a cemented doublet lens with 'f =100mm ，D/f,=1∶5，if the lens has only primary aberration, the aberration coefficient of marginal ray are respectively1 2 3I S ： 0.0100104 -0.0338527 0.0256263find ：1)How large is primary spherical aberration of the lens?2) if primary spherical aberration is less than 4 times of depth of focus, does the aberration satisfy the tolerance of the lens?3) How large is the secondary spectrum of the lens? solution ：1） 1.0''21('312'arctg u Su n L kIkk =-=∑初）δmm L 08979.0('=初）δ2)mm u n m223479.0''42=λ﹥ 0.08979mm, satisfy the tolerance of the lens3) mm f L FCD 052.0'00052.0'==∆11.what called Petzval field curvature? What method can be used to correct field curvature? If the aberration of astigmatism is corrected, does the field curvature is corrected simultaneously?solution ：for thin lens system, with separated positive andnegative lens, make∑=0n φ；or meniscus thick lens ，if astigmatism is corrected , the field curvature is still present 。
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Keywords and concept
1. Magnification
Transverse magnification (Lateral magnification) is defined as the ratio of image size, y', to object size, y.
Axial magnification (longitudinal magnification) is defined as the ratio of a short length, or depth, in the image measured along the axis, to the conjugate length in the object. In the air, it is equal to the square of the transverse magnification. Angular magnification (magnifying power)
The angle made by the incident ray, that made by the refracted ray, and the surface normal at
the point of incidence in a refractive process
obey the expression nsinI=n'sinI' The incident ray, the refracted ray, and the surface normal are all coplanar.
4
5. Total Internal Reflection
normal
Incident ray Denser medium Refracted ray Rarer medium
Surface Critical angle
6. Fermat’ principle
Light takes the path that requires the least time.
Example: Shadows
2
3. The Law of Reflection
A light ray striking a reflecting surface and reflected from it makes equal angles with the
normal to the surface at the point of incidence,
8
Some Concept
1. Vertex is the point where the optic axis intersects the surface. 2. Object distance is the distance from vertex to the object. 3. Image distance is the distance from vertex to the image. 4. Center of curvature is the center of single spherical surface. 5. Meridional plane is a plane containing the optic axis and the object.
10
Some Keywords and concept
1. Principal planes are defined as the loci where
refraction is assumed to occur without
reference as to where it actually does occur.
5
7. Key words and concept Image space extends from infinity on one side to infinity on the other. The same holds for the object space; both completely overlap. Whether a given point is in the object space or the image space depends on whether it is part of a ray before or after refraction. If a given point is in the object space, it is a point object, otherwise it is a point image.
11
4. focal length is the distance from the principal
point to the focal point. 5. A on-axial point object at infinity and the second focal point are conjugate.
English Summary
1. The electromagnetic spectrum
short Wave length long Radio -waves
Gamma X- Ultra V Infra- Micro red -waves Rays Rays violet
380nm 780nm
violet-blue deep-red
The total magnification of a combination of lenses is the product of the magnification of the individual lenses.
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Keywords and concept
2. Nodal points are where no refraction occurs. Whenever the refractive indices on either side of the lens are the same, the nodal points coincide with the principal points. If the refractive indices on the two sides of the lens are different, the N points would move away from the principal planes, toward the side of higher index.
12
Homework ( P18-Example)
13
Some Keywords and concept
1. Graphical construction
Parallel ray Chief ray F Focal ray H H’ F’
The parallel ray is first parallel to the axis and then, after refraction passes through F’. The focal ray passes through F and then is rendered parallel to the axis.
I''=-I
and the incoming ray, the outgoing ray, and the
normal to the surface at the point of intersection
all lie in the same plane.
3
4. The Law of Refraction
7
Some Keywords
1. single spherical surface 2. converge and divergence 3. conjugate (~ points, ~ planes, ~ distance etc) 4. magnification (transverse ~, axial ~, angular ~) 5. Lagrange’s invariant 6. sign convention (vertex, image distance, ) 7. image characteristics (larger or smaller, upright or inverted, real or virtual) 8. Concave and convex mirror
1
Байду номын сангаас
2. Law of Rectilinear propagation
Light as it advances through free space or
through homogeneous, isotropic matter
follows a beeline. (a direct straight course).
The chief ray goes through the principal point without deviation.
14
Some Keywords and concept
2. Newton’s lens equation (x, x' are called Newton’s extrafocal object and image distances, respectively )
9
Homework
1. A goldfish swims 10cm from the side of a
spherical bowl of water of radius 20cm. Where does the fish appear to be? Does it appear larger or smaller? 2. An object is located 2cm to the left of convex end of a glass rod which has a radius of curvature of 1cm. The index of refraction of the glass is n=1.5. Find the image distance. P13-12&14