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1、Contents in this chapterThe imaging properties and features of the mirror and prism systemsThe imaging directions of prism systemsThe conjunction of mirror and prism systems 第1页/共108页第一页，编辑于星期五：十七点 四十七分。4.1 Applications of Mirror and Prism Systems in Optical InstrumentsProperties of coaxial spherica

2、l systems:Advantages: a) Be able to form the image of an object as required b) Images in paraxial region are perfect c) While the object plane is normal to the axis, the image plane will be also normal to the axis, and the image is similar to the object.Defects: Can not change directions object, opt

3、ical system, image locate on a common line第2页/共108页第二页，编辑于星期五：十七点 四十七分。Applications of the mirror and prism systems:a) They can be used to reduce the dimension and weight of an instrument.b) They can be used to change the direction of the image, or make the image inverted. c) They can be used to cha

4、nge the location and direction of the axis, producing a certain periscope height or tilting the axis for an angle. d) With the help of rotating the prism and mirror, the direction of the axis can be changed continuously, enlarging the field of view.第3页/共108页第三页，编辑于星期五：十七点 四十七分。ExamlesExamles第4页/共108

5、页第四页，编辑于星期五：十七点 四十七分。ExamplesExamples名称特性图例平面反射镜具有折转光路的作用，可用于成像、激光和全息系统。曲面反射镜表面可为球面或非球面。具有聚焦和会聚光的作用。可用在光的发射和接受、激光和光纤系统中，也可以和其它反射镜组合形成成像系统。柱面反射镜具有单一方向聚光和聚焦的作用。可用来产生线光源，或单方向扩束和放大图像。第5页/共108页第五页，编辑于星期五：十七点 四十七分。4.2 Imaging properties of mirrorsPAONBDAIIOBpoint A is imaged at point AWe can take incident

6、 ray AO from point O arbitrarily;DAADODAAODODODIADOIAOD9090， The incident AO is arbitrary, the equation has no relationship with the location of point O. All of the extending rays of the reflected ray from point A will meet at one common point A. 1.Imaging of an arbitrary point through a single mirr

7、or第6页/共108页第六页，编辑于星期五：十七点 四十七分。 PODOAAConclusion： a) The image and object points are symmetrical to the mirror. b) The height of the image is equal to that of the object. c) A real object forms a virtual image ，and a virtual object forms a real image.d) A single plane mirror can form ideal images .

8、Is the image similar to the object?第7页/共108页第七页，编辑于星期五：十七点 四十七分。2 Image of object in space reflected by a single mirrorPxyzo y z x oa) The size of the image is equal to that of the object but in different orientation.b) The right hand coordinates in object space will be changed to left hand coordina

9、tes in image space .c) Facing directly to axes z and z separately, when x rotates anticlockwise we can see that x will rotate clockwise. The image satisfying this relationship between the object and image spaces is called mirror image. 第8页/共108页第八页，编辑于星期五：十七点 四十七分。Summery:a) A single plane mirror ca

10、n image any object points ideally in the whole space.b) The image and object points are symmetrical to the mirror. c) The size of the image is equal to that of the object.d) Forming a mirror image, the image formed by a plane mirror is not similar to the object. 第9页/共108页第九页，编辑于星期五：十七点 四十七分。3 Imagin

11、g properties of mirrors systemlForming ideal imageslAn image formed by odd mirrors will be a mirror image and by even mirrors will be one same as the object. Attentions：1. The erected or inverted image has no relationship with the mirror image；2. The object and the mirror image have different shape,

12、 and they are not similar. Generally speaking，we always expect the image is similar to the object, especially in the systems used in military.第10页/共108页第十页，编辑于星期五：十七点 四十七分。4.3 Rotation of MirrorsPNOABIINIBIConclusion：The reflected angle will be 2 if the mirror rotates 2(I+)-2I=21. Rotation of single

13、 mirror第11页/共108页第十一页，编辑于星期五：十七点 四十七分。Advantage：Enlarging the range observedDefect：Errors caused by rotationExample：The plane mirror in Range Finder1. Rotation of single mirror第12页/共108页第十二页，编辑于星期五：十七点 四十七分。2. Rotation of two mirrorsP1P2AO1O2MBI I1I I2 2）（2121222IIII21NOO22121IIIIIn triangle O1O2M，N

14、 the normals of two mirrors meet at point N , In triangleO1O2N，according to the principle of external angle, The angle between the incident and the reflected rays is twice of the angle of two mirrors. 第13页/共108页第十三页，编辑于星期五：十七点 四十七分。2. Rotation of two mirrorsP1P2AO1O2MBI1I2 The rotation direction wil

15、l coincide with that of rotation from P1 to P2 according to the order of reflections on two mirrors.N The rotated angle of the emergent ray will be equal to twice angle of the two mirrors, regardless of the incident rays direction.Application: We can use two mirrors in place of the single mirror in

16、Range Finder.angle mirror，prism.第14页/共108页第十四页，编辑于星期五：十七点 四十七分。第15页/共108页第十五页，编辑于星期五：十七点 四十七分。4.4 Prism and Its Unfolding 1. Advantages and defects of prisms Prism: Optical element which can make use of the reflection in glass to change the direction of raysAdvantages ：The loss of energy is small. I

17、ts hard to break. It is easy to assemble and fix.Defects： Its volume and weight are larger. It has strict requirements for material. It is influenced by circumstance greatly.第16页/共108页第十六页，编辑于星期五：十七点 四十七分。2. Unfolding of the prismmethod of studying the imaging properties of prismsRight-angle Prism M

18、ain section :The plane or section which is perpendicular to each prism. 第17页/共108页第十七页，编辑于星期五：十七点 四十七分。 The method, unfolding the main section of prism along the reflective surface and canceling the reflection and replacing prisms refractions by glass blocks refractions, is called prism unfolding. 1

19、23From Law of reflection, we can get: From the symmetricalrelationship, we can get: So, we can get:第18页/共108页第十八页，编辑于星期五：十七点 四十七分。3.The requirements for prisms(1) After unfolding a prism, the two faces of the glass block must be parallel to each other.第19页/共108页第十九页，编辑于星期五：十七点 四十七分。(2) When a prism

20、locates in converging rays, the axis must be perpendicular to both incident and emergent surfaces. 第20页/共108页第二十页，编辑于星期五：十七点 四十七分。4. Typical examples of unfolding prismsa. Right-angle PrismLocating in parallel raysABCAWhen the prism works in parallel rays the two faces of the glass block unfolded by

21、 the prism must be parallel to each other , then, AB/ACAnd then, 第21页/共108页第二十一页，编辑于星期五：十七点 四十七分。4. Typical examples of unfolding prismsa. Right-angle PrismLocating in parallel raysABCA Face AB needs to be parallel to face AC，then ABC=ACB So the prism needs ABC=ACBThat means triangle ABC must be an

22、isosceles triangle but B and C do not need to equal 45，or A does not need to be a right angle. 第22页/共108页第二十二页，编辑于星期五：十七点 四十七分。When a prism locates in converging rays, both the first and the second surfaces should be perpendicular to the axis.ABCAWhen the axis of this prism is deviated through axis

23、90：第23页/共108页第二十三页，编辑于星期五：十七点 四十七分。When deviating the axis through any angle2BCA290CB If we want the ray to deviate , then the reflective surface must deviate the axis2This kind of prism is called Isosceles prism. If we need the axis of this prism be deviated 45,then,第24页/共108页第二十四页，编辑于星期五：十七点 四十七分。

24、2.Penta Prism第25页/共108页第二十五页，编辑于星期五：十七点 四十七分。45第26页/共108页第二十六页，编辑于星期五：十七点 四十七分。3.Boot Prism45o60第27页/共108页第二十七页，编辑于星期五：十七点 四十七分。4. Cube PrismABCIIEaDSuppose the index of the prism is n，then the refraction angle I isFrom the figure we can see that the diameter D of the rays is： 第28页/共108页第二十八页，编辑于星期五

25、：十七点 四十七分。4. Cube PrismABCIIEaDwe get:If the glass is K9(or BK7)，then n=1.5163，we can get:第29页/共108页第二十九页，编辑于星期五：十七点 四十七分。 In order to enlarge the diameter of the rays, or to reduce the size of the Dove prism for a given diameter of the rays, two Dove prisms can be cemented hypotenuse to hypotenuse

26、which can form the Cube prism .第30页/共108页第三十页，编辑于星期五：十七点 四十七分。Attentions to use Cube PrismlA bundle of rays will be divided into two bundles of rays after entering the prism, then they will be merged together to one bundle of ray after passing through the prism. So the hypotenuses of the two Dove pr

27、isms must be parallel to each other precisely to avoid producing two separated images.lIf the entrance pupil of the rays is circular, the exit pupil of the rays will be divides into two reversed half circles. So the cube prism cannot work in circular rays.lSince the incident and emergent faces are n

28、ot normal to the axis ,the Cube prism can only be used in parallel light rays.第31页/共108页第三十一页，编辑于星期五：十七点 四十七分。11223443第32页/共108页第三十二页，编辑于星期五：十七点 四十七分。4.5 Roof Surfaces and Roof prismsRoof surfaces：Using two right angle surfaces to replace one reflecting surface.Roof prism: Prism which contains foor

29、surfaces.第33页/共108页第三十三页，编辑于星期五：十七点 四十七分。 Roof Surfaces and Roof prismsEffects：The addition of the roof to a prism is to introduce an extra inversion to the image or change the total reflecting number from odd to even, keeping the original axis and image orientation in the main section unchanged. In

30、 this way we can add a reflection and get an image similar to the object.第34页/共108页第三十四页，编辑于星期五：十七点 四十七分。yxz x1 y1 z1yzx x2 y2 z2第35页/共108页第三十五页，编辑于星期五：十七点 四十七分。第36页/共108页第三十六页，编辑于星期五：十七点 四十七分。Requirement for a roof prism: The roof angle must be equal to 90 precisely, otherwise, the emergent rays wi

31、ll be not parallel, producing double images 第37页/共108页第三十七页，编辑于星期五：十七点 四十七分。unfolding of roof prisms第38页/共108页第三十八页，编辑于星期五：十七点 四十七分。4.6 Imaging Property of Parallel Glass Block and Prism Size Calculation 1. Imaging properties of the parallel glass block (1) The final image position AA第39页/共108页第三十九页

32、，编辑于星期五：十七点 四十七分。AALl1 l2Suppose the object distance of the first surface is L1， is the image distance of the second surface, calculate Calculate a ray by using the Gaussian equation for the two surfaces of the parallel glass block one by one:Substitute to the above equation, we get：So the shift of

33、image plane is:第40页/共108页第四十页，编辑于星期五：十七点 四十七分。2. The size of imageALl1 l2 uuConclusion: A parallel glass block only makes the image plane shift a certain distance, having no influence on the imaging property. The shifting value is L-L/n. The emergent ray will be parallel to the incident ray when it

34、passes through a parallel glass block. So, we can get ,In the air,And then,第41页/共108页第四十一页，编辑于星期五：十七点 四十七分。2. The equivalent air thickness of A parallel glass blockALl1 l2P2P1KQKP2=AA=L-L/nKP2=QM=L-L/nNQ=L/nMNL/nAFrom the figure, AQ=l1-L/n=l2=AM The heights of the ray at the two surfaces of the para

35、llel glass block are equal to those of the ray at the two surfaces of the equivalent air thickness. For a parallel glass block whose thickness is L and index is n, L/n is called the equivalent air thickness .第42页/共108页第四十二页，编辑于星期五：十七点 四十七分。lThe distance from the second surface to the image plane is

36、equal to that of by passing through an equivalent air thickness;l The heights of the ray at the two surfaces of the parallel glass block are equal to those of the ray at the two surfaces of the equivalent air thickness ;lThe size of the image formed by the parallel glass block is equal to that forme

37、d by the equivalent air thickness. The equivalences :第43页/共108页第四十三页，编辑于星期五：十七点 四十七分。The nonequivalences：lThe parallel glass block makes the image plane shift;lThere is no shift of image plane formed by the equivalent air thickness;lThe parallel glass block introduces aberrations;lThere are no aberr

38、ations by the equivalent air thickness.第44页/共108页第四十四页，编辑于星期五：十七点 四十七分。2. ApplicationsGiven a thin lens, effective focal length is 100 mm, diameter of the rays is 20 mm, the object is in infinity, diameter of the image is 10 mm. 50mm behind the lens there is a Penta prism which makes the axis deviat

39、e 90. Find out the size of the prism and the position of the image.（n=1.5163）Step 1: Make the corresponding figure of rays.D y10050D1D2Step 2: Calculate the diameter of the first surface . D1=（20+10）/2=15Step 3: Calculate the thicknesses of glass block and equivalent air thickness.L=51.21，e=L/n=33.8

40、第45页/共108页第四十五页，编辑于星期五：十七点 四十七分。Step 5: Calculate diameter of the second surface of the prism.10 510050D1D233.8x62.1121081. 02 .161008 .335010052xDxxStep 6: Calculate the image distance (from the second surface of the prism to the image plane).L2=50-33.8=16.2Homework：No. 2,6 on page 88,8916.2第46页/共1

41、08页第四十六页，编辑于星期五：十七点 四十七分。Example: suppose the diameter of a right angle prism is 10mm，when rotating the prism 45，the emergent ray will be parallel to the incident rays, find out the diameter of the rays.ABC10DA28. 55163. 1, 8 . 0,E,10EnknLkLEDKKN4534. 345sin)28. 510(45sinKNDThe thickness of the glas

42、s block isThe equivalent air thickness is第47页/共108页第四十七页，编辑于星期五：十七点 四十七分。4.7 Determination of Image Orientations for Mirrors and Prisms Intentions:1.Find out the orientations of the image formed by mirrors and prisms. 2.Design a mirrors and prisms system according to the requirements of the orientat

43、ions of the axis and image of the system.第48页/共108页第四十八页，编辑于星期五：十七点 四十七分。Methods of representing the orientations of the object and image in mirror and prism system Take an orthogonal coordinates xyz in object space.x orientation coincides with the incident axis.Y orientation lies in the main sectio

44、n of the prism. Z orientation is normal to the main section. Similarly, in the image space x y z are used to represent the orientations of the image .第49页/共108页第四十九页，编辑于星期五：十七点 四十七分。Methods of determining the image orientations: 1. Determine x orientation: It coincides with the exit axis. 2. Determi

45、ne the y, z orientations.Optic axis section：Main section that coincides with the optic axis. Mirror and prism systems with single main section :All of the main sections of the mirrors and prisms coincide with each other.第50页/共108页第五十页，编辑于星期五：十七点 四十七分。 Mirror and prism systemwith single main section:

46、 No roof surface : z and z have the same orientations .The emergent axis and incident axis Coincide : a) If the number of reflectors is odd, y will be opposite to y ; b) If the number of reflectors is even, y will have the same orientation as y. 第51页/共108页第五十一页，编辑于星期五：十七点 四十七分。 Mirror and prism syst

47、emwith single main section: No roof surface : z and z have the same orientations .The emergent axis and incident axisare reversed :a) If the number of reflectors is odd, y will have the same orientation as y; b) If the number of reflectors is even, y will be opposite to y .第52页/共108页第五十二页，编辑于星期五：十七点

48、 四十七分。Mirror and prism system with single main section (no roof surface)The emergent axis and incident axis Number of reflectorsy and y SignCoincideEvenSame orientation (+)(+)=(+)CoincideOddReverse (+)(-)=(-)ReverseEvenReverse(-)(+)=(-)ReverseOddSame orientation(-)(-)=(+)第53页/共108页第五十三页，编辑于星期五：十七点 四

49、十七分。 After determining the orientations of x and y, we can find out the orientation of z according to total number of reflectors (mirror image or similar image).第54页/共108页第五十四页，编辑于星期五：十七点 四十七分。Mirror and prism system with single main sectionIf there is a roof surface in the system, above rules can a

50、lso be used . However, for the total number of reflectors, the roof surface should be counted twice.第55页/共108页第五十五页，编辑于星期五：十七点 四十七分。 Mirror and prism system with two main sections perpendicular to each other The main sections of prism 1 and prism 3 are parallel, but the main section of prism 2 is ve

51、rtical to those of prism 1 and prism 3. For a prism it can only change the orientation of the coordinate which lies in its main section, and has no influence upon the coordinate whose orientation is normal to the main section. Prism 2 can only change the orientation of z and can not change that of y

52、, also prisms 1 and 3 can only change the orientation of y and do not affect z.第56页/共108页第五十六页，编辑于星期五：十七点 四十七分。 Prism 2 or prisms 1 and 3 all belongs to the systems with single main section and the above rules can be used to find out the orientations of the image. However, we cant simply judge by th

53、e final orientation of emergent axis, but should judge by the actual rotations of optical axis caused by prism 1 and 3. For the axis in this system, after passing through prism 1 the axis rotates 90and after passing through prism 3 the axis again rotates 90,rotating all together 180. 第57页/共108页第五十七页

54、，编辑于星期五：十七点 四十七分。 That means, we can determine the orientation of y according to prisms 1 and 3, and the axis should be considered reversed. Now we can determine the orientation of y according to prisms 1 and 3, the number of reflectors is 2, y is reversed For the orientation of z, for prism 2, the

55、axis is reversed, the number of reflectors is 2, z is reversed. Actually, after determining the one of orientations of either y or z, according to the total number of reflectors, we can determine the coordinate of the object and image space, then find out the other orientation.第58页/共108页第五十八页，编辑于星期五

56、：十七点 四十七分。Note that for the emergent and incident axes, “coincide” and “reverse” are in broad sense. “Coincide” means not only for emergent axis being parallel to the incident axis, but also for the axial deviation angle within 0and 90. “Reverse” means the axial deviation angle greater than 90 If th

57、e deviation angle is just equal to 90, both “coincide” and “reverse” can be used and can get the same result. 第59页/共108页第五十九页，编辑于星期五：十七点 四十七分。1. According to these conditions, two prisms can be used to change the axis 90 twice, and they can be made up by a prism system with a single main section. Fr

58、om the “Handbook of Optical Design” we can find out two types of prisms, 90-1 and 90-2, in which 90-1 is the so-called right angle prism and 90-2 are Penta prism and Boot prism. 2.Since the emergent and incident axes are required to be parallel and the image is required to be inverted to the object,

59、 the total number of reflectors should be odd. That means, the combination should be a 90-1 prism and a 90-2 prism, other combinations are unacceptable. There are two kinds of the combinations.Example: Design a prism system which is made up by two prisms. The system has a 800mm periscope height. The

60、 axis should always lies in one plane and the system is required to produce an inverted image similar to the object. 第60页/共108页第六十页，编辑于星期五：十七点 四十七分。3. since the total number of the reflectors is odd, the image is a mirror image, so one of the reflecting surface should be changed to roof surfaces. In