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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頁/共108頁4.1 Applications of Mirror and Prism Systems in Optical InstrumentsProperties of coaxial spherical systems:Advantages

2、: 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, optical system, image l

3、ocate on a common line第2頁/共108頁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 change the location and direction of the ax

4、is, 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頁/共108頁ExamlesExamles第4頁/共108頁ExamplesExamples名稱特性圖例平面反射鏡具有折轉(zhuǎn)光路的作用，可用于成像、激光和全息系統(tǒng)。曲面反射鏡表面可

5、為球面或非球面。具有聚焦和會(huì)聚光的作用?？捎迷诠獾陌l(fā)射和接受、激光和光纖系統(tǒng)中，也可以和其它反射鏡組合形成成像系統(tǒng)。柱面反射鏡具有單一方向聚光和聚焦的作用?？捎脕懋a(chǎn)生線光源，或單方向擴(kuò)束和放大圖像。第5頁/共108頁4.2 Imaging properties of mirrorsPAONBDAIIOBpoint A is imaged at point AWe can take incident ray AO from point O arbitrarily;DAADODAAODODODIADOIAOD9090， The incident AO is arbitrary, the equat

6、ion 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 mirror第6頁/共108頁 PODOAAConclusion： a) The image and object points are symmetrical to the mirror. b) The h

7、eight 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 . Is the image similar to the object?第7頁/共108頁2 Image of object in space reflected by a single mirrorPxyzo y z x oa) The s

8、ize 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 coordinates 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

9、 image satisfying this relationship between the object and image spaces is called mirror image. 第8頁/共108頁Summery:a) A single plane mirror can 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

10、the object.d) Forming a mirror image, the image formed by a plane mirror is not similar to the object. 第9頁/共108頁3 Imaging 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 erec

11、ted or inverted image has no relationship with the mirror image；2. The object and the mirror image have different shape, 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頁/共108頁4.3 Rotation of MirrorsPNOAB

12、IINIBIConclusion：The reflected angle will be 2 if the mirror rotates 2(I+)-2I=21. Rotation of single mirror第11頁/共108頁Advantage：Enlarging the range observedDefect：Errors caused by rotationExample：The plane mirror in Range Finder1. Rotation of single mirror第12頁/共108頁2. Rotation of two mirrorsP1P2AO1O2

13、MBI I1I I2 2）（2121222IIII21NOO22121IIIIIn triangle O1O2M，N 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頁/共108頁2. Rotation of two mirrorsP1P2A

14、O1O2MBI1I2 The rotation direction will 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 mirr

15、ors in place of the single mirror in Range Finder.angle mirror，prism.第14頁/共108頁第15頁/共108頁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. Its ha

16、rd 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頁/共108頁2. Unfolding of the prismmethod of studying the imaging properties of prismsRight-angle Prism Main section :The plane or

17、section which is perpendicular to each prism. 第17頁/共108頁 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. 123From Law of reflection, we can get: From the

18、symmetricalrelationship, we can get: So, we can get:第18頁/共108頁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頁/共108頁(2) When a prism locates in converging rays, the axis must be perpendicular to both incident and emergent

19、surfaces. 第20頁/共108頁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 the prism must be parallel to each other , then, AB/ACAnd then, 第21頁/共108頁4. Typical examples of unfolding pr

20、ismsa. 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 isosceles triangle but B and C do not need to equal 45，or A does not need to be a right angle. 第22頁/共108頁When a prism locates in con

21、verging rays, both the first and the second surfaces should be perpendicular to the axis.ABCAWhen the axis of this prism is deviated through axis 90：第23頁/共108頁When deviating the axis through any angle2BCA290CB If we want the ray to deviate , then the reflective surface must deviate the axis2This kin

22、d of prism is called Isosceles prism. If we need the axis of this prism be deviated 45,then,第24頁/共108頁2.Penta Prism第25頁/共108頁45第26頁/共108頁3.Boot Prism45o60第27頁/共108頁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

23、 the rays is： 第28頁/共108頁4. Cube PrismABCIIEaDwe get:If the glass is K9(or BK7)，then n=1.5163，we can get:第29頁/共108頁 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 which ca

24、n form the Cube prism .第30頁/共108頁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 prisms must be parallel to each

25、 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 not normal to the axis ,the Cu

26、be prism can only be used in parallel light rays.第31頁/共108頁11223443第32頁/共108頁4.5 Roof Surfaces and Roof prismsRoof surfaces：Using two right angle surfaces to replace one reflecting surface.Roof prism: Prism which contains foor surfaces.第33頁/共108頁 Roof Surfaces and Roof prismsEffects：The addition of

27、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 this way we can add a reflection and get an image similar to the object.第34頁/共108頁yxz x1 y1 z1

28、yzx x2 y2 z2第35頁/共108頁第36頁/共108頁Requirement for a roof prism: The roof angle must be equal to 90 precisely, otherwise, the emergent rays will be not parallel, producing double images 第37頁/共108頁unfolding of roof prisms第38頁/共108頁4.6 Imaging Property of Parallel Glass Block and Prism Size Calculation 1

29、. Imaging properties of the parallel glass block (1) The final image position AA第39頁/共108頁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 b

30、lock one by one:Substitute to the above equation, we get：So the shift of image plane is:第40頁/共108頁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 emerge

31、nt ray will be parallel to the incident ray when it passes through a parallel glass block. So, we can get ,In the air,And then,第41頁/共108頁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 a

32、t the two surfaces of the parallel 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頁/共108頁lThe distance from the second surface to the image p

33、lane is 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 th

34、at formed by the equivalent air thickness. The equivalences :第43頁/共108頁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 aberrations by the

35、equivalent air thickness.第44頁/共108頁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 deviate 90. Find out the size of the prism

36、 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第45頁/共108頁Step 5: Calculate diameter

37、 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頁/共108頁Example: suppose the diameter of a right angle prism is

38、 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 glass block isThe equivalent air thickness is第47頁/共108頁4.7 Determination of Image Or

39、ientations 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 orientations of the axis and image of the system.第48頁/共108頁Methods of representing the orientations of the obj

40、ect 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 section of the prism. Z orientation is normal to the main section. Similarly, in the image space x y z are used to represent the o

41、rientations of the image .第49頁/共108頁Methods of determining the image orientations: 1. Determine x orientation: It coincides with the exit axis. 2. Determine the y, z orientations.Optic axis section：Main section that coincides with the optic axis. Mirror and prism systems with single main section :Al

42、l of the main sections of the mirrors and prisms coincide with each other.第50頁/共108頁 Mirror and prism systemwith single main section: 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 ;

43、b) If the number of reflectors is even, y will have the same orientation as y. 第51頁/共108頁 Mirror and prism systemwith 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 sam

44、e orientation as y; b) If the number of reflectors is even, y will be opposite to y .第52頁/共108頁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 (+)(-)=(-)Rev

45、erseEvenReverse(-)(+)=(-)ReverseOddSame orientation(-)(-)=(+)第53頁/共108頁 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頁/共108頁Mirror and prism system with single main sectionIf there is a

46、roof surface in the system, above rules can also be used . However, for the total number of reflectors, the roof surface should be counted twice.第55頁/共108頁 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 s

47、ection of prism 2 is vertical 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 c

48、an not change that of y, also prisms 1 and 3 can only change the orientation of y and do not affect z.第56頁/共108頁 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 t

49、he 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頁/共108頁 Tha

50、t 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 axis is reversed, the

51、 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頁/共108頁Note that for the emergent and in

52、cident 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 the deviation angle is just equal to 90, both

53、 “coincide” and “reverse” can be used and can get the same result. 第59頁/共108頁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. From the “Handbook of Optical Design” we can find out two types of

54、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, the total number of reflectors should be odd. That means, the co

55、mbination 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 axis should always lies in one plane and the system is required

56、to produce an inverted image similar to the object. 第60頁/共108頁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 this way, four possible systems are shown in Fig. 4.31. Any one of the four systems c

57、an be accepted according to the actual situations. 第61頁/共108頁名稱特性圖例直角棱鏡 用于將光束折轉(zhuǎn)900，可以用于平行和會(huì)聚光路中。道威棱鏡 用于將光束繞光軸轉(zhuǎn)動(dòng)，是一個(gè)像旋轉(zhuǎn)器。只能用于平行光路中。斜方棱鏡 用于將光束產(chǎn)生一定的橫向位移，同時(shí)不使像的方向產(chǎn)生任何變化。可以用于平行和會(huì)聚光路中。五角棱鏡用于將光束折轉(zhuǎn)900和縮短光路，可以用于平行和會(huì)聚光路中。第62頁/共108頁角錐棱鏡 用于將光束沿原方向返回，棱鏡的位置變化不影響返回光束的方向。半五角棱鏡用于將光束折轉(zhuǎn)600，只能用于平行光路中。屋脊直角棱鏡用于將光束折轉(zhuǎn)900和正

58、像?？梢杂糜谄叫泻蜁?huì)聚光路中。色散棱鏡用于將入射光束中不同波長的光在空間上分散開。楔形鏡用于將光束偏轉(zhuǎn)一定的角度。將影響系統(tǒng)的共軸性。第63頁/共108頁Homework：No. 1,3,4,5 on page 88,89第64頁/共108頁4-8 The Prism Rotation Law The method of rotating mirrors and prisms is usually used to enlarge the view angle. The rotation of mirrors and prisms can be used to adjust the axis o

59、f the system or the image direction during the assembly of the system. The prism rotation law can be used to study the problem. 第65頁/共108頁中巴資源衛(wèi)星紅外多光譜掃描儀中巴資源衛(wèi)星紅外多光譜掃描儀 圖1 方案一光學(xué)系統(tǒng)結(jié)構(gòu)示意圖 圖2 方案二光學(xué)系統(tǒng)結(jié)構(gòu)示意圖第66頁/共108頁If a prism works in parallel ray path it is sufficient to consider only the directions of th

60、e image. However, if it does not work in parallel ray path, both the direction and the position should be considered.Suppose is the unit vector that represents the rotating direction and position of the prism. Similarly represents the images rotating direction and position of the prism.: rotating an

61、glen: reflector number PP第67頁/共108頁A special symbol is used to represent the rotation PPPP)(2121PP)(We can easily get the following equations第68頁/共108頁P(yáng)rism Rotation LawSuppose the object space is fixed，if a prism rotates angle around vector ,the image will first rotate angle (-1)n-1around vector ，a

62、nd then rotate angle around vector PPP ) 1( 1PPAn第69頁/共108頁Step 1：Suppose the prism is fixed， the object space rotates angle around the vector ，according to the image properties，if the reflector is odd the image is mirror image, and if the reflector is even the image is similar to the object. That m

63、eans the image space will rotate angle (-1)n-1 around the vector Step 2：Rotate the object space, prism and the image space angle around the vector . Then, the object space keeps unchanged, the prism rotates around the vector , and the image space first rotates angle (-1)n-1around the vector , and th

64、en rotates angle around the vector P ) 1( 1PPAnPPPPP第70頁/共108頁 object space prism image space object space prism image spacestep 1step 1： rotate - unchange rotate(-rotate - unchange rotate(-1)1)n-1n-1Step 2Step 2： rotate rotate rotate rotate rotate rotate Total resultTotal result： unchange rotate ro

65、tate(-1)unchange rotate rotate(-1)n-n-1 1 rotate rotatePPPPPPPP第71頁/共108頁1、Prism rotates around the axis z in the parallel ray path Since the prism works in parallel ray path, we can only determine the direction of the image, without considering the position. ) 1( 1PPAnSuppose there are no roof surf

66、aces, we have zzPzP) 1( 1zzAn第72頁/共108頁If the reflector n is even, we will getIf the reflector n is odd, we will get0 zzA2 zzzAThat means, when the prism rotates angle around axis z，if the reflector is even the image space will be unchanged, and if the reflector is odd the image space will rotate angle 2 around axis z第73頁/共108頁 ) 1( 1PPAnSuppose the prism contains roof surfaces, we havezzPzP) 1( zzAn第74頁/共108頁If the reflector n is even, we will getIf the reflector n is odd, we will get0 zzA2 zzz