Mirror and Prism Systems.ppt

1、Chapter 4 Mirror and Prism Systems,Contents in this chapter,The imaging properties and features of the mirror and prism systems The imaging directions of prism systems The conjunction of mirror and prism systems,4.1 Applications of Mirror and Prism Systems in Optical Instruments,Properties of coaxia

2、l spherical 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

3、 object, optical system, image locate on a common line,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 locati

4、on 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.,Examles,Examples,4.2 Imaging properties of mirrors,P,A,O,N,B,D,A

5、,I,I,O,B,point A is imaged at point A We can take incident ray AO from point O arbitrarily;,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 arbitr

6、ary point through a single mirror,P,O,D,O,A,A,Conclusion： 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

7、images .,Is the image similar to the object?,2 Image of object in space reflected by a single mirror,P,x,y,z,o,y,z,x,o,a) 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 coordinates in image

8、 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.,Summery: a) A single plane mirror can image any object points ideally in th

9、e 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.,3 Imaging properties of mirrors system,Forming ideal images An image forme

10、d 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, and they are not similar.,Generally speaking，we always expect t

11、he image is similar to the object, especially in the systems used in military.,4.3 Rotation of Mirrors,P,N,O,A,B,I,I,N,B,Conclusion：The reflected angle will be 2 if the mirror rotates ,2(I+)-2I=2,1. Rotation of single mirror,Advantage：Enlarging the range observed,Defect：Errors caused by rotation,Exa

12、mple：The plane mirror in Range Finder,1. Rotation of single mirror,2. Rotation of two mirrors,P1,P2,A,O1,O2,M,B,I1,I2,In 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 i

13、s twice of the angle of two mirrors.,2. Rotation of two mirrors,P1,P2,A,O1,O2,M,B,I1,I2,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 mirro

14、rs, regardless of the incident rays direction.,Application: We can use two mirrors in place of the single mirror in Range Finder.,angle mirror，prism.,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 dir

15、ection of rays,Advantages ：The loss of energy is small. Its 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.,2. Unfolding of the prismmethod of studying the imaging properties

16、of prisms,Right-angle Prism,Main section :The plane or section which is perpendicular to each prism.,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.,1,2,

17、3,From Law of reflection, we can get:,From the symmetrical relationship, we can get:,So, we can get:,3.The requirements for prisms,(1) After unfolding a prism, the two faces of the glass block must be parallel to each other.,(2) When a prism locates in converging rays, the axis must be perpendicular

18、 to both incident and emergent surfaces.,4. Typical examples of unfolding prisms,a. Right-angle Prism,Locating in parallel rays,A,B,C,A,When 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/AC,And then,4. Typical exampl

19、es of unfolding prisms,a. Right-angle Prism,Locating in parallel rays,A,B,C,A,Face AB needs to be parallel to face AC，then ABC=ACB,So the prism needs ABC=ACB,That 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.,When a prism

20、 locates in converging rays, both the first and the second surfaces should be perpendicular to the axis.,A,B,C,A,When the axis of this prism is deviated through axis 90：,When deviating the axis through any angle,B,C,A,If we want the ray to deviate , then the reflective surface must deviate the axis,

21、This kind of prism is called Isosceles prism.,If we need the axis of this prism be deviated 45,then,2.Penta Prism,3.Boot Prism,45o,60,4. Cube Prism,A,B,C,I,I,E,a,D,Suppose the index of the prism is n，then the refraction angle I is,From the figure we can see that the diameter D of the rays is：,4. Cub

22、e Prism,A,B,C,I,I,E,a,D,we get:,If the glass is K9(or BK7)，then n=1.5163，we can get:,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 can form the Cube prism .,Attent

23、ions to use Cube Prism A 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 other precisely to avoid producing two

24、 separated images. If 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. Since the incident and emergent faces are not normal to the axis ,the Cube prism can only be used in parallel l

25、ight rays.,1,1,2,2,3,4,4,3,4.5 Roof Surfaces and Roof prisms,Roof surfaces：Using two right angle surfaces to replace one reflecting surface. Roof prism: Prism which contains foor surfaces.,Roof Surfaces and Roof prisms,Effects：The addition of the roof to a prism is to introduce an extra inversion to

26、 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.,y,x,z,x1,y1,z1,y,z,x,x2,y2,z2,Requirement for a roof prism: The roof angle mus

27、t be equal to 90 precisely, otherwise, the emergent rays will be not parallel, producing double images,unfolding of roof prisms,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,A,A,A,A,L,l1,l2,Suppo

28、se 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 image plane is:,2. The size

29、of image,A,L,l1,l2,u,u,Conclusion: 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 passes through a parallel glass block. So, we can get

30、 ,In the air,And then,2. The equivalent air thickness of A parallel glass block,A,L,l1,l2,P2,P1,K,Q,KP2=AA=L-L/n,KP2=QM=L-L/n NQ=L/n,M,N,L/n,A,From the figure, AQ=l1-L/n=l2=AM,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

31、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 .,The distance from the second surface to the image plane is equal to that of by passing through an equivalent air thickness; The heights of the ray at the two sur

32、faces of the parallel glass block are equal to those of the ray at the two surfaces of the equivalent air thickness ; The size of the image formed by the parallel glass block is equal to that formed by the equivalent air thickness.,The equivalences :,The nonequivalences：,The parallel glass block mak

33、es the image plane shift; There is no shift of image plane formed by the equivalent air thickness; The parallel glass block introduces aberrations; There are no aberrations by the equivalent air thickness.,2. Applications,Given a thin lens, effective focal length is 100 mm, diameter of the rays is 2

34、0 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 and the position of the image.（n=1.5163）,Step 1: Make the corresponding figure of rays.,D,y,100,50,D1,D2,Step 2: Calculate the

35、diameter of the first surface . D1=（20+10）/2=15,Step 3: Calculate the thicknesses of glass block and equivalent air thickness. L=51.21，e=L/n=33.8,Step 5: Calculate diameter of the second surface of the prism.,10,5,100,50,D1,D2,33.8,x,Step 6: Calculate the image distance (from the second surface of t

36、he prism to the image plane). L2=50-33.8=16.2,Homework：No. 2,6 on page 88,89,16.2,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.,A,B,C,10,D,A,E,D,K,K,N,45,The thickness

37、 of the glass block is,The equivalent air thickness is,4.7 Determination of Image Orientations for Mirrors and Prisms,Intentions: 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 an

38、d image of the system.,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 section of the prism. Z orientation is normal to the m

39、ain section.,Similarly, in the image space x y z are used to represent the orientations of the image .,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 t

40、he optic axis.,Mirror and prism systems with single main section : All of the main sections of the mirrors and prisms coincide with each other.,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

41、 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.,Mirror and prism systemwith single main section:,No roof surface : z and z have the same orientations .,The emergent axis and incident axis are reversed : a) If

42、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 .,Mirror and prism system with single main section (no roof surface),After determining the orientations of x and y, we can find out the orientation of z according t

43、o total number of reflectors (mirror image or similar image).,Mirror and prism system with single main section,If there is a 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.,Mirror and prism system with

44、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 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

45、 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, also prisms 1 and 3 can only change the orientation of y and do not affect z.,Prism 2 or prisms 1 and 3 all belongs to the systems with single main section and

46、the above rules can be used to find out the orientations of the image. However, we cant simply judge by the 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 rotate

47、s 90and after passing through prism 3 the axis again rotates 90,rotating all together 180.,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 r

48、eflectors is 2, y is reversed For the orientation of z, for prism 2, the 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 an

49、d image space, then find out the other orientation.,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 a

50、xial deviation angle greater than 90 If the deviation angle is just equal to 90, both “coincide” and “reverse” can be used and can get the same result.,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 s

51、ection. From 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 t

52、he object, 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

53、height. The axis should always lies in one plane and the system is required to produce an inverted image similar to the object.,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 poss

54、ible systems are shown in Fig. 4.31. Any one of the four systems can be accepted according to the actual situations.,Homework：No. 1,3,4,5 on page 88,89,4-8 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 c

55、an be used to adjust the axis of the system or the image direction during the assembly of the system.,The prism rotation law can be used to study the problem.,中巴资源卫星红外多光谱扫描仪,图1 方案一光学系统结构示意图,图2 方案二光学系统结构示意图,If a prism works in parallel ray path it is sufficient to consider only the directions of the

56、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

57、gle n: reflector number,A special symbol is used to represent the rotation,We can easily get the following equations,Prism Rotation Law,Suppose the object space is fixed，if a prism rotates angle around vector ,the image will first rotate angle (-1)n-1around vector ，and then rotate angle around vector,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 t