Abstract:
A phase-compensating cube corner retroreflector includes three rear reflecting surfaces. All three rear reflecting surfaces can be coated with a phase-compensating film stack that induces 2nπ phase difference when handling polarized light. So coated, the phase-compensating cube corner preserves the polarization orientation and ellipticity of the incident light. Such a phase-compensating cube corner can be used to improve the accuracy of distance-measuring interferometers. The cube corner directs light to and from other optical elements, including a polarizing beam-splitters, mirrors, and quarter-wave plates.

Description:
CROSS REFERENCE TO RELATED APPLICATIONS  
       [0001]     This application is a divisional of U.S. patent application Ser. No. 10/734,990, filed on Dec. 11, 2003, and incorporated herein by reference. 
     
    
     DESCRIPTION OF RELATED ART  
       [0002]     The conventional cube corner (CC) retroreflector has the property that any ray entering the effective aperture will be reflected and emerge from the entrance/exit face parallel to itself, but with opposite direction of propagation. This property is, within acceptance angle limits, independent of the orientation of the retroreflector. Retroreflectors therefore find frequent application in situations where angular orientation is difficult or impossible to control and where a mirror would therefore be unsatisfactory.  FIGS. 1A and 1B  illustrate a side and a front view of a conventional CC retroreflector  10 .  
         [0003]     Some conventional CC retroreflectors use the principle of total internal reflection (TIR) for its operation. The use of TIR limits the acceptable incident angles for the CC retroreflector. The use of TIR also changes the phase difference between the S and P components so the light exiting the CC retroreflector has a different polarization state, both in orientation and ellipticity, than the light entering the CC retroreflector. Experiments have shown that in solid BK-7 CC retroreflectors using TIR (without any coating), incident linearly polarized light will exhibit about 8.4 degrees of relative rotation of azimuth angle and about 11 degrees of ellipticity. Azimuth is defined as the angle between the major axis of the polarization ellipse to horizontal or vertical reference axes, and tangent of ellipticity is defined as the ratio of minor to major axes of the polarization ellipse.  
         [0004]     The limitation on the incident angle can be removed, at the cost of a reduction in reflective efficiency, by applying a reflective coating to the rear reflecting surfaces (e.g., faces 1, 2, and 3). Thus, some conventional CC refroreflectors have their rear reflecting surfaces coated with silver and then with paint to prevent oxidation of the silver. Nonetheless, the use of the reflecting coating still causes the light exiting the CC retroreflector to have a different polarization state than the light entering the CC retroreflector. Experiments have shown that in solid BK-7 CC retroreflectors with silver coated rear reflecting surfaces, incident linearly polarized light will exhibit about 6.0 degrees of relative rotation of azimuth angle and about 1.0 degree of ellipticity. The paint used to coat the silver is also known to outgas in certain environments and contaminate the surrounding components.  
         [0005]      FIG. 2  illustrates a conventional plane mirror interferometer system  20 . A laser head  22  generates a coherent, collimated light beam consisting of two orthogonally polarized frequency components. One frequency component f A  (e.g., a measurement beam having a P-polarization) enters the interferometer&#39;s measurement path while the other frequency component f B  (e.g., a reference beam having an S-polarization) enters the interferometer&#39;s reference path.  
         [0006]     In the measurement path, a polarizing beam-splitter  24  transmits frequency component f A  to a measurement plane mirror  26  mounted to a moving stage. Since frequency component f A  passes through a quarter-wave plate  28 , the returning polarization is rotated 90 degrees and the newly S-polarized frequency component f A  is reflected by polarizing beam-splitter  24  to CC retroreflector  10 . Retroreflector  10  directs frequency component f A  again to polarizing beam-splitter  24 , which again reflects frequency component f A  to measurement mirror  26 . Again, since frequency component f A  passes through quarter-wave plate  28 , the returning polarization is rotated 90 degrees and the newly P-polarized frequency component f A  is transmitted through polarizing beam-splitter  24  onto a receiver  34 .  
         [0007]     In the reference path, polarizing beam-splitter  24  reflects frequency component f B  to a reference plane mirror  30 . Since frequency component f B  passes through a quarter-wave plate  32 , the returning polarization is rotated 90 degrees and the newly P-polarized frequency component f B  is transmitted through polarizing beam-splitter  24  to CC retroreflector  10 . CC retroreflector  10  directs frequency component f B  through beam-splitter  24  to reference mirror  30 . Again, since frequency component f B  passes through quarter-wave plate  32 , the returning polarization is rotated 90 degrees and the newly S-polarized frequency component f B  is reflected by polarizing beam-splitter  24  coaxially with frequency component f A  onto receiver  34 .  
         [0008]     The importance of maintaining the polarization states through CC retroreflector  10  can be seen in  FIG. 2 . If CC retroreflector  10  changes the orientations of frequency components f A  and f B , then parts of frequency components f A  and f B  would leak through polarizing beam-splitter  24 , which can cause a decrease in the accuracy of the distance measurements.  
         [0009]     A paper by Paul Mauer discloses that a three-layer phase-compensating film stack can be formed on the rear reflecting surfaces of a CC retroreflector to compensate the phase shift due to TIR. “Phase Compensation of Internal Reflection” by Paul Mauer, J. Opt. Soc. Am. 56, 1219 (1966). However, the paper states that although the ellipticity of the light is preserved, the orientation of the light is changed. “Although a coating of this type compensates phase shift on total internal reflection, there remains, in general, a rotation of the plane of polarization when light is returned from a phase-compensated corner cube.” Id. at 1221. Reading this paper, one skilled in the art would be discouraged from using the phase-compensated cube corner in an interferometer system because the paper states that the polarization orientation is changed by the phase-compensated cube corner. As described above, change in the polarization orientation is known to decrease the accuracy of the interferometry system.  
         [0010]     In a series of lecture notes for a UCLA Extension class, Philip Baumeister cites the Mauer paper for describing the use of a three-layer film stack to reduce the differential phase shift upon reflection. “Optical Coating Technologies” by Philip Baumeister, UCLA Extension (1998). In FIG. 1-122 of these lecture notes, Baumeister shows a differential phase shift of approximately 0 degree for a normal angle of incidence when the three-layer film stack is applied to the three rear reflecting surfaces of the cube corner. Reading this paper, one skilled in the art would also read the Mauer paper described above. Thus, for the same reasons described above, one skilled in the art would be discouraged from using the phase-compensated cube corner in an interferometer system because the Mauer paper states that the polarization orientation is changed by the phase-compensated cube corner.  
         [0011]     Thus, what is needed is a CC retroreflector that addresses the relative polarization rotation, ellipticity, reflective efficiency, and contamination problems of a conventional CC retroreflector.  
       SUMMARY  
       [0012]     In one embodiment of the invention, a phase-compensating cube corner retroreflector includes three rear reflecting surfaces. All three rear reflecting surfaces can be coated with a phase-compensating film stack that induces 2nπ phase difference when handling polarized light. So coated, the phase-compensating cube corner preserves the polarization orientation and ellipticity of the incident light. Such a phase-compensating cube corner can be used to improve the accuracy of distance-measuring interferometers. The cube corner directs light to and from other optical elements, including a polarizing beam-splitters, mirrors, and quarter-wave plates. 
     
    
     BRIEF DESCRIPTION OF THE DRAWINGS  
       [0013]      FIGS. 1A and 1B  illustrate a side and a front view of a conventional cube corner retroreflector.  
         [0014]      FIG. 2  illustrates a conventional plane mirror interferometer system.  
         [0015]     FIGS.  3  to  8  illustrate the derivation of the polarization orientation and ellipticity with an uncoated conventional cube corner in one embodiment of the invention.  
         [0016]     FIGS.  9  to  12  illustrate the derivation of the polarization orientation and ellipticity with a phase-compensated cube corner having 0 degree phase difference in one embodiment of the invention.  
         [0017]     FIGS.  13  to  16 A illustrate the derivation of the polarization orientation and ellipticity with a phase-compensated cube corner having 180 degree phase difference in one embodiment of the invention.  
         [0018]      FIGS. 16B and 16C  illustrate the reflected phase from phase-compensated cube corners having 0 degree phase difference in embodiments of the invention.  
         [0019]      FIG. 17A  illustrates a plane mirror interferometer system in one embodiment of the invention.  
         [0020]      FIGS. 17B and 17C  illustrate interferometer systems in embodiments of the invention.  
         [0021]      FIGS. 18A, 18B , and  18 C illustrate a differential interferometer system in one embodiment of the invention.  
         [0022]      FIG. 19  illustrates a single beam interferometer system in one embodiment of the invention.  
         [0023]      FIGS. 20A, 20B , and  20 C illustrate a high-resolution interferometer system in one embodiment of the invention.  
         [0024]      FIG. 21  illustrates a linear interferometer system in one embodiment of the invention. 
     
    
     DETAILED DESCRIPTION  
       [0025]     In one embodiment of the invention, a dielectric multilayer thin film coating is deposited on all reflecting surfaces or only the first and the third reflecting surfaces of a solid BK-7 cube corner. The thin film stack is designed specifically to preserve upon reflection the azimuth and ellipticity angles of the incident polarization. The thin film coating reduces polarization rotation from about 6 degrees (for silver coated cube corner) to about 0.5 degree, and ellipticity from about 1.0 degree to about 0.5 degree for a linearly polarization state. Additionally, since all reflections are accomplished with total internal reflection (TIR), the net irradiance efficiency is near 100%. Furthermore, problems associated with outgassing are alleviated because the rear reflecting surfaces of the cube corner do not need to be painted as the dielectric materials are inert.  
         [0000]     Introduction  
         [0026]     Reflection processes in general induce change to the polarization state of light. Polarization states change on reflection as electric field components undergo relative phase retardation and/or coordinate transformation. The extent of the change depends on the incident angle and the polarization as well as on the optical properties of the surface.  
         [0027]     To be described herein, the polarization effects associated with total internal reflections (TIR) in solid cube corners are considered. Using Stokes parameter representation, a numerical model is created to make predictions and develop ways to preserve ellipticity and azimuth angle of the initial polarization.  
         [0000]     Uncoated TIR Cube Corner  
         [0028]     For an uncoated cube corner, each reflection is TIR with a phase retardation Δ of 45.2 degrees between the S and the P polarization states. For an incident S-polarized beam propagating in negative z direction into face 1 with a face folded edge perpendicular to the horizontal plane as shown in  FIG. 3 , the beam undergoes three reflections and finally emerges from face 3 propagating in the positive z direction. For the first reflection on face 1, the unit normal vector for the plane of incidence is given by: 
 
 N   1 =0.500 i+ 0.866 j+ 0.000 k,   (0.1) 
 
 where i, j and k are unit vectors in the laboratory Cartesian frame of  FIG. 3 . In a Cartesian coordinate frame defined by N 1 , the S-polarization makes an angle of −30 degrees with the y-axis of this frame as shown in  FIG. 4 . The azimuth angle is negative since it is measured counterclockwise from the y-axis. In this rotated frame, the y-axis is perpendicular while the x- and z-axes lie in the plane of incidence of face 1. The incident unit magnitude polarization vector in this frame is represented by its components a 1  and a 2 . The values for components a 1  and a 2  are: 
 
 a   1 =−0.500  (1) 
 
 a   2 =0.866.  (2) 
 
         [0029]     With these values of components a 1 , a 2  and the azimuth ψ(=ψ 0 )=−30 degrees in face 1 coordinate frame and looking into the beam as shown in  FIG. 4 , it is possible to use equation II.12 of Appendix II to calculate the phase difference δ between components a 1  and a 2  of the input beam as follows: 
 
( a   1   2   −a   2   2 )tan 2ψ=2 a   1   a   2 cos δ.  (II.12) 
 
         [0030]     Henceforth, all references to coordinate frames are from the view of looking into a beam. It is important that in equation II.12 the azimuth angle ψ is measured from the y-axis. This is only true when |a 2 |&gt;|a 1 |. Solving for the phase difference δ, it is determined that: 
 
δ=δ 0 =180 deg.  (2.1) 
 
         [0031]     This value for phase difference δ 0  is then added to the relative phase retardation Δ, which is 45.2 degrees for uncoated TIR reflection at 54.7 degrees of incidence, to enable the calculation of ellipticity χ 1  and azimuth angle ψ 1  of the reflected beam from face 1 as follows: 
 
δ 1 =Δ+δ 0 =225.2 deg.  (2.2) 
 
tan α 1   =a   2   /a   1 =−1.732.  (2.3) 
 
 Phase difference δ 1  is post reflection relative phase retardation between components a 1  and a 2 . Additionally due to the reflection being TIR, tan α 1 =tan α 0 . The calculation for azimuth angle ψ 1  proceeds with the reapplication of equation II.12 using the above value for phase difference δ 1  as follows: 
 
tan 2ψ 1 =2 tan α 1 cos δ 1 /(1−tan 2 α 1 ) 
 
ψ 1 =−25.3 deg.  (3) 
 
         [0032]     Next using the Stokes parameter S 3  of equation III.15 in Appendix III, the ellipticity χ 1  is calculated as follows: 
 
 S   3 =sin 2χ={2 tan α/(1+tan 2 α)}sin δ  (4) 
 
χ 1 =sin −1 [{2 tan α 1 /(1+tan 2α 1 )}sin δ 1 ]/2 
 
χ 1 =18.96 deg.  (5) 
 
 In equation 5, the tangent of ellipticity angle χ 1  is the ratio of the minor to the major axis of the elliptical polarization. See equation III.0.b in Appendix III.  FIG. 5  illustrates the reflected polarization state relative to the plane of incidence of face 1. 
 
         [0033]     As was the case for the incident beam, the azimuth angle ψ 1  is negative with respect to the plane of incidence for face 1. The next step is to calculate azimuth angle ψ 1  relative to the plane of incidence of face 2 (ellipticity angle χ 1  remaining unchanged with the rotation of coordinate frame). This is necessary since the reflected polarization from face 2 must be made relative to the plane of that face. To perform the coordinate transformation from face 1 to face 2, it is necessary to first know the unit normal vector for the plane of incidence of face 2. The unit normal vector for the plane of incidence of face 2 is: 
 
 N   2 =0.000 i+ 0.577 j+ 0.816 k.   (5.1) 
 
 Then taking the dot product of normal vectors N 1  and N 2 , the angle φ 12  between planes for faces 1 and 2 is given by: 
 
φ 12 =60 deg.  (5.2) 
 
         [0034]      FIG. 6  illustrates the relative orientation of the two frames of reference where y 1 , x 1  and y 2 , x 2  are coordinate frames for faces 1 and 2 respectively. It is evident from  FIG. 6  that the azimuth angle ψ 12  of the elliptically polarized state with respect to y 2  axis is given by, 
 ψ 12 =ψ 1 +60=60−25.3=34.7 deg.  (5.3)  
 The subscript 12 refers to the azimuth angle relative to frame 2 for the reflected beam from face 1. As shown in  FIG. 6 , the azimuth angle ψ 12  is a positive quantity (whereas azimuth angle ψ 1  was negative). Once azimuth angle ψ 12  and the ellipticity angle χ 1  are determined for y 2 , x 2  frame, the associated components a 1  and a 2  are also calculated. As described earlier with the input polarization for the first reflection, new values for components a 1 , a 2  and phase difference δ 12  are needed in advance of the second reflection. These quantities are determined from the Stokes parameter S 1  of equation III.13 from Appendix III and equation II.12 from Appendix II as follows: 
   S   1 =(1−tan 2 α)/(1+tan 2 α).  (6a)  
 Alternatively, these quantities are determined from the normalized form of equation III.10 from Appendix III as follows: 
   S   1 =cos 2χ cos 2ψ.  (6b)  
 In equation 6b, the azimuth angle ψ is always measured from the x-axis. Therefore, its value equal to: 
 ψ=90−ψ 12 .  (6.1)  
 Solving equation 6a for tan α and combining with equation 6b provide:  
                     tan   ⁢           ⁢     α   12       =       {       (     1   -     cos   ⁢           ⁢   2   ⁢     χ   1     ⁢   cos   ⁢           ⁢   2   ⁢   ψ       )     /     (     1   +     cos   ⁢           ⁢   2   ⁢     χ   1     ⁢   cos   ⁢           ⁢   2   ⁢   ψ       )       }       1   /   2                   =     1.331   .                   (   7   )               
 From equation II.12: 
 cos δ 12 ={(1−tan 2 α 12 )/tan α 12 }tan 2ψ 12 /2  δ 12 =140.2 deg.  (8)  
         [0035]     As a result of measuring the azimuth from frame 2 (given by azimuth angle ψ 12 ), new values are assigned to tan α and δ (see equations 7 and 8). So tan α and δ are inherently dependent on the reference frame in which the azimuth angle ψ is measured. For reflection 2 from uncoated face 2, the relative phase retardation Δ on tan α 12  (or equivalently on components a 1  and a 2 ) is once again 45.2 degrees. Then it follows that the phase difference δ 2  is equal to: 
 
δ 2 =Δ+δ 12 =275 deg.  (8.1) 
 
         [0036]     Using calculation methods developed earlier for reflection 1, and applying equations 3 and 5 respectively for the azimuth angle ψ 2  and the ellipticity angle χ 2 , it is determined that: 
 
tan 2ψY 2 =2 tan α 2 cos δ 2 /(1−tan 2 α 2 ) 
 
ψ 2 =2 36.9 deg.  (8.2) 
 
                     χ   2     =         sin     -   1       ⁡     [       {     2   ⁢           ⁢   tan   ⁢           ⁢       α   2     /     (     1   +       tan   2     ⁢     α   2         )         }     ⁢   sin   ⁢           ⁢     δ   2       ]       /   2                 =       -   2.593     ⁢           ⁢     deg   .                     (   8.3   )             
 
 Not that tan α 2 =tan α 12  for TIR reflection. 
 
         [0037]     The azimuth angle ψ 23  with respect to face 3 is next found by noting that the y 2 , x 2  and y 3 , x 3  frames are relatively rotated by 60 degrees as shown in  FIG. 7 . This is determined from knowing that the unit normal vector for plane of incidence of face 3 is, 
 
 N   3 =0.500 i+ 0.866 j+ 0.000 k.   (8.4) 
 
         [0038]     Hence the dot product of normals N 2  and N 3  yields the angle between x 2 - and x 3 -axes to be 60 degrees. Additionally, x 2  and x 3  axes lie in the plane of incidence for faces 2 and 3 respectively, which is analogous to the fact that previously x 1  and x 2  axes were in the plane of incidences for faces 1 and 2. In this case, however, the azimuth angle ψ 23  is rotated counterclockwise from y 3  axis and is therefore a negative number calculated as follows: 
 
Ψ 23 =ψ 2 −60=−23.1 deg.  (8.5) 
 
         [0039]     Furthermore, tan α 23  is also negative and is calculated using equation 7 as follows: 
 
ψ=90−ψ 23   (8.6) 
 
tan α 23 ={(1−cos 2χ 2 cos 2ψ)/(1+cos 2χ 2 cos 2ψ)} 1/2  
 
tan α 23 =−2.33  (8.7) 
 
         [0040]     Using equation 8, 
 
cos δ 23 ={(1−tan 2 α 23 )/tan α 23 }tan 2ψ 23 /2 
 
δ 23 =172.9 deg.  (8.8) 
 
         [0041]     For the reflection from face 3, tan α 23  and δ 23  from reflection 2 are utilized as inputs to determine the final ellipticity and azimuth angle. Using calculation methods outlined for reflection 1 and 2, it is determined that:  
                     δ   3     =     Δ   +     δ   23                     =     218.1   ⁢           ⁢     deg   .         ,                 (   8.9   )             
 
 where Δ=45.2 degrees and tan α 3 =tan α 23 . Then from equations 3 and 5, 
 
tan 2ψ 3 =2 tan α 3 cos δ 3 /(1−tan 2 α 3 ) 
 
ψ 3 =−19.8 deg.  (8.10) 
 
χ 3 =sin −1 [{2 tan α 3 /(1+tan 2α 3 )}sin δ 3 ]/2 
 
χ 3 =13.3 deg.  (8.11) 
 
         [0042]     Then converting the azimuth angle to laboratory frame as shown in  FIG. 8 ,  
                     ψ   lab     =     30   +       ψ   3     .                   =     10.2   ⁢           ⁢     deg   .                     (   8.12   )             
 
         [0043]     So after three TIR reflections, the initial polarization has transformed from its linear vertical state to one that has rotated 10.2 degrees and with ellipticity of 13.3 degrees. These computed values are in close agreement with experimental results:  
                                               TABLE 1                           Predicted and measured azimuth and ellipticity angles                Predicted (deg.)   Measured (deg.)                            Azimuth   10.2   8.4           Ellipticity   13.3   11.0                      
 
         [0044]     Similar calculation can also be made for P-polarized light.  
         [0000]     Phase Compensated TIR Cube Corner  
         [0045]     Reflection from a TIR cube corner with uncoated faces results in a transformation of the input polarization state from linear to an elliptical state. This is a consequence of non zero phase retardation that occurs on an uncoated surface at the TIR angle of 54.7 degrees (the incident angle for each reflection in a cube corner). The idea of phase compensated coating is to introduce an interference coating stack on the TIR surface with certain phase thicknesses and refractive indices such as to reduce the reflected phase retardation from 45.2 to 0 or 180 degrees.  
         [0046]     In order to produce optical interference within the coating stack, refractive indices for coating materials must be chosen in such a way that the internal transmitted angles be less than their respective critical angles. This way, the phase thicknesses for all layers in the stack will influence the net change in the reflected phase. With the aid of a thin film analysis program, various coating stacks are designed and investigated below.  
         [0047]     Case I: Coating Phase Difference is Zero Degree (i.e. Δ=180 Degrees)  
         [0048]     In one embodiment, the thin film stack on the cube corner induces 0 degree phase difference for each of the three reflections that occur on a cube corner. The term phase difference follows the convention used in thin film calculation whereas phase retardation Δ follows the convention used in ellipsometry. The design is shown in Table 2:  
                                                   TABLE 2                           Zero degree phase difference            Layer#   Material   Index (at 633 nm)   QWOT (nm)                    1   SiO2   1.46   815       2   TiO2   2.45   1066       3   SiO2   1.46   1090       4   TiO2   2.45   1702                  
 
 QWOT is an optical thickness equal to 4*N*T, where N is the refractive index and T is the physical thickness. Please note that the higher TiO2 index is achieved with ion-assist deposition process. Here layer 1 is the first layer on the BK-7 surface.  FIG. 9  illustrates the phase change with the coating of Table 2. 
 
         [0049]     This coating may now be incorporated in the polarization model developed for the analysis of the uncoated TIR cube corner to calculate the polarization state produced from a phase-compensated cube corner. Before continuing, however, it is important to resolve sign convention differences between thin film and polarization analysis. In the thin film convention, coatings that produce zero degree phase difference between the S and the P polarization states do not alter the sense of polarization. For example, incident right circular will remain right circular after reflection. However, the same phenomenon is described in polarization analysis as having flipped in the sense of polarization rotation from right to left circular. This happens because in polarization analysis the coordinate frame is always defined looking into the beam. Therefore, incident right circular polarization (looking into the beam) appears as left circular after reflection when viewing into the beam, even with zero degree reflected phase change. It thus appears, from the point of view of polarization analysis, that the coating induces exactly 180 degrees phase change on reflection between S and P and not zero degree as was predicted by the thin film convention. Even at normal incidence, polarization analysis predicts a change in the handedness of the polarization on reflection from right to left or visa versa, when physically there is no phase difference between S and P at normal incidence.  
         [0050]     With this understanding, the TIR cube corner with the phase compensating coating of Table 2 is hereafter investigated.  
         [0051]     Input Polarization  
         [0052]     As in the earlier example of uncoated cube corner, the input state is S polarized with respect to laboratory frame. Thus, components a1, a2 and phase difference δ 0  have the following values: 
 
 a   1 =−0.500  (8.13) 
 
a 2 =0.866  (8.14) 
 
δ 0 =180.  (8.15) 
 
         [0053]     Reflection 1  
         [0054]     For reflection from face 1: 
 
Δ=180 deg.  (8.16) 
 
tan α=−0.866/0.500=−1.732.  (8.17) 
 
cos(Δ+δ 0 )={(1−tan 2 α)/tan α}tan 2ψ 1 /2  (9) 
 
sin 2χ 1 ={2 tan α/(1+tan 2 α)}sin(Δ+δ 0 )  (10) 
 
 Solving (9) for ψ 1  and (10) for χ 1 , 
 
ψ 1 =30 deg.  (10.1) 
 
χ 1 =0 deg.  (10.2) 
 
         [0055]      FIG. 10  shows the reflected polarization state R 1  from face 1. It is noteworthy in this instance that R 1  lies along the x-axis of frame 2. This important result will be utilized later in on embodiment where the phase compensation coating is only applied to two faces of the cube corner.  
         [0056]     With respect to frame 2, 
 
ψ 12 =ψ 1 +60=90 deg.  (10.3) 
 
ψ=90−ψ 12   (10.4) 
 
                     tan   ⁢           ⁢     α   12       =       {       (     1   -     cos   ⁢           ⁢   2   ⁢     χ   1     ⁢   cos   ⁢           ⁢   2   ⁢   ψ       )     /     (     1   +     cos   ⁢           ⁢   2   ⁢     χ   1     ⁢   cos   ⁢           ⁢   2   ⁢   ψ       )       }       1   /   2                   =   0                 (   11   )             
 cos δ 12 ={( 1 −tan 2 α 12 )/tan α 12 }tan  2 ψ′ 12   /2  
 
δ 12 =90.  (12) 
 
         [0057]     Note that in equation 11, tan α 12 =0 and this implies that |a 2 |&lt;|a 1 |. Therefore, the azimuth angle ψ′ 12  in equation 12 is measured with respect to the x-axis or in this case to the x 2 -axis. The value for ψ′ 12  is equal to, 
 
ψ′ 12 =0 deg.  (12.1) 
 
         [0058]     Reflection 2  
         [0059]     For reflection 2, no change is expected on ellipticity and azimuth angle since R 1  lies on the x 2 -axis with no component of the polarization vector existing on the y 2 -axis. This effect only holds true when the incident polarization to the cube corner (i.e., to face 1) is S or P polarized (with respect to laboratory frame). For any other polarization states, a vector component will exist along the y 2 -axis. 
 
Δ=180 deg.  (12.2) 
 
tan α=tan α 12 =0  (12.3) 
 
cos(Δ+δ 12 )={(1−tan 2 α)/tan α}tan 2ψ′ 2 /2  (13) 
 
sin 2χ 2 ={2 tan α/(1+tan 2 α)}sin((Δ+δ 12 )  (14) 
 
 Solving equation 13 for the azimuth angle ψ′ 2  and equation 14 for the ellipticity angle χ 2 , 
 
ψ′ 2 =0 deg.  (14.1) 
 
χ 2 =0 deg.  (14.2) 
 
 Again note that tan α 12 =0. Therefore the azimuth ψ′ 2  is measured with respect to x 2 -axis. 
 
         [0060]     With respect to frame 3, the polarization vector R 2  forms an angle as shown in FIG.  11 , 
 
ψ 23 =ψ′ 2 +30=30 deg.  (14.3) 
 
         [0061]     In using the S 1  Stokes parameter in equations 6a and 6b, care must be taken by noting that the azimuth angle in these equations are measured from the x-axis. That is to say,  
                   ψ   =     90   -     ψ   23                   =     60   ⁢           ⁢     deg   .                     (   14.4   )                       tan   ⁢           ⁢     α   23       =       {       (     1   -     cos   ⁢           ⁢   2   ⁢     χ   2     ⁢   cos   ⁢           ⁢   2   ⁢   ψ       )     /     (     1   +     cos   ⁢           ⁢   2   ⁢     χ   2     ⁢   cos   ⁢           ⁢   2   ⁢   ψ       )       }       1   /   2                   =   1.732                 (   14.5   )             
 cos δ 23 ={(1−tan 2 α 23 )/tan α 23 }tan 2ψ 23 /2 
 
δ 23 =180 deg.  (14.6) 
 
         [0062]     Reflection 3  
         [0063]     For reflection 3, 
 
Δ=180 deg.  (14.7) 
 
tan α=tan α 23 =1.732.  (14.8) 
 
cos(Δ+δ 23 )={(1−tan 2 α)/tan α}tan 2ψ 3 /2  (15) 
 
sin 2χ 3 ={2 tan α/(1+tan 2 α)}sin(Δ+δ23)  (16) 
 
 Solving equation 15 for the azimuth angle ψ 3  and equation 16 for the ellipticity angle χ 3 , 
 
ψ 3 =−30 deg.  (16.1) 
 
χ 3 =0 deg.  (16.2) 
 
         [0064]     The azimuth angle relative to laboratory frame is given by, 
 
ψ lab =30+ψ 3 =0 deg.  (16.3) 
 
         [0065]     With the azimuth angle ψ lab =0 and ellipticity χ 3 =0, the cube corner with three TIR coatings preserves the state of linear and vertical input polarization.  
         [0066]     Case II: Only Faces 1 and 3 are Coated.  
         [0067]     In one embodiment, only faces 1 and 3 are coated with the coating of Table 2. If the same S-polarization is used as input, then the calculations of the ellipticity χ 1  and azimuth angle ψ 1  up to reflection 2 will be the same as that of Case I. Therefore the Case I results for reflection 1 will be utilized here.  
         [0068]     Reflection 2  
         [0069]     As described in reflection 1 of Case I, the reflected polarization R 1  from face 1 lies in the plane of incidence of face 2 as shown in  FIG. 10 . Thus, regardless of the amount of phase change that occurs from face 2, the reflected state R 2  will always lie in that plane of incidence (i.e., there will be no change in its azimuth angle or ellipticity). Therefore, face 2 could have been left uncoated in Case I and the same end result would have been achieved. This result, however, is only valid for linear S or P input polarization states. 
 
Δ=45.2 deg. (for uncoated surface)  (16.4) 
 
δ 12 =90  (16.5) 
 
tan α=tan α 12 =0  (16.6) 
 
cos(Δ+δ 12 )={(1−tan 2 α)/tan α}tan 2ψ′ 2 /2  (17) 
 
sin 2χ 2 ={2 tan α/(1+tan 2 α)}sin((Δ+δ 12 )  (18) 
 
 Solving equation 17 for the azimuth angle ψ′ 2  and equation 18 for the ellipticity angle χ 2 , 
 
ψ′ 2 =0 deg.  (18.1) 
 
χ 2 =0 deg.  (18.2) 
 
         [0070]     With respect to frame 3 as shown in  FIG. 11 , the reflected state R 2  has 
 
ψ 23 =ψ′ 2 +30=30 deg.  (18.3) 
 
         [0071]     Because these values for the azimuth angle ψ′ 2 , the ellipticity angle χ 2 , and the azimuth angle ψ 23  are identical to corresponding reflection 2 values from Case I, same results are also expected for reflection 3 of this case to that of Case I. From the point of view of vertical linear or horizontal linear input states, the two cases (both Case I and II) produce the same results. However, the same results cannot be produced for circularly polarized input states when only the first and the third faces are coated.  
         [0072]     Case III: Coating Phase Difference is 180 Degrees (i.e., Δ=0 Degree).  
         [0073]     In one embodiment, the thin film stack on the cube corner induces 180 degrees phase difference on reflection. The coating design is shown in Table 3.  
                                 TABLE 3                           180 degree phase difference            Layer#   Material   Index (at 633 nm)   QWOT (nm)               1   TiO2   2.45   1316       2   SiO2   1.46   1279       3   TiO2   2.45   2595                  
 
 Here layer 1 is the first layer on the BK-7 surface.  FIG. 13  illustrates the phase change with the coating of Table 3. 
 
         [0074]     With the 180 degrees phase difference coating on all three faces, the output polarization state from a cube corner is now calculated.  
         [0075]     Input Polarization  
         [0076]     The input state is S polarized with respect to laboratory frame. 
 
 a   1 =−0.500  (18.4) 
 
a 2 =0.866  (18.5) 
 
δ 0 =180 deg.  (18.6) 
 
         [0077]     Reflection 1  
         [0078]     For reflection 1, 
 
Δ=0 deg.  (18.7) 
 
tan α=−0.866/0.500=−1.732.  (18.8) 
 
cos(Δ+δ 0 )={(1−tan 2 α)/tan α}tan 2ψ 1 /2  (19) 
 
sin 2χ 1 ={2 tan α/(1+tan 2 α)}sin(Δ+δ 0 )  (20) 
 
 Solving equation 19 for the azimuth angle ψ 1  and equation 20 for the ellipticity angle χ 1 , 
 
 ψ 1 =−30 deg.  (20.1) 
 
χ 1 =0 deg.  (20.2) 
 
         [0079]      FIG. 14  shows that the reflected R 1  is collinear with the input state. The apparent magnitude difference between S and R 1 , in  FIG. 14 , is only meant for clarity, since for TIR the reflected amplitude is equal to that of the incident.  
         [0080]     With respect to frame 2, 
 
ψ 12 =ψ 1 +60=30 deg.  (20.3) 
 
ψ=90−ψ 12   (20.4) 
 
                     tan   ⁢           ⁢     α   12       =       {       (     1   -     cos   ⁢           ⁢   2   ⁢     χ   1     ⁢   cos   ⁢           ⁢   2   ⁢   ψ       )     /     (     1   +     cos   ⁢           ⁢   2   ⁢     χ   1     ⁢   cos   ⁢           ⁢   2   ⁢   ψ       )       }       1   /   2                   =   1.732                 (   20.5   )             
 cos δ 12 ={( 1 −tan 2 α 12 )/tan α 12 }tan  2ψ   12   /2   ( 20.6 ) 
 
δ 12 =180 deg.  (20.7) 
 
         [0081]     Reflection 2  
         [0082]     For reflection 2, 
 
Δ=0 deg.  (20.8) 
 
tan α=tan α 12 =1.732  (20.9) 
 
cos(Δ+δ 12 )={(1−tan 2 α)/tan α}tan 2ψ 2 /2  (21) 
 
sin 2χ 2 ={2tan α/(1+tan 2 α)}sin((Δ+δ 12 )  (22) 
 
 Solving equation 21 for the azimuth angle ψ 2  and equation 22 for the ellipticity angle χ 2 , 
 
ψ 2 =30 deg.  (22.1) 
 
χ 2 =0 deg.  (22.2) 
 
         [0083]     The angle that R 2  forms relative to the y3-axis is shown in  FIG. 15 . This angle is, 
 
ψ 23 =ψ 2 −60=−30 deg.  (22.3) 
 
ψ=90−ψ 23   (22.4) 
 
                     tan   ⁢           ⁢     α   23       =       {       (     1   -     cos   ⁢           ⁢   2   ⁢     χ   2     ⁢   cos   ⁢           ⁢   2   ⁢   ψ       )     /     (     1   +     cos   ⁢           ⁢   2   ⁢     χ   2     ⁢   cos   ⁢           ⁢   2   ⁢   ψ       )       }       1   /   2                   =     -   1.732                   (   22.5   )             
 cos δ 23 ={( 1 −tan 2 α 23 )/tan α 23 }tan  2ψ   23   /2  
 
δ 23 =180 deg.  (22.6) 
 
         [0084]     Reflection 3  
         [0085]     For reflection 3, 
 
Δ=0 deg.  (22.7) 
 
tan α=tan α 23 =−1.732.  (22.8) 
 
cos(Δ+δ 23 )={(1−tan 2 α)/tan α}tan 2ψ 3 /2  (23) 
 
sin 2χ 3 ={2 tan α/(1+tan 2 α)}sin(Δ+δ 23 )  (24) 
 
 Solving equation 23 for the azimuth angle ψ 3  and equation 24 for the ellipticity angle χ 3 , 
 
ψ 3 =−30 deg.  (24.1) 
 
χ 3 =0 deg.  (24.2) 
 
         [0086]     As shown in  FIG. 16A , the azimuth angle relative to laboratory frame is given by,  
                     ψ   lab     =     30   +     ψ   3                   =     0   ⁢           ⁢     deg   .                     (   24.3   )             
 
         [0087]     This is the exact result that was obtained in Case I, where all sides were coated with zero degree phase difference coating (in Table 2). Therefore, Cases I, II and III all produce the same result for linearly polarized input states.  
         [0088]     Case Iv: Additional Designs of Phase-Compensating Film Stack  
         [0089]     In one embodiment, the thin film stack on the cube corner induces 0 degree phase difference on reflection (i.e., Δ=180 degrees). The coating design is shown in Table 4.  FIG. 16B  illustrates the phase change with the coating of Table 4.  
                                                   TABLE 4                           0 degree phase difference            Layer#   Material   Index (at 633 nm)   QWOT (nm)                    1   MgF2   1.38   715       2   TiO2   2.45   1903                  
 
 Here layer 1 is first layer on the BK-7 surface. 
 
         [0090]     In another embodiment, the thin film stack on the cube corner induces 0 degree phase difference on reflection. The coating design is shown in Table 5. Please note that the lower TiO2 index is a result of a deposition process without ion-assistance.  FIG. 16C  illustrates the phase change with the coating of Table 5.  
                                                   TABLE 5                           0 degree phase difference            Layer#   Material   Index (at 633 nm)   QWOT (nm)                    1   TiO2   2.30   262.5       2   SiO2   1.45   346.5       3   TiO2   2.30   1018.5       4   SiO2   1.45   462       5   TiO2   2.30   850.5                  
 
 Here layer 1 is the first layer on the BK-7 surface. 
 
         [0091]     For linearly polarized input states, the coating designs of Tables 4 and 5 would produce the desired result if all three faces or only the first and the third faces are coated. For circularly polarized input states, the coating designs of Tables 4 and 5 would produce the desired result if all three faces are coated.  
         [0000]     Application to Interferometry Systems  
         [0092]     Plane Mirror Interferometer  
         [0093]      FIG. 17A  illustrates a plane mirror interferometer system  100  in one embodiment of the invention. System  100  is similar to system  20  of  FIG. 2  except that the conventional cube corner  10  has been replaced with a CC retroreflector  102  having phase-compensated interference film stacks  104 . Depending on the embodiment, phase-compensated CC retroreflector  102  can be any of the CC retroreflectors with a phase-compensating film stack deposited on all three reflecting faces or only the first and the third reflecting faces as described above. As CC retroreflector  102  only handles linearly polarized input states, any film stack that induces nπ degree phase difference upon reflection can be used (e.g., 0 degree phase difference and Δ=180 degrees, and 180 degrees phase difference and Δ=0 degree) when all three faces are coated. A film stack that induces 2nπ degree phase difference upon reflection can be used when all three faces or only the first and the third reflecting faces are coated. Whenever nπ and 2nπ are mentioned, n is an integer including 0.  
         [0094]     Although not shown, a 45 degree mirror can be placed in the reference path so reference mirror  30  is parallel to measurement mirror  26 . The 45 degree mirror would be interposed between polarizing beam-splitter  24  and quarter-wave plate  32 . The reference mirror  30  can be mounted to a moving component to provide a differential measurement relative to measurement mirror  26 .  
         [0095]      FIG. 17B  illustrates an interferometer system  120  in one embodiment of the invention. System  120  is similar to system  100  of  FIG. 17A  except that measurement mirror  26  has been replaced with a phase-compensated CC retroreflector  122 . The measurement path has changed as follows.  
         [0096]     In the measurement path, polarizing beam-splitter  24  transmits frequency component f A  to phase-compensated cube corner  122  mounted to a moving stage. Since frequency component f A  passes through a quarter-wave plate  28 , the returning polarization is rotated 90 degrees and the newly S-polarized frequency component f A  is reflected by polarizing beam-splitter  24  to phase-compensated CC retroreflector  102 . Note that after passing through quarter-wave plate  28 , component f A  is circularly polarized when it is incident on CC retroreflector  122 .  
         [0097]     CC retroreflector  102  directs frequency component f A  again to polarizing beam-splitter  24 , which again reflects frequency component f A  to CC retroreflector  122 . Again, since frequency component f A  passes through quarter-wave plate  28 , the returning polarization is rotated 90 degrees and the newly P-polarized frequency component f A  is transmitted through polarizing beam-splitter  24  onto a receiver  34 . Note again that after passing through quarter-wave plate  28 , component f A  is circularly polarized when it is incident on CC retroreflector  122 .  
         [0098]     Depending on the embodiment, CC retroreflector  122  can be any of the CC retroreflectors with a phase-compensating film stack that induces 2nπ degree phase difference upon reflection deposited on all three reflecting faces as described above (e.g., 0 degree phase difference and Δ=180 degrees). This is because CC retroreflector  122  must return a right circularly polarized light as a left circularly polarized light (or vice versa) so that after passing through quarter-wave plate  28 , polarizing beam-splitter  24  would send the light onto the correct path. The handedness of the polarization is defined with the beam coming towards the observer.  
         [0099]      FIG. 17C  illustrates an interferometer system  140  in one embodiment of the invention. System  140  is similar to system  120  of  FIG. 17B  except that CC retroreflector  122  has been replaced with phase-compensated CC retroreflectors  142  and  144 . The measurement path has changed as follows.  
         [0100]     Polarizing beam-splitter  24  transmits frequency component f A  to phase-compensated CC retroreflector  142  mounted to a moving stage. Since frequency component f A  passes through a quarter-wave plate  28 , the returning polarization is rotated 90 degrees and the newly S-polarized frequency component f A  is reflected by polarizing beam-splitter  24  to phase-compensated CC retroreflector  102 . Note that after passing through quarter-wave plate  28 , component f A  is circularly polarized when it is incident on CC retroreflector  142 .  
         [0101]     CC retroreflector  102  directs frequency component f A  again to polarizing beam-splitter  24 , which reflects frequency component f A  to phase-compensated CC retroreflector  144 . Again, since frequency component f A  passes through quarter-wave plate  28 , the returning polarization is rotated 90 degrees and the newly P-polarized frequency component f A  is transmitted through polarizing beam-splitter  24  onto a receiver  34 . Note that after passing through quarter-wave plate  28 , component f A  is circularly polarized when it is incident on CC retroreflector  144 .  
         [0102]     Depending on the embodiment, CC retroreflectors  142  and  144  can be any of the CC retroreflectors with a phase-compensating film stack that causes 2nπ degree phase difference upon reflection deposited on all three reflecting faces as described above. This is because CC retroreflectors  142  and  144  must return a right circularly polarized light as a left circularly polarized light (or vice versa) so that after passing through quarter-wave plate  28 , polarizing beam-splitter  24  would send the light onto the correct path. The handedness of the polarization is defined with the beam coming towards the observer.  
         [0103]     Differential Interferometer System  
         [0104]      FIGS. 18A, 18B , and  18 C illustrate a differential interferometer system  200  in one embodiment of invention. A laser head  202  generates a coherent, collimated light beam consisting of two orthogonally polarized frequency components. One frequency component f A  (e.g., a measurement beam having a P-polarization) enters the interferometer&#39;s measurement path while the other frequency component f B  (e.g., a reference beam having an S-polarization) enters the interferometer&#39;s reference path.  
         [0105]     In the measurement path, a polarizing beam-splitter  204  transmits frequency component f A  to a measurement plane mirror  206  mounted to a moving stage. Since frequency component f A  passes through a quarter-wave plate  208 , the returning polarization is rotated 90 degrees and the newly S-polarized frequency component f A  is reflected by polarizing beam-splitter  204  to a phase-compensated CC retroreflector  210 . CC retroreflector  210  directs frequency component f A  again to polarizing beam-splitter  204 , which again reflects frequency component f A  to measurement plane mirror  206 . Again, since frequency component f A  passes through quarter-wave plate  208 , the returning polarization is rotated 90 degrees and the newly P-polarized frequency component f A  is transmitted through polarizing beam-splitter  204  on a mirror  212 .  
         [0106]     Mirror  212  directs frequency component f A  to a phase-compensated CC retroreflector  214 , which returns frequency component f A  back to mirror  212 . Since frequency component f A  passes through quarter-wave plate  216 , the returning polarization is rotated 90 degrees. Mirror  212  then returns the newly S-polarized frequency component f A  to polarizing beam-splitter  204 , which reflects frequency component f A  to a phase-compensated CC retroreflector  218 . CC retroreflector  218  returns frequency component f A  to polarizing beam-splitter  204 , which reflects frequency component f A  to a receiver  220 .  
         [0107]     In the reference path, polarizing beam-splitter  204  reflects frequency component f B  to phase-compensated CC retroreflector  218 . Phase-compensated CC retroreflector  218  returns frequency component f B  to polarizing beam-splitter  204 , which reflects frequency component f B  to mirror  212 . Mirror  212  directs frequency component f B  to phase-compensated CC retroreflector  214 , which returns frequency component f B  back to mirror  212 . Since frequency component f B  passes through quarter-wave plate  216 , the returning polarization is rotated 90 degrees.  
         [0108]     Mirror  212  directs the newly P-polarized frequency component f B  to polarizing beam-splitter  204 , which transmitted frequency component f B  to a reference mirror  222  that may move. Since frequency component f B  passes through a quarter-wave plate  208 , the returning polarization is rotated 90 degrees and the newly S-polarized frequency component f B  is reflected by polarizing beam-splitter  204  to phase-compensated CC retroreflector  210 . CC retroreflector  210  returns frequency component f B  to polarizing beam-splitter  204 , which again reflects frequency component f B  to reference mirror  222 . Again, since frequency component f B  passes through quarter-wave plate  208 , the returning polarization is rotated 90 degrees and the P-polarized frequency component f B  is transmitted through polarizing beam-splitter  204  onto receiver  220 .  
         [0109]     Depending on the embodiment, phase-compensated CC retroreflector  214  can be any of the CC retroreflectors with a phase-compensating film stack that induces 2nπ degree difference upon reflection deposited on all three reflecting faces as described above. This is because it must return a right circularly polarized light as a left circularly polarized light (or vice versa) so that after passing through quarter-wave plate  216 , polarizing beam-splitter  204  would send the light onto the correct path. The handedness of the polarization is defined with the beam coming towards the observer.  
         [0110]     Depending on the embodiment, phase compensated CC retroreflectors  210  and  218  can be any of the CC retroreflectors with a phase-compensating film stack deposited on all three reflecting faces. The film stack is deposited on all three reflecting faces because components f A  and f B  may strike the reflecting faces of the retroreflectors in different orders. As CC retroreflectors  210  and  218  handle only linearly polarized input states, any film stack that induces nin degree difference upon reflection can be used.  
         [0111]     Single Beam Interferometer System  
         [0112]      FIG. 19  illustrates a single beam interferometer system  300  in one embodiment of the invention. A laser head  302  generates a coherent, collimated light beam consisting of two orthogonally polarized frequency components. One frequency component f A  (e.g., a measurement beam having a P-polarization) enters the interferometer&#39;s measurement path while the other frequency component f B  (e.g., a reference beam having an S-polarization) enters the interferometer&#39;s reference path.  
         [0113]     In the measurement path, a polarizing beam-splitter  304  transmits frequency component f A  to a phase-compensated CC retroreflector  306  mounted to a moving stage. Since frequency component f A  passes through a quarter-wave plate  308 , the returning polarization is rotated 90 degrees and the newly S-polarized frequency component f A  is reflected by polarizing beam-splitter  304  to a receiver  310 .  
         [0114]     In the reference path, polarizing beam-splitter  304  reflects frequency component f B  to a phase-compensated CC retroreflector  312 . Since frequency component f B  passes through a quarter-wave plate  314 , the returning polarization is rotated 90 degrees and the newly P-polarized frequency component f B  is transmitted through polarizing beam-splitter  304  to receiver  310 .  
         [0115]     Depending on the embodiment, phase-compensated CC retroreflectors  306  and  312  can be any of the CC retroreflectors with a phase-compensating film stack that induces 2nπ phase difference upon reflection deposited on all three reflecting faces as described above. This is because CC retroreflectors  306  and  312  must return a right circularly polarized light as a left circularly polarized light (or vice versa) so that after passing through corresponding quarter-wave plates  308  and  314 , polarizing beam-splitter  304  would send the light onto the correct path. The handedness of the polarization is defined with the beam coming towards the observer.  
         [0116]     High-Resolution Interferometer System  
         [0117]      FIGS. 20A, 20B , and  20 C illustrate a high-resolution interferometer system  400  in one embodiment of invention. A laser head  402  generates a coherent, collimated light beam consisting of two orthogonally polarized frequency components. One frequency component f A  (e.g., a measurement beam having a P-polarization) enters the interferometer&#39;s measurement path while the other frequency component f B  (e.g., a reference beam having an S-polarization) enters the interferometer&#39;s reference path.  
         [0118]     In the measurement path, a polarizing beam-splitter  404  transmits frequency component f A  to a measurement plane mirror  406  mounted to a moving stage. Since frequency component f A  passes through a quarter-wave plate  408 , the returning polarization is rotated 90 degrees and the newly S-polarized frequency component f A  is reflected by polarizing beam-splitter  404  to a phase-compensated CC retroreflector  410 . CC retroreflector  410  directs frequency component f A  again to polarizing beam-splitter  404 , which again reflects frequency component f A  to measurement plane mirror  406 . Again, since frequency component f A  passes through quarter-wave plate  408 , the returning polarization is rotated 90 degrees and the newly P-polarized frequency component f A  is transmitted through polarizing beam-splitter  404  on a mirror  412 .  
         [0119]     Mirror  412  directs frequency component f A  to a phase-compensated CC retroreflector  414 , which returns frequency component f A  back to mirror  412 . Mirror  412  then returns frequency component f A  to polarizing beam-splitter  204 , which transmits frequency component f A  to plane mirror  406 . Again, since frequency component f A  passes through quarter-wave plate  408 , the returning polarization is rotated 90 degrees and the newly S-polarized frequency component f A  is reflected by polarizing beam-splitter  404  to phase-compensated CC retroreflector  410 . CC retroreflector  410  directs frequency component f A  again to polarizing beam-splitter  404 , which again reflects frequency component f A  to measurement plane mirror  406 . Again, since frequency component f A  passes through quarter-wave plate  408 , the returning polarization is rotated 90 degrees and the newly P-polarized frequency component f A  is transmitted through polarizing beam-splitter  404  to a receiver  416 .  
         [0120]     In the reference path, polarizing beam-splitter  404  reflects frequency component f B  to a mirror  418 . Since frequency component f B  passes through quarter-wave plate  420 , the returning polarization is rotated 90 degrees and the newly P-polarized frequency component f B  is transmitted through polarizing beam-splitter  404  to phase-compensated CC retroreflector  410 . CC retroreflector  410  returns frequency component f B  to polarizing beam-splitter  404 , which transmits frequency component f B  to mirror  418 . Again, since frequency component f B  passes through quarter-wave plate  420 , the returning polarization is rotated 90 degrees and the newly S-polarized frequency component f B  is reflected by polarizing beam-splitter  404  to mirror  412 .  
         [0121]     Mirror  412  directs frequency component f B  to phase-compensated CC retroreflector  414 , which returns frequency component f B  back to mirror  412 . Mirror  412  then returns frequency component f B  to polarizing beam-splitter  204 , which reflects frequency component f B  to mirror  420 . Since frequency component f B  passes through quarter-wave plate  420 , the returning polarization is rotated 90 degrees and the newly P-polarized frequency component f B  is transmitted by polarizing beam-splitter  404  to phase-compensated CC retroreflector  410 . CC retroreflector  410  returns frequency component f B  to polarizing beam-splitter  404 , which transmits frequency component f B  to mirror  418 . Again, since frequency component f B  passes through quarter-wave plate  420 , the returning polarization is rotated 90 degrees and the newly S-polarized frequency component f B  is reflected by polarizing beam-splitter  404  to receiver  416 .  
         [0122]     Depending on the embodiment, phase-compensated CC retroreflector  414  can be any of the CC retroreflectors with a phase-compensating film stack deposited on all three reflecting faces or only the first and the third faces as described above. The phase-compensating film stack that induces nπ degree phase difference upon reflection can be used when all three faces are coated. A film stack that induces 2nπ degree phase difference upon reflection can be used when all three faces or only the first and the third reflecting faces are coated.  
         [0123]     Depending on the embodiment, phase-compensated CC retroreflector  410  can be any of the CC retroreflectors with a phase-compensating film stack deposited on all three reflecting faces. The film stack is deposited on all three faces because components f A  and f B  each makes two passes through the cube corner and may strike the faces in different orders. As CC retroreflector  410  only handle linearly polarized input states, any phase-compensating interference film that induces nπ phase difference upon reflection can be used.  
         [0124]     Linear Interferometer System  
         [0125]      FIG. 21  illustrates a linear interferometer system  500  in one embodiment of the invention. A laser head  502  generates a coherent, collimated light beam consisting of two orthogonally polarized frequency components. One frequency component f A  (e.g., a measurement beam having a P-polarization) enters the interferometer&#39;s measurement path while the other frequency component f B  (e.g., a reference beam having an S-polarization) enters the interferometer&#39;s reference path.  
         [0126]     In the measurement path, a polarizing beam-splitter  504  transmits frequency component f A  to a phase-compensated CC retroreflector  506  mounted to a moving stage. Phase-compensated retroreflector  506  returns frequency component f A  to polarizing beam splitter  504 , which transmits the frequency component f to a receiver  508 .  
         [0127]     In the reference path, polarizing beam-splitter  504  reflects frequency component f B  to a phase-compensated CC retroreflector  510 . Phase-compensated retroreflector  510  returns frequency component f B  to polarizing beam splitter  504 , which reflects the frequency component f B  to receiver  508 .  
         [0128]     Depending on the embodiment, phase-compensated CC retroreflectors  506  and  510  can be any of the CC retroreflectors with a phase-compensating film stack deposited on all three reflecting faces or only the first and the third reflecting faces as described above. The phase-compensating film stack that induces nπ degree phase difference upon reflection can be used when all three faces are coated. A film stack that induces 2nπ degree phase difference upon reflection can be used when all three faces or only the first and the third reflecting faces are coated.  
       SUMMARY  
       [0129]     With the aid of a numerical model, polarization effects from a TIR cube corner were analyzed. This model predicts significant changes to the azimuth and ellipticity angles of incident polarization from an uncoated cube corner. As a consequence, multilayer interference coatings were designed to produce zero and 180 degrees phase difference between S and P states for TIR reflection. The results indicate that with all or only first and third faces coated, the azimuth and ellipticity angles of the initial polarization were preserved.  
         [0130]     Various other adaptations and combinations of features of the embodiments disclosed are within the scope of the invention. Although the various interferometry systems described above use vertically and horizontally polarized light, the interferometry systems can also use left and right circularly polarized light. Although the various interferometry systems described above show measurement along a single axis, it is understood that these systems could be adopted for multi-axes measurement by replicating the setups or integrating the single-axis measurement components into a multi-axes module. Although specific coating designs for BK-7 cube corners are disclosed, one skilled in the art can modify the coating designs for cube corners made of other materials such as fused silica, SF-10, and etc. Furthermore, one skilled in the art can modify the coating designs to suit other operating wavelengths. Numerous embodiments are encompassed by the following claims.  
         [0000]     Appendix I  
         [0131]     For a monochromatic plane wave E propagating in the z direction with coordinates E x  and E y , 
 
 E   x   =a   1 cos(τ+δ 1 )  I.1.a 
 
 E   y   =a   2 cos(τ+δ 2 )  I.1.b 
 
E z =0,  I.1.c 
 
 where a 1  is the amplitude along the x-axis, δ 1  is the phase along the x-axis (please note this is different from the relative phase difference δ 1  after reflection from the first face described above), a 2  is the amplitude along the y-axis, δ 2  is the phase along the y-axis (please note this is different from the relative phase difference δ 2  after reflection from the second face described above), τ=ωt−kz, ω is angular frequency, t is time, k is wave number (2π/λ), and z is propagation direction. 
 
         [0132]     Rewriting (I.1.a) and (I.1.b), 
 
 E   x   /a   1 =cos(τ+δ 1 )=cos τ cos δ 1 −sin τ sin δ 1 , I.1.d 
 
 E   y   /a   2 =cos(τ+δ 2 )=cos τ cos δ 2 −sin τ sin δ 2 .  I.1.e 
 
         [0133]     To eliminate τ dependence, we multiply each component by sin δ 1 , sin δ 2 , cos δ 1  and cos δ 2 : 
 
( E   x   /a   1 )sin δ 2 =sin δ 2 (cos τ cos δ 1 −sin τ sin δ 1 )  I.2.a 
 
( E   y   /a   2 )sin δ 1 =sin δ 1 (cos τ cos δ 2 −sin τ sin δ 2 )  I.2.b 
 
( E   x   /a   1 )cos δ 2 =cos δ 2 (cos τ cos δ 1 −sin τ sin δ 1 )  I.2.c 
 
( E   y   /a   2 )cos δ 1 =cos δ 1 (cos τ cos δ 2 −sin τ sin δ 2 )  I.2.d 
 
         [0134]     Subtracting (I.2.b) from (I.2.a), 
 
( E   x   /a   1 )sin δ 2 −( E   y   /a   2 )sin δ 1 =cos τ sin(δ 2 −δ 1 )  I.3.a 
 
         [0135]     Subtracting (I.2.d) from (I.2.c), 
 
( E   x   /a   1 )cos δ 2 −( E   y   /a   2 )cos δ 1 =sin τ sin(δ 2 −δ 1 ).  I.3.b 
 
         [0136]     Squaring and adding (I.3.a) and (I.3.b) yields, 
 
( E   x   /a   1 ) 2 +( E   y   /a   2 ) 2 −2( E   x   /a   1 )( E   y   /a   2 )(sin δ 2 sin δ 1 +cos δ 1 cos δ 2 )=sin 2 (δ 2 −δ 1 ) 
 
 and simplifying, 
 
( E   x   /a   1 ) 2 +( E   y   /a   2 ) 2 −2( E   x   /a   1 )(E y   /a   2 )cos(δ 2 −δ 1 )=sin 2 (δ 2 −δ 1 ).  I.4 
 
 Appendix II 
 
         [0137]     In the rotated coordinate system (ξ, η), the components of the electric vector are related by, 
 
 E   ξ   =E   x cos ψ+ E   y sin ψ  II.1.a 
 
 E   η   =−E   x sin ψ+E y cos ψ,  II.1.b 
 
 where ψ is the angle between (ξ, η) and (x, y) coordinate axes. In the (ξ, η) coordinate system, the equation for the electric vector is given by, 
 
 E   ξ   =a *cos(τ+δ ξ )  II.2.a 
 
 E   η   =b *cos(τ+δ η ),  II.2.b 
 
 where a is the amplitude along the ξ-axis, b is the amplitude along the η-axis, and τ=ωt−kz. Then using the results of equation II.2.a and II.2.b, the expressions for the electric vector reduces to, 
 
 E   ξ   =a *cos(τ+δ ξ )  II.2.c 
 
 E   η   =+/−b *sin(τ+δ ξ ).  II.2.d 
 
         [0138]     Now a relationship between the amplitudes a, b of the rotated frame and a 1 , a 2  of the x, y frame is established. First equations I.1.a, I.1.b, II.2.c and (II.2.d) are expanded as follows: 
 
 E   x   =a   1 cos(τ+δ 1 )= a   1 (cos τ cos δ 1 −sin τ sin δ 1 )  II.3.a 
 
 E   y   =a   2 cos(τ+δ 2 )= a   2 (cos τ cos δ 2 −sin τ sin τ 2 )  II.3.b 
 
 E   ξ   =a *cos(τ+δ ξ )= a (cos τ cos δ ξ −sin τ sin δ ξ )  II.3.c 
 
 E   η   =+/−b *sin(τ+δ ξ )=+/− b (sin τ cos δ ξ +cos τ sin δ ξ ).  II.3.d 
 
         [0139]     Combining equation II.1.a and II.1.b with equations II.3.a and II.3.b, 
 
 E   ξ   =a   1 (cos τ cos δ 1 −sin τ sin δ 1 )cos ψ+ a   2 (cos τ cos δ 2 −sin τ sin δ 2 )sin ψ  II.4.a 
 
E η   =−a   1 (cos τ cos δ 1 −sin τ sin δ 1 )sin ψ+ a   2 (cos τ cos δ 2 −sin τ sin δ 2 )cos ψ  II.4.b 
 
 then equating the corresponding terms of equations II.3.c, II.3.d and equations II.4.a, II.4.b: 
 
 For E ξ  equation, 
 
         [0140]     cos τ terms: 
 
 a *cos δ ξ   =a   1 cos δ 1 cos ψ+ a   2 cos δ 2 sin ψ  II.5.a 
 
         [0141]     sin τ terms: 
 
 a *sin δ ξ   =a   1 sin δ 1 cos ψ+ a   2 sin δ 2 sin ψ  II.5.b 
 
 For E, equations, 
 
         [0142]     cos τ terms: 
 
+/− b *sin δ ξ   =−a   1 cos δ 1 sin ψ+ a   2 cos δ 2 cos ψ  II.6.a 
 
         [0143]     sin τ terms: 
 
+/− b *cos δ ξ   =a   1 sin δ 1 sin ψ− a   2 sin δ 2 cos ψ  II.6.b 
 
         [0144]     Now squaring and adding equations II.5.a and II.5.b, 
 
 a   2   =a   1   2 cos 2   ψ+a   2   2 sin 2 ψ+2 a   1   a   2 cos ψ sin ψ cos(δ 2 −δ 1 ),  II.7.a 
 
 and performing the same operations with equations II.6.a and II.6.b, 
 
 b   2   =a   1   2 sin 2   ψ+a   2   2 cos 2 ψ−2 a   1   a   2 sin ψ cos ψcos(δ 2 −δ 1 ).  II.7.b 
 
         [0145]     Adding (II.7.a) and (II.7.b), 
 
 a   2   +b   2   =a   1   2   +a   2   2 .  II.8 
 
         [0146]     Next equations II.5.a and II.6.b are multiplied together and equations II.5.b and II.6.a are also multiplied together as follows:  
                 +     /     -   a         ⁢           ⁢   b   *     cos   2     ⁢     δ   ξ       =         -     a   1       ⁢     a   2     ⁢   sin   ⁢           ⁢     δ   2     ⁢   cos   ⁢           ⁢     δ   1     ⁢     cos   2     ⁢   ψ     +       a   1     ⁢     a   2     ⁢   sin   ⁢           ⁢     δ   1     ⁢   cos   ⁢           ⁢     δ   2     ⁢     sin   2     ⁢   ψ               II   ⁢     .9   .   a                     +     /     -   a         ⁢           ⁢   b   *     sin   2     ⁢     δ   ξ       =         -     a   1       ⁢     a   2     ⁢   sin   ⁢           ⁢     δ   2     ⁢   cos   ⁢           ⁢     δ   1     ⁢     sin   2     ⁢   ψ     +       a   1     ⁢     a   2     ⁢   sin   ⁢           ⁢     δ   1     ⁢   cos   ⁢           ⁢     δ   2     ⁢     cos   2     ⁢   ψ               II   ⁢     .9   .   b               
 
 then adding (II.9.a) and (II.9.b), 
 
 +/−ab=a   1   a   2 sin(δ 1 −δ 2 ) 
 
 or alternatively, 
 
 −/+ab=a   1   a   2 sin(δ 2 −δ 1 ).  II.10 
 
         [0147]     Furthermore, dividing equation II.6.b by equation II.5.a, and dividing equation II.6.a by equation II.5.b yields: 
 
+/− b/a =( a   1 sin δ 1 sin ψ− a   2 sin δ 2 cos ψ)/( a   1 cos δ 1 cos ψ+ a   2 cos δ 2 sin ψ)  II.11.a 
 
+/− b/a =(− a   1 cos δ 1 sin ψ+ a   2 cos δ 2 cos ψ)/( a   1 sin δ 1 cos ψ+ a   2 sin δ 2 sin ψ),  II.11.b 
 
 and equating the right hand side (RHS) of equations II.11.a and II.11.b, 
 
( a   1   2   −a   2   2 )sin 2ψ=2 a   1   a   2 cos(δ 2 −δ 1 )cos 2ψ
 
 or equivalently, 
 
( a   1   2   −a   2   2 )tan 2ψ=2 a   1   a   2 cos δ  II.12 
 
 where δ=δ 2 −δ 1 . 
 
         [0148]     The last relation is derived by multiplying equation II.10 by 2 and then dividing by equation II.8, 
 
−/+2 ab /( a   2   +b   2 )=2 a   1   a   2 sin δ/( a   1   2   +a   2   2 )  II.13 
 
 Appendix III 
 
         [0149]     To derive the expressions for the Stokes parameters, let us recall equations II.8, II.12, and II.13, 
 
 a   2   +b   2   =a   1   2   +a   2   2 .  (II.8) 
 
( a   1   2   −a   2   2 )tan 2ψ=2 a   1   a   2 cos δ  (II.12) 
 
−/+2 ab /( a   2   +b   2 )=2 a   1   a   2 sin δ/( a   1   2   +a   2   2 )  (II.13) 
 
 and define 
 
tan α= a   2   /a   1   (III.0.a) 
 
tan χ=+/− b/a.   (III.0.b) 
 
         [0150]     From equation II.12, we make the following definitions:  
         [0151]     Right Hand Side Definitions (RHS): 
 
 S   1   =a   1   2   −a   2   2   II.1 
 
 S   2 =2 a   1   a   2 cos δ  III.2 
 
         [0152]     Left Hand Side Definition (LHS): 
 
 S   2 =( a   1   2   −a   2   2 )tan 2ψ  III.3 
 
         [0153]     Similarly from equation II.13, we have the following definitions:  
         [0154]     Right Hand Side Definition (RHS): 
 
 S   3 =2 a   1   a   2 sin δ  III.4 
 
         [0155]     Left Hand Side Definition (LHS): 
 
 S   3 =−/+2 ab ( a   1   2   +a   2   2 )/( a   2   +b   2 )  III.5 
 
         [0156]     Next defined is a dependent parameter S 0 . This is defined by combining the squares of the RHS definitions of S 1 , S 2  and S 3 ,  
         [0157]     Right Hand Side Definition (RHS):  
                     S   0   2     =       S   1   2     +     S   2   2     +     S   3   2                   =         (       a   1   2     -     a   2   2       )     2     +       (     2   ⁢     a   1     ⁢     a   2     ⁢   cos   ⁢           ⁢   δ     )     2     +       (     2   ⁢     a   1     ⁢     a   2     ⁢   sin   ⁢           ⁢   δ     )     2                           ⁢     =       (       a   1   2     +     a   2   2       )     2                     S   0     =       a   1   2     +     a   2   2                     III   ⁢   .6             
 
         [0158]     Left Hand Side Definition (LHS): 
 
 S   0   2   =S   1   2   +S   2   2   +S   3   2   III.7 
 
         [0159]     Taking next the LHS definition of S 3 , 
 
 S   3 =−/+2 ab ( a   1   2   +a   2   2 )/( a   2   +b   2 ), 
 
 or alternatively with the substitution of equation III.6, S 3  becomes, 
 
 S   3 =−/+2 abS   0 /( a   2   +b   2 ). 
 
         [0160]     Further, note that tan χ=−/+b/a; and thus S 3  may be expressed as, 
 
 S   3 =2 S   0 tan χ/(1+tan 2 χ)= S   0 sin 2χ
 
 S   3   =S   0 sin 2χ.  111.8 
 
         [0161]     Now to express S 1  and S 2  in terms of the parameters S 0 , ψ and χ, we take the LHS definition of equation III.3 and rewrite it as, 
 
 S   2   =S   1 tan 2ψ.  III.9 
 
         [0162]     Then substituting equations III.8 and III.9 into equation III.7, 
 
 S   0   2   =S   1   2 +( S   1 tan 2ψ) 2 +( S   0 sin 2χ) 2  
 
 and solving for S 1 , 
 
 S   2   =S   0 cos 2ψ cos 2χ.  III.10 
 
         [0163]     Now substitute equation III.10 into equation III.9 to arrive at a new form of S 2 , 
 
 S   2   =S   0 sin 2ψ cos 2χ.  III.11 
 
         [0164]     Summarizing the results, 
 
 S   0   2   =S   1   2   +S   2   2   +S   3   2  
 
 S   1   =S   0 cos 2ψ cos 2χ
 
 S   2   =S   0 sin 2ψ cos 2χ
 
 S   3   =S   0 sin 2χ. 
 
         [0165]     Expressing the Stokes parameters with respect to RHS definitions, 
 
 S   0   =a   1   2   +a   2   2  
 
 S   1   =a   1   2   −a   2   2  
 
 S   2 =2 a   1   a   2 cos δ
 
 S   3 =2 a   1   a   2 sin δ. 
 
         [0166]     These RHS Stokes parameters are then normalize by dividing by S 1 . 
 
s 0 =1 
 
 s   1 =( a   1   2   −a   2   2 )/( a   1   2   +a   2   2 ) 
 
 s   2 =2 a   1   a   2 cos δ/( a   1   2   +a   2   2 ) 
 
 s   3 =2 a   1   a   2 sin δ/( a   1   2   +a   2   2 ). 
 
         [0167]     Then expressing in terms of tan α=a 2 /a 1 , 
 
s 0 =1  (III.12) 
 
 s   1 =(1−tan 2 α)/(1+tan 2 α)  (III.13) 
 
 s   2 =2 tan α cos δ/(1+tan 2 α)  (III.14) 
 
 s   3 =2 tan α sin δ/(1+tan 2 α).  (III.15)