Abstract:
Telecentric projection lenses are disclosed which are precise in terms of their F number and which have a high resolving power and high contrast, namely well corrected spherical aberration, coma, curvature of field and distortion, and are intended for use at aperture efficiency of 100% and at magnification of about 1/10. The telecentric projection lens comprises a first lens group consisting of a positive single lens, a second lens group consisting of a negative single lens and a third lens group consisting of two positive lenses, arranged in this order as viewed from the object field side. The lens of the second group is a double concave lens and one of the positive lenses in the third group has a cemented surface whose center of curvature lies on the side of object field. The focal lengths f 1 , f 2  and f 3  of the first, second and third groups and the focal length f of the whole lens system satisfy the following conditions: 
     
       1.69≦|f.sub.1 /f.sub.2 |≦2.55 
     
     
       -0.33≦f.sub.2 /f≦-0.19 
     
     
       0.41≦f.sub.3 /f≦0.59.

Description:
BACKGROUND OF THE INVENTION 
     1. Field of the Invention 
     The present invention relates to telecentric projection lenses which are intended to use at a magnification of about 1/10 and which are precise in terms of their F number and have their aberrations well corrected. 
     2. Description of the Prior Art 
     In recent years there have been developed and used original scanning methods in which solid image pick-up elements are employed as scanning means for reading apparatus with the solid image pick-up elements are arranged in the image plane as scanning photo-receptor elements to scan the original. 
     To carry out the above known scanning methods, a projection lens must be used to transmit the original image to the solid image pick-up elements and further a color separation prism must be interposed between the projection lens and the focal plane containing the elements to transmit the color signals of the original to the solid image pick-up elements. If a common non-telecentric lens is used as the projection lens in this case, the off-axial rays are obliquely incident upon the color separation prism and shading is caused thereby. To overcome the problem of shading there is used a projection lens having a telecentric property. By using such lens, the principal rays of the incident light pass through the focal point on the side of object field and therefore the principal rays of the exit light including off-axial rays on the side of image field can run in parallel with the optical axis. This is one effective method already known for overcoming the problem of shading caused by the color separation prism. 
     However, projection lenses useful for carrying out the above scanning method have to satisfy many requirements at the same. In general, such a projection lens has to satisfy the following requirements: 
     (1) The F number of the lens should be relatively precise. To carry out a high speed scanning employing solid image pick-up elements it is preferable to increase, as much as possible, the quantity of exposure light to the elements per unit time. On the other hand, it is also preferable to use, as an original illumination lamp, a light source whose illumination is as low as possible. For this reason, the projection lens is required to have a relatively precise F number. 
     (2) For the purpose of reduction in size of the apparatus, namely reduction of the distance between the original surface and the focal plane, the lens should have a wide angle of field. 
     (3) The lens should have a high resolving power because the size of solid image pick-up element is very small which is in the order of 15μ. 
     (4) The lens should be suitable for use with its apperture efficiency for off-axis being 100%. This is because the distribution of light intensity must be uniform over all of the solid image pick-up elements. 
     (5) The lens should be able to project the original surface uniformly. In other words, the distortion of the lens should be of low level. 
     (6) Since a color separation prism must be interposed between the lens and the solid image pick-up elements, the lens should have a long back focal length. 
     SUMMARY OF THE INVENTION 
     Accordingly, it is a general object of the present invention to provide telecentric projection lenses which satisfy the above requirements. 
     It s a more specific object of the present invention to provide such telecentric lenses which are precise in terms of F number and which have high resolving power, as well as contrast, as the result of well corrected spherical aberration, coma, curvature of field and distortion and which are suitable for use at aperture efficiency of 100% and at a magnification of about 1/10. 
     To attain the above objects according to the first embodiment of the present invention, there is provided a projection lens which has a pupil located at the focal point on the side of object field of the whole lens system and comprises a first lens or group composed of a positive single lens or lens elements, a second lens or group composed of a double concave single lens and a third lens group or composed of two positive lenses or lens elements arranged in this order as viewed from the object field side, wherein one of the two positive lenses in the third lens group has a cemented surface whose center of curvature lies on the side of object field, and which projection lens is so designed as to satisfy the following conditions: 
     
         (1) 1.69≦|f.sub.1 /f.sub.2 |≦2.55 
    
     
         (2) -0.33≦f.sub.2 /f≦-0.19 
    
     
         (3) 0.41≦f.sub.3 /f≦0.59 
    
     where, f 1 , f 2  and f 3  are focal lengths of the first, first lens and the second and third lens groups, respectively, and f is the focal length of the whole system. 
     The second embodiment of the invention is directed to a projection lens which has a pupil located at the focal point on the side of object field of the whole lens system and comprises a first lens composed of a positive single lens, a second lens group composed of a double concave singal lens and a third lens group composed of three or four lenses, arranged in this order as viewed from the object field side, wherein any one of the lenses in the third group has a cemented surface, and which projection lens is so designed as to satisfy the following conditions: 
     
         (1&#39;) 1.72≦|f.sub.1 /f.sub.2 |≦2.58 
    
     
         (2&#39;) -0.33≦f.sub.2 /f≦-0.19 
    
     
         (3&#39;) 0.41≦f.sub.3 /f≦0.59 
    
     wherein, f 1 , f 2  and f 3  are focal lengths of the first lens and the second and third lens groups, respectively, and f is the focal length of the whole system. 
     Other and further objects, features and advantages of the present invention will appear more fully from the following description with reference to the accompanying drawings. 
    
    
     BRIEF DESCRIPTION OF DRAWINGS 
     FIG. 1A is a cross-sectional view of the lens described in Example 1; 
     FIG. 1B shows aberrations thereof; 
     FIG. 1C shows transverse aberrations on Gaussian image plane thereof; 
     FIG. 2A is a cross-sectional view of Example 2 and FIG. 2B shows aberrations thereof; 
     FIG. 3A is a cross-sectional view of Example 3 and FIG. 3B shows aberrations thereof; 
     FIG. 4A is a cross-sectional view of Example 4 and FIG. 4B shows aberrations thereof; 
     FIG. 5A is a cross-sectional view of Example 5 and FIG. 5B shows aberrations thereof; 
     FIG. 6A is a cross-sectional view of Example 6 and FIG. 6B shows aberrations thereof; 
     FIG. 7A is a cross-sectional view of Example 7 and FIG. 7B shows aberrations thereof; 
     FIG. 8A is a cross-sectional view of Example 8 and FIG. 8B shows aberrations thereof; 
     FIG. 9A is a cross-sectional view of Example 9 and FIG. 9B shows aberrations thereof; 
     FIG. 10A is a cross-sectional view of Example 10 and FIG. 10B shows aberrations thereof; 
     FIG. 11A is a cross-sectional view of Example 11 and FIG. 11B shows aberrations thereof; 
     FIG. 12A is a cross-sectional view of Example 12 and FIG. 12B shows aberrations thereof; 
     FIG. 13A is a cross-sectional view of Example 13 and FIG. 13B shows aberrations thereof and FIG. 13C shows transverse aberrations on Gaussian image plane thereof; 
     FIG. 14A is a cross-sectional view of Example 14 and FIG. 14B shows aberrations thereof; 
     FIG. 15A is a cross-sectional view of Example 15 and FIG. 15B shows aberrations thereof; 
     FIG. 16A is a cross-sectional view of Example 16 and FIG. 16B shows aberrations thereof; 
     FIG. 17A is a cross-sectional view of Example 17 and FIG. 17B shows aberrations thereof; 
     FIG. 18A is a cross-sectional view of Example 18 and FIG. 18B shows aberrations thereof; 
     FIG. 19A is a cross-sectional view of Example 19 and FIG. 19B shows aberrations thereof; 
     FIG. 20A is a cross-sectional view of Example 20 and FIG. 20B shows aberrations thereof; and 
     FIG. 21A is a cross-sectional view of Example 21 and FIG. 21B shows aberrations thereof. 
    
    
     DESCRIPTION OF PREFERRED EMBODIMENTS 
     Initially, there is described conditions (1) to (3) relating to the projection lens according to the first embodiment of the present invention. 
     By satisfying the condition (1) the spherical aberration can be well corrected while keeping the balance of the refractive powers of the first and second lens groups. The lens of the present invention is used in a telecentric system and the distance between the principal points of the second and third groups is larger than the distance between those of the first and second groups. Therefore, positions at which the paraxial rays pass through the first lens group are greatly spaced apart from the optical axis, which produces a large quantity of spherical aberration. 
     When |f 1  /f 2  | is below the lower limit 1.69, then the refractive power of the first group becomes high and the paraxial rays passing through the surface are intensely refracted in the direction toward the optical axis. As the result of it, a large quantity of negative spherical aberration is produced. On the contrary, when |f 1  /f 2  | exceeds the upper limit of 2.55, the refractive power of the second lens group becomes high and the second group produces such a level of positive spherical aberration which overly compensates the negative spherical aberration produced in the first group. 
     The condition (2) must be satisfied to correct the curvature of field of the lens system. 
     When f 2  /f is larger than the upper limit, -0.19, then Petzval sum is overcompensated and curvature of field is overly currected. To correct it, the refractive power of the second lens group must be increased in absolute value. However, as described hereinafter in connection with the condition (3), if the absolute value of the refractive power of the second group is so increased, then there is produced in the second group a large quantity of distortion which is difficult to correct. 
     On the contrary, when f 2  /f is less than the power limit of -0.33, it becomes difficult to correct Petzval sum of the whole system and thereby undercorrection of the curvature of field is caused. 
     Condition (3) is necessary for correction of the curvature of field and distortion. Since the lens is used in a telecentric system, the positions at which principal rays pass through the third lens group are greatly spaced apart from the optical axis. When f 3  /f is smaller than the lower limit of 0.4, the refractive power of the third lens group becomes high and principal rays passing through the third group are intensely refracted in the direction toward the optical axis. Thereby, a large quantity of distortion is produced. On the contrary, if f 3  /f is larger than the upper limit of 0.59, then curvature of field becomes worse to the extent that it may be hardly corrected. 
     Now, the shape of lens according to the first embodiment of the invention is described in detail. 
     As described above, the lens system of the present invention must be precise in terms of F number. To attain the object, it is most advantageous to effectively correct the spherical aberration of the first lens group which is the group at which paraxial rays are most apart from the optical axis. To this end, the surface of the first lens group on the side of object field is so shaped as to be convexed toward the object field side. 
     In the lens system of the present invention, the lens group which corrects Petzval sum is only the second group. Therefore, the condition of power to the second group becomes severe. For this reason, the negative lens of the second group is shaped as a double concave lens to reduce aberrations in the second group. 
     Since a color separation prism is to be interposed between the image plane and the nearest lens surface to the image field, the lens system is required to have a long back focal length. To this end, the third lens group is composed of two positive lenses and is formed in such manner that the air lens formed by the two positive lenses has a shape of double concave lens. 
     The following examples, Examples 1 to 12 together with FIGS. 1 to 12 illustrate the design of above projection lenses according to the first aspect of the invention and demonstrate the effect of the invention. 
     In the examples, 
     Ri is the radius of curvature of the i-th surface of the lens system; 
     Di is the thickness of air spacing on axis between the i-th surface and the i+1-th surface; 
     ω is angle of field; 
     β is magnification of focus; 
     Ni is the refractive index of the i-th lens to D ray; 
     νi is Abbe&#39;s number of the i-th lens; 
     fi is the focal length of the i-th group; and 
     D0 is the air spacing on axis from the pupil SL to the R1 surface. 
     For all of Examples 1 to 12 the F number is 1:5. 
     EXAMPLE 1 
     
         ______________________________________f = 1 angle of field = 25.2° β = 0.12343______________________________________R1 =  0.3986      D1 = 0.134  N1 = 1.72  ν1 = 50.2R2 = -5.6178      D2 = 0.1653R3 = -0.3958      D3 = 0.0701 N2 = 1.80518                             ν2 = 25.4R4 =  0.4543      D4 = 0.1345R5 = -2.8306      D5 = 0.1384 N3 = 1.697 ν3 = 48.5R6 = -0.3292      D6 = 0.1588 N4 = 1.72825                             ν4 = 28.5R7 = -0.6113      D7 = 0.0181R8 =  0.8726      D8 = 0.1635 N5 = 1.7725                             ν5 = 49.6R9 = -3.6073|f.sub.1 /f.sub.2 | = 2.0161      f.sub.2 /f = -0.2543                  f.sub.3 /f = 0.4971                             DO = 0.0213______________________________________ 
    
     EXAMPLE 2 
     
         ______________________________________f = 1 angle of field = 25.2° β = -0.12343______________________________________R1 =  0.4403      D1 = 0.1682 N1 = 1.72  ν1 = 50.2R2 = -1.8559      D2 = 0.13R3 = -0.4636      D3 = 0.0329 N2 = 1.80518                             ν2 = 25.4R4 =  0.5059      D4 = 0.1492R5 = -1.6321      D5 = 0.2362 N3 = 1.697 ν3 = 48.5R6 = -0.4315      D6 = 0.0414 N4 = 1.72825                             ν4 = 28.5R7 = -0.5964      D7 = 0.0528R8 =  1.0092      D8 = 0.0925 N5 = 1.7725                             ν5 = 49.6R9 = -5.1222|f.sub.1 /f.sub.2 | = 1.723      f.sub.2 /f = -0.296                  f.sub.3 /f = 0.5764                             DO = 0.1764______________________________________ 
    
     EXAMPLE 3 
     
         ______________________________________f = 1 angle of field = 25.2° β = -0.12343______________________________________R1 =  0.3446      D1 = 0.1461 N1 = 1.72  ν1 = 50.2R2 =  4.6621      D2 = 0.1688R3 = -0.3246      D3 = 0.0205 N2 = 1.80518                             ν2 = 25.4R4 =  0.3854      D4 = 0.0696R5 =  1.3077      D5 = 0.2144 N3 = 1.674 ν3 = 48.5R6 = -0.3306      D6 = 0.2108 N4 = 1.72825                             ν4 = 28.5R7 =  0.5451      D7 = 0.0697R8 =  0.9435      D8 = 0.1443 N5 = 1.7725                             ν5 = 49.6R9 =  28.7464|f.sub.1 /f.sub.2 | = 2.3587      f.sub.2 /f = -0.216                  f.sub.3 /f = 0.4485                             DO = 0.0219______________________________________ 
    
     EXAMPLE 4 
     
         ______________________________________f = 1 angle of field = 25.2° β = -0.12343______________________________________R1 =   0.4822      D1 = 0.1456 N1 = 1.72  ν1 = 50.2R2 = -35.0608      D2 = 0.2244R3 =  -0.4207      D3 = 0.0325 N2 = 1.80518                             ν2 = 25.4R4 =   0.6509      D4 = 0.0834R5 =  -1.3008      D5 = 0.1128 N3 = 1.697 ν3 = 48.5R6 =  -0.2572      D6 = 0.1755 N4 = 1.72825                             ν4 = 28.5R7 =  -0.4887      D7 = 0.0625R8 =   0.8924      D8 = 0.1618 N5 = 1.7725                             ν5 = 49.6R9 =  -5.1437|f.sub.1 /f.sub.2 | = 2.1135      f.sub.2 /f = -0.3131                  f.sub.3 /f = 0.4966                             DO =0 0.0399______________________________________ 
    
     EXAMPLE 5 
     
         ______________________________________f = 1 angle of field = 25.2° β= -0.12343______________________________________R1 =  0.3052      D1 = 0.1736 N1 = 1.72  ν1 = 50.2R2 = -1.5021      D2 = 0.089R3 = -0.3001      D3 = 0.0837 N2 = 1.80518                             ν2 = 25.4R4 =  0.3522      D4 = 0.1376R5 =  3.8193      D5 = 0.1725 N3 = 1.697 ν3 = 48.5R6 = -0.5969      D6 = 0.167  N4 = 1.72825                             ν4 = 28.5R7 = -0.7671      D7 = 0.0713R8 =  0.6312      D8 = 0.0988 N5 = 1.7725                             ν5 = 49.6R9 =  4.6414|f.sub.1 /f.sub.2 | = 1.9285      f.sub.2 /f = -0.1904                  f.sub.3 /f = 0.4961                             DO = 0.04______________________________________ 
    
     EXAMPLE 6 
     
         ______________________________________f = 1 angle of field = 25.2° β = -0.12343______________________________________R1 =  0.4186      D1 = 0.0891 N1 = 1.72  ν1 = 50.2R2 = -9.096      D2 = 0.197R3 = -0.4351      D3 = 0.0279 N2 = 1.80518                             ν2 = 25.4R4 =  0.4955      D4 = 0.1493R5 = -1.6863      D5 = 0.1956 N3 = 1.697 ν3 = 48.5R6 = -0.3195      D6 = 0.1151 N4 = 1.72825                             ν4 = 28.5R7 = -0.5561      D7 = 0.0153R8 =  0.9878      D8 = 0.0852 N5 = 1.7725                             ν5 = 49.6R9 = -4.6263|f.sub.1 /f.sub.2 | = 1.9652      f.sub.2 /f = -0.2844                  f.sub.3 /f = 0.5261                             DO = 0.0437______________________________________ 
    
     EXAMPLE 7 
     
         ______________________________________f = 1 angle of field = 25.2° β = -0.12343______________________________________R1 =  0.4165      D1 = 0.1912 N1 = 1.72  ν1 = 50.2R2 = -3.39 D2 = 0.1481R3 = -0.3872      D3 = 0.0785 N2 = 1.80518                             ν2 = 25.4R4 =  0.4866      D4 = 0.1315R5 = -0.7668      D5 = 0.0804 N3 = 1.697 ν3 = 48.5R6 = -0.251      D6 = 0.1176 N4 = 1.72825                             ν4 = 28.5R7 = -0.4241      D7 = 0.0631R8 =  0.7723      D8 = 0.1651 N5 = 1.7725                             ν5 = 49.6R9 = -6.2772|f.sub.1 /f.sub.2 | = 2.0437      f.sub.2 /f = -0.2571                  f.sub.3 /f = 0.4901                             DO = 0.0373______________________________________ 
    
     EXAMPLE 8 
     
         ______________________________________f = 1 angle of field = 25.2° β = -0.12343______________________________________R1 =  0.3482      D1 = 0.1182 N1 = 1.72  ν1 = 50.2R2 = -4.0406      D2 = 0.1449R3 = -0.3748      D3 = 0.0764 N2 = 1.80518                             ν2 = 25.4R4 =  0.3816      D4 = 0.1265R5 =  1.7347      D5 = 0.2061 N3 = 1.697 ν3 = 48.5R6 = -0.4603      D6 = 0.1757 N4 = 1.72825                             ν4 = 28.5R7 = -0.7049      D7 = 0.0618R8 =  0.8119      D8 = 0.1439 N5 = 1.7725                             ν5 = 49.6R9 =  8.081|f.sub.1 /f.sub.2 | = 2.0039      f.sub.2 /f = -0.2247                  f.sub.3 /f = 0.4957                             DO = 0.0344______________________________________ 
    
     EXAMPLE 9 
     
         ______________________________________f = 1 angle of field = 25.2° β = -0.12343______________________________________R1 =  0.4429      D1 = 0.0335 N1 = 1.72  ν1 = 50.2R2 = -4.1849      D2 = 0.204R3 = -0.5629      D3 = 0.0209 N2 = 1.80518                             ν2 = 25.4R4 =  0.4844      D4 = 0.2198R5 = -2.4032      D5 = 0.2335 N3 = 1.697 ν3 = 48.5R6 = -0.4017      D6 = 0.0267 N4 = 1.72825                             ν4 = 28.5R7 = -0.6186      D7 = 0.0515R8 =  1.0314      D8 = 0.0917 N5 = 1.7725                             ν5 = 49.6R9 = -6.7698|f.sub.1 /f.sub.2 | = 1.741      f.sub.2 /f = -0.3205                  f.sub.3 /f = 0.5832                             DO = 0.0585______________________________________ 
    
     EXAMPLE 10 
     
         ______________________________________f = 1 angle of field = 25.2° β = -0.12343______________________________________R1 =  0.3693      D1 = 0.1981 N1 = 1.72  ν1 = 50.2R2 =  2.5301      D2 = 0.1765R3 = -0.2971      D3 = 0.0161 N2 = 1.80518                             ν2 = 25.4R4 =  0.4745      D4 = 0.0643R5 =  1.9365      D5 = 0.1127 N3 = 1.697 ν3 = 48.5R6 = -0.2642      D6 = 0.2305 N4 = 1.72825                             ν4 = 28.5R7 = -0.4824      D7 = 0.07R8 =  0.9622      D8 = 0.1526 N5 = 1.7725                             ν5 = 49.6R9 = -9.8269|f.sub.1 /f.sub.2 | = 2.5728      f.sub.2 /f = -0.2248                  f.sub.3 /f = 0.4199                             DO = 0.0248______________________________________ 
    
     EXAMPLE 11 
     
         ______________________________________f = 1 angle of field = 25.2° β = -0.12343______________________________________R1 =  0.3879     D1 = 0.1531 N1 = 1.72  ν1 = 50.2R2 = -9.0167     D2 = 0.165R3 = -0.3786     D3 = 0.0698 N2 = 1.80518                            ν2 = 25.4R4 =  0.4629     D4 = 0.1477R5 =  21.2811     D5 = 0.1962 N3 = 1.697 ν3 = 48.5R6 = -0.3036     D6 = 0.1701 N4 = 1.72825                            ν4 = 28.5R7 = -0.5721     D7 = 0.052R8 =  0.7612     D8 = 0.1554 N5 = 1.7725                            ν5 = 49.6R9 =  6.2487|f.sub.1 /f.sub.2 | = 2.0849     f.sub.2 /f = -0.2494                 f.sub.3 /f = 0.4834                            DO = -0.2064______________________________________ 
    
     EXAMPLE 12 
     
         ______________________________________f = 1 angle of field = 25.2° β = -0.12343______________________________________R1 =  0.3756      D1 = 0.1325R2 = -3.3599      D2 = 0.145R3 = -0.3905      D3 = 0.0774R4 =  0.4348      D4 = 0.1877R5 = -2.0321      D5 = 0.1835R6 = -0.578      D6 = 0.0618R7 =  0.8196      D7 = 0.1767R8 = -1.6015      D8 = 0.1025R9 = -3.9946|f.sub.1 /f.sub.2 | = 1.9421      f.sub.2 /f = -0.2452                  f.sub.3 /f = 0.5112                             DO = 0.0327______________________________________ 
    
     Ternary aberration coefficients of the above examples are shown in the following table, Table 1. In the table, I is spherical aberration, II is coma, III is astigmatism, P is Petzval sum and V is distortion. 
     
                       TABLE 1______________________________________coefficient   Example______________________________________   Example 1 Example 2 Example 3                               Example 4______________________________________I       0.12406   0.46004   -0.14528                               0.02664II      0.1735    0.1418    0.40244 0.26765III     0.01027   0.03254   0.17935 0.05713P       0.17223   0.27954   0.09523 0.21265V       -0.05443  0.13357   -0.01235                               -0.02325______________________________________   Example 5 Example 6 Example 7                               Example 8______________________________________I       0.31882   0.16961   0.22606 0.32845II      0.16566   0.24766   0.36047 0.19517III     0.13201   0.05117   0.12012 0.09247P       0.13286   0.13695   0.1091  0.24124V       0.10842   -0.0848   -0.32938                               0.04887______________________________________   Example 9 Example 10                       Example 11                               Example 12______________________________________I       0.87371   -0.31711  0.08399 0.08932II      0.18237   0.45304   0.37468 0.19056III     0.00096   0.22948   -0.04541                               -0.00143P       0.30255   0.06921   0.2076  0.21645V       0.05155   0.05192   0.01822 -0.08943______________________________________ 
    
     Hereinafter, description is made of the conditions (1&#39;) to (3&#39;) previously given for the lens according to the second embodiment of the invention. 
     Condition (1&#39;) is necessary for good correction of spherical aberration while maintaining the balance of refractive powers of the first and second lens groups. The lens of the present invention is of telecentric system and the spacing between the principal points of the second and third groups is broader than that between those of the first and second groups. Therefore, positions at which paraxial rays pass through the first group are greatly spaced apart from the optical axis and thereby a large quantity of spherical aberration is produced. If |f 1  /f 2  | is below the lower limit of 1.72, then the refractive power of the first group becomes high and therefore the paraxial rays passing through the surface will be intensely refracted in the direction toward the optical axis so that a large quantity of negative spherical aberration may be produced. On the contrary, when |f 1  /f 2  | is over the upper limit, 2.58, the refractive power of the second lens group becomes high and the second group will produces such a level of positive spherical aberration which may overly compensate the negative spherical aberration produced by the first lens group. 
     Condition (2&#39;) is for correction of curvature of field. When f 1  /f 2  exceeds the upper limit, -0.19, the Petzval sum is overcorrected and the correction of the curvature of field becomes excessive. To overcome the drawback, the absolute value of refractive power of the third group must be increased. However, as described later in connection with condition (3&#39;), by increasing the absolute value of refractive power there is produced in the third lens a large quantity of distortion the correction of which is very difficult. On the contrary, if f 2  /f is smaller than the lower limit, then the correction of Petzval sum of the whole system becomes difficult and curvature of field is undercorrected. 
     Condition (3&#39;) should be satisfied to correct the curvature of field and distortion. This is because the lens of the present invention is of telecentric system and the principal rays pass through the third lens group at positions far away from the optical axis. If f 3  /f is below the lower limit of 0.41, then the refractive power of the third group becomes high and the principal rays passing through the third group are intensely refracted in the direction toward the optical axis. As a result, a large quantity of distortion is produced. On the contrary, when f 3  /f is over the upper limit of 0.59, curvature of field becomes worse to the extent that it is no longer possible to correct the aberration. 
     Now, the shape of the above projection lens according to the second embodiment of the invention is described in detail. 
     As previously described, the projection lens of the present invention is required to be precise in terms of F number. To meet the requirement, it is advantageous that spherical abberation be corrected at the first group in which paraxial rays are most apart from the optical axis of the lens. To this end, the lens surface on the side of object field of the first group is convexed toward the object field side. 
     In this type of lens, Petzval sum is corrected primarily by the power of the second group. Therefore, the condition of power for the second group often becomes very severe and the second group is apt to produce various aberrations. To reduce the aberrations as much as possible, the second lens group is shaped as a double concave lens. 
     Since the lens of the present invention is a telecentric lens, off-axial principal rays should be spaced at a large distance from the optical axis by the third lens group. It is preferred that the nearest lens to the object field of the third group be shaped as a positive meniscus lens whose concaved surface is facing the side of object field. Further, it is preferred that the lens located nearest to the image field of the third group be shaped as a meniscus lens whose concaved surface is facing the image field side. By doing so, the principal rays are made spaced apart from the optical axis by the first mentioned meniscus lens and aberrations are corrected by the concave surface of the second mentioned meniscus lens. In addition, since the lens has no negative power, the requirement of lengthening the back focal length of the whole lens system can be attained at the same time. 
     The following examples, Example 13 to 21 together with FIGS. 13 to 21 illustrate the design of the projection lenses of the above second embodiment of the present invention. In all the following examples, focal length f is standarized to 1, F number to 5.0, angle of field ω to 25.2° and focus magnification β to -0.12343. 
     Again, Ri is the radius of curvature of the i-th surface; 
     Di is the thichness on axis or air spacing on axis between the i-th surface and the i+1-th surface; 
     Ni is the refractive index of the i-th lens to D ray; 
     νi is Abbe&#39;s number of the i-th lens; 
     fi is the focal length of the i-th lens group; and 
     D0 is the air spacing on axis from pupil SL to the R1 surface. 
     EXAMPLE 13 
     
         ______________________________________f = 1 1:5 angle of field = 25.2° β = -0.12343______________________________________R1 =  0.4235      D1 = 0.1596 N1 = 1.72  ν1 = 50.2R2 = -2.934      D2 = 0.156R3 = -0.4137      D3 = 0.0658 N2 = 1.80518                             ν2 = 25.4R4 =  0.5172      D4 = 0.1378R5 = -0.5814      D5 = 0.1784 N3 = 1.60311                             ν3 = 60.7R6 = -0.5646      D6 = 0.0037R7 =  1.7452      D7 = 0.1295 N4 = 1.757 ν4 = 47.9R8 = -1.0784      D8 = 0.0037R9 =  2.934      D9 = 0.1099 N5 = 1.697 ν5 = 48.5R10 = -1.6199      D10 = 0.141 N6 = 1.72825                             ν6 = 28.5R11 =  7.2159      D11 = 0.1119R12 =  0.6834      D12 = 0.1345                  N7 = 1.7725                             ν7 = 49.6R13 =   1.0343|f.sub.1 /f.sub.2 | = 1.895      f.sub.2 /f = -0.2767                  f.sub.3 /f = 0.5216                             DO = 0.0291______________________________________ 
    
     
         ______________________________________f = 1 1:5 angle of field = 25.2° Γ = -0.12343______________________________________R1 =  0.4309      D1 = 0.1372 N1 = 1.72  ν1 = 50.2R2 = -3.3104      D2 = 0.1659R3 = -0.4118      D3 = 0.068  N2 = 1.80518                             ν2 = 25.4R4 =  0.5596      D4 = 0.1133R5 = -0.5297      D5 = 0.1596 N3 = 1.60311                             ν3 = 60.7R6 = -0.4196      D6 = 0.0948R7 =  1.2334      D7 = 0.1282 N4 = 1.697 ν4 = 48.5R8 = -0.7865      D8 = 0.1423 N5 = 1.72825                             ν5 = 28.5R9 = -1.5704      D9 = 0.0168R10 =  0.8342      D10 = 0.1341                  N6 = 1.7725                             ν6 = 49.6R11 =  1.4571|f.sub.1 /f.sub.2 | = 1.8826      f.sub.2 /f = -0.2857                  f.sub.3 /f = 0.5303                             DO =  0.0069______________________________________ 
    
     EXAMPLE 15 
     
         ______________________________________f = 1 1:5 angle of field = 25.2° β = -0.12343______________________________________R1 =  0.4227      D1 = 0.1398 N1 = 1.72  ν1 = 50.2R2 = -3.6537      D2 = 0.1709R3 = -0.402      D3 = 0.0714 N2 = 1.80518                             ν2 = 25.4R4 =  0.5558      D4 = 0.0827R5 = -0.5313      D5 = 0.0414 N3 = 1.76182                             ν3 = 26.6R6 = -1.0604      D6 = 0.1282 N4 = 1.60311                             ν4 = 60.7R7 = -0.414      D7 = 0.1031R8 =  1.335      D8 = 0.2223 N5 = 1.697 ν5 = 48.5R9 = -1.5253      D9 = 0.0315R10 =  0.7952      D10 = 0.1356                  N6 = 1.7725                             ν6 = 49.6R11 =  1.5874|f.sub.1 /f.sub.2 | = 1.904      f.sub.2 /f = -0.2804                  f.sub.3 /f = 0.5184                             DO =  0.0195______________________________________ 
    
     EXAMPLE 16 
     
         ______________________________________f = 1 1:5 angle of field = 25.2° β = -0.12343______________________________________R1 =  0.3957      D1 = 0.1939 N1 = 1.72  ν1 = 50.2R2 =  2.4014      D2 = 0.1921R3 = -0.3159      D3 = 0.0303 N2 = 1.80518                             ν2 = 25.4R4 =  0.5107      D4 = 0.0846R5 = -1.0159      D5 = 0.0546 N3 = 1.60311                             ν3 = 60.7R6 = -0.4331      D6 = 0.0037R7 =  2.5196      D7 = 0.1745 N4 = 1.757 ν4 = 47.9R8 = -0.5692      D8 = 0.0037R9 = -0.6627      D9 = 0.1387 N5 = 1.697 ν5 = 48.5R10 = -0.4418      D10 = 0.1477                  N6 = 1.72825                             ν6 = 28.5R11 = -0.5971      D11 = 0.0856R12 =  0.71      D12 = 0.1196                  N7 = 1.7725                             ν7 = 49.6R13 =   0.7761|f.sub.1 /f.sub.2 | = 2.6519      f.sub.2 /f = -0.2385                  f.sub.3 /f = 0.4206                             DO = 0.0158______________________________________ 
    
     EXAMPLE 17 
     
         ______________________________________f = 1 1:5 angle of field = 25.2° Γ = -0.12343______________________________________R1 =  0.4103      D1 = 0.0989 N1 = 1.72  ν1 = 50.2R2 = -2.1886      D2 = 0.1497R3 = -0.4378      D3 = 0.0595 N2 = 1.80518                             ν2 = 25.4R4 =  0.5473      D4 = 0.1499R5 = -0.5567      D5 = 0.1904 N3 = 1.60311                             ν3 = 60.7R6 = -0.5263      D6 = 0.0036R7 =  5.6929      D7 = 0.1218 N4 = 1.757 ν4 = 47.9R8 = -1.368      D8 = 0.0036R9 =  3.3639      D9 = 0.1172 N5 = 1.697 ν5 = 48.5R10 = -1.539      D10 = 0.1   N6 = 1.72825                             ν6 = 28.5R11 =  35.1532      D11 = 0.0915R12 =  0.6849      D12 = 0.0815                  N7 = 1.7725                             ν7 = 49.6R13 =   1.4832|f.sub.1 /f.sub.2 | = 1.6578      f.sub.2 /f = -0.2941                  f.sub.3 = 0.5815                             DO = 0.02______________________________________ 
    
     EXAMPLE 18 
     
         ______________________________________f = 1 1:5 angle of field = 25.2° β = -0.12343______________________________________R1 =  0.4616      D1 = 0.1423 N1 = 1.72  ν1 = 50.2R2 = -2.5065      D2 = 0.1625R3 = -0.4707      D3 = 0.0522 N2 = 1.80518                             ν2 = 25.4R4 =  0.6353      D4 = 0.1537R5 = -0.5035      D5 = 0.1863 N3 = 1.60311                             ν3 = 60.7R6 = -0.5187      D6 = 0.0037R7 =  3.1351      D7 = 1.1227 N4 = 1.757 ν4 = 47.9R8 = -1.4508      D8 = 0.0037R9 =  2.1422      D9 = 0.1286 N5 = 1.697 ν5 = 48.5R10 = -1.3417      D10 = 0.1032                  N6 = 1.72825                             ν6 = 28.5R11 =  27.4187      D11 = 0.0895R12 =  0.7259      D12 = 0.1182                  N7 = 1.7725                             ν7 = 49.6R13 =   1.2478|f.sub.1 /f.sub.2 | = 1.6798      f.sub.2 /f = -0.3289                  f.sub.3 /f = 0.5736                             DO = 0.0156______________________________________ 
    
     EXAMPLE 19 
     
         ______________________________________f = 1 1:5 angle of field = 25.2° β = -0.12343______________________________________R1 =   0.3092      D1 = 0.1798 N1 = 1.72  ν1 = 50.2R2 =  -1.3737      D2 = 0.0857R3 =  -0.2973      D3 = 0.085  N2 = 1.80518                             ν2 = 25.4R4 =   0.3564      D4 = 0.1288R5 =  -0.5992      D5 = 0.0897 N3 = 1.60311                             ν3 = 60.7R6 =  -0.6701      D6 = 0.0037R7 =   2.6866      D7 = 0.138  N4 = 1.757 ν4 = 47.9R8 =  -0.6575      D8 = 0.0037R9 =  -2.5082      D9 = 0.0833 N5 = 1.697 ν5 = 48.5R10 =  -3.5115      D10 = 0.101 N6 = 1.72825                             ν6 = 28.5R11 = -11.326      D11 = 0.089R12 =   0.5951      D12 = 0.0767                  N7 = 1.7725                             ν7 = 49.6R13 =   2.425|f.sub.1 /f.sub.2 | = 1.9288      f.sub.2 /f = -0.1903                  f.sub.3 /f = 0.496                             DO = 0.0152______________________________________ 
    
     EXAMPLE 20 
     
         ______________________________________f = 1 1:5 angle of field = 25.2° β = -0.12343______________________________________R1 =  0.298      D1 = 0.188  N1 = 1.72  ν1 = 50.2R2 = -1.3058      D2 = 0.0778R3 = -0.2875      D3 = 0.0574 N2 = 1.80518                             ν2 = 25.4R4 =  0.3573      D4 = 0.1482R5 = -0.5748      D5 = 0.0653 N3 = 1.60311                             ν3 = 60.7R6 = -0.4784      D6 = 0.1083R7 =  1.1557      D7 = 0.1496 N4 = 1.697 ν4 = 48.5R8 = -1.383      D8 = 0.1051 N5 = 1.72825                             ν5 = 28.5R9 = -1.6648      D9 = 0.0158R10 =  0.7112      D10 = 0.1017                  N6 = 1.7725                             ν6 = 49.6R11 =  2.7318|f.sub.1 f.sub.2 | = 1.8621      f.sub.2 /f = -0.1903                  f.sub.3 /f = 0.5096                             DO =  0.0099______________________________________ 
    
     EXAMPLE 21 
     
         ______________________________________f = 1 1:5 angle of field = 25.2° β = -0.12343______________________________________R1 =   0.4133      D1 = 0.1413 N1 = 1.72  ν1 = 50.2R2 =  -2.4883      D2 = 0.1479R3 =  -0.4097      D3 = 0.0848 N2 = 1.80518                             ν2 = 25.4R4 =   0.5148      D4 = 0.1134R5 =  -0.5332      D5 = 0.1512 N3 = 1.60311                             ν3 = 60.7R6 =  -0.4298      D6 = 0.1092R7 =   1.1578      D7 = 0.1559 N4 = 1.697 ν4 = 48.5R8 =  -2.0472      D8 = 0.1053R9 =   0.9246      D9 = 0.0956 N5 = 1.7725                             ν5 = 49.6R10 = -24.8952      D10 = 0.1017                  N6 = 1.80518                             ν6 = 25.4R11 =   2.4866|f.sub.1 /f.sub.2 | = 1.8463      f.sub.2 /f =  -0.2722                  f.sub.3 /f = 0.5321                             DO = 0.0079______________________________________ 
    
     Ternary aberration coefficients of the above examples are given in the following table, Table 2 wherein I is spherical aberration, II is coma, III is astigmatism, P is Petzval sum and V is distortion. 
     
                       TABLE 2______________________________________coefficient   Example______________________________________   Example 13             Example 14                       Example 15                               Example 16______________________________________I       0.02847   -0.08302  0.06864 -0.3336II      0.09015   0.1896    0.15269 0.49706III     -0.00516  -0.00769  -0.01428                               0.12913P       0.14767   0.21517   0.19111 0.13869V       0.18112   0.01496   0.0142  -0.0607______________________________________   Example 17             Example 18                       Example 19                               Example 20______________________________________I       0.15339   0.09436   0.13221 0.15643II      0.09323   0.10385   0.2282  0.17737III     -0.01272  0.00036   0.17019 0.07046P       0.25299   0.25626   0.07882 0.11143V       0.14735   0.20489   0.20353 0.12851______________________________________   Example 21______________________________________I       0.0236II      0.15886III     -0.03366P       0.24261V       -0.02899______________________________________