| qlmi |
2006-08-10 13:34 |
关于Effective focal length的问题
ZEMAX中priscription data会给出系统的许多有用的数据，其中包括effective focal length的数据和主平面位置的数据。但是我发现直接给出的effective focal length数据和根据其定义结合主平面的数据算出来的结果不一致，这时怎么回事？ AXBf\ )[
一下是我的Prescription Data数据： lyZ[t P S
GENERAL LENS DATA: A 9\]y%!
>Z-f</v03
Surfaces : 3 p{c+ +P5
Stop : 1 /8](M5X]f
System Aperture : Object Space NA = 0.12 W g \`!T
Glass Catalogs : Schott )}`3haG
Ray Aiming : Paraxial Reference, Cache On &T"X
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X Pupil shift : 0 \ r4Q S
Y Pupil shift : 0 HH`G/(a
Z Pupil shift : 0 YhpNeP{A
Apodization : Gaussian, factor = 1.00000E+000 [,_4#Zz
************************************************************* C B&$tDi
Effective Focal Length : 1.639158 (in air at system temperature and pressure) ea7l:(C
Effective Focal Length : 1.639158 (in image space) ,AX7~;hpq
************************************************************* 9j6# #@{
Back Focal Length : -1.061568 ;ndsq[k>
Total Track : 4.318 k#*tf :R
Image Space F/# : 3.417255 vbXZ Z
Paraxial Working F/# : 2.950535 02|f@b P.
Working F/# : 2.961912 L!fiW`>0G
Image Space NA : 0.1670788 q>JW$8
Object Space NA : 0.12 4&TTP cSt;
Stop Radius : 0.2398355 +aa( YGL
Paraxial Image Height : 0 7J:zIC$u>
Paraxial Magnification : 0 ze5Hg'f
Entrance Pupil Diameter : 0.4796709 4O~E4" ]
Entrance Pupil Position : 0 XnBpL6"T`
Exit Pupil Diameter : 2.511155 `F) Q=
Exit Pupil Position : 7.420935 5|I2
Field Type : Angle in degrees S w<V/t
Maximum Field : 0 >=< qAkk
Primary Wave : 1.31 y]z^e\qc)
Lens Units : Millimeters M@LI (;
Angular Magnification : 0 .'+*>y!
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Fields : 1 Km)5;BQxg
Field Type: Angle in degrees $M `%A
# X-Value Y-Value Weight }4
p3m]
1 0.000000 0.000000 1.000000 JjH#,@'.
E="FE.%A
Vignetting Factors % FN3/iM
# VDX VDY VCX VCY VAN M!KHB r
1 0.000000 0.000000 0.000000 0.000000 0.000000 y -Xd~<*Ia
%j
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Wavelengths : 1 !>%U8 A
Units: Microns st8=1}:&\
# Value Weight 8|\0\Wd;vu
1 1.310000 1.000000 ppwjr
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CARDINAL POINTS: kPh;SCr{
a,B2;4 "
Object space positions are measured with respect to surface 1. '/Y
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Image space positions are measured with respect to the image surface. /U| >
The index in both the object space and image space is considered. >lD*:#o
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Object Space Image Space bzI!;P1 &
W = 1.310000 (Primary) $Vu % 4kq
************************************************************ 6CK WKc
Focal Length : -1.639015 1.639015 (WW*yv.J
************************************************************ D)yCuw{M:
Focal Planes : 0.313078 -1.159568 {i:5XL
Principal Planes : 1.952094 -2.798584 M7$ h
Anti-Principal Planes : -1.325937 0.479447 ;g-L2(T05;
Nodal Planes : 1.952094 -2.798584 3'
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Anti-Nodal Planes : -1.325937 0.479447 3ug|H
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我用*号，标出了需要注意的地方，请高手指教！
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