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    [讨论]如何分析Ghost focus generator的结果 [复制链接]

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    离线sansummer
     
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    只看楼主 倒序阅读 楼主  发表于: 2011-06-27
    例如zemax knowledge base的文章,连接如下:http://www.zemax.com/kb/articles/249/1/How-do-I-Account-for-Energy-Losses-When-Performing-a-Sequential-Ghost-Analysis/Page1.html 0e'@Xo2e  
    6I5LZ^/G9  
    其中使用Ghost focus generator后的结果为: 0bQ"s*K  
    99Nm?$ g  
    Ghost Trace Data
    File : C:\Program Files\ZEMAX - Beta\Samples\Sequential\Objectives\Cooke 40 degree field.zmx ]PH'G>x  
    Title: A SIMPLE COOKE TRIPLET.
    Units are Millimeters.
    Wavelength: 0.550000 μm ge&!GO  
    Type: Double Bounces oHethk  
    The '****' denotes a possible internal focus ukee.:{  
    RMS is the RMS spot radius on axis at the primary wavelength. Is  ( Ji  
    RMS values of 0.00 indicate that an accurate RMS R36A_  
    could not be computed, usually due to ray errors.
    This analysis may be inaccurate if the system is .Ax]SNZ+:A  
    non-rotationally symmetric or uses non-standard surfaces.
    Ghost reflection off surface 2 then 1. (GH002001.ZMX) EHq?yj;  
    Surf     Marginal          F/#          RMS 0*/[z~Z-1  
       1  5.0000E+000     5.737068  3.5355E+000 j;&su=p"  
       2  4.7160E+000     4.595701  3.3570E+000 U ,\t2z  
       1  4.3614E+000     0.990012  3.1277E+000 )3!z2f:e  
       2  2.7155E+000     0.607262  1.9876E+000 Gd[: &h  
       3 -2.2309E+000     0.916280  2.6065E+000 mw${3j~&  
       4 -2.7766E+000     0.514753  0.0000E+000 #t&L}=G{%  
       5 -7.3909E+000     0.888038  0.0000E+000 yzL6oU-{&  
       6 -9.0530E+000     0.823473  0.0000E+000 jQ P2[\  
       7 -3.4681E+001     0.823473  0.0000E+000 M+M\3U  
    Marginal ray height    :       -34.6809 0 SDyE  
    Chief ray height       :       -24.8107 @_"Z]Y ,D0  
    Distance to ghost pupil:       -50.9613 D;oX*`  
    Distance to ghost focus:       -57.1176 vSJ# }&  
    Effective focal length :         8.2347
    >yt8gw0J  
    请问这个结果怎么看? pJ@D}2u(  
    f2M}N  
    其中:Ghost reflection off surface 2 then 1,什么意思? {= T9_c  
    最后的光线高度能用来分析什么?? rn-CQ2{?  
    Distance to ghost pupil: -50.9613 Distance to ghost focus: -57.1176 r )f+j@KF  
    Effective focal length : 8.2347 7MwS[N%#  
    这有代表什么,求高手啊
     
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    离线sansummer
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    只看该作者 1楼 发表于: 2011-06-27
    我现在好像明白一点了,可是这个结果该怎么分析呢?比如这个鬼像发生后的有效焦距8.2347,到底会对成像有多撒影响呢?  s{T6qJ  
    !s[[X5  
    Distance to ghost pupil: -50.9613 Distance to ghost focus到底说明了什么哦
    离线superyu2009
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    只看该作者 2楼 发表于: 2011-11-22
    用CODE V分析更好