| songshaoman |
2020-05-25 15:25 |
在框架结构确定的情况下，基于matlab的消四种像差的三反系统初始结构的求解
%无中间像，焦距输入为负数 i5,kd~%O
function sjr=nfdre(~) scLll ,~
:${HQd +
%系统焦距及各镜间距输入，间距取负正负 :'*~uJrR
, ^f+^^
f=input('f:'); M{hg0/}sUW
d1=input('d1:'); $,Yd>%Y
d2=input('d2:'); K?$^@ N
d3=input('d3:'); cY. bO/&l
_X"N1,0
A=f^2/(d3*d2)-f/d1; B%+T2=&$7
B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); ax5<#3__
C=d3/d2-f/d1; MfQ?W`Kop
M%;hB*9
a1=(-B+sqrt(B^2-4*A*C))/(2*A);%α1 bY QRBi
a2=d3/(a1*f);%α2 Ffta](Z;
b2=a1*(1-a2)*f/d2;%β2 Px`!A EFd[
b1=(1-a1)*f/(d1*b2);%β1 z]D69O b
8k1Dj1@0z
%m$Sp47
%曲率半径 gi
_ 5?$
{
W{]L:
R1=2*f/(b1*b2) Ob& <]
R2=2*a1*f/(b2*(1+b1)) @ xYlS5{
R3=2*a1*a2*f/(1+b2) ocS5SB]8
i5?q,_
A1=b2^3*(a1-1)*(1+b1)^3; &t:Gx<]
B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; 6- B|Y3)B
C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; :\7X}n*&
RcU }}V
A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); (7=!+'T"
B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); d8=x0~7
C2=b2*(a1-1)^2*(1+b1)*(1-b1)^2/(4*a1*b1^2)-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)*(1-b2)^2/(4*a1*a2*b1^2*b2^2)-b2*(a1-1)*(1-b1)*(1+b1)/(a1*b1)-(a2*(a1-1)+b1*(1-a2))*(1-b2)*(1+b2)/(a1*a2*b1*b2)-b1*b2+b2*(1+b1)/a1-(1+b2)/(a1*a2); @>>~CZ`l
!AfHk|
CB=[C1 B1;C2 B2]; Z+. '>
AB=[A1 B1;A2 B2]; oij}'|/Jc
AC=[A1 C1;A2 C2]; 3r."j2$Hs0
TXvI4"&
%非球面系数 9=h'9Wo
k2=-(det(CB)/det(AB)); 91-o}|3v
k3=-(det(AC)/det(AB)); [o+q>|q
k1=(k2*a1*b2^3*(1+b1)^3-k3*a1*a2*(1+b2)^3+a1*b2^3*(1+b1)*(1-b1)^2-a1*a2*(1+b2)*(1-b2)^2)/(b1^3*b2^3)-1 @wo(tf=@P
k2=k2 WQL\y3f5
k3=k3 q0R -7O(
J!pygn O
end NmJWU:W_@
.Blf5b
%有中间像，焦距输入为正数 Y^}Z>
P0 j8- I
function sjr=yfdre(~) 'yG4
LF
T1 Z;r*}
f=input('f:'); Df<xWd2
d1=input('d1:'); \Vy Z
d2=input('d2:'); x QXXC|T
d3=input('d3:');
+:!7L= N#
!()$8
A=f^2/(d3*d2)-f/d1; > $7 v
;Q
B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); _Wq
C=d3/d2-f/d1; Q1
$^v0-)
:^WKT
a1=(-B-sqrt(B^2-4*A*C))/(2*A); yiC^aY=-
a2=d3/(a1*f); h"_;IUZ!
b2=a1*(1-a2)*f/d2; .e=:RkI,
b1=(1-a1)*f/(d1*b2); YS@ypzc/
hYNY"VB
%曲率半径 *%fi/bimG
?Unb?
{,&2
R1=2*f/(b1*b2) 'o+L 41
R2=2*a1*f/(b2*(1+b1)) 6qoyiT%P&
R3=2*a1*a2*f/(1+b2) >dG;w6y'
*4+"Lh.KS
A1=b2^3*(a1-1)*(1+b1)^3; 2Z Mb<b4H
B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; * @v)d[z_
C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; 6S*e xw
'H &2HXw&2
A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); F{Jw ^\
B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); TKY*`?ct
C2=b2*(a1-1)^2*(1+b1)*(1-b1)^2/(4*a1*b1^2)-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)*(1-b2)^2/(4*a1*a2*b1^2*b2^2)-b2*(a1-1)*(1-b1)*(1+b1)/(a1*b1)-(a2*(a1-1)+b1*(1-a2))*(1-b2)*(1+b2)/(a1*a2*b1*b2)-b1*b2+b2*(1+b1)/a1-(1+b2)/(a1*a2); uU <=d
D=m9fFz
CB=[C1 B1;C2 B2]; X=!^] 3zH
AB=[A1 B1;A2 B2]; &d+Kg 0 :
AC=[A1 C1;A2 C2]; A vd
^
E2@65b$
%二次系数 B~JwHwIhA
$.P uK~}
k2=-(det(CB)/det(AB)); P &)xz7wG
k3=-(det(AC)/det(AB)); YoZFwRQU
k1=(k2*a1*b2^3*(1+b1)^3-k3*a1*a2*(1+b2)^3+a1*b2^3*(1+b1)*(1-b1)^2-a1*a2*(1+b2)*(1-b2)^2)/(b1^3*b2^3)-1 ,do v<U[ia
k2=k2 o -{[|/)Tk
k3=k3 *`\Pr
7;sj%U^'l
end
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