[原创]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 [复制链接]

%无中间像，焦距输入为负数 ,iV|^]X3$/  
function sjr=nfdre(~) JHBX'1GQa  
`*^ f =y  
%系统焦距及各镜间距输入，间距取负正负 n'>`2 s  
<;G.(CK@n  
f=input('f:'); I'm.+(1m,  
d1=input('d1:'); 3U^E<H  
d2=input('d2:'); bkTk:-L5:  
d3=input('d3:'); Z$"E|nRN  
/SO 4O|b  
A=f^2/(d3*d2)-f/d1; @p~f*b4H?  
B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); M|xd9kA^  
C=d3/d2-f/d1; RG4T9eZq  
`ZhDoLpH<  
a1=(-B+sqrt(B^2-4*A*C))/(2*A);%α1 i|2CZ  
a2=d3/(a1*f);%α2 hV_bm@f/y  
b2=a1*(1-a2)*f/d2;%β2 7,X5]U&A<x  
b1=(1-a1)*f/(d1*b2);%β1 2NB/&60<  
8cI<~|4_  
XnR9/t  
%曲率半径 EdR1W~JZ  
#l2KJ7AMK  
R1=2*f/(b1*b2) _P,3~ ;  
R2=2*a1*f/(b2*(1+b1)) 4Qwv:4La  
R3=2*a1*a2*f/(1+b2) F/}(FG<'>I  
}&!fT\4  
A1=b2^3*(a1-1)*(1+b1)^3; hhRUC&Y%V  
B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; Qu]F<H*Y|  
C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; <a_ytSoG1  
^N#z&oh  
A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); 4E:kDl*@  
B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); U~H]w,^  
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); J%r$jpd'  
xYSNop3_  
CB=[C1 B1;C2 B2]; , !0-;H.Y  
AB=[A1 B1;A2 B2]; !Z(3dtUy  
AC=[A1 C1;A2 C2]; xQ~}9Kt\  
)/Z% HBn  
%非球面系数 HX}9;O  
k2=-(det(CB)/det(AB)); Oc A;+}>  
k3=-(det(AC)/det(AB)); *e/8
uFX  
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 dK.k,7R  
k2=k2 }*Z *wC  
k3=k3 z"D'rHxy  
s&.VU|=VQ@  
end !I)wI~XF)5  
3pU/Zbb,:  
%有中间像，焦距输入为正数 X
lg0u.  
4Kl{^2  
function sjr=yfdre(~) }:SWgPfc  
';,Rq9-'  
f=input('f:'); O> .gcLA  
d1=input('d1:'); 0*y|k1  
d2=input('d2:'); nI0TvBD  
d3=input('d3:'); +T!7jC(O Q  
6Z,GD  
A=f^2/(d3*d2)-f/d1; nNj<!}HvV  
B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); P80mK-Iyv_  
C=d3/d2-f/d1; lE|Hp  
"1I\~]]  
a1=(-B-sqrt(B^2-4*A*C))/(2*A); "fH"U1Bw  
a2=d3/(a1*f); o%j[]P@4G  
b2=a1*(1-a2)*f/d2; p#A{.6Pa:  
b1=(1-a1)*f/(d1*b2); QP?eKW9 :  
sW#OA\i&  
%曲率半径 N Ftmus  
"Qci+Qq  
R1=2*f/(b1*b2) lX)ZQY:=:  
R2=2*a1*f/(b2*(1+b1)) ZkA05wPZ#  
R3=2*a1*a2*f/(1+b2) BK*Bw,KQ<  
md S`nhb  
A1=b2^3*(a1-1)*(1+b1)^3; Thc"QIk&4  
B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; mu$0x)  
C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; .=`r?#0  
JbR;E`8  
A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); 9jkaEn>m^  
B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); hf('4^  
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); yb4Jsk5%  
oEJYAKN  
CB=[C1 B1;C2 B2]; 
F
<9S,  
AB=[A1 B1;A2 B2]; 09_5niaz[  
AC=[A1 C1;A2 C2]; 6C@W6DR3N  
Q |1-j  
%二次系数 Z23*`yR  
SI"y&[iw  
k2=-(det(CB)/det(AB)); .e Jt]K  
k3=-(det(AC)/det(AB)); j84g6;4Dv  
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 ^.?5!9U  
k2=k2 \""sf{S9  
k3=k3 ]ucz8('  
d&G#3}kOb%  
end

谢谢分享，学习一下 {9_}i#,vR