[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 a9[<^  
function sjr=nfdre(~) <bdyAUeFw  
dYyW]nZ&  
%系统焦距及各镜间距输入，间距取负正负 J v<$*TVS0  
d~<QAh#rG  
f=input('f:'); @*_ZoO7{  
d1=input('d1:'); M@O2 WB1ws  
d2=input('d2:'); 6F`\YSn+  
d3=input('d3:'); kV
1vb  
B+Q+0tw*i  
A=f^2/(d3*d2)-f/d1; NQ!<f\m4n  
B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); ,xj3w#`zaf  
C=d3/d2-f/d1; 
OMd# ^z  
hrT%XJl  
a1=(-B+sqrt(B^2-4*A*C))/(2*A);%α1 M;qb7Mu  
a2=d3/(a1*f);%α2
wA
r~<  
b2=a1*(1-a2)*f/d2;%β2 fN0bIE Y  
b1=(1-a1)*f/(d1*b2);%β1 \ 522,n`  
va>"#;37  
e;;):\p4  
%曲率半径 Tc,$TCF  
<N5rv3 s  
R1=2*f/(b1*b2) ;[caiMA-  
R2=2*a1*f/(b2*(1+b1)) jIZ+d;1  
R3=2*a1*a2*f/(1+b2) 3q CHh  
hpjUkGm5  
A1=b2^3*(a1-1)*(1+b1)^3; A:c]1  
B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; |1i]L@&  
C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; xDLMPo&  
[Gf{f\O  
A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); Q!x`M4   
B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); @%H8"A  
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); _iq2([BpL  
lJ'trYaq7  
CB=[C1 B1;C2 B2]; Ft$^x-d  
AB=[A1 B1;A2 B2]; .}~$1QKS  
AC=[A1 C1;A2 C2]; dQ`:8SK  
[%t3[p<)O  
%非球面系数 4$F:NW,v:)  
k2=-(det(CB)/det(AB)); +]_nbWL(%  
k3=-(det(AC)/det(AB)); ~xE=mg4le  
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 f^u^-l  
k2=k2 _lNC<7+#h  
k3=k3 yFi6jN#~  
Wk,6) jS=}  
end )ZN(2z  
Kny%QB
oiw  
%有中间像，焦距输入为正数 lc"qqt  
ru DP529;  
function sjr=yfdre(~) 33,;i
E  
y3IA '  
f=input('f:'); 1GUqT 9)  
d1=input('d1:'); 9='=-;@/5  
d2=input('d2:'); ?-d Ain1w  
d3=input('d3:'); K Ka c6Zj  
E;xMPK$  
A=f^2/(d3*d2)-f/d1; n+X1AOE[L  
B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); R|$[U  
C=d3/d2-f/d1; [h^f%  
}}s8D>;G~  
a1=(-B-sqrt(B^2-4*A*C))/(2*A); XN4oL[pO  
a2=d3/(a1*f); &7fY_~)B  
b2=a1*(1-a2)*f/d2; {vLTeIxf.G  
b1=(1-a1)*f/(d1*b2); 6TY){Pw  
a6k
(9ZF  
%曲率半径 ;FcExg|k  
8znj~7}#  
R1=2*f/(b1*b2) 0n\^$WY  
R2=2*a1*f/(b2*(1+b1)) Oq[i &  
R3=2*a1*a2*f/(1+b2) {zY`h6d  
x3gwG)Sf  
A1=b2^3*(a1-1)*(1+b1)^3; tK*%8I\s  
B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; jk K#e$7  
C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; >1m)%zt  
c%jsu"  
A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); (LRNU)vD7$  
B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); HgYc@P*b  
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); &|s+KP|d  
[k!-;mi   
CB=[C1 B1;C2 B2]; dFx2>6AZt  
AB=[A1 B1;A2 B2]; T=^jCH &  
AC=[A1 C1;A2 C2]; L7s>su|c(
 
KlY,NSlQ  
%二次系数 -QN1oK@\mE  
HE*7\"9  
k2=-(det(CB)/det(AB)); ht2 f-EKf{  
k3=-(det(AC)/det(AB)); W]Y@WKeT  
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 #-}kG"  
k2=k2 \C!%IR  
k3=k3 ~5FS|[1L  
o`b$^hv{A  
end

谢谢分享，学习一下 l#X=]xQf