[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 wN=;i#  
function sjr=nfdre(~) J2VTo: In  
R=]d%L8  
%系统焦距及各镜间距输入，间距取负正负 Jr;jRe`4c  
Hfym30  
f=input('f:'); #
_i`#d)  
d1=input('d1:'); 9'[ N1Un.=  
d2=input('d2:'); \ZI'|Ad  
d3=input('d3:'); Gl}Qxv#$  
Iu[|<C
x  
A=f^2/(d3*d2)-f/d1; 9?<{_'  
B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); L|hx arJ  
C=d3/d2-f/d1; bBc[bc>R  
V~@^`Gd  
a1=(-B+sqrt(B^2-4*A*C))/(2*A);%α1 #V,R >0"  
a2=d3/(a1*f);%α2 PG~$D];  
b2=a1*(1-a2)*f/d2;%β2 -7 U|a/  
b1=(1-a1)*f/(d1*b2);%β1 PcT?<HU  
tDg}Ys=4K>  
u #w29Pm  
%曲率半径 d5`3wd]]'v  
V)(R]BK{  
R1=2*f/(b1*b2) N$>^g"6o  
R2=2*a1*f/(b2*(1+b1)) yQ,{p@#X8  
R3=2*a1*a2*f/(1+b2) sEJC-$  
#S]ER907  
A1=b2^3*(a1-1)*(1+b1)^3; cyyVg!+  
B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; 4JGtI*%5lq  
C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; TS2ZF{m  
{X(:jAy  
A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); \%]I{  
B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); o1cErI&q"  
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); -i yyn^|  
\R36w^c3  
CB=[C1 B1;C2 B2]; 5,-U.B}  
AB=[A1 B1;A2 B2]; )G^ KDj"  
AC=[A1 C1;A2 C2]; UcOk3{(z$q  
^.dsW0"0  
%非球面系数 1i[FY?6`dh  
k2=-(det(CB)/det(AB)); mpPdG  
k3=-(det(AC)/det(AB)); }X*Riu7gk  
k1=(k2*a1*b2^3*(1+b1)^3-k3*a1*a2*(1+b2)^3+a1*b2^3*(1+b1)*(1-b1)^2-a1*a2*(1+b2)*(1-b2)^2)/(b1^3*b2^3)-1 0Z[oKXm1p  
k2=k2 FKNMtp[`  
k3=k3 8_<4-<}P:  
Y2o?gug  
end p$Ox'A4  
_8Kx6
s%  
%有中间像，焦距输入为正数 VCXJwVb
  
.Asv%p[W  
function sjr=yfdre(~) S}p4iE"n  
MjlP+; !  
f=input('f:'); #]nx!*JNZ  
d1=input('d1:'); i;LXu%3\  
d2=input('d2:'); b2b^1{@h;v  
d3=input('d3:'); M`rl!Ci#  
%?e& WLS  
A=f^2/(d3*d2)-f/d1; \b%kf99  
B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); fFb_J`'ue  
C=d3/d2-f/d1; ]gYz 4OT  
CC#;c1t  
a1=(-B-sqrt(B^2-4*A*C))/(2*A);  J5^'HU3  
a2=d3/(a1*f); bFe+m1Q_  
b2=a1*(1-a2)*f/d2; GrUCZ<S  
b1=(1-a1)*f/(d1*b2); 5R(/Uiv3F  
.C
8PitS  
%曲率半径 Re P|UH  
v?KC%  
R1=2*f/(b1*b2) Q09[[  
R2=2*a1*f/(b2*(1+b1))
E_vq  
R3=2*a1*a2*f/(1+b2) O3I8k\`  
emCM\|NQg&  
A1=b2^3*(a1-1)*(1+b1)^3; 5$V_
Hj  
B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; !%"8|)CAr  
C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; gb
H<]?  
sBr_a5QQ#  
A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); o4|M0  
B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); R8ZK]5{o  
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); ;kY(<{2  
Ney/[3 A  
CB=[C1 B1;C2 B2]; :A/d to  
AB=[A1 B1;A2 B2]; Y;?{|  
AC=[A1 C1;A2 C2];
S:h{2{  
mIK7p6  
%二次系数 eEuvl`&  
zd@m~V  
k2=-(det(CB)/det(AB)); \ExMk<y_&  
k3=-(det(AC)/det(AB)); ,6-:VIHQ  
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 WA+iYLx@H  
k2=k2 R}O_[  
k3=k3 *MKO I'  
vN`klDJgW[  
end

谢谢分享，学习一下 4 o Fel.o