[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 "_=t1UE  
function sjr=nfdre(~) <Ynrw4[)t  
0sCWIGUW  
%系统焦距及各镜间距输入，间距取负正负 . +,{|){c  
?D(aky#cyc  
f=input('f:'); F
zn!  
d1=input('d1:'); >\3N#S"PF  
d2=input('d2:'); 0WyOORuK  
d3=input('d3:'); ZkNet>9  
~(=5`9  
A=f^2/(d3*d2)-f/d1; ,|j\x  
B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); -<e_^  
C=d3/d2-f/d1; Yjl:i*u/  

$4^h>x  
a1=(-B+sqrt(B^2-4*A*C))/(2*A);%α1 {!!8 *ix  
a2=d3/(a1*f);%α2 g[\8s~g,  
b2=a1*(1-a2)*f/d2;%β2 j+>N&.zs
 
b1=(1-a1)*f/(d1*b2);%β1 *?>52 -&b  
X?(R!=a  
;q#Pl!*5  
%曲率半径 !P":z0K4  

1)f <  
R1=2*f/(b1*b2) gJg+ ]-h/  
R2=2*a1*f/(b2*(1+b1)) R3.8Dr0f  
R3=2*a1*a2*f/(1+b2) Dr"PS >.  
|:SIyXGbY  
A1=b2^3*(a1-1)*(1+b1)^3; 6y"T;.FAo  
B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; Qv
)DSl  
C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; ME.!l6lm\  
Wq?vAnLbk  
A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); ,L-V?B(UQ  
B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); zy|h1.gd  
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); Z!4
B=?(  
Nk1p)V SC  
CB=[C1 B1;C2 B2]; #T8PgmR  
AB=[A1 B1;A2 B2]; M<SVH_  
AC=[A1 C1;A2 C2];
0W~1v  
G'wyH[ d/  
%非球面系数 3-
)R'  
k2=-(det(CB)/det(AB)); dl6Ju  
k3=-(det(AC)/det(AB)); ;{0%Vp{  
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 q2;CvoF  
k2=k2 t Q.%f:|  
k3=k3 9"}5jq4*  
kPO+M~+n  
end WtXf~ :R  
aPX'CG4m  
%有中间像，焦距输入为正数 &G?w*w_n  
hdqr~9  
function sjr=yfdre(~) F'@9kdp  
=%B}8$.|  
f=input('f:'); oq,*@5xV2  
d1=input('d1:'); is^5TL%@  
d2=input('d2:'); k4` %.;  
d3=input('d3:'); U? ;Q\=>  
Fu[GQ6{f  
A=f^2/(d3*d2)-f/d1; &3itBQF  
B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); X9C:AGbp  
C=d3/d2-f/d1; |\)Y,~;P  
(
@bq@0g  
a1=(-B-sqrt(B^2-4*A*C))/(2*A); ET%F+  
a2=d3/(a1*f); gj&5>brP  
b2=a1*(1-a2)*f/d2; TiJ \J{  
b1=(1-a1)*f/(d1*b2); 7Q~$&G  
[]?*}o5&>T  
%曲率半径 *=Ma5J.  
aFL<(,~r  
R1=2*f/(b1*b2) n} ]gAX  
R2=2*a1*f/(b2*(1+b1)) eu//Q'W  
R3=2*a1*a2*f/(1+b2) 'vKae  
{J/+KK  
A1=b2^3*(a1-1)*(1+b1)^3; >;}(?+|f  
B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; gRZ!=z[&  
C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; .s|5AC[  
GKG:iR)  
A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); 9j5B(_J^  
B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); xFA`sAucr  
C2=b2*(a1-1)^2*(1+b1)*(1-b1)^2/(4*a1*b1^2)-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)*(1-b2)^2/(4*a1*a2*b1^2*b2^2)-b2*(a1-1)*(1-b1)*(1+b1)/(a1*b1)-(a2*(a1-1)+b1*(1-a2))*(1-b2)*(1+b2)/(a1*a2*b1*b2)-b1*b2+b2*(1+b1)/a1-(1+b2)/(a1*a2); fe}RmnAC  
kc2 8Q2  
CB=[C1 B1;C2 B2]; ^__P;Gr`  
AB=[A1 B1;A2 B2]; -.-@|*5  
AC=[A1 C1;A2 C2]; L\"eE'A  
;)ERxMun  
%二次系数 FR\r/+n:t0  
@[Wf!8_  
k2=-(det(CB)/det(AB)); c57`mOe/b  
k3=-(det(AC)/det(AB)); rY$wC%  
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 8L`wib2  
k2=k2 1\
/~>  
k3=k3 nd5.Py$  
&0{&4,  
end

谢谢分享，学习一下 <W*6=HZ'