[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 2t_g\Q  
function sjr=nfdre(~) Q"D  
5I1YB+$}e  
%系统焦距及各镜间距输入，间距取负正负 89^g$ ac  
Y:O|6%00Y  
f=input('f:'); aaCRZKr  
d1=input('d1:'); #}B1W&\sw  
d2=input('d2:'); IB|6\uKn  
d3=input('d3:'); 4gC(zJ  
A03io8D6  
A=f^2/(d3*d2)-f/d1; <!FcQVH+L  
B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); )
Jk$j  
C=d3/d2-f/d1; ;lb  
Qt{){u
E  
a1=(-B+sqrt(B^2-4*A*C))/(2*A);%α1 YR0AI l:L  
a2=d3/(a1*f);%α2 2^ ]^Yc  
b2=a1*(1-a2)*f/d2;%β2 Z\`SDC  
b1=(1-a1)*f/(d1*b2);%β1 )Cj1VjAg  
T=u"y;&L  
Q"u2<  
%曲率半径 @@K/0:],  
gAorb\
iJ  
R1=2*f/(b1*b2) yUJ#LDW  
R2=2*a1*f/(b2*(1+b1)) /huh}&NNu  
R3=2*a1*a2*f/(1+b2) M^Z=~512g  
-.? @f tY  
A1=b2^3*(a1-1)*(1+b1)^3; IMbF]6%p(  
B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; '}(>s%~  
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|(X_A  
%V]v,  
A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); L5 Cfa-  
B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); Q&{C%j~N  
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); R|8L'H+1x  
As>Og
 
CB=[C1 B1;C2 B2];  rn( drG  
AB=[A1 B1;A2 B2]; ElpZzGj+  
AC=[A1 C1;A2 C2]; %La7);SeY  
%G2g @2  
%非球面系数 $t^Td
<  
k2=-(det(CB)/det(AB)); TA/hj>rV  
k3=-(det(AC)/det(AB)); H $Az,-P  
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 *5oQZ".vA*  
k2=k2 n:wAxU  
k3=k3 Gr&e]M[l  
>Tl/3{V  
end xS%
&l)dT  
'D`lVUB  
%有中间像，焦距输入为正数 zLe(#8G  
~D}fy  
function sjr=yfdre(~) QOUyD;0IW  
`lOW7Z}  
f=input('f:'); ;`pIq-=  
d1=input('d1:'); YHom9&A  
d2=input('d2:'); p<'pqf  
d3=input('d3:'); /KC^x=Xv:  
7J5jf231  
A=f^2/(d3*d2)-f/d1; QnBWZUI  
B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); qonStIP  
C=d3/d2-f/d1; Ru#pJb(R  
mD% qDKI  
a1=(-B-sqrt(B^2-4*A*C))/(2*A); [Q8Wy/o Q  
a2=d3/(a1*f); R'tvF$3=i  
b2=a1*(1-a2)*f/d2; >f
Hu  
b1=(1-a1)*f/(d1*b2); z7XI`MZN^  
[^}bc-9?i  
%曲率半径 '[8w8,v(  
)K]p^lO  
R1=2*f/(b1*b2) $Bc3| `K1v  
R2=2*a1*f/(b2*(1+b1)) `a[fC9  
R3=2*a1*a2*f/(1+b2) Y
KU|D32  
<m~8pM  
A1=b2^3*(a1-1)*(1+b1)^3; Lz_.m  
B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; Q'3tDc<  
C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; r ^*D8  
{g[kn^|  
A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); vs+aUT C\  
B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); 9pj6`5Zn@6  
C2=b2*(a1-1)^2*(1+b1)*(1-b1)^2/(4*a1*b1^2)-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)*(1-b2)^2/(4*a1*a2*b1^2*b2^2)-b2*(a1-1)*(1-b1)*(1+b1)/(a1*b1)-(a2*(a1-1)+b1*(1-a2))*(1-b2)*(1+b2)/(a1*a2*b1*b2)-b1*b2+b2*(1+b1)/a1-(1+b2)/(a1*a2); <>$CYTb  
?o6#i3k#'  
CB=[C1 B1;C2 B2]; zmD7]?|  
AB=[A1 B1;A2 B2]; q'y<UyT6  
AC=[A1 C1;A2 C2]; ucz~y!4L{  
A =Z$H2  
%二次系数 0S>L0qp  
YR/I<m`]}  
k2=-(det(CB)/det(AB)); sl@>
GbnS  
k3=-(det(AC)/det(AB)); o/a2n<4  
k1=(k2*a1*b2^3*(1+b1)^3-k3*a1*a2*(1+b2)^3+a1*b2^3*(1+b1)*(1-b1)^2-a1*a2*(1+b2)*(1-b2)^2)/(b1^3*b2^3)-1 )sK53O$  
k2=k2 HbPn<x^7  
k3=k3 vpT\CjXHZ  
u' kG(<0Y  
end

谢谢分享，学习一下 .w{Y3,dd>