[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 `$oGgz6ZT  
function sjr=nfdre(~) JwWW w1  
9,|{N(N<!  
%系统焦距及各镜间距输入，间距取负正负 -y]\;pbZ0  
fP>_P#gZ  
f=input('f:'); Z7RGOZQ}G  
d1=input('d1:'); yoa"21E$  
d2=input('d2:'); 3a PCi>i!_  
d3=input('d3:'); :e>y= s>  
lJ.:5$2H  
A=f^2/(d3*d2)-f/d1; &ywAzGV{s  
B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); vmdu9"H  
C=d3/d2-f/d1; K +n  
V5sH:A7GJ  
a1=(-B+sqrt(B^2-4*A*C))/(2*A);%α1 taOD,}c|$  
a2=d3/(a1*f);%α2 YT\x'`>Q  
b2=a1*(1-a2)*f/d2;%β2 ']1\nJP[=X  
b1=(1-a1)*f/(d1*b2);%β1 Z$!>hiz2  
[uCW8:e  
kKr|PFz  
%曲率半径 qN((Xz+AZE  
3wZA,Z  
R1=2*f/(b1*b2) g a|RW0  
R2=2*a1*f/(b2*(1+b1)) \jLn5$OW  
R3=2*a1*a2*f/(1+b2) L rV`P)$T  
rT$J0"*=  
A1=b2^3*(a1-1)*(1+b1)^3; 4_qd5K+n"  
B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; eh_{-  
C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; g4USKJ19.  
SrGX4  
A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); 3vRRL  
B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); !f5I.r~  
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); mU.(aLHW  
Zwj\Hz.  
CB=[C1 B1;C2 B2]; F<SMU4]YdG  
AB=[A1 B1;A2 B2]; 1L722I@  
AC=[A1 C1;A2 C2]; ',GWH:B  
CVfQ  
%非球面系数 $s$j
</.q  
k2=-(det(CB)/det(AB)); *Y!'3|T  
k3=-(det(AC)/det(AB)); N8a+X|3]0  
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 Y1RiuJtL  
k2=k2 } :U'aa  
k3=k3 heQ<%NIA"  
UHr0J jQK  
end dIk/vg  
<aps)v
F  
%有中间像，焦距输入为正数 L3[r7 b  
Q/[|
/uNw?  
function sjr=yfdre(~) HPl'u'.Hg  
E__^>=  
f=input('f:'); s5{N+O)~S  
d1=input('d1:'); hE
.NW  
d2=input('d2:'); ["l1\YCi  
d3=input('d3:'); g+>$_s  
3^p<Wx  
A=f^2/(d3*d2)-f/d1; dH4wyd`  
B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); CZ2&9Vb9I  
C=d3/d2-f/d1; Hkq""'Mx+w  
5!
WQ  
a1=(-B-sqrt(B^2-4*A*C))/(2*A); AITV+=sN  
a2=d3/(a1*f); CBaU$`5  
b2=a1*(1-a2)*f/d2; .48Csc-  
b1=(1-a1)*f/(d1*b2); ul0]\(sS:  
f-6hcd@Ca  
%曲率半径 E-Z6qZ^  
i=b<Mz7|  
R1=2*f/(b1*b2) E/x2LYH  
R2=2*a1*f/(b2*(1+b1)) T)%34gN  
R3=2*a1*a2*f/(1+b2) ]T
mx;[D  
U~{fbS3,  
A1=b2^3*(a1-1)*(1+b1)^3; 8@`"ZzM  
B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; !uaV6K  
C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; aLr\Uq,83  
jP*5(*[&y  
A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); 5Fh?YS=  
B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); 5I#L|+  
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); RmXC ^VQ  
Y{c_5YYf  
CB=[C1 B1;C2 B2]; Z}#,E;  
AB=[A1 B1;A2 B2]; J:s^F n  
AC=[A1 C1;A2 C2]; 0*?/s\>PS;  
n_G< /8  
%二次系数 &?~OV:r9  
);4lM%]eb  
k2=-(det(CB)/det(AB)); 8?ig/HSt2  
k3=-(det(AC)/det(AB)); vZEeb j  
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 [x}]sT`#a  
k2=k2 w!$|IC  
k3=k3 S $wx>715  
5sbMp;ZM  
end

谢谢分享，学习一下 {XC1B