[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 p?mQ\O8F  
function sjr=nfdre(~) H0zKL]D'>  
;E8.,#/a  
%系统焦距及各镜间距输入，间距取负正负 R89;<,Ie  
~mqiXr8  
f=input('f:'); U{2[nF  
d1=input('d1:'); )F\tU  
d2=input('d2:'); >AT{\W!N  
d3=input('d3:'); akuV
9S  
o<*H!oyP\  
A=f^2/(d3*d2)-f/d1; v'"0Ya  
B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); _E[zYSo`  
C=d3/d2-f/d1; *(%]|z}]m  
OC$Y8Ofr  
a1=(-B+sqrt(B^2-4*A*C))/(2*A);%α1 E,]G Ek  
a2=d3/(a1*f);%α2 ##gq{hgjb$  
b2=a1*(1-a2)*f/d2;%β2 3|=9aM^x^  
b1=(1-a1)*f/(d1*b2);%β1 N|n"JKw)  
,\y)k}0lH  
^Z:oCTOP  
%曲率半径 @S/jVXA  
35dbDgVz$  
R1=2*f/(b1*b2) }u=-Y'!#]  
R2=2*a1*f/(b2*(1+b1)) ,k*g`OTW  
R3=2*a1*a2*f/(1+b2) *C~O[:6D  
wHSas[4k  
A1=b2^3*(a1-1)*(1+b1)^3; v !~lVv&  
B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; 'N?,UtG R  
C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; <,%qt_ !  
IXp(Aeb  
A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); R<}n?f\#JZ  
B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); <P)vx  
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); 0B0Uay'd_  
]] R*sd*  
CB=[C1 B1;C2 B2]; 50:gk*hy  
AB=[A1 B1;A2 B2]; p`@7hf|hm  
AC=[A1 C1;A2 C2]; yF&"'L  
sb"etc`w%-  
%非球面系数 K5O8G  
k2=-(det(CB)/det(AB)); $"z|^ze  
k3=-(det(AC)/det(AB)); :wn9bCom?M  
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 g5gq{KlU  
k2=k2  ~yQby&s  
k3=k3 ={[
s)G  
eyq8wQT  
end =2[5g!qX  
1wX0x.4d  
%有中间像，焦距输入为正数 $Hr qX?&r  
K3QE>@']  
function sjr=yfdre(~) H1`}3}"  
Uq0GbLjv"  
f=input('f:');  `Pa)H  
d1=input('d1:'); > YHwWf-  
d2=input('d2:'); -e6
~0%X  
d3=input('d3:'); '+6H=Qn  
}P8@\2@=T  
A=f^2/(d3*d2)-f/d1; N"X;aVFs_  
B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); pIKQx5;  
C=d3/d2-f/d1; gxry?':  
HmWU;9Vn+  
a1=(-B-sqrt(B^2-4*A*C))/(2*A);  xZJ r
*  
a2=d3/(a1*f); )Mw<e  
b2=a1*(1-a2)*f/d2; xz1jRI$  
b1=(1-a1)*f/(d1*b2); l+e L:C!  
o"5Bg%H  
%曲率半径 UZ8 vZ  
|a7W@LVYD  
R1=2*f/(b1*b2) 95Q{d'&  
R2=2*a1*f/(b2*(1+b1)) t{\,vI  
R3=2*a1*a2*f/(1+b2) RcQo1  
qJY'"_Q{  
A1=b2^3*(a1-1)*(1+b1)^3; Lq#>N_72W0  
B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; 3z^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; |P{K\;-  
q9GSUkb  
A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); pkd#SY  
B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); 9,h'cf`F  
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); yH\z+A|  
OGgP
~hd  
CB=[C1 B1;C2 B2]; QLx]%
E\  
AB=[A1 B1;A2 B2]; h,:8TMJRRN  
AC=[A1 C1;A2 C2]; 1 J3h_z6/  
p~ `f.q$'  
%二次系数 >EQd;Af  
$O}:*.{(W  
k2=-(det(CB)/det(AB)); $vlc@]~d`&  
k3=-(det(AC)/det(AB)); h{s- e.  
k1=(k2*a1*b2^3*(1+b1)^3-k3*a1*a2*(1+b2)^3+a1*b2^3*(1+b1)*(1-b1)^2-a1*a2*(1+b2)*(1-b2)^2)/(b1^3*b2^3)-1 }O+F#/6  
k2=k2 Ey5E1$w%&  
k3=k3 Tvd=EO  
bERYC|  
end

谢谢分享，学习一下 B>dX
yo