[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 WObvbaK  
function sjr=nfdre(~) U.XNv-M  
j!4et;  
%系统焦距及各镜间距输入，间距取负正负 T[uiPs/xD  
\ 3?LqJ  
f=input('f:'); lR[qqFR  
d1=input('d1:'); %D8ZO0J7H  
d2=input('d2:'); Ikiib WQL+  
d3=input('d3:'); !n$tr  
bKQ"ax>6p  
A=f^2/(d3*d2)-f/d1; <]4i`6{v  
B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); !{)tSipd  
C=d3/d2-f/d1; }!/$M\w  
{R61cD,n  
a1=(-B+sqrt(B^2-4*A*C))/(2*A);%α1 7#U^Dx\yh  
a2=d3/(a1*f);%α2 T G{k0cdOT  
b2=a1*(1-a2)*f/d2;%β2 c`WHNky%j  
b1=(1-a1)*f/(d1*b2);%β1 1~ t{aLPz  
>teOm?@U  
w@c87;c  
%曲率半径 TJ'[--  
(85Fv&a  
R1=2*f/(b1*b2) <Ib[82PU  
R2=2*a1*f/(b2*(1+b1)) [],1lRYI9_  
R3=2*a1*a2*f/(1+b2) * Y7jl#7  
&q~:
~  
A1=b2^3*(a1-1)*(1+b1)^3; uN^=<B?B  
B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; Q~Hh
\Lt  
C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; .G(llA}  
)+"'oY$]}  
A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); Ru>uL@w  
B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); P=+nB*hG  
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); UybW26C;aU  

nF4a-H&Fo  
CB=[C1 B1;C2 B2]; 0}FOV`n  
AB=[A1 B1;A2 B2]; V$icWu  
AC=[A1 C1;A2 C2]; VY"9?2?/  
#8bsxx!s  
%非球面系数 rs*Fy@  
k2=-(det(CB)/det(AB)); M$d%p6Cv  
k3=-(det(AC)/det(AB)); V8wKAj Ux  
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 .tmiQ.  
k2=k2 saAxGG  
k3=k3 #gaQaUjR  
>,gvb5  
end 8kOKwEX  
e)IpPTj#  
%有中间像，焦距输入为正数 ELjK0pE}-  
o|iYd n\  
function sjr=yfdre(~) lR`'e0Lq  
|?,[@z _,  
f=input('f:'); wHv]ViNvXE  
d1=input('d1:'); B
_l{<  
d2=input('d2:'); 'K*AV7>E  
d3=input('d3:'); ',4x$qe  
}Jh: 8BNuP  
A=f^2/(d3*d2)-f/d1; @@g\2Gs  
B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); !W'Ui 9uX  
C=d3/d2-f/d1; ?$Wn!"EC8  
lJs<  
a1=(-B-sqrt(B^2-4*A*C))/(2*A); ~i_Tw#}  
a2=d3/(a1*f); \WrFqm#  
b2=a1*(1-a2)*f/d2; :!f(F9  
b1=(1-a1)*f/(d1*b2); qcJ
ft'>F  
9Sq%s&  
%曲率半径 z6~ H:k1G%  
h~,JdDV8l*  
R1=2*f/(b1*b2) g3sUl&K  
R2=2*a1*f/(b2*(1+b1)) C4#rA.nF|  
R3=2*a1*a2*f/(1+b2) ;t>Z+O%  
[;oCYb$9  
A1=b2^3*(a1-1)*(1+b1)^3; W Qzj[  
B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; qxM
np}O  
C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; 3q:
{1rc  
Y'3}G<'%  
A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); W*:,m8wk  
B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); 0g<K[mPr7  
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); ExXM:1 e26  
sNHSr  
CB=[C1 B1;C2 B2]; !b-bP,q  
AB=[A1 B1;A2 B2]; =}fd6ea(o  
AC=[A1 C1;A2 C2]; HFQR ;9]  
 <n\`d  
%二次系数 B 2.q3T  
 5q<zN  
k2=-(det(CB)/det(AB)); v !Kw< fp|  
k3=-(det(AC)/det(AB)); Lj%{y.Rj  
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 m:O(+Fl  
k2=k2 sK=}E=  
k3=k3 X0+fsf<H}  
RQvVR  
end

谢谢分享，学习一下 1(GHCxA8G