[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 yxCMl.  
function sjr=nfdre(~) u3M`'YCb  
,"N3k(g  
%系统焦距及各镜间距输入，间距取负正负 )7I.N]=  
qX p,d  
f=input('f:'); y'_V/w s  
d1=input('d1:'); vfJ3idvo*w  
d2=input('d2:'); +e0dV_T_>  
d3=input('d3:'); 5:l*Ib:s7  
of+phMev  
A=f^2/(d3*d2)-f/d1; u+z .J4w  
B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); ]{AOh2Z.hv  
C=d3/d2-f/d1; XWUvP  
v?Yd
LR  
a1=(-B+sqrt(B^2-4*A*C))/(2*A);%α1 cXb @H#  
a2=d3/(a1*f);%α2 _H4$$  
b2=a1*(1-a2)*f/d2;%β2 Q(=Vk~v  
b1=(1-a1)*f/(d1*b2);%β1 c#{Ywh  
{+C%D'  
vbRrk($`  
%曲率半径 Sv~YFS :oy  
(o x4K{  
R1=2*f/(b1*b2) BrNG%
%n  
R2=2*a1*f/(b2*(1+b1)) IqhICC1V-  
R3=2*a1*a2*f/(1+b2) 'R4>CZ%jV  
<\1}@?NGC  
A1=b2^3*(a1-1)*(1+b1)^3; QJM-`(  
B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; \~gA+o}Q  
C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; ?418*tXd  
GOYn\N;V2  
A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); L.09\1?.n  
B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); _HWHQF7  
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); 1z})mfsh  
%rpR-}j  
CB=[C1 B1;C2 B2]; (orrX Ez  
AB=[A1 B1;A2 B2]; 93Mdp9v+i  
AC=[A1 C1;A2 C2]; , @%C8Z  
/h?<MI\7V  
%非球面系数 v>#Cg\  
k2=-(det(CB)/det(AB)); c }<*~w;  
k3=-(det(AC)/det(AB)); !X5o7b)  
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 zW"~YaO%C  
k2=k2 I@3Q=14k%  
k3=k3 $ZQlIJZ  

G$;>ueM  
end X'V+^u@W  
]j& FbP)3  
%有中间像，焦距输入为正数 5TXg;v#Z  
//\ds71h  
function sjr=yfdre(~) cT/3yf  
Z|2Eb*  
f=input('f:'); !Oeq G  
d1=input('d1:'); )8pcf`h{  
d2=input('d2:'); b%Wd<N2  
d3=input('d3:'); JQ*D  
tO?21?AD D  
A=f^2/(d3*d2)-f/d1; reJ"r<2  
B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); ~?FK ; (  
C=d3/d2-f/d1; ;EK(b  
q<-%L1kc1  
a1=(-B-sqrt(B^2-4*A*C))/(2*A); HzP.aw4  
a2=d3/(a1*f); yJnPD/i  
b2=a1*(1-a2)*f/d2; =y?#^  
b1=(1-a1)*f/(d1*b2); % 2I  
f"Ost;7zg  
%曲率半径 98BYtxa
 
3H1Pp*PH  
R1=2*f/(b1*b2) y7CWBTH0>  
R2=2*a1*f/(b2*(1+b1)) 8ou e-:/a  
R3=2*a1*a2*f/(1+b2) HDyQzCG,  
g93I+  
A1=b2^3*(a1-1)*(1+b1)^3; N g58/}zO  
B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; 6dF$?I&  
C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; <e'P%tG'  
s>T`l  
A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); $np=eT)  
B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); ppEJs  
C2=b2*(a1-1)^2*(1+b1)*(1-b1)^2/(4*a1*b1^2)-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)*(1-b2)^2/(4*a1*a2*b1^2*b2^2)-b2*(a1-1)*(1-b1)*(1+b1)/(a1*b1)-(a2*(a1-1)+b1*(1-a2))*(1-b2)*(1+b2)/(a1*a2*b1*b2)-b1*b2+b2*(1+b1)/a1-(1+b2)/(a1*a2); V?=TVI*k  
Q`N18I3  
CB=[C1 B1;C2 B2]; dY1J<L}")  
AB=[A1 B1;A2 B2]; 1XG$ z@NN  
AC=[A1 C1;A2 C2];  G]b8]3^  
iDZrK%fl  
%二次系数 y $:yz;  
*]5z^> q;7  
k2=-(det(CB)/det(AB)); !&W|myN^  
k3=-(det(AC)/det(AB)); A 6:Q<  
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 USprsaj  
k2=k2 $)Wb#B  
k3=k3 Z+0?yQ=%  
T$u~E1  
end

谢谢分享，学习一下 lDC}HC