[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 tmUFT  
function sjr=nfdre(~) "Wo,'8{v  
cLVeT  
%系统焦距及各镜间距输入，间距取负正负 -E"GX
 
wU\s; dK  
f=input('f:'); \QE)m<G
Ue  
d1=input('d1:'); }g7]?Ee  
d2=input('d2:'); `n5"0QRd  
d3=input('d3:'); re?s.djT  
r.ZF_^y}+  
A=f^2/(d3*d2)-f/d1; ={>Lrig:l  
B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); X;v$5UKU  
C=d3/d2-f/d1; ::FS/Y]Fg  
zOgTQs"ZH  
a1=(-B+sqrt(B^2-4*A*C))/(2*A);%α1 z7GTaX$d  
a2=d3/(a1*f);%α2 DwHF[]v'  
b2=a1*(1-a2)*f/d2;%β2 #n0Y6Pr  
b1=(1-a1)*f/(d1*b2);%β1 Z_1U9+,  
l>*"mh  
o]:3H8  
%曲率半径 10!wqyj&  
k@ZLg9  
R1=2*f/(b1*b2) Suk  
R2=2*a1*f/(b2*(1+b1)) yeDsJ/L  
R3=2*a1*a2*f/(1+b2) ,to+oSZE  
Urr#N  
A1=b2^3*(a1-1)*(1+b1)^3; FIxF
nh3~  
B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; |sRipWh  
C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; )XcOl7XLN  
5%sE]Y#  
A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); _4^R9Bt  
B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); EF3Cdu{]P  
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); ;4N;D  
;qH
OOT  
CB=[C1 B1;C2 B2]; 7qTE('zt  
AB=[A1 B1;A2 B2]; L|bwZ,M=}?  
AC=[A1 C1;A2 C2]; r{l(O,|e  
.`>y@p!  
%非球面系数 jReXyRmo({  
k2=-(det(CB)/det(AB)); <! )**  
k3=-(det(AC)/det(AB)); (=~&+z  
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 F 8B#}%JE  
k2=k2 orOt>5}b<  
k3=k3 y~()|L[  
yR(x+Gs{]  
end ?QE,;QtpK  
6G=j6gK%P  
%有中间像，焦距输入为正数 8Q_SRwN  
E@7J:|.)R  
function sjr=yfdre(~) Y ')x/H  
kbM3  
f=input('f:'); HRB<Y mP@  
d1=input('d1:'); x c{hC4^V  
d2=input('d2:'); BcI|:qv|  
d3=input('d3:'); +TXX$)3%  
!
.d@L6  
A=f^2/(d3*d2)-f/d1; (<^yqH?  
B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); / X1 x  
C=d3/d2-f/d1; B(M6@1m_  
Ol!ntNhXm  
a1=(-B-sqrt(B^2-4*A*C))/(2*A); F_CYYGZ  
a2=d3/(a1*f); Yk=PS[f  
b2=a1*(1-a2)*f/d2; K4kMM*D  
b1=(1-a1)*f/(d1*b2); 5LOo8xN  
IIbYfPiO  
%曲率半径 YpqrZWvh  
-Z's@'*  
R1=2*f/(b1*b2) %n*-VAfE\  
R2=2*a1*f/(b2*(1+b1)) 8YbE`32  
R3=2*a1*a2*f/(1+b2) EY tQw(!Q  
M3q|l7|9  
A1=b2^3*(a1-1)*(1+b1)^3; z*-2.}&U<  
B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; b9!FC$^J  
C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; 6fw(T.Pe  
0\eIQp  
A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); lv04g} W  
B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); j:VbrR  
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); 13>0OKg`#
 
5k.oW=  
CB=[C1 B1;C2 B2]; jbAx;Xt'=M  
AB=[A1 B1;A2 B2]; .X;3,D[w  
AC=[A1 C1;A2 C2]; 9E6_]8rl  
o,)?!{k}  
%二次系数 #)nSr  
}"|K(hq  
k2=-(det(CB)/det(AB)); ajEjZ6  
k3=-(det(AC)/det(AB)); n^g|Ja  
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 ]iUxp+  
k2=k2 ,LnII  
k3=k3 JT!9\i  
X<I+&Zi  
end

谢谢分享，学习一下 U`2e{>'4t