[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 9dS<^E(ZF  
function sjr=nfdre(~) P"*#mH[W|  
, e^&,5b  
%系统焦距及各镜间距输入，间距取负正负 oF'_x,0  
f2h`bO  
f=input('f:'); zXX=WH  
d1=input('d1:'); N"/J1   
d2=input('d2:'); v7-z<'?s~  
d3=input('d3:'); [A jY~  
YQ? "~[mL  
A=f^2/(d3*d2)-f/d1; 2/o_,k  
B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); `L~gERW#  
C=d3/d2-f/d1; uwI$t[  
Z$*m=]2  
a1=(-B+sqrt(B^2-4*A*C))/(2*A);%α1 8O^<#lh  
a2=d3/(a1*f);%α2 S?Y%
}  
b2=a1*(1-a2)*f/d2;%β2 N>V\  
b1=(1-a1)*f/(d1*b2);%β1 %"[`   
(#e,tu  
y26?>.!  
%曲率半径 ~K$dQb])  
]g] ]\hS  
R1=2*f/(b1*b2) \9t/*%:  
R2=2*a1*f/(b2*(1+b1)) k'6x_ G  
R3=2*a1*a2*f/(1+b2) hqDnmzG  
{!0f.nv  
A1=b2^3*(a1-1)*(1+b1)^3; i<\WRzVT  
B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; $I0&I[_LzK  
C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; '@6O3z_{  
Jb;@'o6  
A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); (e"\%p`  
B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); )L+>^cJI<  
C2=b2*(a1-1)^2*(1+b1)*(1-b1)^2/(4*a1*b1^2)-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)*(1-b2)^2/(4*a1*a2*b1^2*b2^2)-b2*(a1-1)*(1-b1)*(1+b1)/(a1*b1)-(a2*(a1-1)+b1*(1-a2))*(1-b2)*(1+b2)/(a1*a2*b1*b2)-b1*b2+b2*(1+b1)/a1-(1+b2)/(a1*a2); ?3#L?Cq  
'_q&~M{  
CB=[C1 B1;C2 B2]; aM5Hp>'nI  
AB=[A1 B1;A2 B2]; <nvzNXql  
AC=[A1 C1;A2 C2]; qs$w9I  
ZrNBkfe:  
%非球面系数 "?AJ(>wP  
k2=-(det(CB)/det(AB)); +b sc3  
k3=-(det(AC)/det(AB)); '"m-kor  
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 9P1!<6mN\  
k2=k2 %/n#{;c#  
k3=k3 XpH d"(*  
OC6v%@xa  
end \yhj{QS.k  
Xk>YiV",?  
%有中间像，焦距输入为正数 )+ (GE  
% <1&\5f<5  
function sjr=yfdre(~) w~AW( VX  

B<{Yj}..  
f=input('f:'); Oh=
E!  
d1=input('d1:'); Mp`!zwR  
d2=input('d2:'); %#7 ]  
d3=input('d3:'); mS?W+jy%  
U[O7}Nsb"  
A=f^2/(d3*d2)-f/d1; C2;Hugm4  
B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); bQgtZHO  
C=d3/d2-f/d1; Oil~QAd,  
^k2g60]  
a1=(-B-sqrt(B^2-4*A*C))/(2*A); jPIOBEIG  
a2=d3/(a1*f); !d Z:Ih.[{  
b2=a1*(1-a2)*f/d2; ZH)thd9^b  
b1=(1-a1)*f/(d1*b2); g/+|gHq^  
U|~I
JU3-  
%曲率半径 /l` "@  
Mi<l;ZP  
R1=2*f/(b1*b2) m;cgX#k5  
R2=2*a1*f/(b2*(1+b1)) T+j-MR}{\  
R3=2*a1*a2*f/(1+b2) Ake$M^Bz  
h$
)4%Fy  
A1=b2^3*(a1-1)*(1+b1)^3; ,1 ^IFBJ  
B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; @5j3
[e  
C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; x/L(0z  
T@R2H&L  
A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); quL+UFuM  
B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); PiJ>gDx  
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); 9j>LU<Z  
#BW:*$>}  
CB=[C1 B1;C2 B2]; CHckmCgf4  
AB=[A1 B1;A2 B2]; ih=O#f|  
AC=[A1 C1;A2 C2]; 3of
0f{ZTj  
+W+o~BE  
%二次系数 y:YJv x6&4  
~o27~R ]  
k2=-(det(CB)/det(AB)); B-oQj
r-  
k3=-(det(AC)/det(AB)); G2yQHTbl  
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*Ri1   
k2=k2 avb'dx*q>  
k3=k3 #k t+ )>  
ipD/dx.  
end

谢谢分享，学习一下 ebiOR1)sN