[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 ' zz^!@  
function sjr=nfdre(~) |O4LR,{G.w  
rS&"UH?c7  
%系统焦距及各镜间距输入，间距取负正负 yHNx,ra
 
v| Yh]y  
f=input('f:'); SL ) ope  
d1=input('d1:'); aRE%(-5  
d2=input('d2:'); h5^qo ^;g7  
d3=input('d3:'); yh:Wg$qx
 
 J*FUJT  
A=f^2/(d3*d2)-f/d1; N6UPD11}6  
B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); A8oTcX_  
C=d3/d2-f/d1; D8*tzu-  
WkUV)/j  
a1=(-B+sqrt(B^2-4*A*C))/(2*A);%α1 >^N{  
a2=d3/(a1*f);%α2 [8v>jQ)  
b2=a1*(1-a2)*f/d2;%β2 'Tbdo >y  
b1=(1-a1)*f/(d1*b2);%β1 %=[xc?  
-J'0qN!  
CEHtr90P  
%曲率半径 QpI\\Zt6  
U *K6FWqiB  
R1=2*f/(b1*b2) s=[T
,:Z  
R2=2*a1*f/(b2*(1+b1)) }8&?  
R3=2*a1*a2*f/(1+b2) UEeq@ot/4  
}|u>b!7_.  
A1=b2^3*(a1-1)*(1+b1)^3; VV$4NV&`Q  
B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; +L U.QI'  
C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; Xt9vTCox  
Sk7sxy<F'  
A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); gUWW}*\ U  
B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); "OWW -m  
C2=b2*(a1-1)^2*(1+b1)*(1-b1)^2/(4*a1*b1^2)-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)*(1-b2)^2/(4*a1*a2*b1^2*b2^2)-b2*(a1-1)*(1-b1)*(1+b1)/(a1*b1)-(a2*(a1-1)+b1*(1-a2))*(1-b2)*(1+b2)/(a1*a2*b1*b2)-b1*b2+b2*(1+b1)/a1-(1+b2)/(a1*a2); %yPjPUHy  
G5,g$yNs  
CB=[C1 B1;C2 B2]; "J"RH:$v  
AB=[A1 B1;A2 B2]; Rx,5?*b$  
AC=[A1 C1;A2 C2]; 1<;RI?R[9  

<19A=  
%非球面系数 Kv~'*A)d  
k2=-(det(CB)/det(AB)); Z66h  
k3=-(det(AC)/det(AB)); 1G<S'd+N  
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 U~I y),5  
k2=k2 aExt TE  
k3=k3 4H*M^?h\#  
?"-1QG  
end Ou7nk:I@  
>QSlH]M  
%有中间像，焦距输入为正数 0T
2^$^g  
6;Sz^W  
function sjr=yfdre(~) AkAQ%)6qV  
0`KR8# A@  
f=input('f:'); D5"Xjo*  
d1=input('d1:'); LMHiiOs,  
d2=input('d2:'); 3-v&ktD&N'  
d3=input('d3:'); zGaqYbQD  
^I:f4RWo  
A=f^2/(d3*d2)-f/d1; r)|6H"n#]S  
B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); ;Z.sK-NJ4  
C=d3/d2-f/d1; j.kv!;Rj=  
wJF(&P  
a1=(-B-sqrt(B^2-4*A*C))/(2*A); YkF52_^_  
a2=d3/(a1*f); 3g87ir  
b2=a1*(1-a2)*f/d2; ~B\O{5W  
b1=(1-a1)*f/(d1*b2); $bFH%EA.
 
hV}C.- 6h  
%曲率半径 i6P'_  
'37 <+N  
R1=2*f/(b1*b2) UK5u"@T  
R2=2*a1*f/(b2*(1+b1)) 9c{T|+]  
R3=2*a1*a2*f/(1+b2) 7G=Q9^J.H  
^Eif~v  
A1=b2^3*(a1-1)*(1+b1)^3; nSpOTQ  
B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; QB"+B]rV  
C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; UetI4`  
_[h!r;DsG  
A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); j~Mx^ivwj  
B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); sL)7MtNwy  
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); WI1DL&*B@<  
[L=M=;{4  
CB=[C1 B1;C2 B2]; nQ@<[KNd  
AB=[A1 B1;A2 B2]; Yy0U2N[i  
AC=[A1 C1;A2 C2]; U;#G$  
"2ZuI;w  
%二次系数 R
>hL.+l.  
yG2rAG_G&  
k2=-(det(CB)/det(AB)); -_BX\iP{  
k3=-(det(AC)/det(AB)); VE))`?  
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 49=L9:  
k2=k2 rN'8,CV  
k3=k3 C9 j{:&  
g>QN9v})  
end

谢谢分享，学习一下 M'JCT'(X