[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 N/(&&\3  
function sjr=nfdre(~) e(sQgtM6  
vtmvvv  
%系统焦距及各镜间距输入，间距取负正负 o@}Jd0D4  
:*1bhk8~  
f=input('f:'); }{s<!b  
d1=input('d1:'); qjuX16o  
d2=input('d2:'); =F6J%$  
d3=input('d3:'); -[Q%Vv!8  
|iJ37QIM  
A=f^2/(d3*d2)-f/d1; dF"Sz4DY#  
B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); ]*?qaIdqu  
C=d3/d2-f/d1; g9g^zd,  
lf&g *%?1  
a1=(-B+sqrt(B^2-4*A*C))/(2*A);%α1 u9J;OsnHK  
a2=d3/(a1*f);%α2 ;a2TONW   
b2=a1*(1-a2)*f/d2;%β2 k? X7h2  
b1=(1-a1)*f/(d1*b2);%β1 lj(}{O  
.M!HVq47m  
r\Zz=~![<  
%曲率半径 s#>Bwn&b)  
1-4W4"#  
R1=2*f/(b1*b2) zGF_c9X  
R2=2*a1*f/(b2*(1+b1)) <rNCb;  
R3=2*a1*a2*f/(1+b2) GHfsq|*j,Z  
b+,u_$@B  
A1=b2^3*(a1-1)*(1+b1)^3; 1jpcoJ@s  
B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; F.zn:yX5  
C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; &^92z:?  
1d,;e:=j  
A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); \^i/:  
B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); wS9EC}s:Q  
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); zQ&`|kS  
s0WI93+z  
CB=[C1 B1;C2 B2]; @Pb!:HeJE  
AB=[A1 B1;A2 B2]; )Z2HzjE  
AC=[A1 C1;A2 C2]; *%bQp  
uDbz`VpK  
%非球面系数 Q)#+S(TG  
k2=-(det(CB)/det(AB)); 8SR~{  
k3=-(det(AC)/det(AB));
2Ck'A0d  
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 Yjx*hv&?  
k2=k2 %#7Yr(&
 
k3=k3 eX;C.[&7;8  
fL>>hBCqC  
end x8|sdZFxo  
NBU[>P  
%有中间像，焦距输入为正数 v2][gn+58  
B@U;[cO&  
function sjr=yfdre(~) !36jtKdM  
*z&m=G\  
f=input('f:'); U=QfInB  
d1=input('d1:'); vau0Jn%=ck  
d2=input('d2:'); FwKT_XkY  
d3=input('d3:'); '7Q5"M'  
R-5EztmLae  
A=f^2/(d3*d2)-f/d1; ] ;"blB  
B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); /Sy:/
BQ  
C=d3/d2-f/d1; J0K25w  
;w--fqxVl  
a1=(-B-sqrt(B^2-4*A*C))/(2*A); ancs  
a2=d3/(a1*f); %iMRJ}8(7  
b2=a1*(1-a2)*f/d2; 8$4@U;Vh;  
b1=(1-a1)*f/(d1*b2); qD0sD2 x  
p}I,!~}  
%曲率半径 #[=%+*Q  
&iYy  
R1=2*f/(b1*b2) \-i5b  
R2=2*a1*f/(b2*(1+b1)) tj<a , l  
R3=2*a1*a2*f/(1+b2) m$

6u K0  
zI1-l9 o  
A1=b2^3*(a1-1)*(1+b1)^3; !} ~K'1"  
B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; 2vbm=~)$F  
C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; !A@Ft}
FB  
vG;)(.:  
A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); ]N<:6+  
B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); <e)5$Aj  
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); w=Cqv~  
/1O
zX'5f  
CB=[C1 B1;C2 B2]; !5FZxmUup  
AB=[A1 B1;A2 B2]; +OM`c7M:  
AC=[A1 C1;A2 C2]; $=) i{kGS@  
o$disJ  
%二次系数 "eRf3Q7w:  
1T96W :   
k2=-(det(CB)/det(AB)); %z9eVkPI~  
k3=-(det(AC)/det(AB)); U5_1-wV  
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 GBMCw  
k2=k2 01~&H8 =  
k3=k3 PH$fDbC8  
@-9u;aL  
end

谢谢分享，学习一下 02|f@bP.