[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 (W7cQ>  
function sjr=nfdre(~) p+O2:  
PD$gW`V  
%系统焦距及各镜间距输入，间距取负正负 Yy4?|wVl  
(f^K\7HM  
f=input('f:'); <bPn<QI  
d1=input('d1:'); :EISms  
d2=input('d2:'); Zhl}X!:c?\  
d3=input('d3:'); -?A,N,nnX  
8+Y+\XZG  
A=f^2/(d3*d2)-f/d1; IH;+pN  
B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); PKoB~wLH  
C=d3/d2-f/d1; r@_`ob RW;  
\X5>HPB  
a1=(-B+sqrt(B^2-4*A*C))/(2*A);%α1 3Z*'  
a2=d3/(a1*f);%α2 k!gf
t'iU  
b2=a1*(1-a2)*f/d2;%β2 7|A9  
b1=(1-a1)*f/(d1*b2);%β1 SBBDlr^P  
L{K:XiPn  
Rh7unJ
 
%曲率半径 
VO|2  
-saisH6  
R1=2*f/(b1*b2) ;%r#pv~  
R2=2*a1*f/(b2*(1+b1)) k%iZ..  
R3=2*a1*a2*f/(1+b2) l@ +]XyLj  
DwXzmp[qWH  
A1=b2^3*(a1-1)*(1+b1)^3; w<?v78sT  
B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; 6m_whGosi  
C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; gDP\u<2!  
S!0ocS!t  
A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); wl1JKiodg  
B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); k<xiP@b{y  
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); 6e1/h@p\7  
~/hyf]*j  
CB=[C1 B1;C2 B2]; A^PCI*SN[  
AB=[A1 B1;A2 B2]; aB9Pdut  
AC=[A1 C1;A2 C2]; %,u_
`P  
YYrXLt:  
%非球面系数 8y!d^EQ  
k2=-(det(CB)/det(AB)); (>
THN*i  
k3=-(det(AC)/det(AB)); XG fLi  
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 s2t'jIB  
k2=k2 ^c1%$@H  
k3=k3 ;<Dou7=  
:IFTiq5a;  
end rs$sAa*f  
3lc'(ts%  
%有中间像，焦距输入为正数 ~Uw**PT3M  
A&#P=mj  
function sjr=yfdre(~) 5gP#V K  
5}3Q}o#  
f=input('f:'); Z~}=q  
d1=input('d1:'); c?i=6CdD'  
d2=input('d2:'); J]8nbl  
d3=input('d3:'); }-H<wQ&x  
|I7-7d-;/  
A=f^2/(d3*d2)-f/d1; e[fzy0  
B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); k> I;mEV  
C=d3/d2-f/d1; &:g5+([<  
k;v23  
a1=(-B-sqrt(B^2-4*A*C))/(2*A); B mq7w,L.  
a2=d3/(a1*f); L^:+8g  
b2=a1*(1-a2)*f/d2; q^],K'  
b1=(1-a1)*f/(d1*b2); 8UArl
3  
0Y#S2ty  
%曲率半径 s8T}ah!  
CybHr#LBc  
R1=2*f/(b1*b2) /[YH W]  
R2=2*a1*f/(b2*(1+b1)) T)B
yws  
R3=2*a1*a2*f/(1+b2) ^|DI9G(Bs  
24_F`" :-=  
A1=b2^3*(a1-1)*(1+b1)^3; wrq0fHwM  
B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; 7Wg0-{yK4  
C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; M$L1!o1Xf  
"=I ioY  
A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); O@jW&-;  
B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); {.tUn`j6V  
C2=b2*(a1-1)^2*(1+b1)*(1-b1)^2/(4*a1*b1^2)-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)*(1-b2)^2/(4*a1*a2*b1^2*b2^2)-b2*(a1-1)*(1-b1)*(1+b1)/(a1*b1)-(a2*(a1-1)+b1*(1-a2))*(1-b2)*(1+b2)/(a1*a2*b1*b2)-b1*b2+b2*(1+b1)/a1-(1+b2)/(a1*a2); &) 7umdSgi  
'ypJGm  
CB=[C1 B1;C2 B2]; +D|y))fE  
AB=[A1 B1;A2 B2]; 0}]k>ndT  
AC=[A1 C1;A2 C2]; 3IHya=qN  
n|F$qV_p\  
%二次系数 TCJH^gDt  
z5Hz-.  
k2=-(det(CB)/det(AB)); 1,fjdd8OM;  
k3=-(det(AC)/det(AB)); IH?.s k  
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 :`6E{yfM  
k2=k2 e$|g  
k3=k3 l+S08IZ  
;Dg8>  
end

谢谢分享，学习一下 SLRQ3<0W_