[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 gpw,bV  
function sjr=nfdre(~) \s?8}k  
3N[Rrxe2  
%系统焦距及各镜间距输入，间距取负正负 XCyU)[wY  
3\}>nE  
f=input('f:'); 8IkmFXj  
d1=input('d1:'); -e4TqzRr  
d2=input('d2:'); q6PG=9d0B  
d3=input('d3:'); +)hxYLk&I  
6H1;Hl f  
A=f^2/(d3*d2)-f/d1; uv=a}U;  
B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); NkV81?  
C=d3/d2-f/d1; D7r&z?  
S]&:R)#@  
a1=(-B+sqrt(B^2-4*A*C))/(2*A);%α1 @<.ei)
cqb  
a2=d3/(a1*f);%α2 lc8g$Xw3  
b2=a1*(1-a2)*f/d2;%β2 Ht:\ z;cu  
b1=(1-a1)*f/(d1*b2);%β1 ;}Ei #T,D  
[vY#9W"!  
7YxVtN  
%曲率半径 C*}PL  
[Tl66Eyl  
R1=2*f/(b1*b2) n2can  
R2=2*a1*f/(b2*(1+b1)) Pr:\zI  
R3=2*a1*a2*f/(1+b2) DcNp-X40I  
9Fy'L#%  
A1=b2^3*(a1-1)*(1+b1)^3; XzPOqZ`Nv  
B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; 2F+"v?n=\  
C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; \HIBnkj)3n  
k'_f?_PBu  
A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); *u-$$@|y  
B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); \FUMfo^  
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); v6|j.;  
s=]NKJaQH  
CB=[C1 B1;C2 B2]; Qd~z<U l  
AB=[A1 B1;A2 B2]; <S=(
`D  
AC=[A1 C1;A2 C2]; a{L&RRJ  
c9G%;U)  
%非球面系数 w\MWr+4  
k2=-(det(CB)/det(AB)); /@|iI<|  
k3=-(det(AC)/det(AB)); %Z4*;VwQ  
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 ilde<!?  
k2=k2 Q=8YAiCu  
k3=k3 dOKp:|9G  
2F>Y{3&  
end "0uM%*2  
]E-3/r$_cO  
%有中间像，焦距输入为正数 q01 L{~>bz  
>O<a9wz  
function sjr=yfdre(~) >I0;MNX  
Ovc9x\N  
f=input('f:'); =3c?W&:  
d1=input('d1:');
|,@D<  
d2=input('d2:'); &BLCP d  
d3=input('d3:'); YTFU#F  
jan}}7Dly  
A=f^2/(d3*d2)-f/d1; {j9TzR  
B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); Mib.,J~  
C=d3/d2-f/d1; CDM6o!ur3  
{=3A@/vM  
a1=(-B-sqrt(B^2-4*A*C))/(2*A); +HBizJ9K  
a2=d3/(a1*f); &n;*'M  
b2=a1*(1-a2)*f/d2; w") G:K  
b1=(1-a1)*f/(d1*b2); ru@#s2  
a51e~mg Z`  
%曲率半径 CyR`&u  
pe vXixl  
R1=2*f/(b1*b2) #Z1-+X8P  
R2=2*a1*f/(b2*(1+b1)) 4WvW11q8U  
R3=2*a1*a2*f/(1+b2) .*x:  
@KK6JyOTQ  
A1=b2^3*(a1-1)*(1+b1)^3; >lV,K1Z  
B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; P !i_
?M  
C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; #zTy7ZS,0  
4`Ud\Jm[s  
A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); -n.m "O3  
B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); gLX<>|)*  
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); iT :3e%  
y017 B<Ou  
CB=[C1 B1;C2 B2]; 4j=3'Z|  
AB=[A1 B1;A2 B2]; xT"V9t[f  
AC=[A1 C1;A2 C2]; zuLW'a6F-
 
OP1`!P y  
%二次系数 4C=W~6~  
M!J7Vj?Ps  
k2=-(det(CB)/det(AB)); \Ip)Lm0  
k3=-(det(AC)/det(AB)); 3P6O]x<-?  
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  Lto*L X  
k2=k2 H7X-\K 1w  
k3=k3 t,7%| {  
*)Qv;'U=rn  
end

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