[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 e *9c33  
function sjr=nfdre(~) .0ov>4,R  
^ &UezDTS  
%系统焦距及各镜间距输入，间距取负正负 ahJ-T@  
"Sx}7?8AB  
f=input('f:'); .
C HET]  
d1=input('d1:'); 7tZvz `\  
d2=input('d2:'); _Fc :<Ym?  
d3=input('d3:'); dio<?6ZD9P  
_U Q|I|V#  
A=f^2/(d3*d2)-f/d1; *HC[LM  
B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); uGo tXb  
C=d3/d2-f/d1; Uj/m  
LA +BH_t&  
a1=(-B+sqrt(B^2-4*A*C))/(2*A);%α1 A,H|c="  
a2=d3/(a1*f);%α2 r:.ydr@  
b2=a1*(1-a2)*f/d2;%β2 pwIu;:O!?  
b1=(1-a1)*f/(d1*b2);%β1 }Cs.Hm0P  
b '9L}q2m  
'gD,HX  
%曲率半径 {.HFB:<!}  
Q:v9C ^7  
R1=2*f/(b1*b2) \gU=B|W  
R2=2*a1*f/(b2*(1+b1)) 0`H)c) pP  
R3=2*a1*a2*f/(1+b2) kO)+%'L!8  
$ap6Vxjr  
A1=b2^3*(a1-1)*(1+b1)^3; p?
Rq  
B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; r.~^h^c]  
C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; g!![%*' b  
=Rw-@*#l  
A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); .a@>1XO  
B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); pD@2Mt0|]=  
C2=b2*(a1-1)^2*(1+b1)*(1-b1)^2/(4*a1*b1^2)-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)*(1-b2)^2/(4*a1*a2*b1^2*b2^2)-b2*(a1-1)*(1-b1)*(1+b1)/(a1*b1)-(a2*(a1-1)+b1*(1-a2))*(1-b2)*(1+b2)/(a1*a2*b1*b2)-b1*b2+b2*(1+b1)/a1-(1+b2)/(a1*a2); IncHY?ud<  
WVMkLMg8d  
CB=[C1 B1;C2 B2]; ^#0k\f>_  
AB=[A1 B1;A2 B2]; jmkVolz  
AC=[A1 C1;A2 C2];
j34L*?  
=Z~nzyaN  
%非球面系数 {E3329t|'  
k2=-(det(CB)/det(AB)); G&DL)ePu]m  
k3=-(det(AC)/det(AB)); ^{l^Z +b.  
k1=(k2*a1*b2^3*(1+b1)^3-k3*a1*a2*(1+b2)^3+a1*b2^3*(1+b1)*(1-b1)^2-a1*a2*(1+b2)*(1-b2)^2)/(b1^3*b2^3)-1 97L|IZ s)  
k2=k2 qiyX{J7Z  
k3=k3 .$s>b#mO  
"4Bk  
end "I n[= 2w  
<<iwJ U%:  
%有中间像，焦距输入为正数 &X%vp?p  
muZ6}&4  
function sjr=yfdre(~) B{dR/q3;@  
3mCf>qj73  
f=input('f:'); WKDa]({k%  
d1=input('d1:'); kaUEv\T   
d2=input('d2:'); KJQW))%e  
d3=input('d3:'); x<lY&KQ0  
-&UP[Mq  
A=f^2/(d3*d2)-f/d1; R32A2Ml  
B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); -w3KBlo  
C=d3/d2-f/d1; Zk .V   
FcaO-  
a1=(-B-sqrt(B^2-4*A*C))/(2*A); (h$[g"8  
a2=d3/(a1*f); $sa5aUg }  
b2=a1*(1-a2)*f/d2; 3Gyw^_{J  
b1=(1-a1)*f/(d1*b2); O S%  
]A%S&q  
%曲率半径 c1>:|D7w  
<Z:Fnp
 
R1=2*f/(b1*b2) .r4M]1Of  
R2=2*a1*f/(b2*(1+b1)) 5+yT{,(5  
R3=2*a1*a2*f/(1+b2) .`; bQh'!  
3~5%6`  
A1=b2^3*(a1-1)*(1+b1)^3; Hp}dm93T  
B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; 7sglqf>  
C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; ;"T,3JQPn6  
T:?01?m  
A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); 27CVAX ghV  
B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); +ia N[F$  
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); bayDdR4T  
>^|(AzS  
CB=[C1 B1;C2 B2]; miv)R  
AB=[A1 B1;A2 B2]; g$a 5  
AC=[A1 C1;A2 C2]; l+n0=^ Z  
Ph Ep3o&"  
%二次系数 c!20((2|I  
xmp^`^v*  
k2=-(det(CB)/det(AB)); YgfQ{3^I  
k3=-(det(AC)/det(AB)); eJ2$DgB}t  
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 cE SSSH!m  
k2=k2 lQ!)0F  
k3=k3 azT@S=,  
ZBcT@hxm  
end

谢谢分享，学习一下 Dqh rg;