[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 _+By=B.'  
function sjr=nfdre(~) aPelt`  
@6G)(NGD  
%系统焦距及各镜间距输入，间距取负正负 AT1cN1:4?  
{KHI(*r;  
f=input('f:'); i-wRwl4aEF  
d1=input('d1:'); veq3t$sj  
d2=input('d2:'); gttsxOgktH  
d3=input('d3:'); +H3~Infr4f  
Cw(e7K7&  
A=f^2/(d3*d2)-f/d1; MOQ6&C`7q  
B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); u"qu!EY2  
C=d3/d2-f/d1; cIwX sx  
sR9$=91`  
a1=(-B+sqrt(B^2-4*A*C))/(2*A);%α1 CBd%}il  
a2=d3/(a1*f);%α2 )<V!lsUx'-  
b2=a1*(1-a2)*f/d2;%β2 }9\_s*  
b1=(1-a1)*f/(d1*b2);%β1 Nl YFS?5  
/=;,lC  
$^fF}y6N  
%曲率半径 s8,YQ5-  
5Hu[*  
R1=2*f/(b1*b2) #`3Q4  
R2=2*a1*f/(b2*(1+b1)) X&zGgP/  
R3=2*a1*a2*f/(1+b2) ~JT2el2W7p  
)L9eLxI  
A1=b2^3*(a1-1)*(1+b1)^3; fsjLD|?|:  
B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; +1T>Ob;hk  
C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; v>LK+|U  
q.=Q
 
A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); W\>O$IX^e  
B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); ywp_,j9F  
C2=b2*(a1-1)^2*(1+b1)*(1-b1)^2/(4*a1*b1^2)-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)*(1-b2)^2/(4*a1*a2*b1^2*b2^2)-b2*(a1-1)*(1-b1)*(1+b1)/(a1*b1)-(a2*(a1-1)+b1*(1-a2))*(1-b2)*(1+b2)/(a1*a2*b1*b2)-b1*b2+b2*(1+b1)/a1-(1+b2)/(a1*a2); /}@F q  
]z'L1vQl7  
CB=[C1 B1;C2 B2]; #|E#Rkw!  
AB=[A1 B1;A2 B2]; qR cSB  
AC=[A1 C1;A2 C2]; I+ |uyc  
"J,|),Yd  
%非球面系数 Nmx\qJUR(  
k2=-(det(CB)/det(AB)); AHs%?5YTY;  
k3=-(det(AC)/det(AB)); Z~SAlhT  
k1=(k2*a1*b2^3*(1+b1)^3-k3*a1*a2*(1+b2)^3+a1*b2^3*(1+b1)*(1-b1)^2-a1*a2*(1+b2)*(1-b2)^2)/(b1^3*b2^3)-1 {oY"CZ2  
k2=k2 /4Wf\ Zu  
k3=k3 fh`Y2s|:7R  
!f(A9V  
end &C MBTY#u  
,5zY1C==Ut  
%有中间像，焦距输入为正数 Cl3vp_  
R7rM$|n=o  
function sjr=yfdre(~) WILa8"M  
AT I=&O`  
f=input('f:'); dsw^$R}  
d1=input('d1:'); =k<b* 8  
d2=input('d2:'); ;cf$u}+  
d3=input('d3:'); =b$g_+  
D-@6 hWh~  
A=f^2/(d3*d2)-f/d1; uH$hMg  
B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); B)7:*Kj  
C=d3/d2-f/d1; 4e>f}u5  
4O
M ]8I!  
a1=(-B-sqrt(B^2-4*A*C))/(2*A); vfqXHc unj  
a2=d3/(a1*f); gtH^'vFZ  
b2=a1*(1-a2)*f/d2; e/Z{{FP%6  
b1=(1-a1)*f/(d1*b2); BD]J/o  
!KXcg9e  
%曲率半径 ;sA 5&a>!  
L$c 1<7LU  
R1=2*f/(b1*b2) 2n>mISy+  
R2=2*a1*f/(b2*(1+b1)) >AV
9 K  
R3=2*a1*a2*f/(1+b2) x=>dmi3  
xKL(:ePS  
A1=b2^3*(a1-1)*(1+b1)^3; *H/)S5
 
B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; Uot(3p!S6  
C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; ?W ^`Fa)]o  
gAvNm[=wD2  
A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); Tg O]q4  
B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); x3'ANw6E  
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); o!h::j0,~  
RQ|K?^k v  
CB=[C1 B1;C2 B2]; ]NaH *\q  
AB=[A1 B1;A2 B2]; I|*<[/)]y  
AC=[A1 C1;A2 C2]; N@0/
=B[n  
Z5rL.a&  
%二次系数 "xC$Ko _  
&56\@t^  
k2=-(det(CB)/det(AB)); fATn
za  
k3=-(det(AC)/det(AB)); w`boQ_Ir  
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 6@0?~  
k2=k2 m6 M/G  
k3=k3 zLr:zfl  
l{rHXST|  
end

谢谢分享，学习一下 I&8!V)r)