[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 yoQ\lk  
function sjr=nfdre(~) /_)l|<k+V  
EiWd+v,QJQ  
%系统焦距及各镜间距输入，间距取负正负 G(/DtY]  
GKZN}bOm\  
f=input('f:'); :_xh(W+2<  
d1=input('d1:'); /ylc*3e'4  
d2=input('d2:'); ?'eq",c#4N  
d3=input('d3:'); 2/B)O)#ls  
gzf-)J  
A=f^2/(d3*d2)-f/d1; Qqc]aVRF  
B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); d]SYP  
C=d3/d2-f/d1; ZSB?Y1wG  
?qmp_2:WU  
a1=(-B+sqrt(B^2-4*A*C))/(2*A);%α1 +Rwx%=  
a2=d3/(a1*f);%α2 m|O1QM;T  
b2=a1*(1-a2)*f/d2;%β2 j+kC-U;  
b1=(1-a1)*f/(d1*b2);%β1 kccWoU,  
!:3.D,  
\S ."?!U  
%曲率半径 MzkkcQLK  
.WX,Nd3@  
R1=2*f/(b1*b2) /[=E0_t+  
R2=2*a1*f/(b2*(1+b1)) S_b/DO  
R3=2*a1*a2*f/(1+b2) DMn4ll|  
eg<pa'Hw  
A1=b2^3*(a1-1)*(1+b1)^3; 7 ^$;  
B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; TTZe$>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; QR0(,e$Dl  
XRtD< jlA"  
A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); poT&-Ic[  
B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); "& 25D  
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); 3<:jx~y>  
|(%zb\#9  
CB=[C1 B1;C2 B2]; -SCM:j%h  
AB=[A1 B1;A2 B2]; S,{tV=&m]  
AC=[A1 C1;A2 C2]; Am"(+>W21  
S,jZ3^  
%非球面系数 n V&cC  
k2=-(det(CB)/det(AB)); t;NV $!!  
k3=-(det(AC)/det(AB)); ny*i+4Mb  
k1=(k2*a1*b2^3*(1+b1)^3-k3*a1*a2*(1+b2)^3+a1*b2^3*(1+b1)*(1-b1)^2-a1*a2*(1+b2)*(1-b2)^2)/(b1^3*b2^3)-1 vScjq5"p  
k2=k2 -c*\o3)  
k3=k3 IG ~`i I  
M4yI`dr6  
end C! 9}  
-ijC_`>  
%有中间像，焦距输入为正数 `y
iC=$*[  
Dw6fmyJ:  
function sjr=yfdre(~) w@-M{?R  
g)"gw+ZFc  
f=input('f:'); bHE2,;o  
d1=input('d1:'); 3=mr "&]r:  
d2=input('d2:'); v,@F|c?_S  
d3=input('d3:'); kz}R[7  
7[pBUDA  
A=f^2/(d3*d2)-f/d1; ;C.S3}  
B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); bulS&dAX  
C=d3/d2-f/d1; i3$$,W!  
8\.b4FNJ  
a1=(-B-sqrt(B^2-4*A*C))/(2*A); 5GzFoy)j>  
a2=d3/(a1*f); 7gF"=7{-  
b2=a1*(1-a2)*f/d2; 3uWkc3  
b1=(1-a1)*f/(d1*b2); Kn`M4O  
~`ny@WD9  
%曲率半径 ;&b.T}Nf06  
+]Zva:$#`
 
R1=2*f/(b1*b2) i1lBto[  
R2=2*a1*f/(b2*(1+b1)) Ifc]K?  
R3=2*a1*a2*f/(1+b2) *ISZlR\#  
q*7<)VwI  
A1=b2^3*(a1-1)*(1+b1)^3; 3?I;ovsM  
B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; ]}|byo  
C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2;
as yZe  
4<dcB@v  
A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); >Gml4vGK  
B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); I#F!N6;  
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); 8.AR.o  
=@&cHY  
CB=[C1 B1;C2 B2]; 1\*\?\T>_  
AB=[A1 B1;A2 B2];  {^a36i  
AC=[A1 C1;A2 C2]; "TyJP[/  
+ZMls [  
%二次系数 G2bDf-1ew  
*iBTI+"]  
k2=-(det(CB)/det(AB)); )SF}2?7e  
k3=-(det(AC)/det(AB)); PXzsj
.  
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 E>'a,!QPv  
k2=k2 W]M)Q}:Y  
k3=k3 H,fZ!8(A_)  
Q_-_^J  
end

谢谢分享，学习一下 *_a@z1