[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 x._IP,vRx^  
function sjr=nfdre(~) YnnpgR.  
fR_ jYP1  
%系统焦距及各镜间距输入，间距取负正负 xlPUum-o  
9pPb]v,6  
f=input('f:'); 0kN;SSX!  
d1=input('d1:'); .C^1.)  
d2=input('d2:'); &gJKJ=7  
d3=input('d3:'); ,#3}TDC  
%bI(  
A=f^2/(d3*d2)-f/d1; '\%c"?  
B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); `5 py6,  
C=d3/d2-f/d1; Zgp]s+%E  
mv@cGdxu  
a1=(-B+sqrt(B^2-4*A*C))/(2*A);%α1 EtN@ 6xP  
a2=d3/(a1*f);%α2 @|Z:7n6S  
b2=a1*(1-a2)*f/d2;%β2 *8}Y0V\s  
b1=(1-a1)*f/(d1*b2);%β1 1);$#Dlt k  
-q7A\8C  
3L/qU^`  
%曲率半径 [?)=3Pp  
{zoUU  
R1=2*f/(b1*b2) R'a%_sACj>  
R2=2*a1*f/(b2*(1+b1)) "=4`RM  
R3=2*a1*a2*f/(1+b2) 9tZ)#@\  
N;,?k.vU  
A1=b2^3*(a1-1)*(1+b1)^3; SQ#6~zxl  
B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; TJ(PTB;  
C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; Hj ]$  
Ke-Q>sm2Q  
A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); VlKy6PSIg  
B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); #!p=P<4M  
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); \~xI#S@  
8Ml&lfn_8  
CB=[C1 B1;C2 B2]; y e!Bfz>  
AB=[A1 B1;A2 B2]; <4jQbY;  
AC=[A1 C1;A2 C2]; zb9^ii$g  
kTQ:k }%B  
%非球面系数 Tk
s;,C  
k2=-(det(CB)/det(AB)); 0z?b5D;  
k3=-(det(AC)/det(AB)); @k~?h=o\b  
k1=(k2*a1*b2^3*(1+b1)^3-k3*a1*a2*(1+b2)^3+a1*b2^3*(1+b1)*(1-b1)^2-a1*a2*(1+b2)*(1-b2)^2)/(b1^3*b2^3)-1 Du3OmXMk  
k2=k2 zM6yUEg  
k3=k3 ~^/zCPy[w  
Cpa
eo0Oq  
end >Bp%~8f  
L/(e/Jalg  
%有中间像，焦距输入为正数 !- f>*|@  
PpMZ-f@  
function sjr=yfdre(~) m1_?x
U  
/H.QGPr  
f=input('f:'); !8&,GT  
d1=input('d1:'); FzmCS@yA  
d2=input('d2:'); >(z{1'f{  
d3=input('d3:'); J#Fe"  
Sp)KtMV  
A=f^2/(d3*d2)-f/d1; =fMSmn1S  
B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); []Cvma1\  
C=d3/d2-f/d1; X:-X3mV9{  
Pb?H
cg  
a1=(-B-sqrt(B^2-4*A*C))/(2*A); b{
)('C$  
a2=d3/(a1*f); *
lv)9L+0  
b2=a1*(1-a2)*f/d2; O%9Cq}*  
b1=(1-a1)*f/(d1*b2); 4,9AoK)yp  
n? "ti  
%曲率半径 qGkrG38K  
(PGmA>BT  
R1=2*f/(b1*b2) W9 y8dw.  
R2=2*a1*f/(b2*(1+b1)) FcIH<_r  
R3=2*a1*a2*f/(1+b2) m6V
1m0M  
rP ;~<IxEr  
A1=b2^3*(a1-1)*(1+b1)^3; HY#7Ctn3  
B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; %9M; MK  
C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; 1W~-C B>  
+C;ZO6%w  
A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); Y=X"YH|  
B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); OdQ>h$ gZ  
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); 7^sU/3z  
0vG}c5;F  
CB=[C1 B1;C2 B2]; OfTcF_%  
AB=[A1 B1;A2 B2]; *wt yyP@  
AC=[A1 C1;A2 C2]; g|<)J-`Q  
CkoPno  
%二次系数 sxL;o>{  
P@9>4}r$  
k2=-(det(CB)/det(AB)); &_4A6  
k3=-(det(AC)/det(AB)); 5K'EuI)  
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 QXJD'c  
k2=k2 ?f']*pD8  
k3=k3 %fP^Fh  
W3UK[_qK  
end

谢谢分享，学习一下 ?P`wLS^;