[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 _uJV
uCc  
function sjr=nfdre(~) }!=U^A)  
h~fWE  
%系统焦距及各镜间距输入，间距取负正负 L~ 2q1  
Qg>0G%cXU  
f=input('f:'); xx0k$Dqt2I  
d1=input('d1:'); cUsL6y  
d2=input('d2:'); RM#fX^)=  
d3=input('d3:'); ]?# #))RUS  
C Oa.xyp  
A=f^2/(d3*d2)-f/d1; Z8fJ{uOIL  
B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); 1ys(v  
C=d3/d2-f/d1; _k|g@"  
)7w@E$l"  
a1=(-B+sqrt(B^2-4*A*C))/(2*A);%α1 uOFnCy 4  
a2=d3/(a1*f);%α2 R=Ymo.zs6  
b2=a1*(1-a2)*f/d2;%β2 yl]UUBcQ  
b1=(1-a1)*f/(d1*b2);%β1 <N-=fad]  
m r2S!  
_ .!aBy%xf  
%曲率半径 Oj-r;Tt_G}  
}1F6?do3&  
R1=2*f/(b1*b2) -nGwuEngP  
R2=2*a1*f/(b2*(1+b1)) /ISLVp%H  
R3=2*a1*a2*f/(1+b2) 6+)x7g1PL  
X\mz+al>[  
A1=b2^3*(a1-1)*(1+b1)^3; KqK9X  
B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; *(p7NYf1  
C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; !3?yG  
*:[b'D!A  
A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); Vq U|kv  
B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); X?R |x[  
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); Hh@2m\HA  

?CFoe$M  
CB=[C1 B1;C2 B2]; H@4/#V|Uy  
AB=[A1 B1;A2 B2]; i3d y  
AC=[A1 C1;A2 C2]; P
K}vh%  
N;g$)zCV1  
%非球面系数 9 R  
k2=-(det(CB)/det(AB)); ?lyltAxs'  
k3=-(det(AC)/det(AB));  ^ `je
 
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 I5Q~T5Ar  
k2=k2 S QSA%B$<  
k3=k3 _{mJ.1)V;  
? osfL  
end 8UYJye8  
4a?r` '  
%有中间像，焦距输入为正数 s_v}=C^  
s|E%~j[9  
function sjr=yfdre(~) @ce3%`c_  
9GE]<v,_[  
f=input('f:'); pW7kj&a_.  
d1=input('d1:'); mQL8QW[c  
d2=input('d2:'); ;&q]X]bJ  
d3=input('d3:'); Z7oaQ\fR  
ho_4f
Dv  
A=f^2/(d3*d2)-f/d1; ^^eV4Y5`+  
B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); _
\!0t  
C=d3/d2-f/d1; #.b^E3#+  
gLV^Z6eE  
a1=(-B-sqrt(B^2-4*A*C))/(2*A); VT Vm7l  
a2=d3/(a1*f); IX?@~'  
b2=a1*(1-a2)*f/d2; l)H9J]  
b1=(1-a1)*f/(d1*b2); p8_2y~!  
a 1NCVZ  
%曲率半径 #]igB9Cf)w  
n-W?Z'H{r  
R1=2*f/(b1*b2) Z<I[vp6{  
R2=2*a1*f/(b2*(1+b1)) pb6 Q?QG,  
R3=2*a1*a2*f/(1+b2) 69rwX"^  
I IYLA(  
A1=b2^3*(a1-1)*(1+b1)^3; T6;>O`B.r  
B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; _En]@xK3&  
C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; +*<K"H|,  
tfsh!)u?  
A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); 98=XG1sQ@  
B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); C_C$5[~-:  
C2=b2*(a1-1)^2*(1+b1)*(1-b1)^2/(4*a1*b1^2)-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)*(1-b2)^2/(4*a1*a2*b1^2*b2^2)-b2*(a1-1)*(1-b1)*(1+b1)/(a1*b1)-(a2*(a1-1)+b1*(1-a2))*(1-b2)*(1+b2)/(a1*a2*b1*b2)-b1*b2+b2*(1+b1)/a1-(1+b2)/(a1*a2); wD}EW  
BJ"Ay@D*  
CB=[C1 B1;C2 B2]; VGfD;8]z  
AB=[A1 B1;A2 B2]; f7S^yA[[  
AC=[A1 C1;A2 C2]; tG_-;03<`4  
?$
2q P`-  
%二次系数 C9GU6Ao  
+B](5z4  
k2=-(det(CB)/det(AB)); ]gEu.Nth`  
k3=-(det(AC)/det(AB)); vrQFx~ZztH  
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 juR  
k2=k2 $,g 3*A  
k3=k3 5=*@l  
[GOX0}$?  
end

谢谢分享，学习一下 x_]",2 W'