[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 G_4K+ -K  
function sjr=nfdre(~) 0R2S@4%Y  
A52LH,  
%系统焦距及各镜间距输入，间距取负正负 }.fZy&_  
dXf]G6  
f=input('f:'); bUY:XmA  
d1=input('d1:'); yoq\9* ?u^  
d2=input('d2:'); Cy`<^_i  
d3=input('d3:'); WcV\kemf  
Zi2Eu4p l{  
A=f^2/(d3*d2)-f/d1; Mm:a+T  
B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); AaCnTRG  
C=d3/d2-f/d1; \UkNE5  
e{qp!N1!  
a1=(-B+sqrt(B^2-4*A*C))/(2*A);%α1 Xy3g(x]  
a2=d3/(a1*f);%α2 qY*
%p  
b2=a1*(1-a2)*f/d2;%β2 46
Y7HTwE  
b1=(1-a1)*f/(d1*b2);%β1 \S|VkPv  
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z<3}TD  
%曲率半径 Kd AR)EU>  
=DmPPl{  
R1=2*f/(b1*b2) )sY$\^'WY  
R2=2*a1*f/(b2*(1+b1)) MIk #60Ab  
R3=2*a1*a2*f/(1+b2) eY6gb!5u  
$w";*">:0  
A1=b2^3*(a1-1)*(1+b1)^3; rS,*s'G  
B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; 4X(1  
C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; j:de}!wc  
~8Dd<4?F]  
A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); z Et6  
B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); pDq^W@Rq  
C2=b2*(a1-1)^2*(1+b1)*(1-b1)^2/(4*a1*b1^2)-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)*(1-b2)^2/(4*a1*a2*b1^2*b2^2)-b2*(a1-1)*(1-b1)*(1+b1)/(a1*b1)-(a2*(a1-1)+b1*(1-a2))*(1-b2)*(1+b2)/(a1*a2*b1*b2)-b1*b2+b2*(1+b1)/a1-(1+b2)/(a1*a2); b}EYNCw_7S  
RpaA)R,  
CB=[C1 B1;C2 B2]; ^Xt]wl*]+  
AB=[A1 B1;A2 B2]; 
jiw`i  
AC=[A1 C1;A2 C2]; ATXx? b8h  
~PH1|h6  
%非球面系数 w'5dk3$"  
k2=-(det(CB)/det(AB)); K_<lO,[S  
k3=-(det(AC)/det(AB)); g[D,\  
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 wFoR,oXtL/  
k2=k2 LzEE]i  
k3=k3 x9{Sl[2&  
1VPN#Q!  
end yoQ?lh  
U(Hq4D  
%有中间像，焦距输入为正数 .x-Z+Rs{
g  
=vqE=:X6  
function sjr=yfdre(~) u[6`Jr~  
Fm[?@Z&wP  
f=input('f:'); ek0;8Ds9  
d1=input('d1:'); Jb)eC?6O  
d2=input('d2:'); u=ds]XP@  
d3=input('d3:'); +2k|
g2  
7:R{
~|R  
A=f^2/(d3*d2)-f/d1; S9ak '  
B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); tKqCy\-q  
C=d3/d2-f/d1; gYH:EuY,  
XM5;AcD  
a1=(-B-sqrt(B^2-4*A*C))/(2*A); 4{;8 ]/.a  
a2=d3/(a1*f); ][>M<J  
b2=a1*(1-a2)*f/d2; K&"Pm9  
b1=(1-a1)*f/(d1*b2); mG8  
>FMT#x t  
%曲率半径 83 ^,'Z
 
Spr:K,  
R1=2*f/(b1*b2) 2YP"nj#  
R2=2*a1*f/(b2*(1+b1)) ?
` ZGM  
R3=2*a1*a2*f/(1+b2) Y$`hudJ&  
{/|8g(  
A1=b2^3*(a1-1)*(1+b1)^3; 7+Jma!o  
B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; BoiIr[ (  
C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; C6=;(=?C  
krnk%
ug  
A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); oe_[h]Hgl  
B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); 8Q)mmkI\=  
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); ?PxYS%D_L  
TXvt0&-  
CB=[C1 B1;C2 B2]; !+26a*P  
AB=[A1 B1;A2 B2]; $P}]|/Yb  
AC=[A1 C1;A2 C2]; MFdFZkpiV  
4`5Qt=}  
%二次系数 Xy8ie:D  
Vwh&^{Eh  
k2=-(det(CB)/det(AB)); 0|+hm^'_  
k3=-(det(AC)/det(AB)); {pJ@I=q  
k1=(k2*a1*b2^3*(1+b1)^3-k3*a1*a2*(1+b2)^3+a1*b2^3*(1+b1)*(1-b1)^2-a1*a2*(1+b2)*(1-b2)^2)/(b1^3*b2^3)-1 -9<yB  
k2=k2 4;~lpty  
k3=k3 kKk 
|@  
8!fAv$g0  
end

谢谢分享，学习一下 Q_|S^hxQ