[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 9Suu-A  
function sjr=nfdre(~) QB:i/9  
MS(JR  
%系统焦距及各镜间距输入，间距取负正负 PiV7*F4qI.  
}>^Q'BW;65  
f=input('f:'); >e2<!#er|  
d1=input('d1:'); nJRS.xs  
d2=input('d2:'); tx"sH]n  
d3=input('d3:'); ,*4p?|A  
{7!UQrm<  
A=f^2/(d3*d2)-f/d1; Am8x74?  
B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); Eh-n  
C=d3/d2-f/d1; c`lJu_  
=ji1S}e~p  
a1=(-B+sqrt(B^2-4*A*C))/(2*A);%α1 5Zmw} M  
a2=d3/(a1*f);%α2 N=:5eAza  
b2=a1*(1-a2)*f/d2;%β2 OA:%lC!  
b1=(1-a1)*f/(d1*b2);%β1 nA|.t  
M :3u@06a  
hS[yNwD  
%曲率半径 ) \Y7&  
Xi?b]Z  
R1=2*f/(b1*b2) *Nyev]8  
R2=2*a1*f/(b2*(1+b1)) 7'wS\/e4a  
R3=2*a1*a2*f/(1+b2) w;Q;[:y  
7r|(}S  
A1=b2^3*(a1-1)*(1+b1)^3; bX.ja;;   
B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; *A}cL  
C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; QKN<+,h!z>  
o7B[R) 4  
A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); 9Rek4<5  
B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); $?,a
[79  
C2=b2*(a1-1)^2*(1+b1)*(1-b1)^2/(4*a1*b1^2)-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)*(1-b2)^2/(4*a1*a2*b1^2*b2^2)-b2*(a1-1)*(1-b1)*(1+b1)/(a1*b1)-(a2*(a1-1)+b1*(1-a2))*(1-b2)*(1+b2)/(a1*a2*b1*b2)-b1*b2+b2*(1+b1)/a1-(1+b2)/(a1*a2); Ngb(F84H?  
2tROT][J%  
CB=[C1 B1;C2 B2];
Rwr 2gMt7  
AB=[A1 B1;A2 B2]; f84:hXo6  
AC=[A1 C1;A2 C2]; t}v2$<!I  
h._nK\  
%非球面系数 |F.)zC5{  
k2=-(det(CB)/det(AB)); T&86A\D\z  
k3=-(det(AC)/det(AB)); Z~A@o""F  
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 gPAX4'
 
k2=k2 9]t[J_YM  
k3=k3 A2}Rl%+X]6  
r>jC_7  
end >3awn*N  
LqdY Qd51  
%有中间像，焦距输入为正数 .(J
?a"  
8/z3=O&  
function sjr=yfdre(~) 6#j$GH *  
0&ByEN99  
f=input('f:'); SI:U0gUc  
d1=input('d1:'); 7iJ&6=/  
d2=input('d2:'); JQ:Ri  
d3=input('d3:'); AmwWH7,g  
v/*Y#(X  
A=f^2/(d3*d2)-f/d1; E7Cy(LO  
B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); H:p Z-v*  
C=d3/d2-f/d1; B\g]({E  
C"lJl k9g^  
a1=(-B-sqrt(B^2-4*A*C))/(2*A); XC7%vDIt  
a2=d3/(a1*f); Le"oAA#[  
b2=a1*(1-a2)*f/d2; \7"@RHcihB  
b1=(1-a1)*f/(d1*b2); h7s
;m  
2MA]jT  
%曲率半径 Tz2-Bp]h  
WvHw{^(lF  
R1=2*f/(b1*b2) gX{
loG  
R2=2*a1*f/(b2*(1+b1))
u*  
R3=2*a1*a2*f/(1+b2) 1 nvTce  
vzF5xp.  
A1=b2^3*(a1-1)*(1+b1)^3; s:00yQ  
B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; smG>sEp2  
C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; x.1-)\  
Og;-B0,A  
A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); +.y .Mp  
B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); r%DFve:%  
C2=b2*(a1-1)^2*(1+b1)*(1-b1)^2/(4*a1*b1^2)-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)*(1-b2)^2/(4*a1*a2*b1^2*b2^2)-b2*(a1-1)*(1-b1)*(1+b1)/(a1*b1)-(a2*(a1-1)+b1*(1-a2))*(1-b2)*(1+b2)/(a1*a2*b1*b2)-b1*b2+b2*(1+b1)/a1-(1+b2)/(a1*a2); Z ,^9Z  
%!ebO*8q  
CB=[C1 B1;C2 B2]; ~j#~\Ir  
AB=[A1 B1;A2 B2]; 6z,&i  
AC=[A1 C1;A2 C2]; K/ &?VIi`z  
H A}f,),G  
%二次系数 ~.%K/=wK@  
=66Nw(E.  
k2=-(det(CB)/det(AB));
Vtppuu$  
k3=-(det(AC)/det(AB)); gn5)SP8  
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 4/X/>Y1  
k2=k2 Nr2C@FU:0  
k3=k3 :V)lbn\  
E{HY!L[
 
end

谢谢分享，学习一下 I *c;H I