[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 r*$KF!-dg  
function sjr=nfdre(~) C)`/Q(^  
M C>{I3  
%系统焦距及各镜间距输入，间距取负正负 *fl1 =Rfr  
Mb-C DPT  
f=input('f:'); ]RV6(|U4_  
d1=input('d1:'); f"/NY
6  
d2=input('d2:'); | r2'B  
d3=input('d3:'); @qeI4io-n  
7#Mi`W  
A=f^2/(d3*d2)-f/d1; *0Fn C2W1  
B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); n{M!l\1  
C=d3/d2-f/d1; w(VH>t  
CpA=DnZ  
a1=(-B+sqrt(B^2-4*A*C))/(2*A);%α1 )V3G~p=0  
a2=d3/(a1*f);%α2 9+pnpaZB0  
b2=a1*(1-a2)*f/d2;%β2 Bdf3@sbM]  
b1=(1-a1)*f/(d1*b2);%β1 4!D!.t~r  
*5wb8[  
5'@}8W3b  
%曲率半径 {2-w<t  
Hh,\>= ':  
R1=2*f/(b1*b2) /6g*WX2P1  
R2=2*a1*f/(b2*(1+b1)) %h^; "|Z  
R3=2*a1*a2*f/(1+b2) Ta~Ei=d^  
V_h, UYN  
A1=b2^3*(a1-1)*(1+b1)^3; >QCVsX>~  
B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; 8g$pfHt|e  
C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; s3oK[:/  
i9$ -lk  
A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); .#CTL|x  
B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); eR?`o!@y  
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); s+,JwV?b  
x-cg df  
CB=[C1 B1;C2 B2]; *r=6bpi  
AB=[A1 B1;A2 B2]; Y'58.8hl  
AC=[A1 C1;A2 C2]; ^dRB(E}|)  
UkeX">  
%非球面系数 T2GJoJ!  
k2=-(det(CB)/det(AB)); zGA#7W2?0  
k3=-(det(AC)/det(AB)); iWO16=  
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 !M\8k$#"n  
k2=k2 ZxY%x/K  
k3=k3 pFhznH{0  
V4iN2  
end </Ja@%  
]v$VZ'  
%有中间像，焦距输入为正数 J&2cf#  
*$
Bx#0J8  
function sjr=yfdre(~) .*i.Z   
w_PnEJa9  
f=input('f:'); 0;KjP?5  
d1=input('d1:'); FZO}+ P  
d2=input('d2:'); l'm!e'7_  
d3=input('d3:'); 0.qnbDw_  
G,3.'S,7  
A=f^2/(d3*d2)-f/d1; *i?#hTw  
B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); RX\%R  
C=d3/d2-f/d1; [<rV "g  
u8 Q`la  
a1=(-B-sqrt(B^2-4*A*C))/(2*A); 3P N<J  
a2=d3/(a1*f); Sjv_% C$  
b2=a1*(1-a2)*f/d2; ,'Zs")Ydp  
b1=(1-a1)*f/(d1*b2); J4"?D9T3G  
/'O? 8X<  
%曲率半径 A63=$  
2Jm#3zFYz3  
R1=2*f/(b1*b2) }}G`yfs}r  
R2=2*a1*f/(b2*(1+b1)) dv1Y2[  
R3=2*a1*a2*f/(1+b2) lp+Uox  
!59u z4  
A1=b2^3*(a1-1)*(1+b1)^3; b9X"p*'p  
B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; b"k1N9  
C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; [3o^06V8j  
C;>!SRCp  
A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); h6`v%7H?  
B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); wkc)2z   
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); c'O"</  
1UdET#\  
CB=[C1 B1;C2 B2]; )jm!bR`  
AB=[A1 B1;A2 B2]; >>%m,F[  
AC=[A1 C1;A2 C2]; RA^6c
![  
2Kwr
=t  
%二次系数 @^R6}qJ  
S+2we  
k2=-(det(CB)/det(AB)); _D.4=2@|l8  
k3=-(det(AC)/det(AB)); E7mB=bt>=  
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 x`n7D  
k2=k2  r"YOA@  
k3=k3 p.)IdbC`B  
P.#@1_:gC  
end

谢谢分享，学习一下 Tji*\<?