[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 wC*X4 '  
function sjr=nfdre(~) naNghGQ  
pfPz8L.7  
%系统焦距及各镜间距输入，间距取负正负 @)}L~lb[)  
1;iUWU1@  
f=input('f:'); q\%I#1  
d1=input('d1:'); 18Emi<&A  
d2=input('d2:'); ?]5qr?W%  
d3=input('d3:'); OTv)  

\U0'P;em  
A=f^2/(d3*d2)-f/d1; n"8Yv~v*2j  
B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); iow"n$/  
C=d3/d2-f/d1; QV8g#&z  
>mkFV@`  
a1=(-B+sqrt(B^2-4*A*C))/(2*A);%α1 ,: ^u-b|  
a2=d3/(a1*f);%α2 A}w/OA97RO  
b2=a1*(1-a2)*f/d2;%β2 %2h>-.tY  
b1=(1-a1)*f/(d1*b2);%β1 fV~~J2IK  
dWW.Y*339  
GX%g9f!O  
%曲率半径 ]###w;  
HKeK<V  
R1=2*f/(b1*b2) Bp{Ri_&A  
R2=2*a1*f/(b2*(1+b1)) tX[WH\(xI  
R3=2*a1*a2*f/(1+b2) d_CT$  
d'ifLQ\  
A1=b2^3*(a1-1)*(1+b1)^3; R-14=|7a-  
B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; &j6erwaT  
C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; /V By^L:  
1mJHued=6  
A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); s[N@0  
B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); <naz+QK'  
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); 8EY:tzw  
|a@L}m  
CB=[C1 B1;C2 B2]; ,u m|1dh  
AB=[A1 B1;A2 B2]; Ca-j?bb!  
AC=[A1 C1;A2 C2]; [Qr"
cR^  
[hsds\  
%非球面系数 V)4J`xg^  
k2=-(det(CB)/det(AB)); 2qp#N%  
k3=-(det(AC)/det(AB)); JS77M-Ac  
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 Wl4%GB  
k2=k2 .C(tMF]D,  
k3=k3 s^SJY{  
.^.z2 e  
end nFn5v'g  
pk~WrqK}  
%有中间像，焦距输入为正数 w =KPT''!  
>d6|^h'0  
function sjr=yfdre(~) 7Lt)nq-b  
4P0}+  
f=input('f:'); %znc##j)q  
d1=input('d1:'); 2pAW9R#UV-  
d2=input('d2:'); W!<U85-#S  
d3=input('d3:'); &{i{XcqH'  
0$njMnB2l  
A=f^2/(d3*d2)-f/d1; SAz   
B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); KSL`W2}  
C=d3/d2-f/d1; 9FX-1,Jx  
<vP=zk  
a1=(-B-sqrt(B^2-4*A*C))/(2*A); $8FUfJ1@  
a2=d3/(a1*f); /O9EQPm(  
b2=a1*(1-a2)*f/d2; @
XVTU  
b1=(1-a1)*f/(d1*b2); cnLro  
Wjc'*QCPl  
%曲率半径 tVjsRnb{  
d'2A,B~_*  
R1=2*f/(b1*b2) (w{j6).3Dj  
R2=2*a1*f/(b2*(1+b1)) [ 3HfQ  
R3=2*a1*a2*f/(1+b2) 7d vnupLh  
yHGADH0B  
A1=b2^3*(a1-1)*(1+b1)^3;  @8 6f  
B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; ;=N#`l  
C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; q*KAk{kR(v  
n*$ g]G$  
A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); 2?x4vI np;  
B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); a9G8q>h]O  
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); ry!!9Z>9n  
`2snz1>!j  
CB=[C1 B1;C2 B2]; {8aTV}Ha2  
AB=[A1 B1;A2 B2]; Q20%"&Xp]  
AC=[A1 C1;A2 C2]; Y0>y8UV  
D]}G.v1  
%二次系数 rGO8!X 3d  
]?*wbxU0  
k2=-(det(CB)/det(AB)); z:;CX@)*  
k3=-(det(AC)/det(AB)); v:U-6W_)|  
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 HV.t6@\};  
k2=k2 MPV5P^@X  
k3=k3 ^s=8!=A(  
]tD]Wx%  
end

谢谢分享，学习一下 i$@:@&(~Y