[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 3O-vO=D  
function sjr=nfdre(~) w oIZFus  
h*40jZ  
%系统焦距及各镜间距输入，间距取负正负 ig YYkt  
smQl^ 6a  
f=input('f:'); PW5)") z  
d1=input('d1:'); oj{CNa  
d2=input('d2:'); %,~\,+NP  
d3=input('d3:'); U/AiI;Ne  
cNwHY Z'  
A=f^2/(d3*d2)-f/d1; }ssja,;  
B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); G!B:>P|\l  
C=d3/d2-f/d1; l"+8>Mm  
ZCZ@ZN  
a1=(-B+sqrt(B^2-4*A*C))/(2*A);%α1 `C|];mf(#  
a2=d3/(a1*f);%α2 S2\;\?]^~  
b2=a1*(1-a2)*f/d2;%β2 'Nt)7U>oC9  
b1=(1-a1)*f/(d1*b2);%β1 H"UJBO>$  
YU8]W%  
ilK*Xo  
%曲率半径 /
i27F2NQm  
U/kQwrM  
R1=2*f/(b1*b2) vOz1& |;D  
R2=2*a1*f/(b2*(1+b1)) LWTPNp:"{w  
R3=2*a1*a2*f/(1+b2) }Md;=_TP  
Ng!d6]  
A1=b2^3*(a1-1)*(1+b1)^3; 6hd<ys?  
B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; 1|"BpX~D  
C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; F xm:m  
=$)M-;6  
A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); y2jw3R  
B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); =z"
+)N  
C2=b2*(a1-1)^2*(1+b1)*(1-b1)^2/(4*a1*b1^2)-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)*(1-b2)^2/(4*a1*a2*b1^2*b2^2)-b2*(a1-1)*(1-b1)*(1+b1)/(a1*b1)-(a2*(a1-1)+b1*(1-a2))*(1-b2)*(1+b2)/(a1*a2*b1*b2)-b1*b2+b2*(1+b1)/a1-(1+b2)/(a1*a2); d,l?{Ln  
WG6 0  
CB=[C1 B1;C2 B2]; gELG/6l  
AB=[A1 B1;A2 B2]; KgkRs?'z  
AC=[A1 C1;A2 C2]; {]}94T~/k  
ZfqN4  
%非球面系数 [yk-<}#B  
k2=-(det(CB)/det(AB)); 6*>Lud  
k3=-(det(AC)/det(AB)); 7XyCl&Dc:  
k1=(k2*a1*b2^3*(1+b1)^3-k3*a1*a2*(1+b2)^3+a1*b2^3*(1+b1)*(1-b1)^2-a1*a2*(1+b2)*(1-b2)^2)/(b1^3*b2^3)-1 4LB8p7$|a3  
k2=k2 %EVgSF!r  
k3=k3 ^s7!F.OC  
h ':ZF  
end Mhti  
3Y2~HuM  
%有中间像，焦距输入为正数 ZGR5"el!  
0stc$~~v  
function sjr=yfdre(~) HLwMo&*rA  
u)<s*jk  
f=input('f:'); RfTGTz@H  
d1=input('d1:'); 9!uiQ  
d2=input('d2:'); PgK7CG7G  
d3=input('d3:'); _7;:*'>a4  
yTd8)zWq  
A=f^2/(d3*d2)-f/d1; @ G)yz!H  
B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); gHstdp_3  
C=d3/d2-f/d1; f!#!  
@lE'D":?  
a1=(-B-sqrt(B^2-4*A*C))/(2*A); ?BfE*I$\h  
a2=d3/(a1*f); c'eZ-\d{  
b2=a1*(1-a2)*f/d2; sNo8o1Hby  
b1=(1-a1)*f/(d1*b2); jO&*E'pk  
\/Mx|7<  
%曲率半径 aU_Hl+;  
u7[}pf$}  
R1=2*f/(b1*b2) [#q>Aq$11  
R2=2*a1*f/(b2*(1+b1)) qiOJ:'@  
R3=2*a1*a2*f/(1+b2) k[ro[E  
kzRJzJquP  
A1=b2^3*(a1-1)*(1+b1)^3; 6j<!W+~G  
B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; byM-$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; rYr*D[m]  
|sReHt2)d  
A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); _5-h\RB)  
B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); @GFB{ ;=  
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); /!?LBtqy  
/qX?ca1_4^  
CB=[C1 B1;C2 B2]; (*9.GyK  
AB=[A1 B1;A2 B2]; dg24h7|]  
AC=[A1 C1;A2 C2]; m|qktLx  
h0rPMd(K
 
%二次系数 clB K  
6UeYZ g  
k2=-(det(CB)/det(AB)); |]*3
En:  
k3=-(det(AC)/det(AB)); 3O/#^~\'hW  
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 'f-r 6'_ZX  
k2=k2 Fye>H6MU  
k3=k3 _VKI@  
xmvE*q"9]  
end

谢谢分享，学习一下 \l~^dn}