[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 t7l{^d_L  
function sjr=nfdre(~) 7O'.KoMw  
7Q]c=i cg  
%系统焦距及各镜间距输入，间距取负正负 JO`r)_  
gROK4'j6y  
f=input('f:'); e'>q( B  
d1=input('d1:'); JOpH Z?  
d2=input('d2:'); )=sbrCl,C/  
d3=input('d3:'); ' Ut4=@)  
YGC%j  
A=f^2/(d3*d2)-f/d1; R)BXN~dQ  
B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); 'prHXzi(h  
C=d3/d2-f/d1; S\h5 D2G;  
j{YYG|  
a1=(-B+sqrt(B^2-4*A*C))/(2*A);%α1 ~x!up9  
a2=d3/(a1*f);%α2 `:Gzjngc  
b2=a1*(1-a2)*f/d2;%β2 PBnH#zm  
b1=(1-a1)*f/(d1*b2);%β1 DrKB;6  
Jn^b}bk t  
QOo'Iv+EL  
%曲率半径 Vn4wk>b}$2  
&:g:7l]g  
R1=2*f/(b1*b2) ^t5My[R  
R2=2*a1*f/(b2*(1+b1)) uZtN,Un  
R3=2*a1*a2*f/(1+b2) @U18Dj[  
&G\m
cstX  
A1=b2^3*(a1-1)*(1+b1)^3; {='Bd6_=  
B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; !}z'"l4i  
C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; <-!1`@l>  
N=BG0t$  
A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); '1:)q  
B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); 3{$7tck,  
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); M/quswn1  
M&j|5UH%.  
CB=[C1 B1;C2 B2]; 0o=HOCL\  
AB=[A1 B1;A2 B2]; )Q'E^[Ua  
AC=[A1 C1;A2 C2]; \~ChbPnc  
Fs(PVN  
%非球面系数 <'~m1l#2  
k2=-(det(CB)/det(AB)); ^~;ia7V&2  
k3=-(det(AC)/det(AB)); X+aQ 7^"s  
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 :rUMmO-  
k2=k2 k?14'X*7yu  
k3=k3 [|OII!"  
& "&s,  
end W~/d2_|/  
3

NgXM  
%有中间像，焦距输入为正数 t\K (zE  
p0bWzIH  
function sjr=yfdre(~) `y3'v]  
8x U*j  
f=input('f:'); %hsCB .r>|  
d1=input('d1:'); x3=1/#9  
d2=input('d2:'); d fj23+  
d3=input('d3:'); qT@h/Y  
G kjfDY:  
A=f^2/(d3*d2)-f/d1; RW L0@\  
B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); . ,^WCyvq  
C=d3/d2-f/d1; TlS? S+  
CJ\a7=*i  
a1=(-B-sqrt(B^2-4*A*C))/(2*A); )x|;%.8FX7  
a2=d3/(a1*f); NS[eQ_rT  
b2=a1*(1-a2)*f/d2; zl@^[k
m{  
b1=(1-a1)*f/(d1*b2); s$R /!,c  
 l(?B0  
%曲率半径 XP@dg4Z=z  
 vmqa_gU\  
R1=2*f/(b1*b2) ?{S>%P A_B  
R2=2*a1*f/(b2*(1+b1)) KdR4<qVV}  
R3=2*a1*a2*f/(1+b2) N `|A  
@f-X/q]P  
A1=b2^3*(a1-1)*(1+b1)^3; ST*h{:u&A  
B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; N-M.O:p  
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|%41xOsr  
UphTMyn3  
A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); Jj-\Eb?  
B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); OyZR&,q  
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^5'h~  
9(N  
CB=[C1 B1;C2 B2]; 1Z# $X`  
AB=[A1 B1;A2 B2]; OUv<a`0  
AC=[A1 C1;A2 C2]; Z+El(f x  
c@t?R$c  
%二次系数 _Je4&KU  
1>J.kQR^  
k2=-(det(CB)/det(AB)); 0\dmp'j]  
k3=-(det(AC)/det(AB)); l_^OdQ9D  
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 .k,j64 r  
k2=k2 b(*\4n  
k3=k3 J2=4%#R!  
lMFR_g?r  
end

谢谢分享，学习一下 vzIo2,/7