[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 VQ2'a/s  
function sjr=nfdre(~) 8I#^qr5  
7zU~X,  
%系统焦距及各镜间距输入，间距取负正负 \h0+` ;Q  
>zw@!1{1  
f=input('f:'); dN:^RCFzS  
d1=input('d1:'); &t_TLV 8T  
d2=input('d2:'); JX&%5sn(  
d3=input('d3:'); =,qY\@fq  
EKN<KnU%  
A=f^2/(d3*d2)-f/d1; b KDD29  
B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); T/%Y_.NtU  
C=d3/d2-f/d1; C8)s6  
+u5xK  
a1=(-B+sqrt(B^2-4*A*C))/(2*A);%α1 u$ci{<  
a2=d3/(a1*f);%α2 z^4+Un  
b2=a1*(1-a2)*f/d2;%β2 x{So  
b1=(1-a1)*f/(d1*b2);%β1 $$Ibr]$5  
5 /oW/2"  
>Lo'H
}[pF  
%曲率半径 -!M>;M@  
^u0y<kItX  
R1=2*f/(b1*b2) (6[/7e)  
R2=2*a1*f/(b2*(1+b1)) Ic&YiATj  
R3=2*a1*a2*f/(1+b2) U
%#Vz-r  
LtKR15h,  
A1=b2^3*(a1-1)*(1+b1)^3; n<b}6L}  
B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; ,Tx38  
C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; 6e5A8e8"]  
$DnJ/hg;qD  
A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); %X%f0J  
B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); *g:Dg I 2  
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); ~%
`hh9]  
/Zx"BSu
 
CB=[C1 B1;C2 B2]; B !
rb*"[  
AB=[A1 B1;A2 B2]; L7xiq{t`Y  
AC=[A1 C1;A2 C2]; Z$('MQ|Ur  
C+t
|fSJ  
%非球面系数 B7[#z{8'#  
k2=-(det(CB)/det(AB)); gdyWuOxa|  
k3=-(det(AC)/det(AB)); Y[rCF=ZVH  
k1=(k2*a1*b2^3*(1+b1)^3-k3*a1*a2*(1+b2)^3+a1*b2^3*(1+b1)*(1-b1)^2-a1*a2*(1+b2)*(1-b2)^2)/(b1^3*b2^3)-1 ={
6vShG)m  
k2=k2 rF{,]U9`  
k3=k3 [{vX*q 3B  
.;,,{;  
end sdd%u~4,X  
`m"K_\w=/  
%有中间像，焦距输入为正数 Gtv
bm  
'*&V7:  
function sjr=yfdre(~) heb{i5el  
dEX67rUj;  
f=input('f:'); r9D 68*H  
d1=input('d1:'); o(D_ /]'8  
d2=input('d2:'); S~|\bnE  
d3=input('d3:'); (5hUoDr!  
"s`#`'  
A=f^2/(d3*d2)-f/d1; 2<AQ{ c  
B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); ?01ru5ys/o  
C=d3/d2-f/d1; ~0Zy$L/D  
:Z83*SPc  
a1=(-B-sqrt(B^2-4*A*C))/(2*A); !<X/_+G\  
a2=d3/(a1*f); tv]9n8v  
b2=a1*(1-a2)*f/d2;  7(o:J  
b1=(1-a1)*f/(d1*b2); *N6sxFs  
(Bpn9}F-V.  
%曲率半径 DwTVoCC  
Gsm.a  
R1=2*f/(b1*b2) k@>(sXs  
R2=2*a1*f/(b2*(1+b1)) l Yj$3  
R3=2*a1*a2*f/(1+b2) XN3'k[  
XF@34b5(  
A1=b2^3*(a1-1)*(1+b1)^3; 0juP"v$C>  
B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; |a'$v4dCF  
C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; T7Ju7_q}  
:#N]s  
A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); 01]W@\(  
B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); 1@~ 1vsJ  
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); I-Hg6WtB  
#nj;F'O](  
CB=[C1 B1;C2 B2];  Tx'anP  
AB=[A1 B1;A2 B2]; .^ba*qb`{  
AC=[A1 C1;A2 C2]; N6*FlG-  
:/>Zky8,k  
%二次系数 !rZ
Z/M"i  
OU?.}qc<wE  
k2=-(det(CB)/det(AB)); wRX#^;O9?>  
k3=-(det(AC)/det(AB)); h`p=~u +  
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 @v\8+0  
k2=k2 .rlLt5b%  
k3=k3 "837b/>/  
? a/\5`gnN  
end

谢谢分享，学习一下 jZe/h#J)[