[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 gHc1_G]  
function sjr=nfdre(~) <lR:^M[v5<  
%Mj,\J!  
%系统焦距及各镜间距输入，间距取负正负 C)c*s C5N  
'Z#_"s#L  
f=input('f:'); w2xD1oK~o  
d1=input('d1:'); C$0u-Nx8  
d2=input('d2:'); K!ILO  
d3=input('d3:'); cWo>DuW&  
qqo#H O  
A=f^2/(d3*d2)-f/d1; $gnrd~v4e  
B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); z2{y<a9;?  
C=d3/d2-f/d1; SW%}S*h  
kSiyMDY-  
a1=(-B+sqrt(B^2-4*A*C))/(2*A);%α1 OIj.K@Kr  
a2=d3/(a1*f);%α2 c*B< - l<5  
b2=a1*(1-a2)*f/d2;%β2 I'BHNZO5tf  
b1=(1-a1)*f/(d1*b2);%β1 %\HE1d5;  
x2t&Wpvt  
qCI7)L`  
%曲率半径 05{}@tW-  
wbJBGT{sm  
R1=2*f/(b1*b2) QxG^oxU}  
R2=2*a1*f/(b2*(1+b1)) I4%kYp]  
R3=2*a1*a2*f/(1+b2) Ep
O2%|@  
JwxKWVpWv  
A1=b2^3*(a1-1)*(1+b1)^3; lTu&
9)  
B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; u=Ik&^v Wq  
C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; QY4;qA  
qE2V
UEv5Y  
A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); \"$P :Uv  
B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); bK!h{Rr  
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); .'A1Eoo0d  
5qH*"i+|s  
CB=[C1 B1;C2 B2]; c_ncx|dUs  
AB=[A1 B1;A2 B2]; d)V8FX,t  
AC=[A1 C1;A2 C2]; 4v/MZ:%C`  
T8Mqu`$r  
%非球面系数 )EB+(c~E  
k2=-(det(CB)/det(AB)); Ph]b6  
k3=-(det(AC)/det(AB)); qD*y60~]zz  
k1=(k2*a1*b2^3*(1+b1)^3-k3*a1*a2*(1+b2)^3+a1*b2^3*(1+b1)*(1-b1)^2-a1*a2*(1+b2)*(1-b2)^2)/(b1^3*b2^3)-1 Pb;c:HeI/  
k2=k2 6QA`u
*  
k3=k3 MvZa;B  
"~r)_Ko  
end 'WhJ}Uo
\  
d'Bxi"K  
%有中间像，焦距输入为正数 i,^3aZwJ'  
sM MtU@<x  
function sjr=yfdre(~) 9vyf9QE;  
@Q,Q"c2  
f=input('f:'); { rLgyrj$  
d1=input('d1:'); \8$~ i  
d2=input('d2:'); *GoTN  
d3=input('d3:'); $m#^0%  
XX/s@C  
A=f^2/(d3*d2)-f/d1; 
TaD;_)(  
B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); iii|;v]+  
C=d3/d2-f/d1; 4@{?4k-cq  
,DE>:ARZ  
a1=(-B-sqrt(B^2-4*A*C))/(2*A); X uE: dL?  
a2=d3/(a1*f); nl 'MWP  
b2=a1*(1-a2)*f/d2; W9w*=W )Z  
b1=(1-a1)*f/(d1*b2); xE@/8
h  
[5~mP`He  
%曲率半径 6Ot~Q  
;$&\:-6A#  
R1=2*f/(b1*b2) cQhr{W,Un  
R2=2*a1*f/(b2*(1+b1)) -O\i^?lD;
 
R3=2*a1*a2*f/(1+b2) Kw`CN  
F^bY]\-5  
A1=b2^3*(a1-1)*(1+b1)^3; Z[IM\# "  
B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; B2j1GJEO  
C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; \AroSy9  
bD,X.  
A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); u*Xp%vNe  
B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); 2H4vK]]Nl  
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); sq`Xz8u  
\t=0rFV)t  
CB=[C1 B1;C2 B2]; v5'`iO0o  
AB=[A1 B1;A2 B2]; seEo)m`d  
AC=[A1 C1;A2 C2]; )%Fwfb  
_

xv3UzD  
%二次系数 \fTQNF  
.+u b\  
k2=-(det(CB)/det(AB)); V2}\]x'1  
k3=-(det(AC)/det(AB)); la+Cra&xL  
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 :+Ukwno?/  
k2=k2 \wA:58 -j  
k3=k3 ErNYiYLi]  
_|GbU1Hz  
end

谢谢分享，学习一下 oVy{~D=