[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 -`h)$&,  
function sjr=nfdre(~) ydA8wL  
&K#M*B,*p  
%系统焦距及各镜间距输入，间距取负正负 .uZ3odMlx  
N:/D+L
 
f=input('f:'); &U#|uc!+  
d1=input('d1:'); sY&IquK^  
d2=input('d2:'); g*_&  
d3=input('d3:'); |0b`fOS  
013x8!i  
A=f^2/(d3*d2)-f/d1; E{`fF8]K  
B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); 6%_nZvRv  
C=d3/d2-f/d1; !*N@ZL&X  
uo8YP<q  
a1=(-B+sqrt(B^2-4*A*C))/(2*A);%α1 KkbDW3-  
a2=d3/(a1*f);%α2 r`d4e,(  
b2=a1*(1-a2)*f/d2;%β2 \
Gvm9M  
b1=(1-a1)*f/(d1*b2);%β1 [RhO$c$[\  
LU%E:i|  
Bj;'qB
>3  
%曲率半径 ;N0XFjdR  
qo bc<-  
R1=2*f/(b1*b2) dUZ ,m9u  
R2=2*a1*f/(b2*(1+b1)) ?k{?GtSs  
R3=2*a1*a2*f/(1+b2) ;?p>e
'  
VY4yS*y  
A1=b2^3*(a1-1)*(1+b1)^3; _Y;W0Z  
B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; YU'E@t5  
C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; 8(~h"]`!  
/nA{#HY  
A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); bROLOf4S  
B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); \_f(M|  
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); ggR.4&<  
^u ~Q/4  
CB=[C1 B1;C2 B2]; b3, _(;A!  
AB=[A1 B1;A2 B2]; /y}xX  
AC=[A1 C1;A2 C2]; OQJ6e:BGt  
Vt#.eL)Ee  
%非球面系数 Tyx_/pJT  
k2=-(det(CB)/det(AB)); 8&slu{M- t  
k3=-(det(AC)/det(AB)); b8 likP"T  
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 2P0*NQ  
k2=k2 0\P1; ak%  
k3=k3 N`e[:[  
}o`76rDN  
end 37o;;
 
AoxA+.O  
%有中间像，焦距输入为正数 `[ir}+S  
VMWf>ZU  
function sjr=yfdre(~) t%=tik2|7  
$xN|5;+  
f=input('f:'); vr=#3>  
d1=input('d1:'); Lp9E:D->  
d2=input('d2:'); wf<M)Rs|  
d3=input('d3:'); .?$gpM?i  
<)D$51 &0  
A=f^2/(d3*d2)-f/d1; H/M@t\$Dc  
B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); vdwsJPFbc  
C=d3/d2-f/d1; H4+i.*T#  
>4CbwwMA  
a1=(-B-sqrt(B^2-4*A*C))/(2*A); PEZ!n.'S  
a2=d3/(a1*f); E7hY8#G  
b2=a1*(1-a2)*f/d2; 3^yK!-Wp(  
b1=(1-a1)*f/(d1*b2); c]!V'#U  
N;`n@9BF  
%曲率半径 k8zI(5.>
 
UkFC~17P  
R1=2*f/(b1*b2) Qo|\-y-#  
R2=2*a1*f/(b2*(1+b1)) }O p; g^W  
R3=2*a1*a2*f/(1+b2) 4j^ @wV'  
Xsa].  
A1=b2^3*(a1-1)*(1+b1)^3; 5v*\Zr5ha  
B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; f3y=Wxk[  
C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; scV5PUq  
^U/O!GK  
A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); pMM8-R'W-  
B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); 'LDQgC*%  
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); _|`S3}q|d  
?}Y]|c^W  
CB=[C1 B1;C2 B2]; p5*EA x  
AB=[A1 B1;A2 B2]; x]j W<A  
AC=[A1 C1;A2 C2]; Tw<q,O  
GTHt'[t@;  
%二次系数 VUuE T  
6ik$B  
k2=-(det(CB)/det(AB)); w,D+j74e$  
k3=-(det(AC)/det(AB)); Zv{'MIv&v  
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 1_G^w qk  
k2=k2 P.DK0VgY  
k3=k3 ;$Jo+#  
RxQ*  
end

谢谢分享，学习一下 _Z\

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