[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 $N=N(^  
function sjr=nfdre(~) 9n5<]Q(  
z7Eg5rm|QZ  
%系统焦距及各镜间距输入，间距取负正负 6]pX>Xho  
+_xOLiu  
f=input('f:'); \$OF1i@
 
d1=input('d1:'); V-r3-b  
d2=input('d2:'); }~h(w^t  
d3=input('d3:'); o9S+6@  
GMZv RAui  
A=f^2/(d3*d2)-f/d1; SdwS= (e6  
B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); ^e>Wo7r  
C=d3/d2-f/d1; U Gpu\TB  
\3jW~FV  
a1=(-B+sqrt(B^2-4*A*C))/(2*A);%α1 z.VyRBi0  
a2=d3/(a1*f);%α2 4T<Lgb  
b2=a1*(1-a2)*f/d2;%β2 ?8mlZ X9C  
b1=(1-a1)*f/(d1*b2);%β1 8Bq!4uq\5|  
F0UV

o  
v[n7"  
%曲率半径 (V%`k'N7f  
la?Wnw  
R1=2*f/(b1*b2) rf%7b8[v  
R2=2*a1*f/(b2*(1+b1)) ;kJA'|GX  
R3=2*a1*a2*f/(1+b2) 5`RiS]IO]  
?cEskafb>  
A1=b2^3*(a1-1)*(1+b1)^3; .*JA!B  
B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; eBO@7F$  
C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; :BGA.  
rzI|?QaPi  
A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); T!}[yW  
B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); #vqo -y7@  
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); 79yd&5#e?  
=FT98H2*|  
CB=[C1 B1;C2 B2]; .+(
[  
AB=[A1 B1;A2 B2]; 3TZ*RPmFRm  
AC=[A1 C1;A2 C2]; 'iY*
6<xS<  
c$QX)V  
%非球面系数 1I}b|6 `  
k2=-(det(CB)/det(AB)); )CS.F=  
k3=-(det(AC)/det(AB)); eV
cANP  
k1=(k2*a1*b2^3*(1+b1)^3-k3*a1*a2*(1+b2)^3+a1*b2^3*(1+b1)*(1-b1)^2-a1*a2*(1+b2)*(1-b2)^2)/(b1^3*b2^3)-1 %D`,k*X  
k2=k2 NCf"tK'5n  
k3=k3 <(e8sNe  
kzS=g|_  
end #s%-INcR  
mG1IQ!  
%有中间像，焦距输入为正数 sW^a`VM  
KYxBVgJ  
function sjr=yfdre(~) >u(>aV|A  
eb8w~  
f=input('f:'); M#@aB"@J>  
d1=input('d1:'); .\qj;20
W  
d2=input('d2:'); 7gS1~Q4\V2  
d3=input('d3:'); 1]T`n/d V  
y9|K|xO[  
A=f^2/(d3*d2)-f/d1; *X38{rj  
B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); qJ_1*!!91  
C=d3/d2-f/d1; ro[Y-o5Q0  
KZBrE$@%5  
a1=(-B-sqrt(B^2-4*A*C))/(2*A); ;*[9Q'lI*  
a2=d3/(a1*f); \M/6m^zS  
b2=a1*(1-a2)*f/d2; ,s^<X85gp\  
b1=(1-a1)*f/(d1*b2); M'1!<a-Mp  
W];EKj,3W  
%曲率半径 t=#Pya  
il:nXpM!  
R1=2*f/(b1*b2) jRpdft  
R2=2*a1*f/(b2*(1+b1)) cxYfZ4++m  
R3=2*a1*a2*f/(1+b2) !z zW2>  
s/1 #DM"  
A1=b2^3*(a1-1)*(1+b1)^3; =qvZpB7ZZ  
B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; bO/*2oau  
C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; &(xUhX T  
m$$?icA  
A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); 0D(cXzQP  
B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); G;oFTP>o  
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); (a6?s{(  
b]]N{: I  
CB=[C1 B1;C2 B2]; C6&( c  
AB=[A1 B1;A2 B2]; IGAzE(  
AC=[A1 C1;A2 C2]; em]xtya  
DjW$?>  
%二次系数 qU[O1bN  
ZZ)G5ji  
k2=-(det(CB)/det(AB)); 8Vt4HD08  
k3=-(det(AC)/det(AB)); RwTzz] M  
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
}IlP:  
k2=k2 Z#Lx_*p]Q  
k3=k3 J%dJw}  
S9Yt1qb  
end

谢谢分享，学习一下 ^@}#me@