[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 oiP8~  
function sjr=nfdre(~) m&)5QX  
K/;FP'.  
%系统焦距及各镜间距输入，间距取负正负 ,fS}cpV  
iV
X12  
f=input('f:'); r3X|*/  
d1=input('d1:'); FYIzMp.4  
d2=input('d2:'); vJ*IUy  
d3=input('d3:'); u|m>h(O  
5m,{?M`  
A=f^2/(d3*d2)-f/d1; y74Ph:^k  
B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); QG\lXY,  
C=d3/d2-f/d1; "kC uCc  

et|P5%G  
a1=(-B+sqrt(B^2-4*A*C))/(2*A);%α1 ` aTkIo:ms  
a2=d3/(a1*f);%α2 Jd_w:H.  
b2=a1*(1-a2)*f/d2;%β2 5>.)7D%  
b1=(1-a1)*f/(d1*b2);%β1 8>.l4:`  
4^1B'>I  
&Mz3CC6  
%曲率半径 /H+br_D9  
g0ec-  
R1=2*f/(b1*b2) 6Q]c]cCu  
R2=2*a1*f/(b2*(1+b1)) +RexQE  
R3=2*a1*a2*f/(1+b2) HA%%WSuf  
mG[S"?C  
A1=b2^3*(a1-1)*(1+b1)^3; @vWC "W  
B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; *ayn<Vlh`^  
C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; d#:&Uw  
+pU\;x  
A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); }>vf(9sF`  
B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); )/FB73!  
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); W69 -,w/  
l1^/Q~u  
CB=[C1 B1;C2 B2]; XWvT(+J  
AB=[A1 B1;A2 B2]; 4`@]jm  
AC=[A1 C1;A2 C2]; t ZUZNKODW  
G5W6P7-<X  
%非球面系数 Ojj:YLlY>  
k2=-(det(CB)/det(AB)); j""I,$t  
k3=-(det(AC)/det(AB)); P<b.;Oz__-  
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 l!f/0Rx5  
k2=k2 $3BCA)5:  
k3=k3 =ECw'  
X%"
P0P  
end )e.Y"5My  
+}c '4hRv  
%有中间像，焦距输入为正数 y'?|#%D  
IuDg-M[  
function sjr=yfdre(~) 5T,Doxo  
P;[
5#-e  
f=input('f:'); ? lC. Pq  
d1=input('d1:'); dsP|j(y  
d2=input('d2:'); .J' 8d"+  
d3=input('d3:'); |+Z, 7~!  
/0QGU4=  
A=f^2/(d3*d2)-f/d1; fq/F|c  
B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); &fA`Od6l"  
C=d3/d2-f/d1; R?Y#>K  
[ P\3XSR  
a1=(-B-sqrt(B^2-4*A*C))/(2*A); fLK*rK^{"  
a2=d3/(a1*f); v0!>":  
b2=a1*(1-a2)*f/d2; KZO!  
b1=(1-a1)*f/(d1*b2); Saa#Mj`M  
V^aX^;  
%曲率半径 bHcb+TR3  
<tK6+isc  
R1=2*f/(b1*b2) v 1Yf:c  
R2=2*a1*f/(b2*(1+b1)) r{qM!(T  
R3=2*a1*a2*f/(1+b2) E",s]  
9 O| "Ws>{  
A1=b2^3*(a1-1)*(1+b1)^3; "UVqkw,vt  
B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; )t={+^Xe  
C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; u2K{3+r`'  
~rEU83  
A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); NL&(/72V  
B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); #M*h)/d[A  
C2=b2*(a1-1)^2*(1+b1)*(1-b1)^2/(4*a1*b1^2)-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)*(1-b2)^2/(4*a1*a2*b1^2*b2^2)-b2*(a1-1)*(1-b1)*(1+b1)/(a1*b1)-(a2*(a1-1)+b1*(1-a2))*(1-b2)*(1+b2)/(a1*a2*b1*b2)-b1*b2+b2*(1+b1)/a1-(1+b2)/(a1*a2); 7k{Oae\$  
y [e$  
CB=[C1 B1;C2 B2]; uy\<t  
AB=[A1 B1;A2 B2]; %?+A.0]E  
AC=[A1 C1;A2 C2]; [7B:{sH  
X5U.8qI3  
%二次系数 qF=D,Dlz  
=zXpeo&|m  
k2=-(det(CB)/det(AB)); FT73P0!8.  
k3=-(det(AC)/det(AB)); +U&aK dQs  
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 uRG0}>]|U  
k2=k2 (:E_m|00;  
k3=k3 e:{v.C0ez  
Vnuz! 6.  
end

谢谢分享，学习一下 Yo;/7gG>