[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 Cf9{lhE8  
function sjr=nfdre(~) zyhM*eM.7  
Hi$R"O (  
%系统焦距及各镜间距输入，间距取负正负 QkC*om'/!  
n}_}#(a  
f=input('f:'); l5.k2{'  
d1=input('d1:'); 3UaW+@  
d2=input('d2:'); xT]t3'y|-  
d3=input('d3:'); S3l^h4  
Km $o@  
A=f^2/(d3*d2)-f/d1; &a=78Z  
B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); 8lzoiA_9  
C=d3/d2-f/d1; 9TQVgkW  
&l3(+4Sh  
a1=(-B+sqrt(B^2-4*A*C))/(2*A);%α1 fLqjBG]<  
a2=d3/(a1*f);%α2 T- _))  
b2=a1*(1-a2)*f/d2;%β2 
a6j& po  
b1=(1-a1)*f/(d1*b2);%β1 bm 4RR
I  
Ln2FG4{  
4
>i\r  
%曲率半径 &3Yj2
Fw  
l cHf\~  
R1=2*f/(b1*b2) "~0`4lo:Xo  
R2=2*a1*f/(b2*(1+b1)) 2.I|8d[  
R3=2*a1*a2*f/(1+b2) cM3B5Lp  
Jw8?o/1D@  
A1=b2^3*(a1-1)*(1+b1)^3; nj:w1E/R  
B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; `BKb60  
C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; DqT<bNR1*;  
`MCiybl,&P  
A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2);  $8rnf  
B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); Vj9X6u}{  
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); h?p_jI  
N|$9v{ j_  
CB=[C1 B1;C2 B2]; ]t~.?)Ad+2  
AB=[A1 B1;A2 B2]; S'8+jY  
AC=[A1 C1;A2 C2]; :MY=Q]l  
x<M::")5!V  
%非球面系数 d\nXK#)Q  
k2=-(det(CB)/det(AB)); -f&vH_eK  
k3=-(det(AC)/det(AB)); 'mbLK#q  
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<d,9S,)  
k2=k2 e}1Q+h\  
k3=k3 uR :EH.K  
u+r!;-0i  
end wR@>U.XT@  
 Q&xH  
%有中间像，焦距输入为正数 &H%/.4la  
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z&4s@  
function sjr=yfdre(~) #i@f%Bq-  
'z5jnI  
f=input('f:'); U,#x\[3!Jt  
d1=input('d1:'); 4h@,hY1#  
d2=input('d2:'); dh7`eAMY   
d3=input('d3:'); QF/_?Tm4  
(g@\QdH`|  
A=f^2/(d3*d2)-f/d1; k\.9iI'6  
B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); 3?a`@C&x  
C=d3/d2-f/d1; zplv.cf#q  
FHQ`T\fC$@  
a1=(-B-sqrt(B^2-4*A*C))/(2*A); ,M.}Qak^  
a2=d3/(a1*f); nK :YbLdK,  
b2=a1*(1-a2)*f/d2; vvv'!\'#  
b1=(1-a1)*f/(d1*b2); '~&W'='b;  
1#6emMV.`  
%曲率半径 m%`YAD@2z  
Pgr2S I  
R1=2*f/(b1*b2) O*l,&5  
R2=2*a1*f/(b2*(1+b1)) IU"  
R3=2*a1*a2*f/(1+b2) {WrEe7dLy  
PM[_0b  
A1=b2^3*(a1-1)*(1+b1)^3; iVn4eLK^v  
B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; *
)<+u~  
C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; |T""v_q  
,~naKd.ZY  
A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); bPxL+ +  
B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); YUEyGhkMV{  
C2=b2*(a1-1)^2*(1+b1)*(1-b1)^2/(4*a1*b1^2)-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)*(1-b2)^2/(4*a1*a2*b1^2*b2^2)-b2*(a1-1)*(1-b1)*(1+b1)/(a1*b1)-(a2*(a1-1)+b1*(1-a2))*(1-b2)*(1+b2)/(a1*a2*b1*b2)-b1*b2+b2*(1+b1)/a1-(1+b2)/(a1*a2); 1;$XX#7o  
s6 g"uF>k  
CB=[C1 B1;C2 B2]; }8x+F2i  
AB=[A1 B1;A2 B2]; L T$U z  
AC=[A1 C1;A2 C2]; ;zJb("n  
w-pgtO|Us  
%二次系数 s)]jX  
d-_V*rYU  
k2=-(det(CB)/det(AB)); nUP, Yd  
k3=-(det(AC)/det(AB)); CVa>5vt  
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 ~r.R|f]IQ  
k2=k2 &|Duc} t  
k3=k3 6i[Ts0H%<!  
!{.CGpS ]  
end

谢谢分享，学习一下 h+1|.d