[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 h2ou]  
function sjr=nfdre(~) Xa&0j&AH  
MF4B 2d  
%系统焦距及各镜间距输入，间距取负正负 vN`JP`IBx  
kr5'a:F)  
f=input('f:'); Yj>ezFo  
d1=input('d1:'); lgh+\pj  
d2=input('d2:'); (Z at|R.F  
d3=input('d3:'); Ip;;@o&D  
j79$/ Ol  
A=f^2/(d3*d2)-f/d1; uE,j$d  
B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); D P+W*87J  
C=d3/d2-f/d1; F;)qM|7  
*^.OqbO[U  
a1=(-B+sqrt(B^2-4*A*C))/(2*A);%α1 qJ<Ghd`8v  
a2=d3/(a1*f);%α2 ')~[J$qz  
b2=a1*(1-a2)*f/d2;%β2 2)jf~!o)Z  
b1=(1-a1)*f/(d1*b2);%β1 2B=+p83<  
?F@X>zR2  
@ R;o $n  
%曲率半径 ]j^rJ|WTH  
p
"=8{LrO  
R1=2*f/(b1*b2) , @dhJ8/  
R2=2*a1*f/(b2*(1+b1)) #l-/!j  
R3=2*a1*a2*f/(1+b2) {A8w~3F  
X
Q$9E?|=  
A1=b2^3*(a1-1)*(1+b1)^3; >Ms_bfSK  
B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; A>QAR)YP  
C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; EY'48S  
{DbWk>[DkG  
A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); >Gxh=**F  
B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); 1F94e)M)"  
C2=b2*(a1-1)^2*(1+b1)*(1-b1)^2/(4*a1*b1^2)-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)*(1-b2)^2/(4*a1*a2*b1^2*b2^2)-b2*(a1-1)*(1-b1)*(1+b1)/(a1*b1)-(a2*(a1-1)+b1*(1-a2))*(1-b2)*(1+b2)/(a1*a2*b1*b2)-b1*b2+b2*(1+b1)/a1-(1+b2)/(a1*a2); Yjp*T:6  
'n!kqP  
CB=[C1 B1;C2 B2]; 5D`!Tu3  
AB=[A1 B1;A2 B2]; \xexl1_;  
AC=[A1 C1;A2 C2]; |NXe{q7{  
</)QCl'd  
%非球面系数 |bv7N@?e  
k2=-(det(CB)/det(AB)); .Sjg  
k3=-(det(AC)/det(AB)); 3\&I
7o3V  
k1=(k2*a1*b2^3*(1+b1)^3-k3*a1*a2*(1+b2)^3+a1*b2^3*(1+b1)*(1-b1)^2-a1*a2*(1+b2)*(1-b2)^2)/(b1^3*b2^3)-1 CGJ>j}C  
k2=k2 L$ ZZ]?7j  
k3=k3 2U`g[1  
^ lM.lS>)  
end (pkq{: Fs  
.+dego
:  
%有中间像，焦距输入为正数 2N}h<Yd9  
uy oEMT#u  
function sjr=yfdre(~) &=hkB9 ;  
vy1N,8a  
f=input('f:'); Q(ec>+oi  
d1=input('d1:'); :,Ad1(  
d2=input('d2:'); -{s9PZ3~_  
d3=input('d3:'); ^<QF*!  
Ej/P:nB  
A=f^2/(d3*d2)-f/d1; yh"48@L'D  
B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); $BWA=2$  
C=d3/d2-f/d1; QeipfK+me  
Lo^gg#o  
a1=(-B-sqrt(B^2-4*A*C))/(2*A); _cD-E.E%  
a2=d3/(a1*f); |A19IXZ\  
b2=a1*(1-a2)*f/d2; Q804_F F#  
b1=(1-a1)*f/(d1*b2); m005*>IY  
`Fs-z  
%曲率半径 0%>_fMaA  
|n+ `t?L^  
R1=2*f/(b1*b2) ps@{1Rn1  
R2=2*a1*f/(b2*(1+b1)) XZ[3v9?&n  
R3=2*a1*a2*f/(1+b2) 6SN$El 0|G  
WiCJhVF3  
A1=b2^3*(a1-1)*(1+b1)^3; l6
k.`1.In  
B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; %][6TZ}  
C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; b<"LUM*;  
i4r8146D[  
A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); Xe>   
B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); pm k;5 d  
C2=b2*(a1-1)^2*(1+b1)*(1-b1)^2/(4*a1*b1^2)-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)*(1-b2)^2/(4*a1*a2*b1^2*b2^2)-b2*(a1-1)*(1-b1)*(1+b1)/(a1*b1)-(a2*(a1-1)+b1*(1-a2))*(1-b2)*(1+b2)/(a1*a2*b1*b2)-b1*b2+b2*(1+b1)/a1-(1+b2)/(a1*a2); `beU2N  
F@u>5e^6  
CB=[C1 B1;C2 B2]; =y WHm  
AB=[A1 B1;A2 B2]; zvHeoM,  
AC=[A1 C1;A2 C2]; 25/OV"Z  
T+~ _D  
%二次系数 +a!uS0fIJi  
Sx}61
?  
k2=-(det(CB)/det(AB)); R\,qL-Br  
k3=-(det(AC)/det(AB)); t6a$ZN;  
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 E.+BqWZ!  
k2=k2 h$rk]UM/Q  
k3=k3 o1]ZeF  
{BS`v5*
 
end

谢谢分享，学习一下 %r >Y)@$Vt