[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 8\ha@&p  
function sjr=nfdre(~) [IBQvL  
e]>ori 8  
%系统焦距及各镜间距输入，间距取负正负 5i So8*9}  
RMYP"  
f=input('f:'); O9h+Q\0\W  
d1=input('d1:'); )9nElb2  
d2=input('d2:'); IO$z%r7  
d3=input('d3:'); # '|'r+  
J{"kw1Lu  
A=f^2/(d3*d2)-f/d1; C'mL&  
B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); 0Fr1Ku!  
C=d3/d2-f/d1; =([av7  
Y}R}-+bD/  
a1=(-B+sqrt(B^2-4*A*C))/(2*A);%α1 B0UJq./`  
a2=d3/(a1*f);%α2 gQWd&)'muf  
b2=a1*(1-a2)*f/d2;%β2 2(YPz|~W  
b1=(1-a1)*f/(d1*b2);%β1 vR7HF*8  
n]!fO 6kj  
Ju` [m  
%曲率半径 Z0,~V  
LxN*)[Wb  
R1=2*f/(b1*b2) `cB_.&  
R2=2*a1*f/(b2*(1+b1)) xl4=++pu)  
R3=2*a1*a2*f/(1+b2) BNGe exs@  
4jmK].  
A1=b2^3*(a1-1)*(1+b1)^3; -#&kYK#Ph  
B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; TW&DFKK`  
C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; n]CbDbNw7)  
,5k-.Md>2*  
A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); M~T.n)x2  
B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); cd@.zg'sYn  
C2=b2*(a1-1)^2*(1+b1)*(1-b1)^2/(4*a1*b1^2)-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)*(1-b2)^2/(4*a1*a2*b1^2*b2^2)-b2*(a1-1)*(1-b1)*(1+b1)/(a1*b1)-(a2*(a1-1)+b1*(1-a2))*(1-b2)*(1+b2)/(a1*a2*b1*b2)-b1*b2+b2*(1+b1)/a1-(1+b2)/(a1*a2); mXWTm%'[  
wVK*P -C  
CB=[C1 B1;C2 B2]; =H]F`[B=  
AB=[A1 B1;A2 B2];  :S %lv  
AC=[A1 C1;A2 C2]; 1qdZc_x  
FcbM7/  
%非球面系数 m@xi0t  
k2=-(det(CB)/det(AB)); e,1u  
k3=-(det(AC)/det(AB)); zzpZ19"`1  
k1=(k2*a1*b2^3*(1+b1)^3-k3*a1*a2*(1+b2)^3+a1*b2^3*(1+b1)*(1-b1)^2-a1*a2*(1+b2)*(1-b2)^2)/(b1^3*b2^3)-1 7LZ^QC  
k2=k2 B33$ u3d  
k3=k3 ]hw-Bu\{  
0&Gl@4oZ"  
end v@ C,RP9  
Ps[$.h  
%有中间像，焦距输入为正数 2h~-  
m 40m<@  
function sjr=yfdre(~) `i!wq&1g7  
B~WtZ-%%E  
f=input('f:'); ]L_w$ev'  
d1=input('d1:'); &wH:aD  
d2=input('d2:'); %"3 )TN4  
d3=input('d3:'); H
. ,;-  
| FM }  
A=f^2/(d3*d2)-f/d1; #} ,x @]
p  
B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); yOXO)u1n  
C=d3/d2-f/d1; aC=['a>)  
|(IO=V4P  
a1=(-B-sqrt(B^2-4*A*C))/(2*A); Q%ad q-B  
a2=d3/(a1*f); 'JmBh@A  
b2=a1*(1-a2)*f/d2; ?2J?XS>  
b1=(1-a1)*f/(d1*b2); |cbd6e{!  
oh8L`=>&a  
%曲率半径 OmMX$YID  
C9tb\?#  
R1=2*f/(b1*b2) (
'k<XOi  
R2=2*a1*f/(b2*(1+b1)) $Rtgr{ {;"  
R3=2*a1*a2*f/(1+b2) 8^ep/b&|  
|WqOk~)[Z3  
A1=b2^3*(a1-1)*(1+b1)^3; n~0z_;5  
B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; @uleyB  
C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; Rs;15@t@  
D9ufoa&ua  
A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); xh9qg0d  
B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); fZryG  
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); BNm4k7 ]M  
{ShgJ;! Q  
CB=[C1 B1;C2 B2]; AHh#Fx+K  
AB=[A1 B1;A2 B2]; r#876.JK  
AC=[A1 C1;A2 C2]; Fe=8O ^\  
2!&pEqs  
%二次系数 =yiOJyx  
Mhpdaos  
k2=-(det(CB)/det(AB)); ~5aE2w0K
 
k3=-(det(AC)/det(AB)); @"87F{!  
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 |Fln8wB  
k2=k2 b[^{)$(  
k3=k3 ROAI9sW0  
w<nv!e
?  
end

谢谢分享，学习一下 e<+<lj"