[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 8U_{|]M  
function sjr=nfdre(~) jmeRrnC}  
}}<z/zN&^  
%系统焦距及各镜间距输入，间距取负正负 %;D.vKoh  


g&{9VK6.  
f=input('f:'); 4"Hye&O  
d1=input('d1:'); +]Y,
q w  
d2=input('d2:'); )5 R=Z<  
d3=input('d3:'); IH"6? 9nd  
aFLO{tr`  
A=f^2/(d3*d2)-f/d1; QPq7R  
B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); AoyX\iqQ  
C=d3/d2-f/d1; $cZUM}@  
//aF5:Y#  
a1=(-B+sqrt(B^2-4*A*C))/(2*A);%α1 4 uQT5  
a2=d3/(a1*f);%α2 ZzX~&95G  
b2=a1*(1-a2)*f/d2;%β2 "]G\9b)  
b1=(1-a1)*f/(d1*b2);%β1 {`w;39$+  
@/1w4'M  
>+[&3u  
%曲率半径 FJd]D[h  
Km^&<3ch#  
R1=2*f/(b1*b2) 12+>5BA  
R2=2*a1*f/(b2*(1+b1)) [n/hkXa$\  
R3=2*a1*a2*f/(1+b2) znHnVYll(  
QMZ)-ty"  
A1=b2^3*(a1-1)*(1+b1)^3; *<!oHEwkN  
B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; @62Mk},9 c  
C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; '{B!6|"X  
wKsT7c'  
A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); 5>TK^1 :  
B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); 
=xS+5(  
C2=b2*(a1-1)^2*(1+b1)*(1-b1)^2/(4*a1*b1^2)-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)*(1-b2)^2/(4*a1*a2*b1^2*b2^2)-b2*(a1-1)*(1-b1)*(1+b1)/(a1*b1)-(a2*(a1-1)+b1*(1-a2))*(1-b2)*(1+b2)/(a1*a2*b1*b2)-b1*b2+b2*(1+b1)/a1-(1+b2)/(a1*a2); Rx+p
.  
RH|XxH*  
CB=[C1 B1;C2 B2]; aT`%;i^  
AB=[A1 B1;A2 B2]; } GiHjzsR  
AC=[A1 C1;A2 C2]; cA:*V|YV`  
.AU)*7Gh  
%非球面系数 2j&AiD  
k2=-(det(CB)/det(AB)); ;"m ,:5%  
k3=-(det(AC)/det(AB)); jfqopiSi  
k1=(k2*a1*b2^3*(1+b1)^3-k3*a1*a2*(1+b2)^3+a1*b2^3*(1+b1)*(1-b1)^2-a1*a2*(1+b2)*(1-b2)^2)/(b1^3*b2^3)-1 j.@\3'  
k2=k2 lMbAs.!  
k3=k3 f^.AD-  
::/j$bL  
end k Zq!&  
2TA*m{\Hr  
%有中间像，焦距输入为正数 /ID?DtJ  
$-*!pRaVU  
function sjr=yfdre(~) k{Aj^O3gD  
Zp#v Hs  
f=input('f:'); g"> {9YE  
d1=input('d1:'); '3 ^+{=q  
d2=input('d2:'); j+ LawW-  
d3=input('d3:'); +PuPO9jKO@  
Kp,M"Y  
A=f^2/(d3*d2)-f/d1; `9b7>Nn<  
B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); lgtC|kM=  
C=d3/d2-f/d1; s]JF0584  
qC?:*CXH  
a1=(-B-sqrt(B^2-4*A*C))/(2*A); ~7Tc$ "I  
a2=d3/(a1*f); ,c)uX#1  
b2=a1*(1-a2)*f/d2; .uk>QMs1  
b1=(1-a1)*f/(d1*b2); 4tS.G  
=>! Y{: y(  
%曲率半径 RSfzRnhmr  
: 7`[$<~E  
R1=2*f/(b1*b2) BV[5}  
R2=2*a1*f/(b2*(1+b1)) x HhN  
R3=2*a1*a2*f/(1+b2) E3iW-B8u8  
L~NbdaO  
A1=b2^3*(a1-1)*(1+b1)^3; =Fr(9(  
B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; &sL&\+=<(  
C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; ut I"\1hQ  
JQ@E>o7_  
A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); DY?Kfvef  
B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); d|yAs5@  
C2=b2*(a1-1)^2*(1+b1)*(1-b1)^2/(4*a1*b1^2)-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)*(1-b2)^2/(4*a1*a2*b1^2*b2^2)-b2*(a1-1)*(1-b1)*(1+b1)/(a1*b1)-(a2*(a1-1)+b1*(1-a2))*(1-b2)*(1+b2)/(a1*a2*b1*b2)-b1*b2+b2*(1+b1)/a1-(1+b2)/(a1*a2); 2FW\O0U  
y)@[Sl>  
CB=[C1 B1;C2 B2]; L-k@-)98  
AB=[A1 B1;A2 B2]; >M/V oV  
AC=[A1 C1;A2 C2]; f|tjsZxQ  
mA=i)Ga  
%二次系数 /x1![$oC0  
7F{3*`/6  
k2=-(det(CB)/det(AB)); 9c6gkt9eB  
k3=-(det(AC)/det(AB)); 2mGaD\?K  
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 AQiwugs  
k2=k2 ]&pds\  
k3=k3 p ObX42  
O
6G0  
end

谢谢分享，学习一下 ,^. 88<