[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 <g9"Cr`  
function sjr=nfdre(~) 6TH!vuQ1(  
{$M;H+Foh  
%系统焦距及各镜间距输入，间距取负正负 Uw<&Wm`'  
111D3  
f=input('f:'); !Do,>gO  
d1=input('d1:'); *wV[TKaN  
d2=input('d2:'); lr2rQo>  
d3=input('d3:'); LihjGkj\g  
y>\S@I  
A=f^2/(d3*d2)-f/d1; oH2!5;A|  
B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); {5%/T,  
C=d3/d2-f/d1; c0 WFlj9b  
A.Bk/N1G  
a1=(-B+sqrt(B^2-4*A*C))/(2*A);%α1 :Au /2  
a2=d3/(a1*f);%α2 RH~3M0'0  
b2=a1*(1-a2)*f/d2;%β2 Z v0C@r  
b1=(1-a1)*f/(d1*b2);%β1 x"(9II*  
!~
lW3  
J`#`fX  
%曲率半径 !6w{(Rc(C  

N*DhjEU)[  
R1=2*f/(b1*b2) c#b:3dXx9  
R2=2*a1*f/(b2*(1+b1)) Ak~4|w-  
R3=2*a1*a2*f/(1+b2) 2:$ k  
&14W
vAU  
A1=b2^3*(a1-1)*(1+b1)^3; Poa?Ej  
B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; ,M3z!=oIGn  
C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; J<JBdk  
J
fcMca  
A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); Wn!G.(Jq  
B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); GL
KO]y  
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); rdj@u47  
bO49GEUT _  
CB=[C1 B1;C2 B2]; #/j={*-  
AB=[A1 B1;A2 B2]; .9LL+d  
AC=[A1 C1;A2 C2]; +zWrLf_Rc  
]2+g&ox4'  
%非球面系数 >kdM:MK  
k2=-(det(CB)/det(AB)); R V!o4"\]  
k3=-(det(AC)/det(AB)); !W1eUY  
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 UqX1E  
k2=k2 )u@t.)ChAV  
k3=k3 <?$kI>Ot  
F,G,b  
end rbk<z\pc  
^{[`=P'/  
%有中间像，焦距输入为正数 lVeH+"M?  
 SNvb1&  
function sjr=yfdre(~) QJ];L7Hbo  
J(d2:V{h  
f=input('f:'); ,VD6s!(  
d1=input('d1:'); A
|<;  
d2=input('d2:'); W._G0b4}  
d3=input('d3:'); da*9(!OV  
'7nJb6V,0l  
A=f^2/(d3*d2)-f/d1; FU@uH U5fd  
B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); (Cj,\r  
C=d3/d2-f/d1; 
a'[)9:  
J0Four#MD  
a1=(-B-sqrt(B^2-4*A*C))/(2*A); K[?R[  
a2=d3/(a1*f); tE!'dpG5)  
b2=a1*(1-a2)*f/d2; \7E`QY4  
b1=(1-a1)*f/(d1*b2); ~eo^`4O{{  
 3t  
%曲率半径 IYNM
U\s  
0|2%# E  
R1=2*f/(b1*b2) jA2ofC  
R2=2*a1*f/(b2*(1+b1)) ci7~KewJ*  
R3=2*a1*a2*f/(1+b2) ?@a$!_  
F7uhuqA]N  
A1=b2^3*(a1-1)*(1+b1)^3; 'P/taEi=R  
B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; (G5T%[/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; /8p&Qf>lJ1  
yv${M u  
A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); e<'U8|}hc{  
B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); `2x34  
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); ;a{rWz1Wm  
Ck(.N  
CB=[C1 B1;C2 B2]; #J.u  
AB=[A1 B1;A2 B2]; #D:RhqjK  
AC=[A1 C1;A2 C2]; K%<GU1]-]  
X'2Gi  
%二次系数 #5
d8?n  
$Z7:#cZ Y  
k2=-(det(CB)/det(AB)); P7 8uq  
k3=-(det(AC)/det(AB)); m9g^ -X  
k1=(k2*a1*b2^3*(1+b1)^3-k3*a1*a2*(1+b2)^3+a1*b2^3*(1+b1)*(1-b1)^2-a1*a2*(1+b2)*(1-b2)^2)/(b1^3*b2^3)-1 /$OIlu  
k2=k2 3^H/LWx`{]  
k3=k3 kt<@H11  
")\ *2d  
end

谢谢分享，学习一下 -~]^5aa5n