[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 3 %DA{  
function sjr=nfdre(~) 4c2P%X( C  
wB9IP{Pf  
%系统焦距及各镜间距输入，间距取负正负 >=U$s@  
07(E/A]  
f=input('f:'); V2g$"W?3  
d1=input('d1:'); ]2h[.qa  
d2=input('d2:'); wW%I <M  
d3=input('d3:'); E*(Q'p9C  
EHzZ9zH\  
A=f^2/(d3*d2)-f/d1; `b\4h/~  
B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); IC}zgvcW  
C=d3/d2-f/d1; lg8~`96  
'Fonn  
a1=(-B+sqrt(B^2-4*A*C))/(2*A);%α1 ;Ly4Z*!2  
a2=d3/(a1*f);%α2 bzJKoxU  
b2=a1*(1-a2)*f/d2;%β2 ~B=\![  
b1=(1-a1)*f/(d1*b2);%β1 2$\f !6p  
LL[+QcH  
hJ}G5pX  
%曲率半径 G x,D'H'  
+vU.#C_2  
R1=2*f/(b1*b2) SbGp  
R2=2*a1*f/(b2*(1+b1)) fLD9RZ8_  
R3=2*a1*a2*f/(1+b2) 66|lQE&n  
Rw/G =zV@2  
A1=b2^3*(a1-1)*(1+b1)^3; 9&d BL0  
B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; il#rdJ1@t  
C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; Q'8v!/"}p{  
(vI7qD_  
A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); 2f:'~ P56  
B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); r p@=
  
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); #5H@/o8!s=  
RtI
c:ym  
CB=[C1 B1;C2 B2]; 7Ru0
>4B  
AB=[A1 B1;A2 B2]; ~f@<]  
AC=[A1 C1;A2 C2]; 3R{-\
ZMd  
NGze: gPmO  
%非球面系数 >|iy= Zn%'  
k2=-(det(CB)/det(AB)); ;8T<L[ ^U  
k3=-(det(AC)/det(AB)); WEsH@ [  
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 .Z^g 7 *s  
k2=k2 "|Pl(HX  
k3=k3 #ERn 8k  
H!Od.$ZIX  
end sW]n~
kTt'  
bkM$ Qo  
%有中间像，焦距输入为正数 ~Fx[YPO,  
uZYeru"w  
function sjr=yfdre(~) S1B
/ClKWq  
%bimcRX#W  
f=input('f:'); 0)/214^&  
d1=input('d1:'); )F~_KD)7jJ  
d2=input('d2:'); Y{O&-5H^|  
d3=input('d3:'); 1
z`,*eD7  
zJsoenU  
A=f^2/(d3*d2)-f/d1; 6 %=BYDF  
B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); AzV5Re8M  
C=d3/d2-f/d1; s{IoL_P
JP  
7d<v\=J
}  
a1=(-B-sqrt(B^2-4*A*C))/(2*A); cpy"1=K~M  
a2=d3/(a1*f); kDz
.{Ih  
b2=a1*(1-a2)*f/d2; W+eN%w5  
b1=(1-a1)*f/(d1*b2); ~"wD4Ue  
4ku/3/6  
%曲率半径 e"2QV vB  
OP&[5X+Y  
R1=2*f/(b1*b2) 68!]q(!6F  
R2=2*a1*f/(b2*(1+b1)) 7 dzE"m  
R3=2*a1*a2*f/(1+b2) VsmL#@E  
dL
\8^L  
A1=b2^3*(a1-1)*(1+b1)^3; MC
CZh{uo  
B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; K@i*Nl  
C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; 3(&.[o Z  
xL BG}C  
A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); %b?Pasf.  
B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); }ot _k-  
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); K+}Z6_:  
|]q=D1/A  
CB=[C1 B1;C2 B2]; O<3,n;56Z  
AB=[A1 B1;A2 B2]; n~ql]Ln  
AC=[A1 C1;A2 C2]; [s/@z*,M1  
q>[% C5  
%二次系数 \PFx# :-c  
O)Qz$  
k2=-(det(CB)/det(AB)); Lm?*p>\Q  
k3=-(det(AC)/det(AB)); 93J)9T  
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 "uCQm '  
k2=k2 `[KhG)Y7t  
k3=k3 V!P3CNK  
_y9P]@Q7%  
end

谢谢分享，学习一下 a%!XLyq