| songshaoman |
2020-05-25 15:25 |
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
%无中间像，焦距输入为负数 ;7Okyj6EP
function sjr=nfdre(~) ?Vc0)
MQ` %``
%系统焦距及各镜间距输入，间距取负正负 ]-:6T0JuS
k! 3 cq )
f=input('f:'); VRbQdiZ{
d1=input('d1:'); {x {H$ f
d2=input('d2:'); &94W-zh
d3=input('d3:'); &R O7{,`
gn)R^
A=f^2/(d3*d2)-f/d1; pOA !#Aj)
B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); ()\jCNLT
C=d3/d2-f/d1; .'T 40=7
Kkv<"^H
a1=(-B+sqrt(B^2-4*A*C))/(2*A);%α1 -V5w]F'
a2=d3/(a1*f);%α2 OJ1tV% E
b2=a1*(1-a2)*f/d2;%β2 P!e= b-T
b1=(1-a1)*f/(d1*b2);%β1 1nI^-aQ3
1e}8LH7
IQnIaZ
%曲率半径 `}fw1X5L
C:i|-t e
R1=2*f/(b1*b2) U>F{?PReA?
R2=2*a1*f/(b2*(1+b1)) ~<)vKk
R3=2*a1*a2*f/(1+b2) HB iBv-=,
->gZ)?Fqy
A1=b2^3*(a1-1)*(1+b1)^3; gU;&$
B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; |=K_F3aJ
C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; hXB|g[zT
2K{6iw"h
A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); qP1FJ89H
B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); 41V e }%
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_pw
R_N:#K.M
CB=[C1 B1;C2 B2]; QDTNx!WL
AB=[A1 B1;A2 B2]; gl7|H&&xV
AC=[A1 C1;A2 C2]; Y)|N"f;
2 7A!\pn
%非球面系数 %d;ezY '2
k2=-(det(CB)/det(AB)); -VT+O+9_A
k3=-(det(AC)/det(AB)); u:dx;*
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 9 OT,TpA
k2=k2 GP a`e
k3=k3 /*rhtrS)
k'3Wt*i
end t ^ SzqB
Z(GfK0vU
%有中间像，焦距输入为正数 (zcLx;N
](jFwxU
function sjr=yfdre(~) yj_4gxJ\
tTanW2C
f=input('f:'); !L2 4+ $
d1=input('d1:'); W+ =o&V
d2=input('d2:'); A1i!F?X
d3=input('d3:');
V
9;[M;
*rh,"Zo
A=f^2/(d3*d2)-f/d1; $8~e}8dt|
B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); jZ*WN|FK?
C=d3/d2-f/d1; k G0Yh2;#
$E!J:Y=
a1=(-B-sqrt(B^2-4*A*C))/(2*A); vVxD!EL
a2=d3/(a1*f); |`/TBQz:r
b2=a1*(1-a2)*f/d2; cr;`Tl~}s
b1=(1-a1)*f/(d1*b2); gm"#:< )
f 2YLk
%曲率半径 R.9V,R5
SPkn 3D6
R1=2*f/(b1*b2) >Q kP7Kb
R2=2*a1*f/(b2*(1+b1)) 98XVa\|tl
R3=2*a1*a2*f/(1+b2) fS&6
??&<k
A1=b2^3*(a1-1)*(1+b1)^3; 3G}AH E4
B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; # ]Y>KX2HG
C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; K }$&:nao
E%N2k|%8d_
A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); 4Y \wnwI
B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); RP4Ku9 hk
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); LW) H"6v
Vr.Y/3N&'
CB=[C1 B1;C2 B2]; *#
{z 3{+
AB=[A1 B1;A2 B2]; V^s0fWa
AC=[A1 C1;A2 C2]; qQ]]~F
}f} } A =
%二次系数 PJ 4(}a
i5}4(sV
k2=-(det(CB)/det(AB)); 9LJZ-/Wq
k3=-(det(AC)/det(AB)); ;]2s,za)qs
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 Ol_q{^
k2=k2 AnBJ(h
k3=k3 8 .>/6M
Z ~
end
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