| 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 0 z]H=
function sjr=nfdre(~) $/5\Hg1
y HlQKI
%系统焦距及各镜间距输入，间距取负正负 i_l{#*t
)C{20 _
f=input('f:'); I&gd"F _v}
d1=input('d1:'); fo`R=|L[
d2=input('d2:'); 8bs' Ek{'o
d3=input('d3:'); C&%NO;Ole
ja/wI'J<
A=f^2/(d3*d2)-f/d1; 1^V.L+0s]
B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); >&R@L KP
C=d3/d2-f/d1; 2Ub-ufkU
5} ur,0{
a1=(-B+sqrt(B^2-4*A*C))/(2*A);%α1 XP:fL
NpQ
a2=d3/(a1*f);%α2 v&7<f$5
b2=a1*(1-a2)*f/d2;%β2 BY HyqpP9
b1=(1-a1)*f/(d1*b2);%β1 WPlf8* -fQ
/0Qo (
OI78wG
%曲率半径 o"z;k3(i$7
Qp) ?wny4
R1=2*f/(b1*b2) 0 R `>F">
R2=2*a1*f/(b2*(1+b1)) ^,vFxN- -q
R3=2*a1*a2*f/(1+b2) A
#m _w*
"^ BA 5
A1=b2^3*(a1-1)*(1+b1)^3; f7!48,(fB
B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; rz-61A) _
C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; { D|ST2:E
x
_d
A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); ADB)-!$xoi
B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); &DHIYj1 i
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}|<*!4zUQ
z 8%qCq
CB=[C1 B1;C2 B2]; c; -N RvVb
AB=[A1 B1;A2 B2]; Eyk:p nKJb
AC=[A1 C1;A2 C2]; BD}%RTeWKq
? u".*!%
%非球面系数 J(maJuY
k2=-(det(CB)/det(AB)); w`+-xT%
k3=-(det(AC)/det(AB)); ) R5j?6}xF
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 *=v%($~PK6
k2=k2 nhC8Tq[m
k3=k3 MZcvr 9y
ydY 7 :D
end t0v >J9
[q_62[- X
%有中间像，焦距输入为正数 qdKqc,R1{
Ie=gI+2
function sjr=yfdre(~) ahCwA}
\v<S:cTf
f=input('f:'); ht>/7.p]
d1=input('d1:'); iycceZ
d2=input('d2:'); yD.(j*bMK;
d3=input('d3:'); >hq{:m
q@XJ,e1A
A=f^2/(d3*d2)-f/d1; *icaKy3
B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); LikCIO
C=d3/d2-f/d1; _y>drvg
:V#xrH8R
a1=(-B-sqrt(B^2-4*A*C))/(2*A); 3vAP&i'I
a2=d3/(a1*f); 5!$sQ@#}D
b2=a1*(1-a2)*f/d2; )\2KDXc
b1=(1-a1)*f/(d1*b2); F0zaA
VYh/ URU>
%曲率半径 z[R
dM#L
Ex*{iJ;\
R1=2*f/(b1*b2) ;V?(j 3b[
R2=2*a1*f/(b2*(1+b1)) 6@FhDj2X
R3=2*a1*a2*f/(1+b2) }aXS MxCd
$^czqA-&
A1=b2^3*(a1-1)*(1+b1)^3; :AqtPV'
B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; D8 PC;@m
C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; Bj><0
cNF
O\Z!7UQ$
A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); ; !t?*
B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); \dE{[^.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); ;~[}B v
s~B)xYmyB'
CB=[C1 B1;C2 B2]; UGgo;e
AB=[A1 B1;A2 B2]; }2m>S6""A
AC=[A1 C1;A2 C2]; c'Ibgfx%m
t98S[Z(-%+
%二次系数 p W 5D!z
?Ov~\[) F
k2=-(det(CB)/det(AB)); n6Uf>5
k3=-(det(AC)/det(AB)); _ nxu8g]
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 N `fFYO
k2=k2 u(f;4`
k3=k3 QXL .4r%
P 0hr=/h4
end
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