| 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 :q~qRRmjBe
function sjr=nfdre(~) MjE.pb
jd l 1Q<Z
%系统焦距及各镜间距输入，间距取负正负 |`s}PcV
mW1Sd#0
f=input('f:'); Ol9U^
d1=input('d1:'); FFbMG:>:
d2=input('d2:'); q?8MKf[N
d3=input('d3:'); v6oPAqj ,r
G#5Cyu<r!
A=f^2/(d3*d2)-f/d1; @S?.`o
B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); /}2
bsiJT
C=d3/d2-f/d1; T =3te|fv
-6~*:zg,
a1=(-B+sqrt(B^2-4*A*C))/(2*A);%α1 &[*_ -
a2=d3/(a1*f);%α2 7E Y~5U/4
b2=a1*(1-a2)*f/d2;%β2 7oF`Os+U
b1=(1-a1)*f/(d1*b2);%β1 nX5*pTfjL3
#i ?@S$
ce2d)FG}e
%曲率半径 (J.(Fl>^
qS& PMQ"$
R1=2*f/(b1*b2) .`Z{ptt>
R2=2*a1*f/(b2*(1+b1)) >1pD'UZIy7
R3=2*a1*a2*f/(1+b2) 90sM S]a
B_hob
A1=b2^3*(a1-1)*(1+b1)^3; E}k#-+u<S4
B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; Sx;zvc
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[J,j+f<
?M Kf=! w
A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); K blOP{I
B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); K rMIJA4>
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); }S*6+4
^ eM=h
CB=[C1 B1;C2 B2]; )@eBe ^
AB=[A1 B1;A2 B2]; PC \Xm,,
AC=[A1 C1;A2 C2]; lU Uq|Qr
+D?d)lK
%非球面系数 {Hp?rY@
k2=-(det(CB)/det(AB)); n,PHfydqX
k3=-(det(AC)/det(AB)); &M>S$+I
n
k1=(k2*a1*b2^3*(1+b1)^3-k3*a1*a2*(1+b2)^3+a1*b2^3*(1+b1)*(1-b1)^2-a1*a2*(1+b2)*(1-b2)^2)/(b1^3*b2^3)-1 "oz @w'rG
k2=k2 ,Lv} Xku
k3=k3 (ZShh y8g
=#BeAsFfO
end 82$By] Y9
b8b-M]P-=
%有中间像，焦距输入为正数 $k'f)E
$nn~K
function sjr=yfdre(~) JTx}{kVO
U&u7d$AN P
f=input('f:');
dZ%b|CUb
d1=input('d1:'); ljiq +t T
d2=input('d2:'); H kg@M?(
d3=input('d3:'); H5&>Eny
|o!<@/iH=
A=f^2/(d3*d2)-f/d1; tH_# q"@)
B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); EHzZ9zH\
C=d3/d2-f/d1; `b\4h/~
~%<PEl|
a1=(-B-sqrt(B^2-4*A*C))/(2*A); M<ad>M
a2=d3/(a1*f); 2CmeO&(Qf*
b2=a1*(1-a2)*f/d2; gKYn*
b1=(1-a1)*f/(d1*b2); o8s&n3mY}y
~B= \![
%曲率半径 UD9h5PgT
dtF6IdAf
R1=2*f/(b1*b2) 2Cr+Z(f
R2=2*a1*f/(b2*(1+b1)) RLKO 0 #
R3=2*a1*a2*f/(1+b2) dtg Ja_
8U;!1!+
7)
A1=b2^3*(a1-1)*(1+b1)^3; aLsGden|
B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; vi5~ Rd`
C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; spl*[ d
s &.Z;X
A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); qL'3MY.!
B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); 0I#<-9&d-
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); PN.6BJvu
3zl!x
CB=[C1 B1;C2 B2]; i44:V R|
AB=[A1 B1;A2 B2]; >#q|Pjv]
AC=[A1 C1;A2 C2]; 9723f1&Vd
?BLOc;I&a
%二次系数 3YLnh@-
JQtH },T r
k2=-(det(CB)/det(AB)); plf <O5'
k3=-(det(AC)/det(AB)); ;8T<L[ ^U
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 WEsH@
[
k2=k2 .Z^g
7 *s
k3=k3 J BwTmOv Q
!3b%Q</M H
end
|
|