| 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 OQh 36BM
function sjr=nfdre(~) AOaf ,ZF
8
uXNf)?MpA
%系统焦距及各镜间距输入，间距取负正负 Hvq< _&2
oLn| UWe_
f=input('f:'); (;T; ?v`-
d1=input('d1:'); I fZaK([
d2=input('d2:'); ;6 1m
d3=input('d3:');
1Nk}W!v
Ffm Q$>S
A=f^2/(d3*d2)-f/d1; o+O\VNW
B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); vAJfMUlP
C=d3/d2-f/d1; [21tT/
A_%}kt
(6
a1=(-B+sqrt(B^2-4*A*C))/(2*A);%α1 b['TRYc=:
a2=d3/(a1*f);%α2 *0R=(Gy
b2=a1*(1-a2)*f/d2;%β2 {Pg7IYjH
b1=(1-a1)*f/(d1*b2);%β1 i
c]f o
_pNUI { De
Drlt xI)
%曲率半径 Y#6@0Nn[G
xL>0&R
R1=2*f/(b1*b2) U &Ay3/
R2=2*a1*f/(b2*(1+b1)) 7Hpsmfm
R3=2*a1*a2*f/(1+b2) va; d[D,
wrn[q{dX
A1=b2^3*(a1-1)*(1+b1)^3; _jZDSz|Y b
B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; X5U! 25d]
C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; 2.&v{gq
jVRd[
A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); ^B& Z
B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); `bT{E.(T
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); YQN=.Wtc
.(S,dG0P
CB=[C1 B1;C2 B2]; 6XQ )Q)
AB=[A1 B1;A2 B2]; 317Buk
AC=[A1 C1;A2 C2]; nfDPM\FFD
:M3l#`4Q
%非球面系数 8d) F #
k2=-(det(CB)/det(AB)); rP`\ <}a.
k3=-(det(AC)/det(AB)); Y+?bo9CES!
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 $z mES tcm
k2=k2 Ky nZzR
k3=k3
5Ll[vBW
:`
~b&Oz)
end =,@SZsM*B
+Kq> r|;
%有中间像，焦距输入为正数 7FDraEr#f
1C$^S]v%a
function sjr=yfdre(~) Z^fF^3x
}(tGjx]
f=input('f:'); Tz*5;y%4
d1=input('d1:'); !)9zH
d2=input('d2:'); ',!#?aGV
d3=input('d3:'); ao-C9|2>NU
#Y18z5vo
A=f^2/(d3*d2)-f/d1; 6:EO
B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); I$ mOy{/#
C=d3/d2-f/d1; p[o2F5 T2
[
objdQ U`
a1=(-B-sqrt(B^2-4*A*C))/(2*A); <![T~<.
a2=d3/(a1*f); r>)\"U#
b2=a1*(1-a2)*f/d2; [U jbox
b1=(1-a1)*f/(d1*b2); MJg^
QVM
Q49 |,ou[H
%曲率半径 "Z{^i3 gN
g{ J3Ba
R1=2*f/(b1*b2) FD@ ! z
:
R2=2*a1*f/(b2*(1+b1)) }dXL= ul
R3=2*a1*a2*f/(1+b2) ttw@nv%
@
|;_
yAL
A1=b2^3*(a1-1)*(1+b1)^3; So8P 8TCK
B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; u\E.H5u27
C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; Xl aNR+
O.$<B f9
A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); df:,5@CJ8
B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); m|7g{vHVV
C2=b2*(a1-1)^2*(1+b1)*(1-b1)^2/(4*a1*b1^2)-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)*(1-b2)^2/(4*a1*a2*b1^2*b2^2)-b2*(a1-1)*(1-b1)*(1+b1)/(a1*b1)-(a2*(a1-1)+b1*(1-a2))*(1-b2)*(1+b2)/(a1*a2*b1*b2)-b1*b2+b2*(1+b1)/a1-(1+b2)/(a1*a2); ' qM 3.U
q/3}8 BJ
CB=[C1 B1;C2 B2]; ^Ue.9#9T&g
AB=[A1 B1;A2 B2]; FCe503qND$
AC=[A1 C1;A2 C2]; SUVr&S6Nk
iK#{#ebAoW
%二次系数 ry<
P LRN
;% !?dH6
k2=-(det(CB)/det(AB)); =_1" d$S&
k3=-(det(AC)/det(AB)); ~xJD3Qf
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 B#DV<%GPl
k2=k2 4Ek<
5s[
k3=k3 -12v/an]L7
d}=p-s.GA
end
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