| 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 kl!wVLE
function sjr=nfdre(~) $
i)bq6
!l?.5Pm])
%系统焦距及各镜间距输入，间距取负正负 H(c72]@Vg
GG064zPq7
f=input('f:'); k %e^kej
d1=input('d1:'); r`y ezbG
d2=input('d2:'); i$G;f^Z!Y
d3=input('d3:'); C
(n+SY^
u@'0Vk0zGH
A=f^2/(d3*d2)-f/d1; EKEjv|_)
B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); l:0s2
C=d3/d2-f/d1; 6oa azB^L
RpWTpT1
a1=(-B+sqrt(B^2-4*A*C))/(2*A);%α1 3`d}~v {
a2=d3/(a1*f);%α2 0]KraLu"N
b2=a1*(1-a2)*f/d2;%β2 4'j
sDcs
b1=(1-a1)*f/(d1*b2);%β1 n~"$^Vr
4(e59ZgY
Q$& sTM
%曲率半径 ~CB [9D=
P./V6i<:
R1=2*f/(b1*b2) d2oh/j6`TA
R2=2*a1*f/(b2*(1+b1)) x";w%
R3=2*a1*a2*f/(1+b2) T{V/+ RM
5$DHn ]
A1=b2^3*(a1-1)*(1+b1)^3; PWh^[Rd)
B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; 4c
oJRqf=
C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; B\J^=W+`
3G.5724,
A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); *D o/+[Ae
B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); F(4?tX 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); \k;`}3 uO
/V3=K Y`_J
CB=[C1 B1;C2 B2]; ul%h@=n
AB=[A1 B1;A2 B2]; w%WF-:u7|
AC=[A1 C1;A2 C2]; 5MFxo63
F'g Vzf
%非球面系数 JnHo 9K2.
k2=-(det(CB)/det(AB)); >fH=DOz$&
k3=-(det(AC)/det(AB)); Gl1jxxd
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 O: @}lK+H
k2=k2 ^TMJ8` e
k3=k3 vN 2u34
[: xiZ
end &]
\X]p
iX0iRC6f
%有中间像，焦距输入为正数 da1]mb=4 5
d,8mY/S> w
function sjr=yfdre(~) $P1O>x>LIL
nz9DLAt
f=input('f:'); :2njp%
d1=input('d1:'); zBO (`=|
d2=input('d2:'); gp$ Ucfu'
d3=input('d3:'); XU+<?%u}z
SIb Dj[s
A=f^2/(d3*d2)-f/d1; AuR$g7z
B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); D;UV&.$'v
C=d3/d2-f/d1; kdb(I@6
A
AHt218
a1=(-B-sqrt(B^2-4*A*C))/(2*A); }I@L}f5N
a2=d3/(a1*f); "V&+7"Q
b2=a1*(1-a2)*f/d2; 9-] i.y
b1=(1-a1)*f/(d1*b2); <hwy*uBrD
]xguBh ]
%曲率半径 rP!#RzL
*^([ ~[
R1=2*f/(b1*b2) %,hV [[ @.
R2=2*a1*f/(b2*(1+b1)) 54rk C/B>
R3=2*a1*a2*f/(1+b2) `vc
"Q/
1SrJ6W @j[
A1=b2^3*(a1-1)*(1+b1)^3; e^ lWR] v
B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; c)@> zto#
C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; S1/`th
fX
jG5Tv
A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); ['%69dPh
B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); ~4?9a(>3
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)Nve
&-5_f* {
CB=[C1 B1;C2 B2]; LN_ xq&.
AB=[A1 B1;A2 B2]; : >itXD!
AC=[A1 C1;A2 C2]; P ~PIMkt
E*?<KZe"
%二次系数 5=e@d:Sz
[! %![E
k2=-(det(CB)/det(AB)); S$2b>#@UJ
k3=-(det(AC)/det(AB)); F 4Y @
B
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 )"+( butI&
k2=k2 3 =W!4
k3=k3 ={hX}"*D
<L&m4O#|
end
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