Surface Profile Refraction index Thickness
0 object n0 = 1 t0 = −50mm
1 z1 = r12/f1 n1 = 1.5 t1 = 7mm
2 z2 = t1+ r22/f2 n2 = 1 t2 = 5mm 3 z3 = t1+ t2+ r32/f3 n3 = 1.7 t3 = 9mm
4 Eq. (4.22) n4 = 1 t4 = 90mm
5 image not apply not apply
Table 4.1: The image is located at z = 7 + 5 + 9 + 90 = 111mm.
Figure 4.2: The focuses of first parabolic surfaces are with given by f1 = −50, f2 = −50 and f3 = −90.
where,
ϑ3 = −[f n3+n
2
4(r3℘r,3+τ ℘z,3)]±√
[f n3+n24(r3℘r,3+τ ℘z,3)]2−(n24−n23)[n24(r23+τ2)−f2] (n24−n23)
τ =z3−P4 i=1ti
f =−t0n0+P4
i=1niti−n0
√
r21+(z1−t0)2−n1
√
(r2−r1)2+(z2−z1)2−n2
√
(r3−r2)2+(z3−z2)2
ϑ2 =(2√ f3
√
f3℘2z,2+r2℘r,2℘z,2+℘2r,2(t1+t2−z2)−2f3℘z,2−r2℘r,2)/(℘2r,2),
r3 =ϑ2℘r,2+r2,
z3 =ϑ2℘z,2+z2,
ϑ1 =(2√ f2
√
f2℘2z,1+r1℘r,1℘z,1+℘2r,1(t1−z1)−2f2℘z,1−r1℘r,1)/(℘2r,1),
r2 =ϑ1℘r,1+r1,
z2 =ϑ1℘z,1+z1,
z1 =r12/f1.
(4.25)
Surface Profile Refraction index Thickness
0 object n0 = 1 t0 = −40mm
1 z1 = r12/f1 n1 = 1.5 t1 = 8mm
2 z2 = t1+ r22/f2 n2 = 1.7 t2 = 4mm 3 z3 = t1+ t2+ r32/f3 n3 = 1.55 t3 = 9mm
4 Eq. (4.24) n4 = 1 t4 = 90mm
5 image not apply not apply
Table 4.2: The image is located at z = 8 + 4 + 9 + 90 = 111mm.
Figure 4.3: The first three surfaces are parabolas with focuses given by f1 = 50, f2 = −80 and f3 = 50.
It is essential to mention that the reader is left to try to find the case when the object is in menus infinite, or the image is at infinity. With the method used throughout chapters 2, 3 and 4, it is just a routine procedure.
On-axis stigmatic reflective telescope
5.1 Introduction
The generalisation of the on-axis stigmatic lens in an arbitrary number of lenses can also be applied in refractive-reflective telescopes. The refractive-reflective telescopes are telescopes that at the input have a system of lenses, and then they have a reflective surface, typically a mirror.
In this chapter, we are going to get the equations that describe a stigmatic refractive-reflective telescope. The problem in concrete is given an arbitrary number of lenses how must be the shape of the last mirror of the refractive-reflective telescope to get an optical system free of spherical aberration.
5.1.1 Mathematical model
The purpose is to solve the following conjecture: Given a sequence by (N − 1) segments of optical material with refractive indices ni and axial thicknesses ti delimited by N ra-dially symmetric aspherical surfaces (zN, rN),(zN −1, rN −1),...,(z1, r1), where ri is the local radial variable at each plane zi. There exists a unique additional aspheric mirror given by (zN +1, rN +1) that eliminates the spherical aberration produced by the whole optical system, and places the output image in a plane located at a distance tI from the centre of the aspheric mirror. Like in chapter 4, all the parameters mentioned above are defined by the user except (zN +1, rN +1) which is the unknown surface to determine.
The refraction index between the object and the first surface is constant, and it is given by n0. The origin of the coordinate system is located at the centre of the first surface, i.e.
z1(r1 = 0) = 0, as can be seen in Fig. 5.1
Like the chapter 4, the course of a light ray traveling from the point (zi, ri) to the point (zi+1, ri+1) inside the i-th optical segment is given by the unit vector,
vˆi = [℘r,i, ℘z,i] = [ri+1− ri, zi− zi−1]
p(ri+1− ri)2+ (zi+1− zi)2, (5.1) where ℘r,iand ℘z,iare the Cartesian components of the unit vector ˆvi, i.e. its direction cosines with ℘2r,i + ℘2z,i = 1. Each time a light ray crosses an interface, its direction is modified according to the Snell law.
61
Figure 5.1: Telescopic system composed by N lenses with a aspherical corrective mirror (zN +1, rN +1) such as the system is on-axis stigmatic.
The cosine directors refracted by i-th surface are in terms of previous cosine directors.
We have already computed the cosine directors, refracted by the i-th surface in chapter 4. We recall them,
℘r,i = ni−1r0i[(ri− ri−1) ri0+ (zi− zi−1) zi0]
nip(ri−1− ri)2+ (zi−1− zi)2[(r0i)2+ (zi0)2] (4.5)
− zi0
r
1 − n
2
i−1[(ri−ri−1)r0i+(zi−zi−1)zi0]2
n2i](ri−1−ri)2+(zi−1−zi)2][(ri0)2+(z0i)2] p(r0i)2 + (zi0)2 , and,
℘z,i = ni−1zi0[(ri− ri−1) r0i+ (zi− zi−1) zi0]
nip(ri−1− ri)2+ (zi−1− z1)2[(ri0)2+ (zi0)2] (4.6)
+ ri0
r
1 − n
2
i−1((ri−ri−1)r0i+(zi−zi−1)z0i)2
n2i[(ri−1−ri)2+(zi−1−zi)2][(r0i)2+(z0i)2] p(r0i)2+ (z0i)2 ,
where zi0 = dzi(r1)/dr1and ri0 = dri(r1)/dr1 are the derivatives of zi and rirespect to r1. Solution
To discover the mentioned mirror, we are going to use the Fermat principle and Snell’s law.
We have already constructed the relation between the refractive surfaces and Snell’s law in the last section. So, let’s focus on the Fermat principle and Snell’s law. Let take as our reference ray, the axial ray, and compare its optical path with the path of any other ray.
The Fermat principle predicts that for an on-axis stigmatic refractive-reflective tele-scope, the optical paths between of an axial ray a non-axial ray must be equal.
The optical path of the axial ray is just the sum of the thicknesses multiplied by the respective refractive index. Notice that the distance from the centre of the mirror to the image tI should have a negative sign since the light is moving backwards.
For the path of a non-axial ray, it can be expressed as the sum of square roots. Some of the square roots should show the distance between one surface to following the refractive surface. The other square roots are the distance from the point object to the first surface and from the mirror to the point image. See Fig. 5.1.
Let us first to consider that the incoming rays come from a point source located at the optical axis at z = t0 < 0. Thus, applying the Fermat’s principle, the comparison of the optical paths traversed by both trajectories gives,
N
X
i=1
niti− t0n0+ nN(−tI) = nN v u u
trN +12 + −
N
X
i=1
ti+ zN +1− tI
!
2
+
N
X
i=2
ni−1p
(ri− ri−1)2+ (zi− zi−1)2 + nNp
(rN +1− rN)2+ (zN +1− zN)2 + n0
q
r12+ (z1− t0)2. (5.2) Now, to recover the parallelism of the input rays, we let the distance t0 to tend to −∞ in Eq. (5.2).a The limit gives,
−
N
X
i=2
ni−1p
(ri− ri−1)2+ (zi− zi−1)2+
N
X
i=1
niti+ tI(−nN) − n0z1
= nN
p(rN +1− rN)2+ (zN +1− zN)2+ nN
v u u
trN +12 + −
N +1
X
i=2
ti+ zN +1− tI
!
2, (5.3) replacing rN + ϑN℘r,N and zN + ϑN℘z,N is above’s equation,
−
N
X
i=2
ni−1 q
(ri− ri−1)2+ (zi− zi−1)2+
N
X
i=1
niti+ tI(−nN) − n0z1
= nN q
(rN + ϑN℘r,N − rN)2+ (zN + ϑN℘z,N − zN)2
+ nN v u u
t(rN + ϑN℘r,N)2 + −
N +1
X
i=2
ti+ zN + ϑN℘z,N − tI
!2
. (5.4)
where ϑN =p(rN +1− rN)2+ (zN +1− zN)2.
aPlease notice that we do not put the limit when t0→ −∞ in zN +1, rN +1. We avoid it consciously since we are not going to put examples when is t0is finite. In this way, without expressing the limit, the notation seems cleaner.
Let’s assign variables to all the terms of the above’s expression without the unknowns zN +1and rN +1, so,
f ≡ −
N
X
i=2
ni−1 q
(ri− ri−1)2+ (zi− zi−1)2 (5.5)
+
N
X
i=1
niti+ tI(−nN) − n0z1, and,
τN ≡ −
N
X
i=2
ti− tI, (5.6)
replacing Eqs. (5.6) and (5.6) in Eq. (5.4), f = nNϑN + nN
q
(rN + ϑNVr,N)2+ (ϑNVz,N + τ )2. (5.7) Now, if we solve the above’s equation for αN directly we get the radius rN +1and the saggita zN +1of the aspheric output mirror. The analytic solution of the system is given by
(zN +1= zN + ϑN℘z,N,
rN +1= rN + ϑN℘r,N. (5.8)
where,
ϑN = f2− n2N(r2N + τ2)
2nN (nN(rN℘r,N + τ ℘z,N) + 1). (5.9) Eq. (5.8) is the principal outcome of this chapter. It gives the analytic representation of the shape of the aspheric mirror to eliminate the spherical aberration generated by all the preced-ing surfaces. The validity of Eq. (5.8) requires that rays inside the system do not cross each other, i.e. preserve their relative locations. Note that the course cosines of the last refractive surface know all the preceding refractions experimented by the rays since the user defines all the preceding refractive surfaces.
Eq. (5.8) has the data of the optical path of every ray and its direction cosines. Thus, to apply it, every refractive surface must be expressed in terms of the rays refracted by the preceding surfaces. In section 4.2.1 of chapter 4. we already showed this procedure.