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Geometric optics interpretation for rainbow scattering
of a chiral sphere
Zhensen Wu (吴振森)1,2,*, Qingchao Shang (尚庆超)1, Tan Qu (屈檀)1,
Zhengjun Li (李正军)1, and Lu Bai (白璐)1
1School of Physics and Optoelectronic Engineering, Xidian University, Xi’an 710071, China
2Collaborative Innovation Center of Information Sensing and Understanding at Xidian University, Xi’an 710071, China
*Corresponding author: wuzhs@mail.xidian.edu.cn
Received September 11, 2015; accepted October 27, 2015; posted online December 7, 2015
Research on light scattering from a large chiral sphere shows that the rainbow phenomenon is different from that
of an isotropic sphere. A chiral sphere with certain chirality generates three first-order rainbows. In this Letter,
we present a geometric optics interpretation for the phenomenon and make a calculation of the rainbow angles.
The ray traces inside the sphere are determined by the reflection and refraction laws of light at the achiral–chiral
interface and the chiral–achiral interface. The calculated rainbow angles achieve good agreements with those
obtained by the analytical solutions. The effects of chirality and the refractive index of the sphere on rainbow
angles are analyzed.
OCIS codes: 160.1585, 080.0080, 290.5850.
doi: 10.3788/COL201513.121602.
The rainbow phenomenon of an isotropic sphere can be
described properly by using both Lorenz–Mie theory
and geometric optics[1–4]. Related research has been applied
in measurements of particle characteristics such as size
and velocity[5]. Chiral media, also known as the “optical
active media”at optical frequencies, have been researched
a lot in the past several decades due to their different
optical and electromagnetic properties[6]. Research on light
scattering from large-sized chiral spheres[7,8]has shown
that for a linearly polarized wave incidence, there are three
first-order rainbow structures. The phenomenon is nu-
merically analyzed based on analytical solutions[9].
In this Letter, we present a geometric optics model to
compute the rainbow angles of a chiral sphere based on
the optical properties of chiral media. A light wave prop-
agates in chiral media in circularly polarized modes, which
is different from the way light travels in general isotropic
media. For a chiral medium, there are two different wave
numbers for different handednesses of polarized light at
one frequency. According to the electromagnetic descrip-
tion of a chiral medium, the wave numbers of a right-
handed circularly polarized (RCP) wave and a left-handed
circularly polarized (LCP) wave in a chiral medium are
kR¼k0ðncþκÞand kL¼k0ðnc−κÞ, respectively. k0is
the wave number of the light wave in vacuum, κis the chi-
rality parameter of the medium, and ncis a parameter
related to the permittivity and the permeability of the
medium. Thus, it can be considered that there is a refrac-
tive index for an RCP wave, which is nR¼ncþκ; for an
LCP wave, it is nL¼nc−κ. In the figures, the RCP rays
are denoted by the red lines, and the LCP rays are denoted
by the blue lines. For convenience, all media considered in
this Letter are lossless media.
Before we research the ray traces in a chiral sphere, an
introduction to the reflection and refraction of light at the
achiral–chiral interface and the chiral–achiral interface is
necessary. Much study was devoted to it several years
ago[10–13]. Consider a light ray impinging on an achiral–
chiral interface with an incidence angle θi, as shown in
Fig. 1(a).n0is the refractive index of the isotropic medium
and kiis the wave number of the incident light. There will
be two transmitted rays in the chiral medium. One is an
RCP ray with refracted angle θRand the other is an LCP
ray with refracted angle θL. The continuity of the tangential
components of the wave vectors in the two regions leads to
n0sin θi¼nRsin θR¼nLsin θL;(1)
z
O
i
θ
L
θ
ˆ
LL
kr
ˆ
RR
kr
0
n
,
RL
nn
R
θ
x
ˆ
ii
kr
ˆ
ir
kr
chiral media
general
isotropic media
i
k
L
k
R
k
(a)
O
rR
θ
t
k
L
k
R
k
0
n
,
RL
nn
rL
θ
general
isotropic media
chiral media
0
κ
>
ˆ
Ri
kr
ˆ
RrR
kr
ˆ
LrL
kr
x
ˆ
tt
kr
t
θ
z
i
θ
(b)
Fig. 1. Light interaction with chiral interface. (a) Achiral–chiral
interface. (b) Chiral–achiral interface.
COL 13(12), 121602(2015) CHINESE OPTICS LETTERS December 10, 2015
1671-7694/2015/121602(4) 121602-1 © 2015 Chinese Optics Letters
from which follow the two refracted angles:
sin θR¼n0∕nRsin θi;sin θL¼n0∕nLsin θi:(2)
At the chiral–achiral interface, suppose that an RCP ray
illuminates the interface with an incident angle θi.There
will be two reflected rays in the chiral medium and one
transmitted ray in the general isotropic medium, as shown
in Fig. 1(b). Similarly, the reflected angles and the
refracted angle can be obtained by:
nRsin θi¼nRsin θrR ¼nLsin θrL ¼n0sin θt:(3)
It can be seen that, the reflected angle of the reflected RCP
ray is equal to the incident angle θrR ¼θi, while the
reflected angle of the reflected LCP ray is obtained by
sin θrL ¼nR∕nLsin θiand the refracted angle of the
transmitted ray is obtained by sin θt¼nR∕n0sin θi.In
the same way, if the incident light is LCP, there will be
a reflected RCP ray and a reflected LCP ray, and the
reflected angle of the reflected LCP ray will equal the
incident angle. The reflected angle of reflected RCP ray
can be obtained by sin θrR ¼nL∕nRsin θrL.
According to the geometric optics model for rainbows of
isotropic spheres[3], the first-order rainbow is formed by
transmitted rays after one internal reflection of the inci-
dent light ray. Similarly, to explain first-order chiral
sphere rainbows and calculate the rainbow angles, we need
to determine the transmitted rays after one internal reflec-
tion. According to the illustration of the light interaction
with the chiral interface in Fig. 1, it can be inferred that a
linearly polarized ray incident at the surface of a chiral
sphere generates a refracted RCP ray and a refracted
LCP ray inside the sphere. Each refracted ray inside
the sphere generates two reflected rays after one internal
reflection. Thus, there are four total transmitted rays
outside the sphere after one internal reflection.
A sketched map of the light rays interacting with a chi-
ral sphere is illustrated in Fig. 2. The two refractive indices
of the chiral sphere are nRand nL. The refractive index of
the surrounding medium is n1. In the figure, A, B, C, D, E,
F, and G denote points of intersections of rays and the
sphere’s surface. O denotes the center of the sphere.
Figure 2(a) illustrates ray AC (the refracted RCP ray) and
its following traces. Figure 2(b) illustrates ray AB (the
refracted LCP ray) and its following traces. In Fig. 2(a),
AC generates a reflected RCP ray and a reflected LCP
ray, respectively denoted by CG and CE. Then, CG gen-
erates transmitted ray RR outside the sphere, and CE
generates transmitted ray RL. In Fig. 2(b), AB generates
reflected RCP ray BF and reflected LCP ray BD, produc-
ing transmitted rays LR and LL, respectively.
By applying the reflection and refraction laws of light at
the chiral interface presented above, ray traces in the chi-
ral sphere can be determined. In Fig. 2, the incident angle
is denoted by θi; the refracted angle of the refracted RCP
ray, i.e., ∠OAC, is denoted by θR, and the refracted angle
of the refracted LCP ray, i.e., ∠OAB, is denoted by θL.
According to Eq. (1), the three angles satisfy the following
relations: n1sin θi¼nRsin θR¼nLsin θL. Inside the
sphere, ∠OCA ¼∠OAC ¼θR. Thus, considering the
internal reflection at point C, the incident angle is θR.
According to Eq. (3), the two reflected angles follow:
∠OCG ¼θR,∠OC E ¼θL. Then at point G and point
E, the incident angles are θRand θL, respectively. Sub-
sequently, the corresponding refracted angles are found
to be both equal to θi. Thus, rays RR and RL can be
determined. Rays AB, BF, BD, LR, and RL in Fig. 2(b)
can be determined similarly.
With the above angles determined, the directions of the
final transmitted rays can be computed. The transmitted
directions are specified by the deviation angles of the
transmitted rays from the positive z-axis, denoted by
θRR,θRL ,θLR, and θLL. According to the illustrations in
the figure, these angles can be readily derived using the
following:
θRR ¼2θiþπ−4θR;(4)
θRL ¼2θiþπ−2θR−2θL;(5)
θLR ¼2θiþπ−2θR−2θL;(6)
θLL ¼2θiþπ−4θL;(7)
where θR¼arcsinðn1∕nRsin θiÞand θL¼arcsinðn1∕
nLsin θiÞ.
By comparing ray traces in Figs. 2(a) and 2(b), we have
∠AOC ¼∠BOF,∠COE ¼∠AOB, thus obtaining
•
O
Incident light
i
θ
i
θ
i
θ
24
LL i L
θθπθ
=+−
i
θ
z
1
n
,
RL
nn
i
θ
L
θ
R
θ
L
θ
222
LR i R L
θθπθθ
=+− −
LCP
L
θ
L
θ
R
θ
LL
LR
A
B
D
F
C
RCP
θ
LL
θ
LR
(b)
O
Incident light
i
θ
i
θ
z
1
n
,
RL
nn
RL
RR
R
θ
R
θ
R
θ
•
L
θ
R
θ
L
θ
222
RL i R L
θθπθθ
=+− −
24
RR i R
θθπθ
=+−
i
θ
i
θ
θ
RR
θ
RL
A
C
G
E
i
θ
B
RCP
LCP
L
θ
(a)
Fig. 2. Light interaction with the chiral sphere. (a) AC and its
following traces. (b) AB and its following traces.
COL 13(12), 121602(2015) CHINESE OPTICS LETTERS December 10, 2015
121602-2
∠AOC þ∠COE ¼∠AOB þ∠BOF;(8)
which implies that point E and point F are superposed.
From Eqs. (5) and (6), it can be seen that θRL ¼θLR;
therefore, RL and LR are also superposed. This means
that the number of final transmitted rays after one inter-
nal reflection is three, instead of four. The traces of a
linearly polarized incident ray impinging on a chiral
sphere are depicted in Fig. 3.
Taking a further step past Eqs. (4)–(7), the transmitted
directions of the three transmitted rays are derived as
functions of the incident angle:
θRR ¼2θiþπ−4 arcsinðn1∕nRsin θiÞ;(9)
θLL ¼2θiþπ−4 arcsinðn1∕nLsin θiÞ;(10)
θRL∕LR ¼2θiþπ−2 arcsinðn1∕nRsin θiÞ
−2 arcsinðn1∕nLsin θiÞ:(11)
As the incident angle varies from 0 to π∕2, the three
deviation angles, θRR,θLL , and θRL , pass through their
own extrema, which form the first-order rainbows. For
θRR and θLL, the extrema can be readily determined by
the following expressions:
θext
RR;LL ¼2θRR;LL
iþπ−4 arcsinðn1∕nR;Lsin θRR;LL
iÞ;
(12)
θRR
i¼arcsin
4
3
−
1
3ðn1∕nRÞ2
s;(13)
θLL
i¼arcsin
4
3
−
1
3ðn1∕nLÞ2
s;(14)
where θext
RL, the extremum of θRL , can be computed by
numerical methods.
The extrema of the three deviation angles, θext
RR,θext
LL , and
θext
RL, construct the three rainbow angles of a chiral sphere.
From Eqs. (9)–(11), it can be seen that formulations of the
rainbow angles for RR and LL are the same as that of a
general isotropic sphere. Both AC and CG are RCP rays.
For traces including incident light, AC, CG, and RR, the
chiral sphere can be regarded as an isotropic sphere with
refractive index nR. For traces including incident light,
AB, BD, and LL, the chiral sphere can be regarded as
an isotropic sphere with refractive index nL. It seems that
θext
RR and θext
LL are just the rainbow angles of an isotropic
sphere with refractive index nRand nL, respectively. How-
ever, the rainbow angle θext
RL is determined by both nRand
nL. Besides, note that according to Eq. (14), nR;L<2n1is
a necessary condition for the rainbow phenomenon of a
chiral sphere.
In order to examine the model, we calculate the rainbow
angles by using the above expressions and compare them
with the rainbow structure calculated by analytical solu-
tions. Figure 4presents the rainbow structures of a chiral
sphere in Ref. [8], along with the rainbow angles calculated
by the geometric model in this Letter. The three rainbows
are named the “right rainbow,”“middle rainbow,”and
“left rainbow,”according to their relative positions. The
radius of the sphere is 500 times the light wavelength,
and nc¼1.33. The chirality parameter is κ¼0.08, and
surrounding medium is the vacuum. Thus, we have nR¼
1.41, nL¼1.25, and n1¼1.0. The rainbow angles, i.e., the
extrema of θRR,θRL , and θLL based on Eqs. (9)–(11), are
computed as: θext
RR ¼147.93°, θext
RL ¼136.50°, and θext
LL ¼
124.10°. As we know, there is little difference between
the peak angle of the rainbow structure and the rainbow
angle. It can be seen that the three rainbow angles com-
puted by the geometric model agrees well with the analyti-
cal solutions. As light propagates inside the chiral sphere
in circularly polarized modes, rainbows occur in both the
E-plane (ϕ¼0°) and the H-plane (ϕ¼90°), which are dif-
ferent from those of an isotropic sphere. We also calculate
the rainbow angle of an isotropic sphere (κ¼0.0), which
denoted by θIso in Fig. 4(b).θIso ¼137.48° shows that the
•
O
Incident light
i
θ
i
θ
z
1
n
,
RL
nn
LCP
RCP
LL
RL
LR
RR
A
B
C
D
()EF
G
Fig. 3. Final ray traces in the chiral sphere.
105 110 115 120 125 130 135 140 145 150 155
0
2
4
6
8
10
12
14
16
18
Left rainbow
Middle rainbow
σ/λ
2
(x10
−5
)
Scattering angle θ (φ=0
ο
)
Mie
θ
ext
RR
θ
ext
RL
θ
ext
LL
124.10
o
136.50
o
147.93
o(a)
Right rainbow
105 110 115 120 125 130 135 140 145 150 155
0
2
4
6
8
10
12
14
16
18
Left rainbow
(b)
Mie
θ
ext
RR
θ
ext
RL
θ
ext
LL
θ
Iso
147.93
o
137.48
o
136.50
o
124.10
o
σ/λ
2
(
x
10
−5
)
Scattering angle
θ (φ=90
ο
)
Right rainbow
Middle rainbow
Fig. 4. Rainbow angles compared with analytical solutions.
a¼500λ,εr¼1.7689, μr¼1.0 (i.e., nc¼1.33), and κ¼0.08.
(a) E-plane. (b) H-plane.
COL 13(12), 121602(2015) CHINESE OPTICS LETTERS December 10, 2015
121602-3
rainbow angle of the isotropic sphere is not equal to that of
the middle rainbow of the chiral sphere.
It is convenient to use the geometric model to analyze
the variation of the rainbow angles. They can be com-
puted by formulas directly. Figure 5presents the variation
of the three rainbow angles of a chiral sphere with the
chirality parameter calculated by the geometric model,
compared with the peak angles of the rainbow structures
presented in Ref. [9]. It can be seen that the rainbow angles
calculated by the geometric model in this Letter achieve
good agreement with those from the analytical work. Note
that the rainbow angle of the middle rainbow varies
slightly with the chirality. In fact, the result is readily
understandable if we note that the two refractive indices
for a chiral medium are nR¼ncþκfor the RCP wave and
nL¼nc−κfor the LCP wave. For a larger chirality
parameter, the difference between nRand nLis larger.
It seems that the left rainbow angle, θext
LL , is more sensitive
to the variation of the chirality parameter than the
other two.
Figure 6shows the variation of the rainbow angles with
the parameter ncof the chiral sphere. The chirality param-
eter is κ¼0.05; therefore, nR;L¼nc0.05. ncvaries from
1.1 to 1.94 with a step of 0.01. It can be seen that as nc
increases, all three rainbow angles increase, while the
differences between them get smaller. The curves in the
figure calculated by the geometric optics model cover large
angle range, from 60° to 180°. However, according to the
analytical solutions, a rainbow only occurs in a certain
angle range, usually 90°–160°. The rainbow intensity
gradually decreases and vanishes eventually when their
positions move in a forward or backward direction.
In conclusion, we perform a primary analysis of the rain-
bow angles of a chiral sphere using a geometric optics
model. The key point of the model is that a ray incident
on the chiral sphere splits into two transmitted rays inside
the sphere and a ray inside the sphere generates two
reflected rays after the internal reflection. Based on the
model, rainbow angles can be computed directly from
formulas, instead of finding peak angles from the angular
distributions of the scattering intensity. It is convenient to
evaluate the relationship between the rainbow angles and
parameters of the sphere. Further work can be done to
study the geometric optics model for light scattering from
a chiral sphere.
This work was supported by the National Natural Sci-
ence Foundation of China (Nos. 61172031, 61308025,
61475123, and 61571355) and the Fundamental Research
Funds for the Central Universities.
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Fig. 5. Variation of rainbow angles with chirality.
1.11.21.31.41.51.61.71.81.9
60
80
100
120
140
160
180
Rainbow Angle θ (degrees)
Refractive index nc
θext
RR
θext
RL
θext
LL
θIso
κ=0.05
Fig. 6. Effects of refractive index of the chiral sphere.
COL 13(12), 121602(2015) CHINESE OPTICS LETTERS December 10, 2015
121602-4
