Geometrical optics
�
Consider the wave equation for the electric field propagating in a
source-free, homogeneous medium
�2 E � k 2 E � 0
k 2 � � 2� �
��E � 0
�
Luneberg-Kline high frequency approximation:
�
E � r ; � � � exp �� � jk� � r � �� �
m �0
Em � r �
� j� �
m
� � r � : real function of position
Geometrical optics
13
Geometrical optics
�
We substitute this equation into the wave equation, and
separate different orders in 1/j�
�
0th order:
�� � �� � ��
�
2
�1
Eikonal equation
1st order terms:
� 2�
E0 � 0
�� � � � � E 0 �
2
�� � E0 � 0
Geometrical optics
14
Geometrical optics
�
Higher order terms
� 2�
c 2
E m � � E m �1
�� � � � � E m �
2
2
� � � E m � c � � E m �1
c�
Geometrical optics
1
��
15
Geometrical optics
�
Let us focus on the eikonal equation. It tells us that the gradient
of �, which is a vector normal to surfaces of constant phase,
always has a constant length (unity)
Constant phase
surfaces
�
Consider now the surfaces of constant �
(or phase) and the curves perpendicular to
��
these surfaces.
�
At any point on these curves, the tangential
vector to the curve is normal to a constant
� surface
Curves normal to
constant phase
surfaces
Geometrical optics
16
Geometrical optics
�
Imagine that we parametrize these curves in the 3D space:
r � r (� ) � � x (� ), y (� ), z (� ) �
�
d r (� )
eˆ� � d � �
d r (� )
d�
�
r (� )
Unit vector tangential to the curve
d r (� )
d�
d r (� ) d r (� )
�
d�
d�
Length along the curve:
l12 �
�2
�
�1
Geometrical optics
z
eˆ�
y
x
d r (� )
d�
d�
17
Geometrical optics
�
Imagine that we parametrize these curves by changing to a
different parameter s such that
s � s (� )
�
ds
d r (� )
�
d�
d�
r (s)
z
Actually, s is the length measured along
the line (with respect to a reference)
dr
dr d �
dl �
d� �
ds � ds
d�
d � ds
l12 �
s2
� ds � s
2
y
x
� s1
s1
Geometrical optics
18
Geometrical optics
�
Constant phase
surface
With this new parameter:
r (s)
r � r ( s ) � � x ( s ), y ( s ), z ( s ) �
dr
dr d�
�
�1
ds
d � ds
z
eˆs
dr ( s )
eˆ s �
ds
y
x
�
At each point in space we have a unit vector eˆs which is
along such a curve and normal to a surface of constant �
� eˆ s � � �
Geometrical optics
Because the gradient had a unit length!
19
Geometrical optics
�
Look at the change of the unit
Constant phase
surface
tangential vector along the curve
r (s)
� eˆ s
� � eˆ s � � � eˆ s
�s
�
z
eˆs
y
Next use the relation
x
� � a � b � � � a � � � b � � b � � � a � a � �� � b � � b � �� � a �
� � � a � a � � 2 � a � � � a � 2a � �� � a �
Geometrical optics
20
Geometrical optics
� eˆ s
�0
�s
�
It follows that
�
This means that the curves perpendicular to the surfaces of
constant phase are straight lines: they have no curvature
�
These lines are called rays
Geometrical optics
21
Geometrical optics
�
Starting from a certain wave front corresponding to �=�1, look at
the change in � along each ray:
s2
s2
s1
s1
� 2 � � 1 � � eˆ s � � � ds � � 1 � � eˆ s � eˆ s ds � � 1 � � s 2 � s1 �
�
� 2 ��1 � l
Thus, another surface of constant �
can be constructed but starting from
�=�1 and drawing normal rays each
with the same length l � s 2 � s1
�
l
�1
Note: each ray is normal to both
surfaces
Geometrical optics
22
Geometrical optics
�
Now let us look at the 1st order equations which we write as
� E 0 � 2�
�
E0 � 0
�s
2
eˆ s � E 0 � 0
�
From 1st Maxwell equation it follows for the lowest order
magnetic field that
eˆ s � E 0 � � H 0
�
Locally, this is like a TEM plane wave propagating along s
(along the ray). We neglect higher order terms.
Geometrical optics
23
Geometrical optics
�
Note that the wave fronts (surfaces of constant �) are, in general,
curved surfaces. The electric and magnetic fields are tangent to
such a surface, and normal to the propagation direction which is
along the ray.
�
The amplitude of the electric field
vector changes in space. But its
eˆs
E0
H0
polarization does not change
along a ray (follows from the ray
equation for the electric field)
Geometrical optics
24
Geometrical optics
Example: plane waves along z:
�
� � z , eˆ s � zˆ , � 2� � 0 �
�
�E 0
� 0 � E 0 : constant
�z
Example: cylindrical waves
E0 � �0 �
�E 0
1
1
E0 � 0 � E0 � � � �
�
�
� � � , eˆ s � ρˆ , � � �
��
2�
�
� / �0
2
�
Spherical waves:
� � r , eˆ s � rˆ , � 2� �
Geometrical optics
E 0 � r0 �
�E 0 1
2
�
� E0 � 0 � E0 �r � �
r
r
r / r0
�r
25
Geometrical optics
�
In general, however, a surface may be none of these
�
To see how one can still solve the ray equation note that
� E 0 � 2�
�
E0 � 0
�s
2
�
� 2� � � � eˆ s
Central ray
Consider small portions of two constant �
surfaces around a central ray, and “cut”
by walls of rays around the central ray.
�
Length of all ray segments:
� 2 � � 1 � s 2 � s1 � l
Geometrical optics
rays
26
Geometrical optics
�
In this volume in space, let us use the divergence theorem
� � � eˆ dV � � eˆ
s
v
�
s
� nˆ dS
S
Contribution from side walls is zero, only
nˆ � eˆs
the top and bottom surfaces contribute
� � � eˆ dV
s
� � 2 � �1
v
�
�2
l
If the volume is small:
� � � eˆs dV � � � eˆs � dV � � � eˆs
v
Geometrical optics
v
l
� � 2 � �1 �
2
�1
nˆ � �eˆs
27
Geometrical optics
�
Consequently:
nˆ � eˆs
�2
� 2 � 2 � �1 �
� � eˆ s � lim l � 0 �
�
� l � 2 � �1 �
�
l � s2 � s1
Recall that
�1
� 1 � � ( s1 ), � 2 � � ( s 2 ), l � s 2 � s1
� � eˆ s �
Geometrical optics
1 d�
� ds
nˆ � �eˆs
�: Small element of the surface
around the central ray
28
Geometrical optics
�
The ray equation now becomes:
�E 0
� ( s1 )
1 d�
�
E 0 � 0 � E 0 ( s ) � E 0 ( s1 )
�s
�(s)
2 � ds
�
To know the behavior of the field along each ray (with respect
to a reference point), we have to know how a small area
around the ray, on the constant phase surface, changes with s
Geometrical optics
29
Geometrical optics
�
We define a system of curvilinear coordinates in space, where s
is defined along the rays, an u and v are defined by curves on
the constant phase surfaces
�
Two points on a ray have different s
s
values, but the same u and v values
�
Let us restrict ourselves to orthogonal
coordinate systems where lines along s
(constant u,v), along u (constant s,v) and
v
u
along v (constant u,s) are perpendicular to
each other at each point
Geometrical optics
30
Geometrical optics
�
What is the area of a small segment determined by du,dv?
s
(u0 , v0 � dv, s2 )
(u0 , v0 , s2 )
d� u ,2
� 2 � d � u ,2 d � v ,2
d� v ,2
(u0 � du, v0 , s2 )
(u0 , v0 , s1 )
d� u ,1
Geometrical optics
�1 � d � u ,1d � v ,1
d � v ,1
31
Geometrical optics
�
To find the electric field we need to
� ( s 2 ) / � ( s1 )
know
�
eˆs � u0 , v0 �
Consider the ratio d � u ,2 / d � u ,1 . If the
two rays passing through the endpoints
eˆs � u0 � du, v0 �
d� u ,2
of the line segments lie in the same
�u ,1 � s2 � s1
plane, the calculation of the ratio is easy
d � u ,2
d � u ,1
�
�u ,1 � s2 � s1
d� u ,1
�u ,1
�u ,1
Center of curvature
Geometrical optics
32
Geometrical optics
�
But for an arbitrary coordinate
system there is no guarantee that
eˆs � u0 , v0 � eˆv
the two ray cross each other. They
eˆu
eˆs � u0 � du, v0 �
may not lie on the same plane.
�
The only way to ensure this is that
��eˆs � u0 � du, v0 � � eˆs � u0 , v0 � �� � eˆu � u0 , v0 � � 0
� �eˆs ˆ � ˆ
� �u � es � � eu � 0 at u0 , v0
�
�
�eˆs
� eˆv � 0 at u0 , v0
�u
Geometrical optics
33
Geometrical optics
�
Similarly, we demand that
eˆs � u0 , v0 �
� �eˆs ˆ � ˆ
� �v � es � � ev � 0 at u0 , v0
�
�
eˆv
eˆu
�eˆs
� eˆu � 0 at u0 , v0
�v
�
eˆs � u0 , v0 � dv �
�v ,1 � s2 � s1
It is always possible to find such a
coordinate system. Then the line segments
along u and v lie in planes passing through
the ray.
�v ,1
Center of curvature
Different from the other one!
Geometrical optics
34
Geometrical optics
�
Now, in this coordinate system we
need to know the distances to the
eˆs
point of coincidence of the rays for
eˆv
eˆu
the reference surface (s1) and from
there we get
� 2 d � u ,2 d � v ,2 � �u ,1 � s2 � s1 �� � v ,1 � s2 � s1 �
�
��
��
��
�
��
�1 d � u ,1d � v ,1 �
�u ,1
�v ,1
��
�
�
�u ,1
�v ,1
These distances are the principal
radii of curvature of the surface at
the point of coincidence with the ray
Geometrical optics
35
Geometrical optics
�
eˆ s
The principal radius of curvature of the
surfaces can also be expressed as
�1
�1
d � u ,1 � d� u �
d � u ,1 �eˆs
�u ,1 �
�
�
� �
d� u
du � du �
du �u
d � u ,1
�1
d � v ,1 � d� v �
d � v ,1 �eˆs
�
�v ,1 �
�
� �
d� v
dv � dv �
dv �v
d � v,1
�
�1
du
� hu ,
Geometrical optics
d � v ,1
� u ,1
Note that:
d � u ,1
d � u ,1
d � v ,1
dv
� hv
Metric coefficients!
d� u
� v ,1
d� v
36
Geometrical optics
�
Alternatively, if instead of u and v we use the length parameter
along these curves, then
d � u �eˆs
�u �
du �u
Geometrical optics
�1
�eˆs
�
�� u
�1
d � v �eˆs
�v �
dv �v
�1
�eˆs
�
�� v
�1
37
Geometrical optics
�
Now we know how a wave propagates in space in the ray
approximation. But happens when it hits an object? (We
restrict ourselves to perfectly conducting surfaces)
�
The reflected wave can also be represented in terms of fields
along rays:
E s � r ; � � � E 0, s � r � exp �� � jk� s � r � ��
� E 0, s
� 2� s
E 0, s � 0
�
�s
2
Geometrical optics
eˆ s � E 0 , s � 0
38
Geometrical optics
�
But how can we find the ray directions? We can use the short
wave length approximation: reflection from a conductive
surface happens as if the wave is locally reflected from a plane
surface tangent to the true surface. Then the Snell’s reflection
law is employed.
Geometrical optics
39
Geometrical optics
�
Now, along each reflected ray again the polarization does not
change and the amplitude changes according to the radii of
the curvature of the reflected wave fronts
�
If we know the polarization and radii of curvature directly after
reflection, we know on every point along the ray
�
Then, to write down the electric field at any point, we should
first see which ray (or rays) pass through that point, calculate
the electric field (polarization and amplitude) along the rays,
and add them up
Geometrical optics
40
Geometrical optics
�
How can we calculate the polarization along the reflected ray?
�
Again use the plane approximation: view the incident ray as a
plane wave (locally), and a plane tangent to the surface (at the
point of incidence) as the reflecting plane
�
Normal
We can now define the
polarizations in terms of the
normal to the local tangent plane
ki
ks
and the plane of reflection, as in
case of a plane wave
Tangent plane
Geometrical optics
41
Geometrical optics
�
eˆt ,2
Use local system of coordinates defined by
kˆi � nˆ
�
kˆ � nˆ
i
unit vectors normal and tangent to local
eˆt ,1 � eˆt ,2 � nˆ
tangent plane
E i � eˆt ,1 � eˆt ,1 � E i � � eˆt ,2 � eˆt ,2 � E i � � nˆ � nˆ � E i �
�
Reflected field:
E r � � eˆt ,1 � eˆt ,1 � E i �
� eˆt ,2 � eˆt ,2 � E i � � nˆ � nˆ � E i �
Ei
Er
n̂
ks
ki
eˆt ,1
eˆt ,2
Tangent plane
Geometrical optics
42
Geometrical optics
�
Can be written as E r � R � E i
eˆt ,1 � eˆt ,2 � nˆ
eˆt ,2
ˆˆ
R � � eˆ t ,1eˆt ,1 � eˆ t ,2 eˆ t ,2 � nn
kˆi � nˆ
�
kˆ � nˆ
i
�
Er
The polarization of the reflected
wave is thus expressed in terms
of the dyadic reflection matrix R
which itself depends on direction
of incidence and the (local) unit
normal to the surface
Ei
n̂
ks
ki
eˆt ,1
eˆt ,2
Tangent plane
Geometrical optics
43
Geometrical optics
�
So far the polarization, but how can we calculate the radii of
curvature of the reflected surface of constant phase?
�
This is a complicated problem in 3D. Let us restrict ourselves to
2D. (For 3D case see Balanis, Advanced Electromagnetic
Engineering, chapter 13)
�
Now, in 2D, consider a wave
impinging on a curved, perfectly
conducting surface
�
Each ray is reflected according to
Snell’s law
Geometrical optics
44
Geometrical optics
�
For the 2D problem consider a
narrow tube of rays hitting the
�i
conducting surface. Locally,
assume the radius of curvature of
�r
the incident rays to be � i .
�
The local radius of curvature of the
�a
surface is � a . Thus, locally, we
can view the surface as part of a
circular cylinder with the radius � a
Geometrical optics
45
Geometrical optics
�
If follows that
sin � � i
��i
1
�
�
� i � l sin �1 � l sin �1
�� r
sin � � r
1
�
�
� r � l sin � 2 � l sin � 2
�
��i
If can be shown that
�1 � � / 2 � � i � � � / 2
� i�
�i
�l
�1 �1�
� 2 � 2�
�� r
�2 � � / 2 � � i � �� / 2
�1� � � / 2 � � i� � � � / 2
� 2� � � / 2 � � i� � � � / 2
Geometrical optics
��
46
Geometrical optics
�
��i
We have
� � r � � � i � �1 � � 2� � � 2 � �1�
� 2��
�
Hence:
1
1
2��
2
�
�
�
� r � i � l cos � i � a cos � i
� i�
�i
�l
�1 �1�
� 2 � 2�
�� r
��
Geometrical optics
47
Geometrical optics
�
Thus we have all the elements to compute the electric field
along any reflected ray
�
Considering any point in space, we should trace the reflected
rays which pass through that point, compute the electric field
using the properties of the surface and the field on the
corresponding incident rays, and add up the results
�
Note that multiple reflections should also be included (reflected
ray may be again reflected by another part of the surface, etc)
�
This constitutes the geometrical optics approximation
Geometrical optics
48
Geometrical optics
�
Limitations:
•
Ray theory just an approximation (first order terms in KlineLuneberg expansion)
•
In case of caustics (points or lines where infinite rays pass through
a single point or line, e.g. in a dish reflector) the theory breaks
down: adding up the fields leads to infinity
•
Scattering from surfaces was also approximated by expressions
which are (strictly speaking) only valid for a perfectly flat surface. If
the surface changes rapidly (compared to wavelength) this
approximation is not valid.
Geometrical optics
49
Extestion of Geometrical optics
�
One of these limitations (scattering from bends or surfaces of
small radius of curvature) can be lifted to a certain extent by
extending the theory
�
The known exact results from structures such as wedges,
cylinders, and spheres can be used (in high frequency
approximation) to calculate the diffracted field in addition to the
usual incident and reflected fields
Geometrical optics
50
Extension of Geometrical optics
�
Consider a surface edge (or rapid bend): it can be considered
as the tip of a wedge. We know that in addition to incident and
reflected fields we also have a diffracted field which behaves
as if its source is on the tip of the wedge
diffracted
incident
reflected
Sharp edge
Geometrical optics
51
Extension of Geometrical optics
�
So, as in case of scattering from a slowly varying surface, we
consider the scattering to be a local phenomena
�
We 1st find the ray which hits the wedge tip, and consider that
as the incoming incident plane wave in the wedge problem
Sharp edge
Geometrical optics
�
Wedge
52
Extension of Geometrical optics
�
Next, to use the language of rays, we view the wedge tip as the
source of infinite many rays
Diffracted rays
Incident ray
Geometrical optics
�
�
Wedge
Wedge
53
Extension of Geometrical optics
Diffracted rays
Incident ray
�
To find the field along each
�0
ray, we again use the result of
�
a wedge for the diffracted field:
�
Wedge
EzTM , diff ( � ,� ) �
1
� j� � �
exp � �
exp � � jk � � �V� � V� �
�
�0
� 4 � 2k �
H zTE ,diff ( � , � ) �
1
� j� � �
exp � �
exp � � jk � � �V� � V� �
�
�0
� 4 � 2k �
V� � VB �� � �0 � , VB �� � �0 � �
Geometrical optics
sin �� 2 /� 0 �
cos �� 2 /� 0 � � cos ��� �� � �0 � /� 0 ��
54
Extension of Geometrical optics
�
These results are in line with the
ray theory equations as well: for a
�0
cylindrical wave surfaces of
constant phase for the diffracted
rays are cylindrical and the fields
drop as 1 /
�
�
�
Wedge
�
Only the amplitude depends on
the angle of a refracted ray
Geometrical optics
55
Extension of Geometrical optics
�
Remarks:
•
We considered the 2D case, in the 3D case we get a cone of
refracted rays since we have to consider the z-component of
the wave-vector of the incident and diffracted waves
•
In 3D the length of the surface edge does not have to be
infinite: the theory is still applicable since the diffraction is
considered to be local
•
To apply ray theory we replaced the incoming ray by a plane
wave hitting the edge: this is strictly speaking not correct. One
has to include the curvature of the incoming rays as well
Geometrical optics
56
Extension of Geometrical optics
�
Remarks (continued):
•
After diffraction, the rays may again be reflected or even
diffracted by other parts of the surface, this should be taken into
account
•
For a more comprehensive account see for instance:
Balanis, Advanced Electromagnetic Engineering, chapter 13
Geometrical optics
57
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