EM scattering 9 new - Sharifee.sharif.edu/~emscattering_ms/Lecture 9B.pdfGeometrical optics 30...
Transcript of EM scattering 9 new - Sharifee.sharif.edu/~emscattering_ms/Lecture 9B.pdfGeometrical optics 30...
Geometrical optics 13
Geometrical optics
� Consider the wave equation for the electric field propagating in a
source-free, homogeneous medium
0� � �E
2 2k � �� �
� Luneberg-Kline high frequency approximation:
2 2 0k� � �E E
� � � � � �� �0
; exp mm
m
jkj
� ��
�
�
� �� �� ��E r
E r r
� � : real function of position� r
Geometrical optics 14
Geometrical optics
� We substitute this equation into the wave equation, and
separate different orders in 1/j�
� 0th order:2
1� � �� �� � � �
� 1st order terms:
� �2
0 0 02
��
�� �� � �E E
0 0�� � �E
Eikonal equation
Geometrical optics 15
Geometrical optics
� Higher order terms
� �2
212 2m m m
c�� �
�� �� � � �E E E
1m mc� �� � � � �E E
1c
���
Geometrical optics 16
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)
� 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
Constant phase surfaces
��
Curves normal to constant phase surfaces
Geometrical optics 17
Geometrical optics
� �( ) ( ), ( ), ( )x y z� � � �� �r r
� Unit vector tangential to the curve
� Imagine that we parametrize these curves in the 3D space:
x
y
z
( )�r
( ) ( )
ˆ( ) ( ) ( )
d d
d dd d d
d d d
�
� �� �� � �� � �
� ��
r r
er r r
ˆ�e
� Length along the curve: 2
1
12
( )dl d
d
�
�
��
�� �
r
Geometrical optics 18
Geometrical optics
( )s s ��
� Actually, s is the length measured along
the line (with respect to a reference)
� Imagine that we parametrize these curves by changing to a
different parameter s such that
x
y
z
( )sr
( )ds d
d d
�� ��
r
2
1
12 2 1
s
s
l ds s s� � ��
d d ddl d ds ds
d d ds
��
� �� � �
r r
Geometrical optics 19
Geometrical optics
� With this new parameter:
x
y
z
( )sr
1d d d
ds d ds
��
� �r r
� �( ) ( ), ( ), ( )s x s y s z s� �r r
( )ˆs
d s
ds�r
e
ˆse
� At each point in space we have a unit vector which is
along such a curve and normal to a surface of constant �
ˆse
Constant phase surface
ˆs �� � �e Because the gradient had a unit length!
Geometrical optics 20
Geometrical optics
� Look at the change of the unit
tangential vector along the curve
� �ˆˆ ˆs
s ss
�� ��
�e
e e
x
y
z
( )sr
ˆse
Constant phase surface
� Next use the relation
� � � � � � � � � �� � � �� � �� � � � � � � � �a b a b b a a b b a
� � � � � �2 2� � � � �� � � � �a a a a a a
Geometrical optics 21
Geometrical optics
ˆ0s
s
��
�e
� This means that the curves perpendicular to the surfaces of
constant phase are straight lines: they have no curvature
� These lines are called rays
� It follows that
Geometrical optics 22
Geometrical optics
� Starting from a certain wave front corresponding to �=�1, look at
the change in � along each ray:
� Thus, another surface of constant �
can be constructed but starting from
�=�1 and drawing normal rays each
with the same length
� Note: each ray is normal to both
surfaces
� �2 2
1 1
2 1 1 1 2 1ˆ ˆ ˆs s
s s s
s s
ds ds s s� � � � �� � �� � � � � � �� �e e e
2 1 l� �� �
1�
l
2 1l s s� �
Geometrical optics 23
Geometrical optics
� Now let us look at the 1st order equations which we write as
20
0 02s
�� �� �
�E
E
0ˆ 0s � �e E
� From 1st Maxwell equation it follows for the lowest order
magnetic field that
0 0ˆs �� �e E H
� Locally, this is like a TEM plane wave propagating along s
(along the ray). We neglect higher order terms.
Geometrical optics 24
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.ˆ
se
0E
0H
� The amplitude of the electric field
vector changes in space. But its
polarization does not change
along a ray (follows from the ray
equation for the electric field)
Geometrical optics 25
Geometrical optics
� Example: plane waves along z:
2 00ˆ ˆ, , 0 0 : constantsz
z� �
�� � � � � � �
�E
e z E
� Example: cylindrical waves
� � � �0 02 00 0
0
1 1ˆ ˆ, , 0
2 /s
�� � � �
� � � � ��
� � � � � � � � ��
EEe ρ E E
� Spherical waves:
� � � �0 02 00 0
0
2 1ˆ ˆ, , 0
/s
rr r
r r r r r� �
�� � � � � � � � �
�EE
e r E E
Geometrical optics 26
Geometrical optics
� In general, however, a surface may be none of these
� To see how one can still solve the ray equation note that
20
0 02s
�� �� �
�E
E 2 ˆs�� � � � e
rays
� 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 2 1s s l� �� � � �
Central ray
Geometrical optics 27
Geometrical optics
� In this volume in space, let us use the divergence theorem
ˆ ˆ ˆs s
v S
dV dS� � � �� ��e e n
� Contribution from side walls is zero, only
the top and bottom surfaces contributeˆ ˆ
s�n e
ˆ ˆs� �n e
2�
1�2 1ˆs
v
dV� � � � � �� e
� If the volume is small:
� �2 1ˆ ˆ ˆ2s s s
v v
ldV dV� � � � � � � � �� �� �e e e
l
Geometrical optics 28
Geometrical optics
� Consequently: ˆ ˆs�n e
ˆ ˆs� �n e
2�
1�� Recall that
2 10
2 1
2ˆ lims l l�
� �� � �� � � � �� � �� �e
2 1l s s� �
1 1 2 2 2 1( ), ( ),s s l s s� � � � � � � �
1ˆs
d
ds
�� � �
�e �: Small element of the surface
around the central ray
Geometrical optics 29
Geometrical optics
� The ray equation now becomes:
0 10 0 0 1
( )10 ( ) ( )
2 ( )
sds s
s ds s
� ��� � � �
� � �E
E E E
� 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 30
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
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
along v (constant u,s) are perpendicular to
each other at each point
s
u
v
Geometrical optics 31
Geometrical optics
� What is the area of a small segment determined by du,dv?
,2ud�
s
,2vd�
,1ud�,1vd�
1 ,1 ,1u vd d� � � �
2 ,2 ,2u vd d� � � �0 0 2( , , )u v s
0 0 1( , , )u v s
0 0 2( , , )u v dv s�
0 0 2( , , )u du v s�
Geometrical optics 32
Geometrical optics
� To find the electric field we need to
know
� Consider the ratio . If the
two rays passing through the endpoints
of the line segments lie in the same
plane, the calculation of the ratio is easy
,2ud�
,1ud�
,2 ,1/u ud d� �
2 1( ) / ( )s s� �� �0 0ˆ ,s u ve
� �0 0ˆ ,s u du v�e
,1u�
,1 2 1u s s� � �
,2 ,1 2 1
,1 ,1
u u
u u
d s s
d
��� �
��
�
Center of curvature
Geometrical optics 33
Geometrical optics
� But for an arbitrary coordinate
system there is no guarantee that
the two ray cross each other. They
may not lie on the same plane.
� The only way to ensure this is that
� �0 0ˆ ,s u ve
� �0 0ˆ ,s u du v�eˆ
ue
� � � � � �0 0 0 0 0 0ˆ ˆ ˆ, , , 0s s uu du v u v u v� � � �� �� �e e e
0 0
ˆˆ ˆ 0 at ,s
s u u vu
�� �� � �� ��� �
ee e
0 0
ˆˆ 0 at ,s
v u vu
�� �
�ee
ˆve
Geometrical optics 34
Center of curvatureDifferent from the other one!
Geometrical optics
� Similarly, we demand that � �0 0ˆ ,s u ve
� �0 0ˆ ,s u v dv�e
0 0
ˆˆ ˆ 0 at ,s
s v u vv
�� �� � �� ��� �
ee e
ˆue
ˆve
0 0
ˆˆ 0 at ,s
u u vv
�� �
�ee
� 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. ,1v�
,1 2 1v s s� � �
Geometrical optics 35
Geometrical optics
� Now, in this coordinate system we
need to know the distances to the
point of coincidence of the rays for
the reference surface (s1) and from
there we get
ˆue
ˆveˆ
se
,2 ,2 ,1 2 1 ,1 2 12
1 ,1 ,1 ,1 ,1
u v u v
u v u v
d d s s s s
d d
� �� �
� �� �� � � ��� � � �� �� �� �� � �� �
� �
� � ,1u�,1v�
� These distances are the principal
radii of curvature of the surface at
the point of coincidence with the ray
Geometrical optics 36
Geometrical optics
� The principal radius of curvature of the
surfaces can also be expressed as
1 1,1 ,1 ,1
,1
ˆu u uu su
u
d d dd
d du du du u
��
�
� ��� �� � �� � �� �
� � � e
,1ud �
ˆse
,1u� ,1v�� Note that: ud�
vd�
1 1,1 ,1 ,1
,1
ˆv v vv sv
v
d d dd
d dv dv dv v
��
�
� ��� �� � �� � �� �
� � � e,1vd �
,1 ,1,u vu v
d dh h
du dv� �
� �Metric coefficients!
Geometrical optics 37
Geometrical optics
� Alternatively, if instead of u and v we use the length parameter
along these curves, then
11ˆ ˆu s su
u
d
du u�
��� �� �
� ��
�
e e11ˆ ˆv s s
vv
d
dv v�
��� �� �
� ��
�
e e
Geometrical optics 38
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:
� � � � � �0,; exps s sjk� �� �� �� �E r E r r
20,
0, 02
s sss
�� �� �
�E
E 0,ˆ 0s s� �e E
Geometrical optics 39
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 40
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 41
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
Tangent plane
iksk
� We can now define the
polarizations in terms of the
normal to the local tangent plane
and the plane of reflection, as in
case of a plane wave
Geometrical optics 42
Geometrical optics
� Use local system of coordinates defined by
unit vectors normal and tangent to local
tangent plane
Tangent plane
ik
sk� Reflected field:
n̂
,1ˆte ,2ˆ
te
,2
ˆ ˆˆ
ˆ ˆi
t
i
��
�
k ne
k n
,1 ,2ˆ ˆ ˆt t� �e e n
iE
rE� � � � � �,1 ,1 ,2 ,2
ˆ ˆ ˆ ˆ ˆ ˆi t t i t t i i� � � � � �E e e E e e E n n E
� �� � � �
,1 ,1
,2 ,2
ˆ ˆ
ˆ ˆ ˆ ˆ
r t t i
t t i i
� � �
� � � �
E e e E
e e E n n E
Geometrical optics 43
Geometrical optics
� Can be written as
Tangent plane
ik
skn̂
,1ˆte ,2ˆ
te
iE
rE
,1 ,1 ,2 ,2ˆ ˆ ˆ ˆ ˆ ˆ r i t t t t� � � � � �E R E R e e e e nn
,2
ˆ ˆˆ
ˆ ˆi
t
i
��
�
k ne
k n,1 ,2ˆ ˆ ˆt t� �e e n
� 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
Geometrical optics 44
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 45
Geometrical optics
� For the 2D problem consider a
narrow tube of rays hitting the
conducting surface. Locally,
assume the radius of curvature of
the incident rays to be .
� The local radius of curvature of the
surface is . Thus, locally, we
can view the surface as part of a
circular cylinder with the radius
i�
a�
a�
i�
a�
r�
Geometrical optics 46
Geometrical optics
� If follows that
i��
r��
l� 1� �
2� �
1 1
2 2
sin1
sin sin
sin1
sin sin
i i
i
r r
r
l l
l l
� �� � �
� �� � �
� �� �� �
� �� �� �
� If can be shown that
��
1
2
/ 2 / 2
/ 2 / 2i
i
� � � �� � � �
� � � �� � � �
i� �i�
1�
2�
1
2
/ 2 / 2
/ 2 / 2i
i
� � � �� � � �
� �� � � �� �� � � �
Geometrical optics 47
Geometrical optics
� We have
i��
r��
l� 1� �
2� �
1 2 2 1
2r i� � � � � �
�
� �� � � � � � �
� �
� Hence:
��
i� �i�
1�
2�1 1 2 2
cos cosr i i a il
�� � � � �
�� � �
�
Geometrical optics 48
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 49
Geometrical optics
� Limitations:
• Ray theory just an approximation (first order terms in Kline-
Luneberg 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 50
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 51
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
incidentreflected
Sharp edge
diffracted
Geometrical optics 52
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 edgeWedge�
Geometrical optics 53
Extension of Geometrical optics
� Next, to use the language of rays, we view the wedge tip as the
source of infinite many rays
Wedge�
Wedge�
Diffracted rays Incident ray
Geometrical optics 54
Extension of Geometrical optics
� To find the field along each
ray, we again use the result of
a wedge for the diffracted field:
Wedge�
Diffracted rays Incident ray
� � � � � �� � � �
20
0 0 20 0 0
sin /,
cos / cos /B BV V V
� �� � � �
� � � � � �� � � � �� �� �� �
� �� �,
0
1( , ) exp exp
4 2TM diffz
jE jk V V
k
� �� � �
� � � �� �� � � �� �� �
� �� �,
0
1( , ) exp exp
4 2TE diffz
jH jk V V
k
� �� � �
� � � �� �� � � �� �� �
0�
�
Geometrical optics 55
Extension of Geometrical optics
� These results are in line with the
ray theory equations as well: for a
cylindrical wave surfaces of
constant phase for the diffracted
rays are cylindrical and the fields
drop as
� Only the amplitude depends on
the angle of a refracted ray
Wedge�
0�
�
1 / �
Geometrical optics 56
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 57
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
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