Chapter 3 Plane EM Waves and Lasers in Distinct Media
3-1 Normal Incidence at a Plane Conducting Boundary
aˆ ni   zˆ , aˆ nr   zˆ

Ei ( z )  xˆEi 0 e  j1z
Incident plane wave: 
,
E
H i ( z )  yˆ i 0 e  j1z


Reflective wave: E r ( z )  xˆE r 0 e j1z ,



Total field: E1 ( z )  Ei ( z )  E r ( z )



∵ Conducting boundary, ∴ E1 ( z  0)  0  E r 0   Ei 0  E r ( z )   xˆEi 0 e  j1z



E1 ( z )  Ei ( z )  E r ( z )   xˆj 2 E10 sin 1 z


E
1
H r ( z )  aˆ nr  Er ( z )  yˆ i 0 e  j1z   


Ei 0
1
1
H 1 ( z )  H i ( z )  H r ( z )  yˆ 2  cos 1 z
1

Standing wave:
Zero (null) of E1 at β1z=-nπ
Maximum of H1 at z=-nλ/2
Zero (null) of H1 at β1z=-(2n+1)π/2
Maximum of E1 at z=-(2n+1)λ/4
Eg. A y-polarized plane wave, f=100MHz, amplitude=6mV/m, propagates +x direction and


impinges normally on a conducting plane at x=0. (a) Write E i , H i ,
(b) Determine the location nearest to the conducting plane when
(Sol.)   2f  2  108 , 1 

c

2
, 1 
3




E r , H r , E1 , and H 1 .

E1 =0.
0
 120 , aˆ ni  xˆ , aˆ nr   xˆ
0
2
2
2

j
x
4

j
x


j
x
3
1
10
3
3
3
(a) Ei ( x)  yˆ 6  10 e
, H i ( x)  aˆ ni  E i ( x)  zˆ
, E r ( x)   yˆ 6  10 e 3 ,
e
1
2
2



2
10 4  j 3 x 
, E1 ( x)  Ei ( x)  E r ( x)   yˆj12  10 3 sin(
x) ,
H r ( x)  zˆ
e
3
2



2x
3
10 4
2
  , x  
H 1 ( x)  H i ( x)  H r ( x)  zˆ
cos(
x) (b)
3
2

3
Eg. A right-hand circularly plane wave represented by E ( z )  E 0 ( xˆ  jyˆ )e  jz
impinges normally on a perfectly conducting wall at z=0. (a) Determine the
polarization of the reflected wave. (b) Find the induced current on the
conducting wall.



(Sol.) (a) Er ( z)  ( xˆErx  yˆEry )e jz , Ei (0)  E r (0)  0  Erx=-E0, Ery=jE0,

∴ E r ( z )  ( xˆE 0  yˆjE0 )e jz is left-hand circularly-polarized and –z-propagated.





(b) J  aˆ n  ( H 1  H 2 )  aˆ ni  H 1 ( H 2  0), aˆ ni  zˆ, aˆ nr   zˆ

 E



1
E
1
H i ( z )  aˆ ni  Ei  0 ( xˆj  yˆ )e  jz , H r ( z )  aˆ nr  Er  0 ( xˆj  yˆ )e jz  H 1 ( z )  E0 ( xˆj  yˆ )(e  jz  e jz )
0
0
0
0
0

E
J ( z )  0 ( xˆ  yˆj )(e  jz  e jz )
0
3-2 Oblique Incidences at a Plane Conducting Boundary
Case 1 Perpendicular Polarization (TE):
aˆ ni  xˆ sin  i  zˆ cos  i , aˆ nr  xˆ sin  r  zˆ cos  r

Incidence plane wave: Ei ( x, z )  yˆ Ei 0 e  j1 ( x sini  z cosi )


E
1
H i ( x, z )  [aˆ ni  Ei ( x, z )]  i 0 ( xˆ cos  i  zˆ sin  i )e  j1 ( x sini  z cosi )
1
1



∵ E1 ( x,0)  Ei ( x,0)  E r ( x,0)  0 , ∴ Er0=-Ei0, θr=θi

E r ( x, z )   yˆ E i 0 e  j1 ( x sin i  z cos  i ) , H ( x, z )  1 [aˆ  E ( x, z)]  Ei 0 ( xˆ cos   zˆ sin  )e  j ( x sin  z cos  )
r
nr
r
i
i
1
1
1
i
Case 2 Parallel Polarization (TM):
aˆ ni  xˆ sin  i  zˆ cos  i , aˆ nr  xˆ sin  r  zˆ cos  r

E
Incident plane wave: H i ( x, z )  yˆ i 0 e  j1 ( x sini  z cosi )
1


Ei ( x, z )  1 [aˆ ni  H i ( x, z )]  Ei 0 ( xˆ cos  i  zˆ sin  i )e  j1 ( x sini  z cosi )



∵ E1 ( x,0)  Ei ( x,0)  E r ( x,0)  0 , ∴ Er0=-Ei0, θr=θi


E
E r ( x, z )   Ei 0 ( xˆ cos  i  zˆ sin  i )e  j1 ( x sini  z cosi ) , H r ( x, z )  yˆ i 0 e  j1 ( x sini  z cosi )
1
i
Eg. A uniform sinusoidal plane wave Ei ( x, y )  yˆ10e  j ( 6 x 8 z ) in air is incident on
a perfectly conducting plane at z=0. (a) Find the frequency and wavelength of the


wave. (b) Determine the incident angle. (c) Find E r ( x, z ) and H r ( x, z ) of the
reflected wave.
(Sol.) TE case:
9
(a)  2  6 2  8 2        3 10 9 , f  3 10 ,    .
2
vp
(b) tan  i 
5
3
  i  37
4



(c) Er  10  Er ( x, z)   yˆ10e  j (6 x8 z ) , H r  1 aˆ nr  E r ( x, z )   1 ( xˆ 4  zˆ3 )e  j ( 6 x 8 z )
0
12
3-3 Normal Incidence at a Plane Dielectric Boundary

Ei ( z )  xˆEio e  j1z

Incident plane wave:  
Eio  j1z , Reflected wave:
ˆ
H
(
z
)

y
e
i

1

5
5

 E r ( z )  xˆE ro e j1z

E ro j1z

 H r ( z )   yˆ  e
1


 Et ( z )  xˆEto e  j 2 z

Transmitted wave:  
Eto  j 2 z
 H t ( z )  yˆ  e
2

Continuity of tangential field component at z=0
  1

Er 0  2
Ei 0
 Ei 0  E r 0  Et 0

1   2
 Ei (0)  E r (0)  Et (0)



1
Et 0  
 H i (0)  H r (0)  H t (0)  ( Ei 0  E r 0 )  
 E  2 2 E
2
 1
 t 0  2  1 i 0
E
H
  1
Reflection coefficient: Γ= r 0   r 0  2
Ei 0
H i 0  2 1
Transmission coefficient: τ=
Et 0
2 2

Ei 0  2 1
 Et ( z )  xˆEi 0 e  j 2 z

and 1+Γ=τ  
E i 0 e  i  2 z
ˆ
H
(
z
)

y
 t
2

Case 1 Γ>0 (η2>η1): Maximum at 2β1zmax=-2nπ or zmax=-nλ1/2=-nπ/β1
Minimum at 2β1zmin=-(2n+1)π or zmin=-(2n+1)λ1/4=-(2n+1)π/2β1
Case 2 Γ<0 (η2<η1): Max at zmax=-(2n+1)λ1/4. Min at zmin=-nλ1/2
Standing wave ratio (SWR): S 
E max
E min

1 
1 
 
S 1
S 1


Eg. What condition E r  Et
occurs in normal incident case? And Standing
wave ratio =?
1   2 (not reasonable )
1 
 2  1
2 2



(Sol.)
3
1
1, S
 2  1  2  1
1 
1  3 2     ,  

2
2
3-4 Oblique Incidences at a Plane Dielectric Boundary
Case 1 Perpendicular Polarization (TE):
aˆ ni  xˆ sin  i  zˆ cos i ,
aˆ nr  xˆ sin  r  zˆ cos r ,
aˆ nt  xˆ sin  t  zˆ cos t .
Incident plane wave:

Ei ( x, z )  yˆEio e  j1 ( x sini  z cosi )

,
Eio

 j1 ( x sin i  z cos  i )
ˆ
ˆ
H
(
x
,
z
)

(

x
cos


z
sin

)
e
i
i
i

1


E r ( x, z )  yˆE ro e  j1 ( x sin r  z cos r )

Reflected wave:  
,
E ro
 j1 ( x sin r  z cos  r )
ˆ
ˆ
H
(
x
,
z
)

(
x
cos


z
sin

)
e
r
r
r

1


Et ( x, z )  yˆEto e  j 2 ( x sint  z cost )

Transmitted (refractive) wave:  
,
Eto
 j 2 ( x sin t  z cos  t )
ˆ
ˆ
H
(
x
,
z
)

(

x
cos


z
sin

)
e
t
t
t

2

Eiy ( x,0)  Ery ( x,0)  Ety ( x,0)
Continuity of tangential field components at z=0  
H ix ( x,0)  H rx ( x,0)  H tx ( x,0)
Ei 0 e  j1x sini  E r 0 e  j1x sini  Et 0 e  j 2 sint

1
E
 j1 x sin i
 E r 0 cos  r e  j1x sin r )   t 0 cos  t e  j 2 x sint
 ( Ei 0 cos  i e
2
 1
1x sin i  1x sin r  2 x sin t , Ei 0  Er 0  Et 0

1
Et 0
 ( Ei 0  Er 0 ) cos i   cos t
2
 1
 sin  t  1 n1 v p 2  2

 
 , i  r

 sin  i  2 n 2 v p1  1
and 1+Γ⊥=τ⊥.

E

/
cos



/
cos

E
2

/
cos

  r 0  2
t
1
i
2
t
,    t0 
  E
 2 / cos  t   1 / cos  i
E i 0  2 / cos  t   1 / cos  i
i0

Note: Snell’s Law holds only in case of lossless media.
Brewster angle: No reflection occurs when sin  B  
if ε1=ε2, μ1≠μ2
1  1 2 /  2 1
1

2
1  ( 1 /  2 )
1  ( 1 /  2 )
(Proof) Γ⊥=0   2 cos  i  1 cos  t ,
cos  t  1  sin  t  1 
2
n1
2
n2
2
sin 2  i 
1   1 2 /  2  1
1 /  1
2
.
cos  i 
cos  i  sin  B  
1
1  ( 1 /  2 ) 2
2 /  2

Eg. A uniform plane wave Ei ( x, z )  yˆ 6e  j ( 4 x 3 z ) in medium 1 (4ε0,μ0) is incident
on a plane of medium 2 (64ε0/9,μ0) at z=0. (a) Is it in perpendicular polarization
or parallel polarization? (b) Find the frequency and wavelength of the wave in
medium 1. (c) Write down the reflected angle and the refractive angle of the
transmitted wave. (d) Write down the unit propagation vectors of the incident
wave â ni , the reflected wave â nr , and the transmitted wave â nt , respectively. (e)

Compute the reflection and the transmission coefficients. (f) Find E r ( x, z ) and



H r ( x, z ) of the reflected wave. (g) Find E t ( x, z ) and H t ( x, z ) of the
transmitted wave.
(Sol.) (a)  yˆ  xz  plane ,  Perpendicular polarization (TE)
(b)  1  4 2  3 2  5 
(c)
2

   0 4 0 
2
c
 
2
15
 10 8
, f 
5
4
sin  i
2 4
4
3
3

  sin  t  sin  i  , cos  t 
5
sin  t
1 3
4
5
(d) aˆ ni 
4 xˆ  3 zˆ
4 xˆ  3 zˆ
3 xˆ  4 zˆ
, aˆ nr 
, aˆ nt 
5
5
5
(e) 1 
0
0
 45
 60 ,  2 
(64 0 / 9)
4 0
 
 2 / cos  t  1 / cos  i
2 2 / cos  t
18
7

  ,  
 2 / cos  t  1 / cos  i 25
 2 / cos  t  1 / cos  i
25

42
(f) E r ( x, z )    yˆ 6e  j ( 4 x 3 z )   yˆ e  j ( 4 x 3 z )
25


1
7
H r ( x, z )  (aˆ nr  E r ) 
(3xˆ  4 zˆ )e  j ( 4 x 3 z )
1
1250
(g) In medium 2,  2 
15
64
20
 108   0   0 
2
9
3
16

108  j ( 4 x  3 z )
Et ( x, z )     yˆ 6e  j 2 ( x sint  z cost )  yˆ
e
25
16


 j ( 4 x z )
1
27
4
3
3
H t ( x, z )  ( aˆ nt  Et ) 
(  xˆ  zˆ )e
2
125
5
5
Case 2 Parallel Polarization (TM):
aˆ ni  xˆ sin  i  zˆ cos i , aˆ nr  xˆ sin  r  zˆ cos r
aˆ nt  xˆ sin  t  zˆ cos t
Incident plane wave:
Ei 0  j1 ( x sini  z cosi )

H i ( x, z )  yˆ  e
,
1


E ( x, z )  E ( xˆ cos   zˆ sin  )e  j1 ( x sini  z cosi )
i0
i
i
 i

E r ( x, z )  E ro ( xˆ cos  r  zˆ sin  r )e  j1 ( x sin r  z cos r )

Reflected wave:  
,
E r 0  j1 ( x sin r  z cos r )
ˆ
H
(
x
,
z
)


y
e
r

1


Et ( x, z )  Eto ( xˆ cos  t  zˆ sin  t )e  j 2 ( x sint  z cost )

Transmitted wave:  
.
Et 0  j 2 ( x sint  z cost )
ˆ
H
(
x
,
z
)

y
e
t

2

Continuity of tangential field components at z=0 
sin  t  1 v p 2  2



sin  i  2 v p1 1
E r 0  2 cos  t   1 cos  i

( Ei 0  E r 0 ) cos  i  Et 0 cos  t
||  E   cos    cos 
cos  t


i0
2
t
1
i
, and 1  ||   || 

Et 0
1
cos  i
2 2 cos  i
  E t 0 
 ( Ei 0  E r 0 )  
||
2
 1

E i 0  2 cos  t   1 cos  i
Brewster angle: No reflection occurs when sin  B|| 
If μ1=μ2   B||  tan 1
1   2  1 /  1 2
1  ( 1 /  2 )
2

1
1  ( 1 /  2 )
2
n
 tan 1 ( 1 ) .
1
n2
Eg. (a) Find θB|| and its corresponding angle of transmission. (b) A TE plane wave
is incident from air to water at θi=θB||. Find Γ⊥ and τ⊥.
sin  B||
1
 81 ,  t  sin 1 (
(Sol.) (a) εr of water is 80.  B||  sin 1
)  6.38
1  1 / 80
80
120
(b) 1  120 , 1 / cos i  2410 ,  2 
 40.1 ,  2 / cos t  40.4
80
 
40.4  2410
2  40.4
 0.967 ,   
 0.033
40.4  2410
40.4  2410

Eg. A uniform plane wave Ei ( x, z )  ( 3xˆ  4 zˆ )e  j (8 x 6 z ) in air is incident on a
plane of medium (16ε0/9,μ0) at z=0. (a) Is it in perpendicular polarization or
parallel polarization? (b) Find the frequency and wavelength of the wave in
medium 1. (c) Write down the refractive angle θt of the transmitted wave. Is θi
equal to the Brewster angle? (d) Write down the unit propagation vectors of the
incident wave â ni , the reflected wave â nr , and the transmitted wave â nt ,
respectively. (e) Compute the reflection and the transmission coefficients. (f) Find




E r ( x, z ) and H r ( x, z ) of the reflected wave. (g) Find E t ( x, z ) and H t ( x, z ) of
the transmitted wave.


1
(Sol.) (a) H i  ( aˆ ni  E )   yˆ 5e  j ( 8 x 6 z ) , ∴ Parallel polarization (TM)
0
(b) 1  8 2  6 2  10 
  3  10 9  f 
15

2

   0 0   

5
m
 10 8
4
(c)  i  sin 1 ( ), sin  B 
5
1
1 (
1
)
2

4
,  i   B
5
sin  i
2 4
4
4 3 3
3

  sin  t      t  sin 1 ( ) , cos  t 
5
sin  t
1 3
5 4 5
5
(d) aˆ ni 
(e) 1 
4 xˆ  3 zˆ
3 xˆ  4 zˆ
, aˆ nr  0 , aˆ nt 
5
5
0
0
 cos t  1 cos  i
 0,
 120 ,  2 
 90  ||  2
 2 cos t  1 cos  i
0
(16 0 / 9)
2 2 cos  i
3

 2 cos  t  1 cos  i 4


(f) Er ( x, z )  H r ( x, z )  0
 || 
(g) In medium 2,  2    0 
16
40
0 
9
3
32

 j (8 x  z )
9
 j 2 ( x sin  t  z cos  t )
3
Et ( x, z )   ||  (3 xˆ  4 zˆ )e
 ( xˆ  3zˆ )e
4
32
32


 j (8 x  z )
1
1
18
1  j (8 x  3 z )
3
ˆ
ˆ
ˆ
H t ( x, z ) 
(a nt  Et ) 
(  y )e
 y
e
2
90
5
25
Snell’s Law:
sin  t 1 n1 v p 2  2

 r1
and θr=θi




 1 
sin  i  2 n2 v p1 1
2
 r2
The reflection coefficient: Γ=
Er0
E
and the transmission coefficient: τ= t 0
Ei 0
Ei 0
 2 / cos  t  1 / cos  i n1 cos  i  n 2 cos  t sin(  t   i )

   / cos    / cos   n cos   n cos   sin(    )

2
t
1
i
1
i
2
t
t
i

2

/
cos

2
n
cos

2
cos

sin
t
2
t
1
i
i
 




 2 / cos  t  1 / cos  i n1 cos  i  n 2 cos  t
sin(  t   i )
for perpendicular polarization (TE)
 2 cos  t  1 cos  i n1 / cos  i  n 2 / cos  t tan( t   i )

||   cos    cos   n / cos   n / cos   tan(   )

2
t
1
i
1
i
2
t
t
i

2

cos

2
n
/
cos

2
cos
 i sin  t
2
i
1
t
 


||

 2 cos  t  1 cos  i n1 / cos  i  n 2 / cos  t sin(  i   t ) cos( i   t )

for parallel polarization (TM)
The media are called right-handed materials in case ε, μ,
and n are positive. All the natural media are right-handed.
However, some artificial materials, such as a few photonic
crystals, may be left-handed. It implies that the effective
values of ε, μ, and n are negative. The refractive angle θt, as
well as θi, is positive in case of obliquely-incident light passing the interface between
the distinct right-handed media. That is, the incident light and the transmitted light are
the opposite sides of the normal line of the interface.
Example of an obliquely-incident laser beam passing a slab made of left-handed
medium:
Eg. A light ray is incident from air obliquely on a transparent sheet of thickness d
with an index of refraction as shown in the figure. The angle of incidence is θi.
Find (a) θt, (b) the distance l1 at the point of exit, and (c) the amount of the
lateral displacement l2 of the emerging ray.
sin  t 1
sin  i
(Sol.) (a)
   t  sin 1 (
)
sin  i n
n
(b)  1  d tan  t 
(c)  2  12  d 2  sin( i  t ) 
d sin  i
n 2  sin 2  i
d 2 sin 2 i
d sin i cos i
 d 2  [sin i cos t  cos i sin t ]  d sin i 
2
2
n  sin i
n2  sin 2 i
Eg. A uniform plane wave in medium 1 having a refractive index n1 is incident on
a plane interface at z=0 with medium 2 having a refractive index n2 (<n1) at the
critical angle. Let Ei0 and Et0 denote the amplitudes of the incident and refracted
electric fields, respectively. (a) Find the radio Et0/Ei0 for perpendicular
polarization. (b) Find the ratio Et0/Ei0 for parallel polarization.

(Sol.) θi=θc, θt=π/2, cosθt =0. (a) τ⊥=2, (b)  ||  2( 2 )
1
3-5 Total Reflection and Critical Angle (θc)
In case of ε1>ε2 (or n1>n2):
Define critical angle as θc  sin 1 (
2
n
)  sin 1 ( 2 ) ,
1
n1
1
sin  i  1
2
While θi>θc  sin  t 
 cos  t  1  sin 2  t   j
1
sin 2  i  1
2


aˆ nt  xˆ sin  t  zˆ cos t , E t and H t  e  j 2 z cost  e  j 2 x sint = e 2 z  e  j2 s x  0 as
z→∞, where  2   2
2 2

sin  t  1 and  2 x   2 2 sin  i
1
1
Eg. An electromagnetic wave from an underwater source with perpendicular
polarization is incident on a water-air interface at θi=20°. Using μr=1 and εr=81
for fresh water, find (a) critical angle θc, (b) reflection coefficient Γ⊥, (c)
transmission coefficient τ⊥, and (d) attenuation in dB for each wavelength into
the air.
(Sol.)  2   0  120 , 1 
(a)
 c  sin 1 (
(d)  2   2 
1
)  6.38 ,
81
40
,  i  20 , cos t  1  sin 2  t  1  (9 sin 20) 2
3
(b)    2 / cos  t  1 / cos  i  e j 0.66 , (c)    1    1.89e j 0.33 ,
 2 / cos  t  1 / cos  i
1 2
sin  i  1 .
2
Attenuation per wavelength is -20log e 2 22 =159dB, where λ2=
0
81

0
9
Eg. ε of water at optical frequency =1.75ε0, find d at a distance under water yields
an illuminated circular area of a radius r=5m.
(Sol.)  c  sin 1
加水前
(by C. H. Yang)
1
 49.2 , d tan 49.2  5  d  4.32m
1.75
加水後
Eg. For preventing interference of waves in neighboring fibers and for
mechanical protection, individual optical fibers are usually cladded by a material
of a lower refractive index, as shown in the figure, where n1>n2. (a) Express the
maximum angle of incident θa in terms of n0, n1 and n2 for meridional rays
incident on the core’s end face to be trapped inside the core by total internal
reflection. (Meridional rays are those that pass through the fiber axis. The angle
θa is called the acceptance angle, sin(θa) is the numerical aperture (NA) of the
fiber.)
n
n

(Sol.) sin   0 sin  a ,       c  sin 1 ( 2 )
n1
2
n1
sin  c 
n2
2
n1
2
n sin 2 
n2
 sin   cos   1  0 2 a
n1
n1
2
n0 sin 2  a
 1
n1
n1  n2
2
2
, sin  a 
2
2
, where n0=1
n0
Eg. Determine εr1 such that light is confined within the rod.
(Sol.) sin 1  sin  c 
sin  t 
1
 r1
1
 r1
, 1 
 sin  i …(2)  1 

2
  t  cos  t 
sin 2  i
 r1

1
 r1
1
 r1
…(1)
  r1  1  sin 2  i
Eg. Assuming εr=4 for glass, calculate the percentage of the incident light power
reflected back by the prism.
1
 30  45 ,  air  120 ,  prism  60
4
2  60
2
2  120
4

 ,  prism air 

120  60 3
120  60 3
(Sol.)  c  sin 1
 air prism
2
 total
2
2
4
64

 12  12 

 79%
3
3
81
Note: The depth of focus (DOF) of an optical imaging system is usually defined as
DOF=0.5λ2/NA2.
3-6 Normal Incidence at Three-layer Dielectric Interfaces
In medium 1,

E1  xˆ ( Ei 0 e  j1z  E r 0 e j1z )

, Er 0  0  Ei 0
1

 j1 z
j1 z
ˆ
H

y
(
E
e

E
e
)
1
i
0
r
0

1

In medium 2,

E 2  xˆ ( E 2  e  j 2 z  E 2  e j 2 z )

1

  j 2 z

 E 2 e j 2 z )
H 2  yˆ  ( E 2 e
2


 E3  xˆE3  e  j3 z


In medium 3,  
E3  j3 z
ˆ
H

y
e
 3

3

∵ Continuity of tangential field components at z=0 & z=d,
∴ E1(0)=E2(0), H1(0)=H2(0), E2(d)=E3(d), H2(d)=H3(d)
 cos  2 d  j 2 sin  2 d
  j 2 tan  2 d
Define Z 2 (0)   2  3
=2  3
 2  j3 tan  2 d
 2 cos  2 d  j 3 sin  2 d
 Effective reflection coefficient Γ0=
Er 0
H
Z (0)  1
  r0  2
Ei 0
H i 0 Z 2 (0)  1
Eg. If no reflection occurs, find the relation among d, η1, η2, and η3.
0  0 
(Sol.)
Z 2 (0)  1
  2 ( 3 cos  2 d  j 2 sin  2 d )  1 2 cos  2 d  j3 sin  2 d )
Z 2 (0)  1
  1 3
 3 cos  2 d  1 cos  2 d
1  3
 2
 2
or 
 2 sin  2 d  1 3 sin  2 d cos  2 d  0
sin  2 d  0
Case 1:
Case 2:
 2  1 3 or n 2  n1 n3


(2n  1)
 2 , n  0,1,2,...
d 
4

1   3 or n1  n3


n 2
, n  0,1,2, . . .
d 
2

(Quarter-wave impedance transformer)
(Half-wave impedance transformer)
Eg. A transparent dielectric coating is applied to glass (μr=1, εr=4) to eliminate
the reflection of red light λ0=0.75μm. (a) Determine the required dielectric
constant and thickness of the coating. (b) If violet light λ0=0.42μm is shone
normally on the coating, what percentage of the incident power will be reflected?
(Sol.) (a) 1   0  120 , 3  60
(2n  1)
(2n  1) 0.75
120 120
 2  1 3 

  r  2  n2  2 , d=
2 

, n=0,1,2,…
4
4
2
r
2
3-7 Optical Theory of Multi-Layer Films
Ei  E r  E 2  E 2 '

Normal incidence: At z=0: 
,
n1 Ei  n1 E r  n2 E 2  n2 E2 '
 E2 e  jk2d  E2 ' e jk2d  E3
z=d: 
 jk2 d
 n2 E2 ' e jk2d  n3 E3
n2 E2 e

 1   1  Er
cos( k 2 d )
  
( )  


n1   n1  Ei
 jn2 sin( k 2 d )
 j sin( k 2 d ) 
   1   ( E3 )
n2
 n  Ei
cos( k 2 d )   3 
1
1  1 
Or,    
   M      , where k2=2πn2/λ, Γ=Er/Ei, τ=E3/Ei

n1   n1 
n3 
 j sin( k 2 d ) 

cos( k 2 d )

 =  M 11 M 12 
Optical transfer matrix: M 
n2

 M 21 M 22 

jn
sin(
k
d
)
cos(
k
d
)
2
2
2


M n  M 12 n3 n1  M 21  M 22 n3
2n1
   11 1
and  
M 11n1  M 12 n3 n1  M 21  M 22 n3
M 11n1  M 12 n3 n1  M 21  M 22 n3
 j sin[ k 2 d cos( i )] 

cos[ k 2 d cos( i )]

 , where p=n2cosθi
Oblique incidence: M=
p


 jp sin[ k 2 d cos( i )] cos[ k 2 d cos( i )] 
(TE), or p=n2/cosθi (TM)
Anti-reflecting film:  =0 and d=λ/4n2  M11=M22=0, M12=-j/n2, M21=-jn2
 M12n3n1=-jn3n1/n2=M21=-jn2  n2= n1 n3
Effective optical transfer matrix of n-layer film: Mt=M1M2M3…Mn.
High-reflectance film: A stack of N alternate quarter-wave layers
of high index nh and low index nl materials

0
M 

 jnl

 j 
0
nl   
 
0   jnh
( n h nl ) 2 N  1
→1.
( n h nl ) 2 N  1
 j 
0
nh   
 
0   jnl
 j 
0
nl   
 
0   jnh
  nh N

0 
 j
( n )

nh   =  l
 nl N 


0
( )
0 

nh 
Eg. Determine the effective reflectances of an eight-layer stack (N=4) and
thirty-layer stack (N=15) of ZnS (nh=2.3) and Mg2F (nl=1.35).
(2.3 1.35) 30  1 2
(2.3 1.35)8  1 2
2
2
(Sol.) (a)   [
]  0.9409 , (b)   [
]  0.999
(2.3 1.35)8  1
(2.3 1.35) 30  1
Eg. The relation between the reflectivity and the incidence angle.
3-8 Fabry-Perot Resonators
Fabry-Perot resonator: Lightwave is resonant between two parallel plates.
Path difference between two successive rays:
d
d

 cos 2  2d cos
cos  cos
Total output E-field:
Et  Ei tt' Ei tt' rr ' e j  Ei tt' r 2 r ' 2 e j 2   
where δ=2kdcosθ=
4nd cos 
0
Ei tt'
,
1  rr ' e j
is the optical phase
difference, and d is the thickness.
Total output intensity Fabry-Perot resonator:
It 
I iT 2
1  Re j
2
=
I iT 2
(1  R) 2  4 R sin 2 ( 2)
,
where
R=rr’,
T=tt’,
and
R+T+A(absorption)=1.
Transmittance of Fabry-Perot resonator:
It
T2
1
,
where


2
(1  R) 1  F sin 2 ( 2)
Ii
4R
F=
is called the coefficient of finesse.
(1  R) 2
It is a measurement of the sharpness of the
interference fringes. Note that F becomes larger
as well as R increases.
I
4nd cos 
c
N
T2
 2 N (Or, f=

( t ) max 
occurs if  
).
2
0
 0 2nd cos 
Ii
(1  R)
I
4nd cos
c
N
T2
 N (Or, f=

( t ) min 
occurs if  
).
2
0
 0 4nd cos 
Ii
(1  R)
∴ Fabry-Perot Resonator can be used as an optical spectral analyzer.
Free spectral range in frequency: f N 1  f N 
c
2nd cos 
Natural cleavage planes: Semiconductor lasers usually utilize the natural cleavage
planes as the both parallel mirrors of the Fabry-Perot resonator.
Miller indices: If a plane crossing (k,0,0), (0,h,0) and (0,0,l), then (k,h,l) is called
Miller indices of the plane. Note: ( k ,h,l) denotes (-k,h,l), (k, h ,l) denotes (k,-h,l), etc.
Note: GaAs’s natural cleavage plane is (1,1,0)-plane. Si’s and Ge’s natural cleavage
plane are (1,1,1)-plane.
3-9 Introduction to Lasers
Stimulated emission: A photon induces an electron to
fall down from a higher-energy level to a lower-energy
level. And then it generates another photon with the
same wavelength and the same phase.
Basic characteristics of lasers:
1. Alignment.
2. Small broadening angle.
3. Single wavelength.
4. High Coherence.
Lasing conditions:
1. Population inversion: A certain higher-energy level
has more electrons than a lower-energy level.
Eg. The energy-level system of a ruby laser: The
population inversion occurs between the metastable
state and the ground state.
2. Pumping systems: Utilizing current driving or
other methods to pump electrons from a lower-energy level into a higher-energy
level.
3. Optical resonators: The laser light traverses back and forth within an optical
resonator to generate the stimulated emission.
Application of lasers:
Laser Pointer
Laser Pickup Head
Laser Sight/Collimation
Laser knife in surgery
Laser acupuncture
Laser acupuncture machine
Laser scanner and its applications
3-10 General Optical Resonators and Laser Modes
General laser resonator: Two mirrors with radii R1
and R2, separation L.
L
L
Stability condition: 0  (1  )(1  )  1
R1
R2
If a laser medium is situated within an unstable
resonator, no stable laser can emit.
Laser modes: TEMmn mode if m+1 spots in the horizontal
direction and n+1 spots in the vertical direction.
Gaussian beam (TEM00 mode): Fundamental laser mode in an optical resonator or a
lenslike medium.



2 1 
2
2
2
2
2
 Ek E 0 ( 2 
 2  k ) E  ( t  2  k 2 ) E  0 .
r r z
r
z
2
Let E=ψ(x,y,z)e-jkz be a scalar field  e  jkz  t  
2

[ ' e  jkz  jke  jkz )  k 2e  jkz  0 ,
z
where ψ’=dψ/dz. Assume kψ’>>ψ”<<k2ψ  ▽t2ψ-2jkψ’=0.
Let ψ=E0exp{-j[P(z)+kr2/2q(z)]}, where r2=x2+y2.
Substitute ψ=E0exp{-j[P(z)+kr2/2q(z)]} into ▽t2ψ-2jkψ’=0
 -(k/q) 2r2-2j(k/q)- k2r2(1/q)’-2kP’=0
 (1/q)2+(1/q)’=0 and P’=-j/q.
Let 1/q=s’/s  s=az+b, q=q0+z, and P’=-j/(z+q0)  P=-jln(1+z/q0).
Set q0=jπw02n/λ and k=2πn/λ  ψ=exp{-j[-jln(1+z/q0)+kr2/2(z+q0)]} 
 w0 2 n 2 

w0
1
jk 
2
2
] ,
 exp  j[kz   ( z)]  ( x  y )  [ 2 
] , where R(z)= z 1  [
E(x,y,z)= E0
z 
w( z)
w ( z ) 2 R( z ) 



1
1
j
z
z 


w(z)= w0  1  [ 2 ]2  , η(z)= tan 1 ( 2 ) , and
.
2
q
(
z
)
R
(
z
)

nw
(
z
)

w
n

w
n
0


0
At z=0, waist: w(0)=w0, R(0)=∞, and η(0)=0  E(x,y,z=0)=E0exp[-(x2+y2)/w02]

].
Beam divergent angle of Gaussian beam: θbeam= tan 1 [
w0 n
Beam Shaper: Reshape laser beam and intensity distribution.
Transformation of the Gaussian beam propagating through an optical element—
ABCD law: q2=
Aq1  B
Cq1  D
ABCD matrix: It can be utilized
to describe a ray propagating
through an optical element
 rout   A B   rin 
r '   C D r ' 
  in 
 out  
A
For cascade of n optical elements,  t
Ct
Bt   An
=
Dt  C n
Bn   A2
…
Dn  C 2
1
A B 
Case 1 Ray passes a planar dielectric interface: 
 = 0
C D  
B2 
D2 
 A1
C
 1
B1 
D1 
0
n1 
n2 
(Proof) rout=rin, r‘out/r‘in=tanθout/tanθin≒sinθout/sinθin=n1/n2
1
A B 

 = 0
C D  
0
n1 
n2 
 A B  1 d 
n
Case 2 Ray passes a dielectric of thickness d: 
=
C D  0 1 
(Proof) rout≒rin+(d/n)tanθin= rin+(d/n)r‘in, r‘out= r‘in
 A B  1 d 
 A B  1 d 
n  . If n=1, 

=
.

=
C D  0 1 
C D  0 1 
 A B   11
Case 3 Ray passes a thin lens of focal length f: 
 = 
C D   f
0

1

(Proof) rout=rin, r‘out=tanθout≒tanθin-rin/f=r‘in-rin/f
 A B   11

 = 
C D   f
0

1

1
A B n  n
Case 4 Ray passes a spherical dielectric interface: 
1
= 2
C D   n2 R
0
n1  , where
n2 
R>0 if the surface is concave; R<0 if the surface is convex.
(Proof) rout=rin, r‘out=tanθout≒sinθout
≒(n1/n2)sinθin+(1-n1/n2)rin/R≒(n1/n2)tanθin+(1-n1/n2)rin/R
1
A B n  n
=(n1/n2)r’in+(1-n1/n2) rin/R  
1
= 2
C D   n2 R
0
n1 
n2 
 A B   1 0
Case 5 Ray is reflected by a spherical mirror: 
 =   2 1
C
D


  R
Eg. Obtain the effective focal length of the thin lens
system.
0
 1 0
 1
1 1
1 2
1 
(Sol.) M1=M7=  1
, M2=M6= 
,
M
=
,
M
=M
=

4
3
5
0 1 

 5 1
 10 1


0 1
 At
C
 t
Bt 
=M7M6M5M4M3M2M1  Ct=-1/f, f=6.3376mm
Dt 
Eg. Show that lens-maker’s formula
1
1 1
 (n  1)(  ) ,
f
R1 R2
where R1 or R2>0 if the surface is convex to right; otherwise, R1
or R2<0.
A
(Proof)  t
Ct

Bt   1
= 1  n
Dt   R
 2
1
1 1
 (n  1)(  )
f
R1 R2
0

n

 1
 n 1
 nR
1

0 
1
1  = 1  n 1  n

n   R2
R1
0  1
  1
1 = 
  f
0

1

Eg. Consider a system of two closely-contacted thin lenses, of focal lengths f1 and
 1
f2, respectively. The ABCD matrix is  1
 f
 2
Hence the effective focal length f is expressed by
0

1

 1
 1
 f
 1
0 
1
0
1
= 1

1   (  ) 1 .
  f1 f 2

1
1
1


.
f
f1 f 2
Eg. A Gaussian beam propagated a
distance d1 in free space and passed a
thin lens of focal length f. And then it
traversed a distance d2 in free space.
Determine the ABCD matrix.
A
(Sol.)  t
Ct
Bt  1 d 2   1
 1
=
Dt  0 1   f

0

1

 d2
1
1 d1  
f
0 1  =  1

  
 f
d1 d 2 
f 

d1

1

f
d1  d 2 
Theorem The stable condition of an optical
resonator, composed of two spherical mirrors of
curvature radii R1 and R2, separation L, is
L
L
0  (1  )(1  )  1 .
R1
R2
(Proof) One round trip:
2L

1
1
0
1
0





A B  2
1 L   2
1 L 
R1



=


C D  = 
1
1

2
2
2L

  R2  0 1   R1  0 1  
 (1  )
 R2 R1
R2
2L

)

R1

 2L
2L
2L 
 (1  )(1  )
R2
R1
R2 
L( 2 
Characteristic polynomial is λ2-(A+D)λ+AD-BC=λ2-(A+D)λ+1=0.
Stable resonator if  =1 and λ is imaginary: (A+D) 2-4≦0   2  2 
 0 4
4 L 4L 4 L2
 
2
R1 R2 R1 R2
L
L
L L
L2
4 L 4 L 4 L2
 
 4  0  1  
 1  0  (1  )(1  )  1 .
R1
R2
R1 R2 R1 R2
R1 R2 R1 R2
Eg. Show that the Fabry-Perot resonator is a stable resonator.
L
L
(Sol.) R1=R2=∞, (1  )(1  ) =1 fulfills the stable condition.
R1
R2
Eg. Consider two resonators: (a) R1=5cm, R2=10cm, L=20cm. (b) R1=5cm,
R2=10cm, L=3cm. Which is a stable resonator?
L
L
(Sol.) (a) (1  )(1  ) =(-3)(-1)=3>1, ∴ It is not a stable resonator!
R1
R2
(b) 0< (1 
L
L
)(1  ) =(0.4)(0.7)=0.28<1, ∴ It is a stable resonator!
R1
R2
Eg. A Gaussian beam is focused by a
lens. The waist of Gaussian beam is
2w1, which is located on the surface
of a lens of focal length f. Determine
the dimension of the waist 2w3 of the
output beam and its location l.
1
1
j
j
(Sol.) At plane 1, R1=∞,
=


2
q1 R1 nw1
nw1 2
B1   1
= 1
D1   f

A
A q  B1
At plane 2, q2= 1 1
and  1
C1q1  D1
C1
b=λ/πw12n. At plane 3, q3=
A
A2 q 2  B2
and  2
C 2 q 2  D2
C 2
0
a  jb

1  q2= a 2  b 2 , where a=1/f,

B 2  1 l 
=
D2  0 1
a
jb
l  2
2
1
1
j
j
a b
a  b2
 q3= q2+l 
=
=


2
a
b
q3 R3 nw3
2
2
nw3 2
[ 2

l
]

(
)
a  b2
a2  b2
2
 l=
a
=
a  b2
2
1 (
f
f
<f and w3=
w1 2 n
)
2
f / w1 n
1  ( f / w1 n) 2
2
≠0
1° ∵ l<f, ∴ the realistic focal length of a Gaussian beam is
less than the ideal focal length of the lens.
2° ∵ In case of very short λ, w3=
f / w1 n
1  ( f / w1 n)
2
 λ, ∴
2
we must develop the short-λ laser to read data stored in the high-density optical disk.
Otherwise, the long-λ laser can not be focused within a small range to detect a small
pit. Thus conventional VCD/DVD utilizes red (650nm) but blu-ray DVD utilizes blue
light (405nm). By the same reason, we must utilize short-wavelength light to fabricate
the masks of the IC patterns comprising tiny devices.
3-11 Gratings and Photonic Crystals
Grating: Periodical structure for optical
diffraction.
Photonic Crystals: Arrays of optical scatterers.
Eg. Comparison between 3 types of 90° bent
photonic crystal waveguides. (by K. –Y. Lee, C. –C.
Tsai 蔡佳辰, T. –C. Weng 翁宗誠, Y. –L. Kuo 郭奕
麟, C. –W. Kao 高智偉, K. –Y. Chen 陳奎元, and
Y. –J. Lin 林宇仁)