Wave Incidence

Wave Incidence
Electrical and Computer Engineering Dept.
Electrical

Light
Light traveling in air encounters the
water; another medium.

Wave incidence
Wave incidence
For many applications, like fiber optics,
app
fi
opt
it’s necessary to know what happens
to a wave when it meets a different
medium
medium.
• How much is transmitted?
transmitted
• How much is reflected back?reflected

We will look at
We will look at…
Normal incidence
ave arrives at 0o from normal
•

Standing waves

. Oblique incidence
ave
•
•
•

arrives at another angle
Snell’s Law and Critical angle
Parallel or Perpendicular
Brewster angle

eflection at Normal
eflection at Normal Incidence
x
Et

Ei
ak

Hi

Ht

Incident wave

ak

Transmitted wave

Er

y
z=0akr

Hr

Reflected wave

z

Now in terms of equations …


Incident wave

Ei
Hi


 1 z
ˆ
Eis ( z )  Eio e x

ak

Incident wave


Eio  1z
 1 z
ˆ
ˆ
H is ( z )  H io e y 
ey

1

Reflected wave


Er

It’s traveling along –z axis
akr


ˆ
E rs ( z )  E ro e  1 z x

Hr

Reflected wave


Ero  1z
 1z
ˆ
ˆ
H rs ( z)  H roe (y)  
ey

1

Transmitted wave
Et


 2 z
ˆ
Ets ( z )  Eto e x

Ht

ak

Transmitted wave


Eto  2 z
 2 z
ˆ
ˆ
H ts ( z )  H to e y 
ey

2

The total fields
The total fields
At medium 1 and medium 2



E1  Ei  Er



H1  H i  H r



E2  Et


H2  Ht

Tangential components must be
continuous
continuous at the interface



Ei (0)  Er (0)  Et (0)



H i (0)  H r (0)  H t (0)

Define
Define
Reflection coefficient,
Reflection



Note:

Ero  2  1


Eio  2  1
Transmission
Transmission coefficient,

Eto
2 2


Eio  2  1

•1+ 



•Both are
dimensionless
dimensionless
and may be
complex
• 0≤||≤1

PE 10.8
PE 10
Hz
Hz uniform plane wave Eis=10e-jz ax in
e space is incident normally on a large
ne, lossless dielectric slab (z>0) having
4o and =o.
:
Answer:
e reflected wave Ers and -3.33 ej1z x V/m,
reflected
e transmitted wave Ets.
transmitted
-j2z
6.67 e

x V/m

where 2 = 21 = 200 /3

Case 1:
Case 1:
Medium 1 = perfect dielectric
1=0

Medium 2 = perfect conductor
perfect
2=∞
Halla impedancias int.Refleccion,
Refleccion,
Transmisión
Y campos

 2  0,
  1,  0
ˆ
E1s  2 jEio sin 1 z x
ˆ
E1  2 Eio sin 1 z sin t x

http://www.phy.ntnu.edu.tw/java/waveSuperposition/waveSuperpo

The
The EM field forms a
Standing Wave on medium 1
ˆ
E1  2 Eio sin 1 z sin t x
|E1|

1  0
2Eio

z

Minima
Minima @ - 1 z  0,  , 2
Maxima @ - 1 z 
z



n



 35
2

n1

,

2

,

2

n  1,3,5

Conducting
material
material
 

Standing Wave Applets
Standing Wave Applets
http://www.phy.ntnu.edu.tw/java/waveSuperposi
tion/waveSuperposition.html
http://www.ngsir.netfirms.com/englishhtm/StatW
ave.htm
http://www.physics.smu.edu/~olness/www/03fall
1320/applet/pipe-waves.html

Case 2:
Case 2:
Medium 1 = perfect dielectric 1=0Medium 2 = perfect dielectric 2=0
If
If  2  1 ,
E1s  Eis  Ers
 j z
 j z
  0,
 Eoi (e
 e
)
 and  are real.
 E e  j z (1  e  2 j z )
1

1

1

1

oi

z max  0, 2 ,4 ,6 ...
z max  0,  ,2 ,3 ,...

(2n  1)

z max

n1


1
2

(2n  1)

n

n  0,1,2,3

Standing waves due to reflection
Standing waves due to reflection
Ei Er  Eoi (e  j1z  e  j1z )  Eoi e  j1z (1  e 2 j1z )
|E 1 |

Lossless Medium 1

1  0

Eio (1+||)

 2  1

z

2  0

z max

n

n1


1
2

n  0,1,2,3

Lossless
Medium 2

Case 3:
Case 3:
Medium 1 = perfect dielectric 1=0
Medium 2 = perfect dielectric 2=0

If  2  1 ,
  0, and  are real.
z max
z

(2n  1)
(2n  1)1
...