Equation for reflection.pdf

Equation for reflection

1. 13. Fresnel's Equations for Reflection
and Transmission Incident, transmitted, and reflected beams
Boundary conditions: tangential fields are continuous
Reflection and
transmission
coefficients
The "Fresnel Equations"
Brewster's Angle
Total internal reflection
Power reflectance
and transmittance
Augustin Fresnel
1788-1827

2. Posing the problem
What happens when light, propagating in a
uniform medium, encounters a smooth interface
which is the boundary of another medium (with a
different refractive index)?
k-vector of the
incident light
boundary
n
incident
n
transmitted
First we need to
define some
terminology.

3. Definitions: Plane of Incidence and
plane of the interface
Plane of incidence(in this
illustration, the yzplane) is the
plane that contains the incident
and reflected k-vectors.
x
y
z
Plane of the interface (y=0, the xzplane) is the plane that
defines the interface between the two materials

4. Definitions: “S”and “P”polarizations
2. “P” polarization is the parallel
polarization, and it lies parallel
to the plane of incidence.
1. “S” polarization is the perpendicular polarization, and it sticks upout of the plane of incidence
The plane of the interface(y=0)
is perpendicular to this page.
Here, the plane of
incidence(z=0) is the
plane of the diagram.
x
y
z
IR
T
A key question:which way is the E-field pointing?
There are two distinct possibilities.

5. reflected light
reflecting medium
Definitions: “S”and “P”polarizations
The amount of reflected (and transmitted) light is
different for the two different incident polarizations.
Note that this is a different use of the word “polarization”
from the way we’ve used it earlier in this class.

6. n
i
n
t
i
k
r
k
t
k

i

r

t
E
i
B
i
E
r
B
r
E
t
B
t
Interface
x
y
z
Beam geometry for
light with its electric
field sticking up out of
the plane of incidence
(i.e., out of the page)
We treat the case of s-polarization first:
the xz plane (y = 0)
AugustinFresnel was the first to do this calculation (1820’s). Fresnel Equations—Perpendicular E field

7. n
i
n
t
i
k
r
k
t
k

i
r

t
E
i
B
i
E
r
B
r
E
t
B
t
Interface
Boundary Condition for the Electric
Field at an Interface: s polarization
x
y
z
In other words,
The Tangential Electric Field is Continuous
So: E
i(y= 0) + E
r
(y= 0) = E
t(y= 0)
The component of
the E-field that lies in
the xz plane is
continuous as you
move across the
plane of the interface.
Here, all E-fields are
in the z-direction,
which is in the plane
of the interface.
(We’re not explicitly writing
the x, z, and t dependence,
but it is still there.)

8. Boundary Condition for the Magnetic
Field at an Interface: s polarization
n
i
n
t
i
k
r
k
t
k

i

r

t
E
i
B
i
E
r
B
r
E
t
B
t
Interface
x
y
z

i

i
*It's really the tangential B/

, but we're using

i

t

0
–B
i(y= 0) cos

i
+ B
r
(y= 0) cos

r
= –B
t(y= 0) cos

t
The Tangential Magnetic Field* is Continuous
In other words,
The total B-field in the
plane of the interface is
continuous.
Here, all B-fields are in
the xy-plane, so we take
the x-components:

9. Reflection and Transmission for
Perpendicularly Polarized Light
Ignoring the rapidly varying parts of the light wave and keeping
only the complex amplitudes:
00 0
00 0

cos( ) cos( ) cos( )



ir t
iirr tt
EEE
BB B


0000
00 00
:
( )cos( ) ( )cos( )
Substituting for using



tirt
ir i i tr i t
EEEE
nE E nE E


00 0
()cos()cos()


ir i i tt t
nE E nE


00
/( / ) / . But and 
ir
BEcn nEc


Substituting into the second equation:

10. Reflection & Transmission Coefficients
for Perpendicularly Polarized Light

00 00
00
()cos()()cos() :
cos( ) cos( ) cos( ) cos( )
iriitrit
ri i t t ii i t t
nE E nE E
En n En n


 

 
 
Rearranging yields


00
/ 2 cos( )/ cos( ) cos( )
tiiiiitt
tEE n n n



 
00
/, is transmission coefficient Analogously, the ,
ti
EE




00
/ cos( ) cos( ) / cos( ) cos( )
riiittiitt
rEE n n n n



  
00
/ Solving for yields treflection coefficient he :
ri
EE
These equations are called the Fresnel Equationsfor
perpendicularlypolarized (s-polarized) light.

11. n
i
n
t
i
k
r
k
t
k

i

r

t
E
i
B
i
E
r
B
r
E
t
B
t
Interface
×
Fresnel Equations—Parallel electric field
x
y
z
Beam geometry
for light with its
electric field
parallelto the
plane of incidence
(i.e., in the page)
Note that the reflected magnetic fi eld must point into the screen to
achieve for the reflected wave. The x with a circle
around it means “into the screen.”
EB k


Note that Hecht
uses a different
notation for the
reflected field,
which is confusing!
Ours is better!
This leads to a
difference in the
signs of some
equations...
Now, the case of P polarization:

12. Reflection & Transmission Coefficients
for Parallel Polarized Light
These equations are called the Fresnel Equationsfor
parallelpolarized (p-polarized) light.




|| 0 0
/ cos( ) cos( ) / cos( ) cos( )
riittiitti
rEE n n n n


  


|| 0 0
/ 2 cos( )/ cos( ) cos( )
tiiiitti
tEE n n n


 
Solving for E
0r
/ E
0i
yields the reflection coefficient, r
||
:
Analogously, the transmission coefficient, t
||
= E
0t
/ E
0i
, is
For parallel polarized light,
B
0i

B
0r
= B
0t
and
E
0i
cos(

i) + E
0r
cos(

r
) = E
0t
cos(

t)

13. To summarize…
||
cos( ) cos( )
cos( ) cos( )



itti
itti
nn
r
nn




||
2cos()
cos( ) cos( )


ii
itti
n
t
nn



2cos()
cos( ) cos( )



ii
iitt
n
t
nn



cos( ) cos( )
cos( ) cos( )




iitt
iitt
nn
r
nn




s-polarized light: p-polarized light:
And, for both
polarizations:
sin( ) sin( )

iitt
nn


plane of incidence
incident
wave
transmitted wave
interface
plane of incidence
incident
wave
transmitted wave
interface
E-field vectors are red.
k vectors are black.

14. Reflection Coefficients for an
Air-to-Glass Interface
Incidence angle,

i
Reflection coefficient, r1.0
.5
0
-.5
-1.0
r
||
r
┴
0° 30° 60° 90°
The two polarizations are
indistinguishable at = 0°
Total reflection at = 90°
for both polarizations.
n
air
1 < n
glass
1.5
Brewster’s angle
r
||
=0!
Zero reflection for parallel
polarization at:
“Brewster's angle”
The value of this angle
depends on the value of
the ratio n
i/n
t:

Brewster
= tan
-1
(n
t/n
i)
Sir David Brewster
1781 - 1868
For air to glass
(n
glass
= 1.5),
this is 56.3°.

15. Incidence angle,

i
Reflection coefficient, r1.0
.5
0
-.5
-1.0
r
||
r
┴
0° 30° 60° 90°
Brewster’s
angle
Total internal
reflection
Critical
angle
Critical
angle
Total internal reflection
above the "critical angle" 
crit
sin
-1
(n
t
/n
i)

41.8° for glass-to-air
n
glass
> n
air
(The sine in Snell's Law
can't be greater than one!)
Reflection Coefficients for a
Glass-to-Air Interface

16. http://www.ub.edu/javaoptics/docs_applets/Doc_PolarEn.html
The obligatory java applet.

17. Reflectance (R)
RReflected Power / Incident Power
rr
ii
IA
IA

Because the angle of incidence = the angle of reflection,
the beam’s area doesn’t change on reflection.
Also, nis the same for both incident and reflected beams.
A= Area
2
00
0
2
c
In E






i
w
i
n
i
n
t

r
w
i
2
R
r

So:
since
2
02
2
0

r
i
E
r
E

18. Transmittance (T)
tt
ii
IA
IA

A= Area
2
00
0
2
c
In E





cos( )
cos( )
tt t
ii i
Aw
Aw



t

i
w
i
w
t
n
i
n
t
If the beam
has width w
i:
2
00
2
0
02
2
2
00
0
0
2
2
tt
tt t tt t tt
ii i iiii i
ii
c
nE
nE w
I
Awnw
Tt
c
I
AwnwnE w
nE




 
  




The beam expands (or contracts) in one dimension on refraction.
since
2
02
2
0
t
i
E
t
E






2
cos
cos
tt
ii
n
Tt
n












TTransmitted Power / Incident Power

19. Reflectance and Transmittance for an
Air-to-Glass Interface
Note that it is NOT true that: r+ t= 1.
But, it is ALWAYS true that: R + T = 1
Perpendicular polarization
Incidence angle,

i
1.0
.5
0
0° 30° 60° 90°
R
T
Parallel polarization
Incidence angle,

i
1.0
.5
0
0° 30° 60° 90°
R
T
Brewster’s
angle

20. Perpendicular polarization
Incidence angle,

i
1.0
.5
0
0° 30° 60° 90°
R
T
Reflectance and Transmittance for a
Glass-to-Air Interface
Parallel polarization
Incidence angle,

i
1.0
.5
0
0° 30° 60° 90°
R
T
Note that the critical angle is the same for both polarizations.
And still, R + T = 1

21. Reflection at normal incidence, 
i
= 0
2

ti
ti
nn
R
nn





2
4
ti
ti
nn
T
nn


When

i
= 0, the Fresnel
equations reduce to:
For an air-glass interface ( n
i
= 1and n
t
= 1.5),
R= 4% and T= 96%
The values are the same, whichever
direction the light travels, from air to
glass or from glass to air.
This 4% value has big implications
for photography.
“lens flare”

22. Windows look like mirrors at night
(when you’re in a brightly lit room).
One-way mirrors (used by police to
interrogate bad guys) are just partial
reflectors (actually, with a very thin
aluminum coating).
Disneyland puts ghouls next to you in
the haunted house using partial
reflectors (also aluminum-coated one-
way mirrors).
Smooth surfaces can produce pretty
good mirror-like reflections, even
though they are not made of metal.
Where you’ve seen Fresnel’s Equations in action

23. Optical fibers only
work because of total
internal reflection. Fresnel’s Equations in optics
R= 100%
R= 90%
Laser medium
0% reflection!
0% reflection!
Many lasers use Brewster’s
angle components to avoid
reflective losses: