Download Fresnel equations

The Fresnel equations
ki
kr
Ei
Bi
Er
qi qr
Br
ni
Interface
qt
Et
Bt
nt
kt
Augustin Fresnel
Augustin Fresnel (1788-1827)
did experiments to establish the
wave theory and derived expressions
for the fraction of a light wave
reflected and transmitted by a flat
interface between two media with
different refractive indices. These
expressions now carry his name.
Augustin Fresnel
“S” and “P” polarizations
“S” polarization is the perpendicular polarization, and it sticks up out of the
plane of incidence
y
Plane of incidence (z = 0)
is the plane that contains
the incident and reflected
k-vectors.
z
x
Plane of the interface (x = 0)
(perpendicular to page)
“P” polarization is the parallel polarization, and it lies parallel to the plane
of incidence.
Fresnel equations – perpendicular polarized light (s)
We would like to compute the fraction of a light wave reflected and
transmitted by a flat interface between two media with different refractive
indices. Fresnel was the first to do this calculation. Here we do a somewhat
simpler derivation than given in the text.
ki
kr
Ei
Bi
Er
qi qr
Br
Interface
Beam geometry
for light with its
electric field perpendicular to the
plane of incidence
(i.e., out of the page)
y
ni
z
qt
Et
Bt
nt
kt
x
Boundary condition for the electric field
As you will learn or remember from Physics 320:
The tangential component of E to the interface is always
continuous, and the normal component of E to the interface is
only continuous in the absence of a surface charge.
ki
In other words, for our
configuration the total Efield is continuous because
all E-fields are in the zdirection, which is in the
plane of the interface (xz).
Bi
Ei
Er
qi qr
y
x
z
kr
Br
ni
Interface
qt
Et
Bt
Hence:
nt
kt
Ei ( x, y  0, z, t )  Er ( x, y  0, z, t )  Et ( x, y  0, z, t )
Ignoring the rapidly varying parts of the light wave
and keeping only the complex amplitudes:
E0t  E0i  E0r
Boundary conditions for the magnetic field y
Also: The normal component of B to the interface is always
continuous and the tangential component of B to the interface is
continuous in the absence of a surface current and assuming
i = r = t = 0.
Hence, in the absence of a
surface current, the total Bfield in the plane of the
interface is continuous.
ki
qi
Bi qi qi qr
Interface
Here, all B-fields are
in the xy-plane, so we
take the x-components:
Er
Ei
z
kr
Br
qt
Et
Bt
x
ni
nt
kt
 Bi ( x, y  0, z, t ) cos qi  Br ( x, y  0, z, t ) cos qr   Bt ( x, y  0, z, t ) cos qt
Fresnel equations - perpendicular polarized light (s)
Ignoring the rapidly varying parts of the light wave and keeping
only the complex amplitudes:
 B0i cos qi  B0r cos qr   B0t cosqt
but: qi  qr
  B0r  B0i  cosqi   B0t cosqt
E
E
E

n
 ni ( E0 r  E0i ) cos qi  nt E0t cos qt
v c/n
c
Substitute for E0t  E0i  E0r  ni ( E0r  E0i ) cosqi  nt ( E0r  E0i ) cosqt
also: B 
Rearranging: E0r  ni cosqi  nt cosqt   E0i  ni cosqi  nt cosqt 
Solving for E0 r / E0i yields the reflection coefficient r
Fresnel equations for perpendicular polarized light
 E0 r 
ni cos qi  nt cos qt
r  
 
 E0i  ni cos qi  nt cos qt
Amplitude reflection
coefficient
Similarly,
from previous slide: ni ( E0 r  E0i ) cos qi  nt E0t cos qt
Substitute for E0r  E0t  E0i  ni ( E0t  E0i  E0i ) cos qi  nt E0t cos qt
Solving for E0t / E0i yields the transmission coefficient t
 E0t 
2ni cos qi
t  
 
 E0i  ni cos qi  nt cos qt
Amplitude transmission
coefficient
Fresnel equations for parallel polarized light (p)
In a similar fashion the Fresnel equations for parallel polarized
light can be derived. Here we give only the result:
 E0 r  ni cos qt  nt cos qi
r||  
 
 E0i  ni cos qt  nt cos qi
Amplitude reflection
coefficient
 E0t 
2ni cos qi
t||  


E
 0i  ni cos qt  nt cos qi
Amplitude transmission
coefficient
Notational simplification using Snell's law
Using Snell’s law:
ni sin qi  nt sin qt
,
the Fresnel's equations can be written as:
sin qi  qt 
r  
sin qi  qt 
2sin qt cos qi
t  
sin qi  qt 
tan qi  qt 
r 
tan qi  qt 
2sin qt cos qi
t 
sin qi  qt  cos qi  qt 
A note of caution: the literature is not standardized and all possible sign
variations can be found. In case you are unsure, relate back to the specific
field direction from which the equations were derived.
Reflection & transmission coefficients
for an air-to-glass interface
nair 1 < nglass 1.5
(external reflection:
ni < nt)
Note:
• Total reflection at qi = 90°
for both polarizations
• Zero reflection for parallel
polarization at qP=56.3°
“Brewster's angle”
This polarization by
reflection is exploited in
numerous optical devices.
1
0
1
0
45
qi  degrees 
q P  56.3o
90
Near glancing
incidence, the
floor looks
mirrorlike
Try this with your copy of
Hecht. Hold the book
horizontally at the level
of the middle of your
eye and face a bright
source
 you will see a nice
reflection of the source
Reflection coefficients for a glass-to-air interface
nglass  1.5 > nair  1
(internal reflection: nt < ni)
1
Total internal reflection
above the "critical angle"
here: qt = p/2 and all light is
0
reflected back into the medium
(The sine in Snell's Law
can't be > 1!).
 nt
qc  arcsin 
 ni

o
  41.8

This phenomenon of total
internal reflection has many
practical applications in optics.
1
0
45
qi  degrees 
Zero reflection for parallel polarization at
Brewster's angle: q'P=33.7°. Here the
reflected light is linearly polarized in a
plane perpendicular to the incident plane.
90
Brewster windows
• Polarization by reflection via Brewster's angle is exploited in
numerous optical devices
Courtesy Thorlabs Inc.
Brewster windows are uncoated substrates which are used at Brewster's
angle. When used in a laser cavity, a Brewster window causes the laser
output to be polarized and there is no reflection.
R = 100%
0% reflection!
Laser medium R = 90%
0% reflection!
  c
I   n 0  E0
 2 
Transmittance
Transmitted Power I t At
T 

Incident Power
I i Ai
If the beam
has width wi:
wi
qi
ni
nt
qt
wt
2
A = Area
At wt cos(qt )


Ai wi cos(qi )
The beam expands in one dimension on refraction.
2
  0c 
nt
E0t
2


n
E


I t At  2 
wt
nt wt 2
t
0 t wt
T



t


2
2
I i Ai   0c 
 wi  ni E0i wi ni wi
n
E
 i 2  0i



 nt cos qt  2
T 
t
 ni cos qi 
The transmittance is also
called the transmissivity
since
E0t
E0i
2
2
 t2
  0c 
I  n
 E0
 2 
Reflectance
R
Reflected Power I r Ar

Incident Power
I i Ai
wi
ni
nt
qi qr
A = Area
wi
Because the angle of incidence = the angle of reflection,
the beam area doesn’t change on reflection.
Also, n is the same for both incident and reflected beams.
E0t
I r Ar
R

I i Ai
E0i
2
2
 r2
 R  r2
The reflectance is also called the reflectivity
2
Energy
conservation
Ii A cos qi  I r A cos qr  I t A cos qt
1
2
qi  q r , ni  nr , and I  nc 0 E0
2
ni E02i cos qi  ni E02r cos qi  nt E02t cos qt
2
 E0 r   nt cos qt  E0t 
1 
 


E
n
cos
q
E
i  0 i 
 0i   i
2

1 R T
Neglecting absorption in the medium, you can calculate T from knowledge
of R. It can also be shown that this relation holds for the individual
polarization components:
R T 1
R  T  1
Energy
conservation
So you can find perpendicular and parallel
polarization components of T from
knowledge of R:
2
 E0 r   ni cos qi  nt cos qt 
R  Ro  r  
 

E
n
cos
q

n
cos
q
i
t
t 
 0i    i
2

2
 E0 r   ni cos qt  nt cos qi 
R  Ri r  
 

 E0i   ni cos qt  nt cos qi 
2
2
2
note: in your manual,
the numerator is:
nt cos qi  ni cos qt
which doesn’t make a
difference due to the
square
Reflectance and transmittance for an
air-to-glass interface
Perpendicular polarization
1.0
Parallel polarization
1.0
T
T
.5
.5
R
R
0
0
0°
30°
60°
90°
Incidence angle, qi
0°
30°
60°
Incidence angle, qi
Note that:
R T 1
R  T  1
qB
90°
Reflectance and transmittance for a
glass-to-air interface
Perpendicular polarization
1.0
Parallel polarization
1.0
R
.5
R
.5
T
T
0
0°
30°
0
60°
90°
Incidence angle, qi
qC
0°
30°
60°
Incidence angle, qi
Note that:
R T 1
R  T  1
qC
90°
Reflection and transmission at normal
incidence
When qi = 0, the incident plane becomes undefined, and any distinction
between parallel and perpendicular components vanishes:
2
4 nt ni
 nt  ni 
R  R  R  
 and T  T  T 
2
n

n
n

n
 t i
 t i
For an air-glass interface (ni = 1 and nt = 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. The 4% has big implications for photography
lenses. It can be significantly reduced using anti reflection (AR)
coatings (later).
4% back reflection
This back reflection is of concern if you are dealing with complicated lens
systems containing 10 or 20 of these interfaces.
Polarization by reflection
A wave reflecting and
refracting at an interface
The incoming wave
drives the bound
electrons, and they in turn
re-radiate according to
the dipole radiation
pattern.
Brewster's angle revisited
From Fresnel equations, the
reflection coefficient for parallelpolarized light goes to zero for
Brewster's angle incidence:
q p  qt  90o
 qt  90o  q p
ni
qt
ni sin(q p )  nt sin(qt )
ni sin(q p )  nt sin(90  q p )
 nt cos(q p )
nt
tan(q p ) 
ni
qp qp
nt
Brewster's angle can be found from Snell's
law:
qp +qt = 90°
 nt 
q p  tan  
 ni 
1
Stick model by Robert W. Wood
This behavior can be explained from Fresnel equations, but perhaps an
easier method is to employ the stick model first proposed by Robert W.
Wood:
Consider a stick of wood in place of the electric vector. If the wood
impacts a water surface at an angle, the stick slides into the water and is
not reflected. However, if the stick lands parallel to the water surface, it
can bounce back. Because in nature, horizontal surfaces are almost
exclusively encountered, it will be a horizontal vibration that is reflected.
To minimize or cut glare, Polaroid sunglasses are designed to remove the
horizontal vibrations and transmit the electric vector that is vibrating
vertically.
Electron oscillators and Brewster’s law
We can understand Brewster's angle considering the dipole radiation pattern.
q p Brewster's angle
the reflected beam
although completely
polarized is rather weak.
http://physics.bu.edu/~duffy/
semester2/c27_brewster.html
Electron oscillators vibrate under the influence of the refracted (transmitted)
wave. When the reflected beam makes a right angle with the transmitted
beam, the reflected beam vanishes due to the scattered dipole emission
pattern.
Glare is horizontally polarized
Puddle reflection viewed
through polarizer that
transmits only horizontally
polarized light
Puddle reflection viewed
through polarizer that
transmits only vertically
polarized light
Light reflected
into our eyes
from the puddle
reflects at about
Brewster's Angle.
So parallel
(i.e., vertical)
polarization sees
zero reflection.
http://www.colorado.edu/physics/2000/applets/polarized.html
Polarizer sunglasses transmit only vertically polarized light. That's why I use
them for fly-fishing so can see through the water and spot the fish.
Sir David Brewster
• Brewster’s law named after Sir David
Brewster
• Professor of physics at St. Andrews
University
• Inventor of the kaleidoscope, lenticular
stereoscope, binocular camera,
polyzonal lens, polarimeter, and
lighthouse illuminator.
Sir David Brewster
(1781-1868)
http://www.brewstersociety.com/brewster_bio.html