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Lecture 11
Chapter 4
Fresnel Equations cont.
 Total internal reflection and evanescent waves
 Optical properties of metals
 Familiar aspects of the interaction of light and matter
Total internal reflection
nt
ni
C
ni >nt
total internal reflection
Critical angle C : the incident angle
for which t is 90o (ni<nt)
nisin(C) = ntsin(90o)
Critical angle
(for total internal reflection)
 C  sin 1
nt
ni
Since t cannot exceed 90o, there will be no transmitted beam
For i > C light is completely reflected: total internal reflection
Fresnel Equations: total internal reflection
Case ni > nt (glass to air), internal reflection
At some incidence angle
(critical angle c) everything is
reflected (and nothing
transmitted).
=0
It can be shown that for any
angle larger than c no
waves are transmitted into
media: total internal
reflection.
Note:
Component normal to the
plane of incidence
experiences no phase shift
upon reflection when
ni > nt
The Evanescent Wave
Problem with total internal reflection:
with only two waves it is impossible to satisfy the boundary conditions
Consequence:
• There must be transmitted wave even for total internal reflection
• It cannot, in average, carry energy across the interface
Solution:
There is an evanescent wave that propagates along the surface whose
amplitude drops off as it penetrates the less dense medium
evanescent wave
beam splitter
(frustrated total internal reflection)
microscope
Total internal reflection: example
Can the person standing on the edge of the pool be
prevented from seeing the light by total internal
reflection ?
1) Yes
2) No
“There are millions of light ’rays’ coming
from the light. Some of the rays will be
totally reflected back into the water,
but most of them will not.”
Exercise: right angle prism
45o
Idea: use total internal reflection to construct
a mirror with 100 % reflecting efficiency
Design: right angle prism
Will it work ?
Solution:
Angle of incidence is 45o. It must be larger than critical angle
nglass = 1.5
nair = 1
nt
1 1
 C  sin
 sin
 41.8o
1.5
ni
Conclusion: it will work
1
Right angle prism: applications
A periscope
Binoculars
Fiber Optics
Optical fibers use TIR to transmit light long distances.
They play an ever-increasing role in our lives!
Design of optical fibers
Core: Thin glass center of the fiber that carries the light
Cladding: Surrounds the core and reflects the light back into the core
Buffer coating: Plastic protective coating
ncore > ncladding
Propagation of light in an optical fiber
Light travels through the
core bouncing from the
reflective walls. The walls
absorb very little light from
the core allowing the light
wave to travel large
distances.
Some signal degradation occurs due to imperfectly constructed glass
used in the cable. The best optical fibers show very little light loss -- less
than 10%/km at 1.550 m.
Maximum light loss occurs at the points of maximum curvature.
Fiber optics: applications
Applications:
Signal transmission: computers, phones etc.
Laser surgery
Endoscope
Fiber optics: applications
Decorative
Optical properties of metals
Metal: sea of ‘free’ electrons.
E = E0 cos t
conductivity


Electrons will move under E - electric current: J  E
Ideal conductor:  =
, and J is infinite. No work is done to
move electrons - no absorption
Real conductor:  = finite. Electrons are moving against force absorption is a function of .
Optical properties of metals


damping
Assumption: medium is continuous, J  E





2
2
2
2
 E  E  E
 E
E
Maxwell eq-ns lead to:
 2  2   2  
2
x
y
z
t
t
Due to damping term solution leads to
complex index of refraction: n~  nR  inI
Wave equation:
Rewrite using exp:
y
metal
 

E  E0 cost  ky   E0 cos t  n~y / c 
 
E  E0 e
split real and imaginary terms
 i t n y / c n y / c
i t  n~y / c 
R
I
 E0 e
  n y / c i t n y / c 
R
E  E0 e I e
   n y / c
E  E0e I cos t  nR y / c 
x
Metals: absorption coefficient
   n y / c
E  E0e I cos t  nR y / c 
I
amplitude decays exponentially
Intensity is proportional to E2:
I  y   I 0 e

n I y / c 2
 I 0 e  2 n I y / c
Intensity of light in metal:
I  y   I 0e y
y
metal
absorption coefficient:   2nI / c
Intensity will drop e times after beam propagates y=1/:
1/ - skin or penetration depth
Example: copper at 100 nm (UV): 1/=0.6 nm
at 10,000 nm (IR): 1/=6 nm
Metals: dispersion
p 
2
It can be shown that for metals: n    1   
 
p - plasma frequency
2
For  < p n is complex, i.e. light intensity drops exponentially
For  > p n is real, absorption is small - conductor is transparent
Example: Critical wavelengths, p = c/p
Lithium
Potassium
Rubidium
155 nm
315 nm
340 nm
Metals: reflection
Normal incidence:
2

nR  1  nI2
R
nR  12  nI2
nI depends on conductivity. For dielectrics nI is small (no absorption)
Light: wavelength and color
Typically light is a mixture of EM waves at many frequencies:
 



Enet   Ei   E0i cos  i t  ki  r   i

i

i
Power of waves of each wavelength forms a spectrum of EM
radiation
I()
Sun spectrum:
Mixture of all wavelengths
is perceived by people as
‘white’ light.
How do we see colors?
Scattering and color
Water is transparent, vapor is white: diffuse reflection from droplets
White paint: suspension of colorless particles (titanium oxide etc.)
Scattering depends on difference in n between substances: wet
surfaces appear darker - less scattering
Oily paper - scatters less
The Eyeball
There are four kind of ‘detectors’ of light.
They are built around four kinds of organic molecules that can
absorb light of different wavelength
Color vision - three kinds of ‘cones’, B&W - ‘rods’
The eye’s response to light and color
•The eye’s cones have three receptors, one
for red, another for green, and a third for blue.
How film and digital cameras work
Most digital cameras interleave
different-color filters
The Eye: a digital camera?
Brain interprets each combination
of three signals from R, G and B
receptors (cones) as a unique color
Signal
R G
25 0
98 70
65 80
25 35
There are ~120 million receptors in your eye
Equivalent to 120 Megapixel digital camera!
color
B
0
0
20
60
red
yellow
green
blue
The eye is poor at distinguishing spectra.
Because the eye perceives intermediate colors, such as orange and
yellow, by comparing relative responses of two or more different
receptors, the eye cannot distinguish between many spectra.
The various yellow spectra below appear the same (yellow), and the
combination of red and green also looks yellow!
RGB vision
Suppose we think that
light is yellow.
What wavelength is it?
R
98
G
80
B
0
yellow
Is  = 560 nm ?
Lets mix two light waves at 650 nm and 530 nm in proportion 1.9:0.73
R=25×1.9 + 70×0.73  98
G= 0×1.9 + 95×0.73  80
It will be indistinguishable
from yellow!
RGB: additive coloration
By mixing three wavelengths we can reproduce any color!
Primary colors for additive mixing: Red, Green, Blue
Complimentary colors - magenta, cyan, yellow (one of the
primaries is missing)
Computer monitors
LCD display
CMY: subtractive coloration
Use white light and absorb some spectral components
Primary colors for additive mixing: Cyan, Magenta, Yellow
Cyan - absorbs red
Yellow - absorbs blue
Magenta - absorbs green
Any picture that is to be seen in ambient white light can be painted
using these three colors.
Color printer: uses CMYK - last letter stands for Black (for better
B&W printing)
Dyes: molecules that absorb light at certain wavelengths in visible
spectral range (due to electronic transitions)
Practical Applications of Fresnel’s Equations
Windows look like mirrors at night (when you’re in a brightly lit room).
Indoors
Outdoors
RIin
Iin
TIout
Iin >> Iout
TIin
Iout
RIout
R = 8% T = 92%
One-way mirrors (used by police to interrogate bad guys) are just
partial reflectors (aluminum-coated), and you watch while in the dark.
Disneyland puts ghouls next to you in the haunted house using partial
reflectors (also aluminum-coated).
Practical Applications of Fresnel’s Equations
Lasers use Brewster’s angle components to avoid reflective losses:
R = 100%
0% reflection!
Laser medium R = 90%
0% reflection!