Download - Lecture11 Ch4 total internal reflection.ppt
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Chapter 4
Fresnel Equations cont.
Lecture 11
Total internal reflection and evanescent waves Optical properties of metals Familiar aspects of the interaction of light and matter
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Total internal reflection
nisin(C) = ntsin(90o)
Critical angle C : the incident angle for which t is 90o (ni<nt)
C
ni
nt
ni >nt
Since t cannot exceed 90o, there will be no transmitted beamFor i > C light is completely reflected: total internal reflection
total internal reflection
Critical angle(for total internal reflection)
i
tC n
n1sin
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Fresnel Equations: total internal reflectionCase ni > nt (glass to air), internal reflection
At some incidence angle(critical angle c) everything is reflected (and nothing transmitted).
It can be shown that for any angle larger than c no waves are transmitted into media: total internal reflection.
=0
Note:Component normal to the plane of incidence experiences no phase shift upon reflection when ni > nt
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The Evanescent WaveProblem with total internal reflection: with only two waves it is impossible to satisfy the boundary conditionsConsequence:• There must be transmitted wave even for total internal reflection• It cannot, in average, carry energy across the interfaceSolution:There is an evanescent wave that propagates along the surface whose amplitude drops off as it penetrates the less dense medium
evanescent wave
(frustrated total internal reflection)beam splitter
microscope
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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.”
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Exercise: right angle prism
Idea: use total internal reflection to construct a mirror with 100 % reflecting efficiencyDesign: right angle prism
Will it work ?
Solution:
45o
Angle of incidence is 45o. It must be larger than critical angle
nglass = 1.5nair = 1
o
i
tC n
n 8.415.1
1sinsin 11
Conclusion: it will work
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Right angle prism: applications
A periscope Binoculars
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Optical fibers use TIR to transmit light long distances.
Fiber Optics
They play an ever-increasing role in our lives!
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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
Design of optical fibers
ncore > ncladding
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Propagation of light in an optical fiber
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.
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.
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Fiber optics: applicationsApplications:Signal transmission: computers, phones etc.Laser surgeryEndoscope
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Fiber optics: applicationsDecorative
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Optical properties of metals
Metal: sea of ‘free’ electrons.
Electrons will move under E - electric current: EJ
conductivity
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 .
E = E0 cos t
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Optical properties of metals
Assumption: medium is continuous, EJ
Maxwell eq-ns lead to:tE
tE
zE
yE
xE
2
2
2
2
2
2
2
2
damping
Due to damping term solution leads to complex index of refraction: IR innn ~
x
y metal
cyntEkytEE /~coscos 00
Wave equation:
Rewrite using exp: cyncynticynti IReEeEE //0
/~0
split real and imaginary terms
cynticyn RI eeEE //0
cynteEE RcynI /cos/
0
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Metals: absorption coefficient
cynteEE RcynI /cos/
0
amplitude decays exponentially
Intensity is proportional to E2:
cyncyn II eIeIyI /20
2/0
yeIyI 0
Intensity of light in metal:
cnI /2 absorption coefficient:
y
I
metal
Intensity will drop e times after beam propagates y=1/: 1/ - skin or penetration depth
Example: copper at 100 nm (UV): 1/=0.6 nmat 10,000 nm (IR): 1/=6 nm
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Metals: dispersion
It can be shown that for metals: 2
2 1
pn
p - plasma frequency
For < p n is complex, i.e. light intensity drops exponentiallyFor > p n is real, absorption is small - conductor is transparent
Example: Critical wavelengths, p = c/p
Lithium 155 nmPotassium 315 nmRubidium 340 nm
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Metals: reflection
22
22
11
IR
IR
nnnnR
nI depends on conductivity. For dielectrics nI is small (no absorption)
Normal incidence:
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Light: wavelength and colorTypically light is a mixture of EM waves at many frequencies:
i
iiiii
inet rktEEE cos0
Power of waves of each wavelength forms a spectrum of EM radiation
Sun spectrum:Mixture of all wavelengths is perceived by people as ‘white’ light.
I()
How do we see colors?
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Scattering and color
Water is transparent, vapor is white: diffuse reflection from dropletsWhite paint: suspension of colorless particles (titanium oxide etc.)Scattering depends on difference in n between substances: wet surfaces appear darker - less scatteringOily paper - scatters less
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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 wavelengthColor vision - three kinds of ‘cones’, B&W - ‘rods’
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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.
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How film and digital cameras work
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Most digital cameras interleave different-color filters
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The Eye: a digital camera?
There are ~120 million receptors in your eyeEquivalent to 120 Megapixel digital camera!
Brain interprets each combination of three signals from R, G and B receptors (cones) as a unique color
Signal colorR G B25 0 0 red98 70 0 yellow65 80 20 green25 35 60 blue
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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!
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RGB vision
Suppose we think that light is yellow.What wavelength is it?
R G B98 80 0 yellow
Is = 560 nm ?
Lets mix two light waves at 650 nm and 530 nm in proportion 1.9:0.73R=25×1.9 + 70×0.73 98G= 0×1.9 + 95×0.73 80
It will be indistinguishable from yellow!
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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)
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Computer monitors
LCD display
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CMY: subtractive colorationUse white light and absorb some spectral componentsPrimary colors for additive mixing: Cyan, Magenta, Yellow
Any picture that is to be seen in ambient white light can be paintedusing these three colors.Color printer: uses CMYK - last letter stands for Black (for better B&W printing)
Cyan - absorbs red
Magenta - absorbs green
Yellow - absorbs blue
Dyes: molecules that absorb light at certain wavelengths in visible spectral range (due to electronic transitions)
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Windows look like mirrors at night (when you’re in a brightly lit room).
Practical Applications of Fresnel’s Equations
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).
Indoors Outdoors
Iin
RIin
Iout
RIoutTIout
TIin
Iin >> Iout R = 8% T = 92%
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Lasers use Brewster’s angle components to avoid reflective losses:
Practical Applications of Fresnel’s Equations
R = 100%R = 90%Laser medium
0% reflection!
0% reflection!