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

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

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

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

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

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

Right angle prism: applications

A periscope Binoculars

Optical fibers use TIR to transmit light long distances.

Fiber Optics

They play an ever-increasing role in our lives!

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

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.

Fiber optics: applicationsApplications:Signal transmission: computers, phones etc.Laser surgeryEndoscope

Fiber optics: applicationsDecorative

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

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

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

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

Metals: reflection

22

22

11

IR

IR

nnnnR

nI depends on conductivity. For dielectrics nI is small (no absorption)

Normal incidence:

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?

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

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’

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?

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

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 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!

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 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)

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%

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!