Fresnel Equations Cont

Lecture 11

 Total internal reflection and evanescent waves  Optical properties of metals  Familiar aspects of the interaction of light and matter Fresnel Equations: phases

Case ni < nt (air to glass) external reflection

If nt > ni , then r< 0  E  =0 r   0r    E   0i  =0 Amplitude changes to negative, equivalent to 180o phase shift at surface = = Keep in mind for later: Component normal to the plane of incidence experiences 1800 phase shift upon reflection when

ni < nt Fresnel Equations: Brewster angle

Case ni < nt (air to glass) external reflection For light polarized so that E is in the plane of incidence there will be no reflection! (Brewsters =0 angle, or polarization angle)

  tan1 nn =0 Bti 

= Note: at Brewsters angle = reflected and transmitted rays form right angle Total internal reflection

Case ni > nt (glass to air) internal reflection

n Critical angle C : the incident angle t o for which t is 90 (nin i t (for total internal reflection)

1 nt total internal reflection C  sin ni

o Since t cannot exceed 90 , 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). It can be shown that for any

=0 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 microscope (frustrated total internal reflection) Total internal reflection: example

Can the person standing on the edge of the pool be prevented from seeing the light by total internal reflection ? “There are millions of light ’rays’ coming from the light. Some of the rays will be 1) Yes 2) No totally reflected back into the water, but most of them will not.” Exercise: right angle prism Idea: Can we use total internal reflection to 45o 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

n = 1.5 n 1 glass   sin1 t  sin1  41.8o n = 1 C air ni 1.5

Conclusion: it will work 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! 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 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   Assumption: medium is continuous, J  E damping       2E 2E 2E 2E E Maxwell eq-ns lead to:      x2 y 2 z2 t 2 t

Due to damping term solution leads to y metal complex index of refraction: n~  n  in  R I x     ~ Wave equation: E  E0 cos t  ky E0 cost  ny / c

split real and imaginary terms   ~  Rewrite using exp: i tny / c i tnR y / cnI y / c E  E0e  E0e     nI y / c i tnR y / c E  E0e e   nI y / c E  E0e cost  nR y / c  Metals: absorption coefficient   E  E enI y / c cost  n y / c 0 R I amplitude decays exponentially

2 Intensity is proportional to E : y

2  nI y / c 2nI y / c I y  I0 e  I0e metal Intensity of light in metal:

y I y  I0e 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   2 2  p  It can be shown that for metals: n  1     

p - plasma frequency

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 155 nm Potassium 315 nm Rubidium 340 nm Metals: reflection

n 12  n2 Normal incidence: R  R I 2 2 nR 1  nI 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 it  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 color R G B 25 0 0 red 98 70 0 yellow 65 80 20 green 25 35 60 blue There are ~120 million receptors in your eye Equivalent to 120 Megapixel digital camera! 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: 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)