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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 cost 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: 2nI / 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 12 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!