Week5 - Where can you get polarized light ? A. Reflection...

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How the waves of the various colors add at your eye determines colors that you see. This is possible because waves can be added and subtracted. In phase Out of phase The same principle is behind the rich colors seen from light reflected off a CD disk that can be modeled by multiple thin films. -filtrJ Ftr.t4 tilf€Kf€n€il tE Ba& 4 -la Hoa b aro-. e.rpla:n ? 1rr* oil tlt*g. sap bebbltt 'I (msi/e'n a' fliln sf ,r,,;l*^ h,1 h, .nV rU t-+2: 3->4, n;l ai* - 6 = w, .Silm coN adry 1 ?ta:o tf8,4/t'7ivu-Jt@EEffi 2L- nh Znt,= cnaA, ? 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Everyday Physics Questions to ask: • What is a CD/DVD disk made up of ? • How is information stored ? • If you put a fingerprint smudge on a CD, will it work ? • How is a Laser used to play and record on CDs/DVDs ? • Why is the sound or video from optical media free of noise ? • If you damage a spot on the CD, will it still work ? • How should you clean your valuable CDs and DVDs ? Digital Recording Sound – is produced by fluctuations in the air density (or pressure) • Represent air pressure fluctuations as electrical current. • Measure magnitude of current. • Convert current measurements to binary. • Use these binary values to represent sound. Structure of CDs(compact disks) • CDs/DVDs are typically 120 mm (4.72 in) in diameter and 12 mm (0.05 in) thick. CDs: One side is clear and smooth Other side contains layers: a thin film of aluminum protective lacquer printed label • Thin* aluminum reflective layer has tiny pits that are detected by a 780-nm laser beam. Pits are about 0.83 μm wide. 1.63 μm apart • 20,000 spiralling tracks of “pits” and “lands” on a CD. * So thin that you can see through it with a bright light The Physics 1. Shine a laser* light on CD. 2. Reflect laser light from the optical surface. 3. Measure reflected intensity to obtain information CD PhotoElectric Detector A Laser Diode (source of light) Diode Laser Source Schematic of Optical System *A laser is a source of light with almost pure wavelength and frequency. A cheap source is a semiconductor diode laser. Let’s look at it with some more detail. 1. Light waves from the laser gets reflected from “lands”. The different waves of light travel almost same distances and arrive at the same time at the detector, producing a bright intensity at the detector (Zero phase difference). 2. If a ridge is encountered, light going into the ridge travel more distance and get delayed. This results in non-zero phase difference and will produce a zero intensity at the detector. No delays produce bright intensity at the detector. “1” The delay from light reflected from the ridges produce zero intensity at the detector. “0” Intensity of light Land or Pit “1” or “0” Compare the wavelength of laser beam light with pit size: CD: λ ~ 0.780 μm (air) DVD: λ ~ 0.635 μm (air) λ ~ 0.503 μm (plastic) λ ~ 0.410 μm (plastic) pit size ~ 0.830 μm pit size ~ 0.40 μm • Note that the pit size is close to the wavelength of diode laser. If they were different by a factor of 10, the diode laser will be unable to detect the pits. • Historically, the wavelength of commercialized lasers determined the sizes of the pits, and therefore, the amount of information that can be stored in CDs. • More recently, blue lasers – which have shorter wavelengths have come to the market. With such a laser, more information can be stored in a CD/DVD since one can make pits even smaller. Unfortunately, though in the market, blue lasers are still expensive to commercialize and you will pay a premium for devices based on it. % eAhp. gtFFQr4u[r o^) inlu{uorr4, fi.dfs ptvda*r/- t4/'^ oat/Lg str;lCe a- brn'+- r e/ge * +- olass of QC: Otrc.a- hcUrd e gen ?le:fnc glrebo bardaa/ b? alte*r,utin ,bnk t bnrtf bo",ats - nfi b","rp/d:4 bA II% ofrb:, T nilt of,Lt AcM,etTrL Cwrv;d*^ sl# , P"y.iad.!- orrlo a. sl.or-n ' lfi,=f?.- ll_tn i^-l I Ylll v '1' I irrfdrt g, Vfita* & tt"* nbfrr*tcc- crr,r4^E a44" ? -' Agk (lVt^ ! A Diffraction: Bending of light around corners Area behind a geometric shadow is bordered by interference fringes Diffraction is useful in materials science Light: Particle or Wave ? Predictions of geometric and wave optics Diffraction: Origins and Classes Single-slit Diffraction – produced by interference from secondary wavelets from point sources within the slit width. Fresnel diffraction – “near-field” Fraunhofer diffraction – “far-field” (this is simpler to analyze! ) Modeling Diffraction Minima given by: (a/2) sin θ = +/- (λ/2) in general by: (a/n) sin θ = +/- (λ/n) \,Vr',ryt's 41,* (W WM /LAtt ? srhg/c- sl,{ 7,4'"-an) Irl t,a. | | | -t .rrI | 74 l.a ttl tat- llLl ITT | -l-t, \{> L €o, 2 soutzcg v ig c"' z €Kp.A eLottn-,, Taat d,tle"r* + D snn$ u Da* ft/"t6c whc*u I s,ug a f / 4 W. "ilL t, ',*llo €rtg* 4, t,n4 " r'Z drl hr,rre- din thL sr,t ,mzllL 5r. . . 5s, 4t^{ a, da"rJa 5rn -b . rr4 A- At Cm. + l,lzt ! r, ) Qd+lr,,.Aw A**'4o,.^" A tnfi sia.€- -e t'4 gt'4 lflltr/ tfr'fr I sln 4 4, i L 'a- ^'!a Qu -L Sk1z- L a- Minima (dark fringe): sin θ = +/- (mλ/a) Calculating the Intensity of Single Slit Diffraction Radius = arclength/β In the limit of many sources, trail of phasors circular sector can get the “chord” of circular sector Ep = Eo sin (β/2) (β/2) Ep = amplitude of the resultant E field at point P = geometrically the “chord” of a trail of phasors I = Io sin (β/2) 2 (β/2) β = (2π/λ) a sin θ The Double Slit Interference Pattern with Diffraction This is the familiar interference fringes for Young’s double slit experiment. We had always assumed narrow slits. Slit separation = 4 x slit width d = 4a i.e. slit is “narrow” But what if slits are not narrow ? Then we must factor in diffraction ! Diffraction “modulates” the two-slit interference pattern. I = Io cos2 (φ/2) sin (β/2) (β/2) 2 two-slit interference intensity φ = (2π/λ) d sin θ β = (2π/λ) a sin θ diffraction factor (envelope) ...
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