PHY122 Physics for the Life Sciences II

Transcription

1 PHY122 Physics for the Life Sciences II Lecture 16 Waves and Interference HW 10 is due Sunday, 6 Nov. at 8:00 pm Make-ups for Labs 3,4,5 MUST be done this week (or else! As you all know since Day 1 of PHY 122.) Note: Clicker Channel 21 11/01/2011 Lecture 16 1

2 Electromagnetic Spectrum Frequency and wavelength: f λ = c (in vacuo: m/s exactly!) The visible part of the EM spectrum is only small, from 400 nm < λ< 800 nm different wavelengths are seen as different COLORS from short λ to long: violet, blue, green, yellow, orange, red larger λ (lower f): INFRARED, followed by MICROWAVES: 0.1 mm < λ < 1 m, RADIOWAVES: λ > 1 m: UHF, VHF, MW, LW, VLF, shorter λ (higher f): ULTRAVIOLET: 1 nm < λ < 400 nm, SOFT/HARD X-RAY: 1 pm < λ < 1 nm, and GAMMA: λ < 1 pm, where instead of wavelength one usually uses photon energy in kev or MeV or GeV or TeV (or higher!) to denote the radiation (k=10 3, M=10 6, G=10 9, T=10 12 ) 11/01/2011 Lecture 16 2

3 LiveWeb2 transverse and longitudinal waves 11/01/2011 Lecture 16 3

4 Slide 3 LiveWeb2 LiveWeb, 11/1/2011

5 Coherence and Monochromaticity Monochromatic (= one color ) waves are by definition waves that have a single unique wavelength λ ( 1 frequency) examples: light of a single color (λ red =700 nm), radio waves of a single FM station (λ = c/(91.1 MHz)), a sound tone (λ=(344 m/s)/(1000 Hz)), etc. Waves will interfere : waves will add wherever waves from two or more sources come together: this is the important principle of SUPERPOSITION e.g., sound-pressure amplitudes from two sources add; for waves meeting on a rope: the amplitudes add as function of time; for EM waves: the electric and magnetic field vectors each add everywhere interference is typically not noticed because many frequencies are present in typical wave trains or wave packets if monochromatic waves meet, AND if they are COHERENT (i.e., they oscillate with a stable difference in PHASE), a stable interference pattern will emerge 11/01/2011 Lecture 16 4

6 LiveWeb3 constructive & destructive interference of transverse slinky wave packets (pulses) 11/01/2011 Lecture 16 5

7 Slide 5 LiveWeb3 LiveWeb, 11/1/2011

8 When waves meet an obstacle parallel (water) wave fronts meeting a small (size ~λ) opening and a large (size >>λ) opening in a barrier waves approach from the top of the picture small opening large opening circular wave fronts after the opening; no strong shadow regions straight wave fronts after the opening with minimal side ripple ; little/no energy in shadow regions 11/01/2011 Lecture 16 6

9 Similarly for Light Waves Light waves spread out behind a narrow slit (width ~ λ): DO THE DEMO using a caliper for the slit d = 100 µm λ green 550 nm = 0.55 µm 11/01/2011 Lecture 16 7

10 Huygens Principle: Consider a particular wave front (e.g., one crest of a wavetrain) moving in the direction shown... Each point on a wave front serves as a point from which circular wave fronts ( wavelets ) originate. The wave front at t later in time is tangential to all the circular wavelets at that time wavelets all at a time t later and so forth 11/01/2011 Lecture 16 8

11 LiveWeb1 animation of two-circular-wave interference 11/01/2011 Lecture 16 9

12 Slide 9 LiveWeb1 LiveWeb, 11/1/2011

13 Interference: 2 different point sources A B Note: the angleθ of the lines of maxima (and minima) changes with changing distance d between the sources A and B 11/01/2011 Lecture 16 10

14 A B d Example: Interference of 2 different sources r r θ λ P Q maxima (twice the amplitude) minima (dead spot) Interference of two coherent monochromatic (point) sources A and B (How to make them coherent?): coherent: wave fronts in phase phase difference in P: ϕ I I cos 11/01/2011 wave fronts (E=E m ) ( ) y φ=k r= (2π/λ) BP AP ; here: path diff: BP AP = 8λ 6λ= 2λ= r x φ= k r= (2π/λ) 2λ = 4π CONSTRUCTIVE interference maxima: path diff. = d sinθ m,max = mλ, m=0,±1,±2,.. path diff BQ AQ = λ = 0.5λ= r φ= k r= (2π/λ) 0.5λ = π DESTRUCTIVE interference minima: d sinθ m,min =(m+½)λ, m=0,±1,±2,.. Intensity pattern: alternating bright and dark lines (proof later): = ( (2 π λ) r ) π d sinθ = λ P 0 = I Lecture cos I0 cos 2 11

15 How do we make 2 coherent sources of light? 11/01/2011 Lecture 16 12

16 Answer: Do what we did before Light waves spread out behind a narrow slit (width ~ λ): d = 100 µm λ green 550 nm = 0.55 µm 11/01/2011 Lecture 16 13

17 TWICE! 11/01/2011 Lecture 16 14

18 Young s 2-slit expt: 2 coherent sources from 1 Thomas Young Powerful evidence for the wave nature of light 11/01/2011 Lecture 16 15

19 11/01/2011 Interference from 2 slits: trig + algebra Path Difference r (or the Phase Difference φ) determines the effect: defn: angular wavenumber =2 / units of m -1 Path difference of waves from A, B in P: r = BK = d sinθ Phase difference of waves from A, B in P: φ=(k)(bk)=(2π/λ)(bk) because AK CP AP BP =BK; for CP>>AB: BK d sinθ A Intensity MAXIMUM in P: BK=mλ d sinθ m =mλ, m=0,±1,±2, d C Intensity MINIMUM in P: BK=(m+½)λ BK=d sinθ m =(m+½)λ if P is on a screen at distance L from C: y/l = BK/AK BK/d (for small θ) y BK L/d maxima: y m = m (λl/d), m=0,±1,±2, ; mimima: y m = (m+½) (λl/d) thus the interference pattern on the screen consists of alternating bright and dark lines (if A and B are slits), separated by y =λl/d e.g.: λ=600 nm, L=1 m, d=600 µm y=1 mm Intensity pattern: alternating bright and dark bands (proof later): ϕ (2 π λ ) BK 2 π d sinθ 2 π d I P = I0 cos 2( 2 ) I0 cos sinθ θ λ θ 2 = I 2( ) = I 0 cos 0 cos = I0 cos 2 λ sin θ y/ L B θ K θ r=bk=d sinθ=λ φ/(2π) L Lecture P y O π d = y λl

20 Summary Double-Slit Experiment angles under which maxima occur on the screen: λ θ = m d m d positions of maxima on the screen: λ = L tanθ m = m d y m with m=0, ±1,±2, ±3, Diffraction see later! I 0 y Intensity pattern: Intensity I see later! = I 0 cos ϕ 2( ) (or m = ±, ±, ±, ±, L for minima!) L 11/01/2011 Lecture 16 17

21 Examples Two antenna masts, 10 m apart, are emitting at 91.1 MHz. Find the angle with the bisector of the antenna baseline for which there is a A first minimum in θ d C intensity ( 1 ) 1 λ 3 λ 5 λ minima for: d sinθ m = m + λ sin θ 2 m = ±, ±, ±, L 2 d 2 d 2 d 8 λ c 3 10 sinθ = ± = ± = ± = ± θ = ± = ± d f 2d note the dependence on θ and on λ: for differentλthe minima and maxima will occur at differentθ! B K θ BK=d sinθ =λ φ/(2π) L P y O 11/01/2011 Lecture 16 18

22 λ maxima (twice the amplitude) minima (dead spots) Interference by Same- Frequency Sound Waves A B P y x Interference of two synchronous, equalfrequency, equalamplitude sound sources, ignoring reflections from walls, floor, ceiling, etc Maximum in P: AP BP = mλ, m=0,1,2, ; Minimum in P: AP BP = (m+½)λ 11/01/2011 Lecture 16 19

23 2-slitActivPhysics applet 11/01/2011 Lecture 16 20

24 The Behavior of Light in Dielectrics We know the speed of EM waves in a dielectric is reduced compared to the speed in vacuum: v = dielectric c = κ c n index of refraction phase change 180 = path length increase by λ/2 11/01/2011 Lecture 16 21

25 Why does this happen? 11/01/2011 Lecture 16 22

26 11/01/2011 Interference in Thin Transparent Films Reflection of normal incidence light off a thin layer of oil (n=1.45) on water (n=1.33): ray A, under reflection off a material of larger index of refraction, undergoes a phase shift φ=π otherwise φ=0 identical to a wave on a rope that meets a more massive piece of the rope or a fixed end: the reflected wave is inverted (i.e. a phaseshift of π) the electric field in particular: Erefl, a = E Refracted waves have zero phase shift Q: dominant color of the reflected light? maxima for: φ AB =(2π/λ oil )(2t) +π = m2π λ air,m = n oil λ oil = 4n oil t/(2m+1), m=0,1,2, λ air,0 = 2204 nm (infrared; not visible) λ air,1 = 735 nm (near infrared; not visible) λ air,2 = 441 nm (violet-blue; visible) λ air,3 = 315 nm (near UV; not visible) inc, a n n a a n + n n oil = 1.45 Lecture b b A B n water =1.33 n air =1.00 t=380 nm from reflection of ray A at larger angles, larger-λ colors dominate Only thin layers reflect colors selectively; thick layers reflect all colors (many maxima)

27 Summary Thin Film Interference Speed inside a dielectric: v dielectric =c/n λ dielectric =λ vacuum /n Extra phase change of 180 =π if n inc =n 1 <n 2 =n refract remember: pulse on a rope meeting a fixed end or a loose end Constructive interference if: λ 2t = m m = 0,1, 2, L n plus an EVEN number of phase changes (180 =π) from reflection ELSE: ( 1 ) λ 2t = m + m = 0,1, 2, L 2 n plus an ODD number of phase changes (180 =π) from reflection 11/01/2011 Lecture 16 24