Physics 116. Nov 3, Lecture 21 Wave optics. R. J. Wilkes 11/3/11 1

Transcription

1 Physics 116 Lecture 21 Wave optics Nov 3, 2011 R. J. Wilkes 11/3/11 1

2 Announcements 3 clickers have quiz data logged, but no registration: If one of these is yours, see me pls Exam 2 next Monday: same procedures as last time Practice exam posted Thursday, in class Friday YOU bring bubble sheet, pencil, calculator Practice questions for Exam #2 are posted (in slides directory), I will go over solutions in class tomorrow Kyle will have special office hour: Monday, 11:30, B442

3 Lecture Schedule (up to exam 2) 11/3/11 Today 3

4 Quiz: #8 Today, almost all large astronomical telescopes are reflectors, because with a mirror instead of a lens A.! Since refraction is not involved, there is no color aberration B.! There is only one surface to grind and polish, which is easier and cheaper C.!Both of the above D.!None of the above 11/3/11 4

5 Physical (wave) optics So far, we have been using simple ray-tracing methods to understand behavior of light on a macroscopic scale! Analyze optical systems with diagrams and lens/mirror eqns.! Works as if light were a stream of particles (rays)! Accurate for system lengths from cm up to size of Universe On a microscopic scale (few mm down to wavelength of light), ray optics does not work: must employ the wave nature of light! Take into account the physical properties of waves: superposition and interference effects We ve already discussed this: brief review follows! Wave optics applies to behavior of light in and around Narrow apertures (mm or smaller) Arrays of apertures (gratings, meshes, screens) Edges of barriers (if we look closely enough) Reflections in thin (few wavelengths thick) films 5

6 From mid-october: Interference patterns Imagine two kids paddling their hands in a pond in unison Waves (ripples) propagate outward, and overlap There will be lines of reinforcement, and lines of cancellation (constructive and destructive interference) lines of reinforcement, where peaks overlap (constructive interference) Halfway between are lines of destructive interference 11/3/11 6

7 From mid-october: Interference = defining property of waves If you see interference effects, you are looking at waves! Case study: Isaac Newton thought light was a stream of particles Newton s Opticks (1687) explained all observations at the time Thomas Young (120 years later) observed interference effects with light Only waves could do that Wave theory of light replaced Newton s Important point to review: Interference depends on phase relationship of overlapping waves Phase relationship depends on distance from source Recall: Phase at distance D from source = 2! (D/") but sin/cos repeat every cycle, so all that matters is where we are relative to start of latest cycle: fraction of a cycle Phase at distance D from source = 2! [ fractional part of (D/") ] Example: fractional part of is /3/11 7 7

8 From mid-october: Phase relationships Example: source of sound has "=1.5m Point A is 2.5 m away, B is 5.5 m away, C is 7 m away Phase values are To make this acoustic example describe light waves, just change wavelength units from meters to nanometers! A B C A, B, C are all 2/3 along a cycle, so sound waves have the same phase at all these points D / " (distance, in units of wavelengths) 11/3/11 8 8

9 From mid-october: light Example: 2 sound sources with same f Solid line: R = one wavelength, dashed line: R=! wavelength 1 2 B C A These could be closely spaced pinholes in an opaque screen Point A: sources 1 and 2 are in phase, so we get constructive interference Points B and C: 1 and 2 are 180 deg out of phase, so destructive interference 11/3/11 9 9

10 Another example reworked: Instead of loudspeakers, let s consider radio antennas! spaced 3 km apart, in phase: same electrical signal goes to both Observer 4km away, parallel to antenna 1, finds constructive interference (signal is double intensity of a single antenna s) What f is being transmitted? D 1 = 4km D 2 = ( 3km) 2 + ( 4km) 2 = 5km D 2! D 1 = n" for n = 1, " = D 2! D 1 = 1km f = c 300,000 km / s = " 1km = 300kHz 4km 3km 1 2 5km This is just the lowest possible frequency to give constructive interference. Any integer multiple (harmonic) of this f will also produce constructive interference at the observer s location: then n > 1 Notice: this is just the stereo loudspeaker example, with meters changed to km and speed of sound in air changed to speed of light 11/3/11 10

11 Coherence One big difference between acoustics and optics: ordinary light sources have billions of independent radiators! In acoustics, we only discussed simple cases of one source (or 2 sources in phase )! Light bulb = tungsten atoms, each emitting light on its own schedule! Incoherent light source: jumble of wavelengths and phases! To have constant phase relationships, to observe interference effects, we need to either View a tiny part of an incoherent source (use slits, pinholes) Have a source of coherent light (all atoms emit in synchrony)! Simple to do today: use a laser! This option did not exist until 1960s! 11/3/11 Broad spectrum of wavelengths mixed together Monochromatic (only 1 wavelength is produced) 11

12 Constructive vs destructive interference We see interference effects when waves overlap (superposition) after travelling different paths from their source For EM waves that are coherent (phase relationship at their source is fixed, and does not vary with time)! Constructive interference if path difference is zero or an integer number of wavelengths! Destructive interference if path difference is an odd integer number of half-wavelengths (Even number of half-wavelengths = integer number of full wavelengths) 1 2 Constructive if:! 2!! 1 = m" m = 0,1,2 and # Destructive if:! 2!! 1 = m! 1 & $ % 2' ( " m = 1,2, 3 # # alternative statement:! 2!! 1 = n & & $ % 2' ( " n = 1, 3,5 $ % ' ( l 1 l 2 observer (For phase relationships between these extremes, we get intermediate intensities) 11/3/11 12

13 Young s 2-slit experiment Some call it the most important experiment in physics! Fundamental impact on quantum theory as we shall see Make a mask with 2 pinholes or narrow slits, closely spaced! What s narrow? What s close? On the scale of visible light wavelength: fraction of mm Illuminate with plane waves (optical equivalent of equally spaced water waves at a beach, with all wave peaks parallel)! How do we get plane waves of visible light? Use a laser! How did Thomas Young manage in 1804? Used a pinhole to view a tiny patch of lamp surface Partially coherent light pattern is partially washed out 2-slit interference pattern in laser light The waves interfere to form bright and dark fringes on a viewing screen some distance away We can simulate this with water waves in a tank Slits = barrier with 2 holes plane waves

14 Huygens principle Christiaan Huygens (English pronunciation: /"ha#$%nz/), Dutch, Contemporary of Newton, advocate of wave theory of light Proposed a mechanism for wave propagation: Each point on a light wavefront acts like a point source of light! Produces expanding spherical wavelets! Superposition of wavelets from all points on original wavefront produces the next wavefront! Repeat Picture shows wavelets from 3 points imagine every point adding in its contribution: spherical wavelets Today, Maxwell s equations tell us how EM waves propagate, but Huygens Principle provides a handy model to help us picture interference phenomena. plane wave

15 2-slit experiment in detail Illuminate a pinhole or slit with light from a distant, coherent source! Nearby source would give spherical waves, distant source = plane waves! Coherent = from a laser*, or by using another pinhole to restrict area of light source used * Lasers supply plane waves at any distance Place a mask with 2 slits in the beam of light! Slit spacing d should be small (not many wavelengths) Place a screen far away (many wavelengths)! Rays* from each slit are nearly parallel when they arrive * We ll talk about rays only to get path lengths then we ll compare to wavelengths We get a bright fringe on screen parallel to midpoint! Central bright fringe equal path lengths from the 2 slits! Constructive interference Where is the next bright fringe?! At what angle do we get next constructive interference?! Path difference from the 2 slits must be 1 or more wavelengths!! = "!! = 0

16 2-slit experiment in detail Calculate the angle to the next bright fringe: 1 2! d l 1 l 2 Slit mask d sin!! (angle to next bright fringe) 90! Ray to next bright fringe Ray to slit midpoint! 2!! 1 = d sin" Constructive if d sin" = m#, m = 0, ± 1, ± 2, ± 3 Ray to next bright fringe Ray to slit midpoint On this scale, the screen is miles away!! = "!! = 0! Central fringe for m=1, the next bright fringe on either side for +1, etc! Dark fringes = destructive interference: #!! = m " 1 & $ % 2' ( ) m = ±1, ± 2, ± 3