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Interference of light • Interference of light waves similar to interference of water waves – two different waves arrive at the observation point – the total influence is the sum of the two wave amplitudes at each time and at each point in space • High frequency of light has important consequences – Cannot follow the fast cycling of the field – Detectors measure the effect of many oscillations – Only interference that persists over many periods is observable • Only two-beam interference is discussed in this module – Multiple beam interference will be treated in next module April 03 LASERS 51 What does an optical power meter measure? • de Amplitu =A Electron in atom Field wiggles electron causing it to escape from atom. Becomes electrical current • Electron is released due to wiggling – Current from detector doesn’t go up and down with field – Responds to the amplitude not phase • Fast detectors can measure changes in 100 psec (10-10sec) – This includes 60,000 cycles of the field – Unlike radio waves where individual cycles can be measured • The irradiance (power per unit area) is proportional to A2, the square of the amplitude April 03 LASERS 51 Power measurement with interfering waves • Detector measures square of amplitude of resultant field (field obtained by adding waves at detector) – Two 1000 Watt beams impinging on detector will give a zero power reading if out of phase and 4000 Watts if in phase!!! – Where does the power go? – somewhere else • If the relative phase of the fields is not constant (e.g. incoherent light) then interference effects go away April 03 – This makes observation of interference difficult LASERS 51 Two point sources create interference pattern • Along red lines – Crests of two sources always coincide – Valleys of two sources always coincide – Net disturbance has twice the amplitude •Along green lines –Crest of one wave always occurs with valley of the other –Water is undisturbed along these lines Interference of two spherical waves source 1 crests of wave • “Frozen” in time – where two crests coincide, amplitude crests intersect is double that of a single source – where two troughs lines of constructive coincide, amplitude interference is negative and twice as deep as a single source troughs intersect source 2 April 03 Along indicated arrows waves have twice amplitude of a single source Constructive interference LASERS 51 Interference of two spherical waves (cont) source 1 • “Frozen” in time crest wave 1 intersects trough wave 2 lines of destructive interference – If a crest of one wave coincides with a trough of another wave, there is no net disturbance crest wave 2 intersects trough wave 1 Along indicated arrows waves have zero amplitude! source 2 April 03 Destructive interference LASERS 51 Optical path difference (OPD) source 1 source 2 d1 d2 OPD = d1-d2 observation point • • Optical path length determines how long it takes light to travel from the source to the observation point – Phase at an instant of time depends on optical path length • Optical path difference determines the phase difference at observation point between light from two sources • Understanding interference given the OPD is easy • Finding the OPD in many cases is very complicated geometry April 03 LASERS 51 amplitude_ (to tal) amplitude (wave 2) amplitude (wave 1) OPD = 0, in-phase waves 1 Wave at P due to source 1 0 -1 1 0 -1 2 1 0 -1 -2 April 03 Wave at P due to source 2 source 1 d1 Observation point, P source 2 d2 d1=d2 • • Optical path from either source to observing point is the same • Resulting wave 2x amplitude, same phase as either component • Intensity is four times that of either source LASERS 51 amplitude (wave 2) amplitude (wave 1) OPD = λ/2, 180° out of phase 1 0 due to source 1 -1 1 0 -1 2 2 amplitude (to tal) Wave at P source 1 1 0 -1 April 03 Wave at P due to source 2 d1 Observation Point, P source 2 d2 • d1=d2+λ/2 • Optical path difference (OPD)=λ/2 – Phase difference=180° • Resultant amplitude zero Conservation of energy works! Energy “missing” due to destructive interference is redistributed to regions of constructive interference LASERS 51 amplitude (wave 2) amplitude (wave 1) Superposition of waves 90°source out 1of phase source 2 1 0 -1 d1 d2 1 0 -1 2 amplitude (to tal) d1=d2+λ/4 1 0 -1 -2 April 03 observation point • • Path length from source 1 and 2 not the same – Optical path difference (OPD)=λ/4 – Phase difference=90° • Result: same frequency, amplitude 1.4x, phase different from either wave • Intensity 2 times single source intensity LASERS 51 More about superposition • To find interference pattern you need to know the OPD Interference observed on card – Often difficult to calculate. laser Calculation not needed to beam understand interference. Diverging lens – Changing OPD by λ, 2λ, 3λ, etc. Glass plate doesn’t change interference – Constructive interference: OPD=0, λ, 2λ, 3λ, etc. (integral number of wavelengths) – Destructive interference: OPD=λ/2, 3λ/2, 5λ/2, etc. (half-integral number of wavelengths) • Interference between waves with unequal amplitudes – OPD=half-integral number of waves • incomplete cancellation • Dark fringes are not completely dark – When OPD=integral no. of waves April 03 • intensity less than four times that of a single wave LASERS 51 amplitude (wave 1) Coherence 1 Wave due to source 1 has random phase jumps 0.5 0 -0.5 -1 -1.5 amplitude (wave 2) 1.5 Wave due to source 1 has different random phase jumps 1 0.5 0 -0.5 -1 amplitude (to tal) 1.5 Observation Point, P d2 • – On a femtosecond timescale interference still occurs but it is not observable 1 0.5 0 -0.5 -1 -1.5 April 03 d1 source 2 • No interference observed -1 5 2 -2 source 1 Superposition has some regions of constructive other regions of destructive interference • If phase jumps in each source are the same interference returns LASERS 51 Young’s double-slit experiment • First demonstration of wave nature of light – Thomas Young, 1802 – not accepted until Fresnel’s work 12 years later point source • Division of wavefront opaque screen with two slits April 03 – single wavefront from point source strikes both slits simultaneously – source is nearly monochromatic – point source must be nearly same distance from each pinhole LASERS 51 Young’s experiment—fringe spacing Essentially two sources, Huygen’s principle D d δ=OPD L slits OPD calculation easy if: Observation point screen • Assume L is very large compared to D or d • By (approximately) similar triangles Photo of pattern on screen D δ = L d • Constructive interference (bright fringes) if δ is 0 or an integral number of wavelengths L D = nλ where n = 0,1,2, L d Spacing between bright fringes is λL/d April 03 LASERS 51 Two slit interference pattern • Interference of two slits gives a sinusoidal variation in intensity • This pattern can be modulated by an overall intensity pattern Intensity – This pattern is charcteristic of all two beam interference – Due to diffraction as in next module, or uniformity of illumination in other cases April 03 Position on screen If intensity at low points don’t go to zero (unequal illumination of slits or partial coherence) fringes are indistinct or fuzzed out (low visibility) LASERS 51 Interference using nonmonochromatic sources • Each wavelength produces interference fringes screen – spacings are different • At center n=0, all wavelengths have a bright fringe • For larger n, fringes become colored, red on outside blue on slits inside • Finally for larger n fringes become completely washed out White-light interference can only be observed for OPD << λ April 03 Fringes in red light Fringes in blue light LASERS 51 Young’s fringes - large source • For a large source each point on the source produces a set of fringes point 2 • Fringes are shifted relative to each other • Waves from each point interfere add point 1 “incoherently” – Interfere with themselves only • A light field incident on the set of slits is said to have spatial coherence if an interference pattern is produced on the screen April 03 screen slits Fringes from point 1 Fringes from point 2 LASERS 51 Why do fringes from two different points wash out? • Atoms excited by some energy source • Excited atoms decay at random times – average lifetime (~10-8 sec) – each emission results in very short burst (wavepacket) – no phase relationship Atoms between different packets in gas • Emission between different points, or even same point at different times are incoherent April 03 Wave packets emitted by each atom at random times in random directions LASERS 51 Interference between incoherent sources a m plitude (wa ve 1) 1.5 1 0.5 0 -0.5 -1 a m plitude (wa ve 2) -1.5 1.5 1 0.5 0 -0.5 -1 -1.5 2 1.5 a m plitude (tota l) 1 0.5 0 -0.5 -1 -1.5 -2 April 03 • Shown is superposition of a randomly restarted wave with a perfect sine wave • sudden jumps represent termination of light from one atom and start of light from another • Superposition sometimes in phase sometimes out of phase – average over many cycles shows no enhancement by interference LASERS 51 a m plitude (wa ve 1) 1.5 1 0.5 0 -0.5 -1 a m plitude (wa ve 2) -1.5 1.5 1 0.5 0 -0.5 -1 -1.5 2 1.5 a m plitude (tota l) 1 Coherence length of a laser • Because of feedback, laser light has longer “memory” of its phase • Nevertheless, the phase drifts and eventually goes out of phase 0.5 0 -0.5 -1 -1.5 -2 April 03 LASERS 51 amplitude (wave 2) amplitude (wave 1) 1.5 1 Superposition of waves with different frequencies • May be from two different lasers or two different atoms with different Doppler shifts • Resultant varies periodically in amplitude as the two waves go in and out of phase • Hetrodyne detection of fm signal 0.5 0 -0.5 -1 -1.5 1.5 1 0.5 0 -0.5 -1 -1.5 2 1.5 amplitude (total) 1 0.5 0 -0.5 -1 -1.5 -2 “A photon can only interfere with itself.”-P. Dirac April 03 LASERS 51 Fringe visibility • Perfectly coherent sources of High equal intensity give maxima visibility of intensity 4x intensity of one source and zero intensity minima • Coherent source of unequal Low intensities give fringes with visibility maxima of I1+I2+2 √(I1I2) and minima of I1+I2-2 √(I1I2) • Partially coherent sources give fringes with lower maxima and higher minima depending on the degree of coherence • Completely incoherent sources give no fringes (intensity everywhere = 2x intensity of one source) I max − I min Visibility ≡ I max + I min LASERS 51 April 03 Lloyd’s mirror point source image of point source mirror Interference fringes in overlap region • Light reflected from mirror interferes with light directly from point source • Considered as interference between the point source and its image, this is almost identical to Young’s fringes – one significant difference is that center fringe is dark due to phase change on reflection April 03 LASERS 51 Division of amplitude Michelson interferometer • Source may also be a non- expanded laser point source or even laser beam an extended source – compensator plate required in this case • Tilting one mirror produces straight-line fringes • Often used for optical testing April 03 mirror 2 beamsplitter mirror 1 observation screen LASERS 51 Interference from wedged plates input wave • Waves from two surfaces interfere at observation point – OPD determined only by path between reflecting surfaces – other parts of path are common • Interference from other d surfaces occurs with laser sources wave reflected α – With incoherent sources OPD for other surfaces is greater than wave reflected OPD = 2d + λ / 2 from second surface coherence length phase change on reflection – Small gap, nearly d = αx for small angles monochromatic source needed OPD = 2αx + λ / 2 to observe interference with incoherent sources Bright fringes when OPD=integral number λ 5λ 9λ of waves – Spatial coherence not needed x= , , , etc 4α 4α 4α LASERS 51 April 03 x from first surface Newton’s rings monochromatic, extended light source eye part under test OPD, optical path difference High quality optical flat • Common optical testing technique – optical flat may be replaced by a curved test plate • Optical path difference varies with shape of tested part – radius of part, as well as size of defects can be measured – dark fringe in center due to phase change on reflection April 03 LASERS 51 Antireflection (AR) coatings Air, index na Glass, index ng Coating, index nc Reflection coefficient of uncoated glass R= (ng − na ) 2 (ng + na ) 2 • Reflection from air/coating interface interferes with reflection from coating/glass interface – If nc=√(ng*na) and the coating is λ/(4nc) thick the two reflections cancel completely for one wavelength – For ng=1.5 and na=1.0 this requires nc=1.225 (unknownium) – a 1/4-wave coating of a relatively low index material MgF(n=1.38) can be applied to obtain a reflectance of 1.4% (at one wavelength) – to get lower reflectance or more than one wavelength multiple LASERS 51 April 03layers must be used Shear plate interferometer shear Front surface • For a plane input beam, reflection Back surface fringes are straight lines reflection perpendicular to tilt direction – Spacing of fringes depends on tilt angle and wavelength input beam Top view Tilt Side view April 03 • Working out the OPD for any other given wavefront is complicated • Shear can be measured using shadow of object in the input beam LASERS 51 Shear-plate interferometer fringe patterns • Horizontal lines indicates a plane wave – Infinite radius of curvature • Tilted straight lines indicates a spherical wavefront – Tilts one direction for converging wave, opposite direction for diverging wave • Deviation from straight lines means wavefront is not plane or spherical – i.e. aberrations April 03 LASERS 51