Superposition of waves (review) PHYS 258 Fourier optics SJSU Spring 2010 Eradat

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1 Superposition of waves (review PHYS 58 Fourier optics SJSU Spring Eradat

2 Superposition of waves Superposition of waves is the common conceptual basis for some optical phenomena such as: Polarization Interference Diffraction What happens when two or more waves overlap in some region of space. How the specific properties of each wave affects the ultimate form of the composite disturbance? Can we recover the ingredients of a complex disturbance? PHYS 58 Spring SJSU Eradat Superposition

3 Linearity and superposition principle ψ(, rt ψ(, rt The scaler 3D wave equation = r V t is a linear differential equation (all derivatives apper in first power. So any linear combination of its solutions ψ(, rt = Ciψi(, rt is a solution. Superposition principle: resultant disturbance at any point in a medium is the algebraic sum of the separate constituent waves. We focus only on linear systems and scalar functions for now. At high intensity limits most systems are nonlinear. Example: power of a typical focused laser beam = ~ W / cm n i= compared to sun light on earth ~ W / cm. Electric field of the laser beam triggers nonlinear phenomena. PHYS 58 Spring SJSU Eradat Superposition 3

4 Superposition of two waves Two light rays with same frequency meet at point p traveled by x and x [ ω ε ] [ ω α ] [ ω ε ] [ ω α ] E = E sin t ( kx + = E sin t+ E = E sin t ( kx + = E sin t+ Where α = ( kx + ε and α = ( kx + ε Magnitude of the composite wave is sum of the magnitudes at a point in space & time or: E = E + E = E t+ where sin ( ω α E = E + E + E E cos( α α and tanα = sinα + E sinα E cosα + E cosα The resulting wave has same frequency but different amplitude and phase. E E cos( α α is the interference term δ α α is the phase difference. E PHYS 58 Spring SJSU Eradat Superposition 4

5 Phase difference and interference π δ α α = ( kx+ ε ( kx + ε = ( x x + ( ε ε = δ+ δ λ Total phase difference between the two waves has two different origins. ( a δ = ε ε phase difference due to the initial phase of the waves. Waves with constant initial phase difference are said to be coherent. ππ b δ = nx ( x = kλ is pahse difference due to the Optical Path λ Difference or OPD Λ= n x ( x Waves in-phase: δ α α =, ± π, ± 4 π,... then E is mximum Waves out of phase: δ α α =± π, ± 3 π,... then E is minimum Waves in-phase interfere constructively E = Emax = ( E + E Waves out of phase interfere destructively E = E = ( E E If E Λ = λ = E then E = ( E and E = max min x x λ min number of waves in medium= number of waves in vacuum PHYS 58 Spring SJSU Eradat Superposition 5

6 Addition of two waves with same frequency Superposition of two waves 3 Superposition of two waves Amplitude - Amplitude Distance x -6 Distance x -6 Superposition of two waves Superposition of two waves Amplitude Amplitude Distance.5 3 x -6 PHYS 58 Spring SJSU Eradat Superposition Distance.5 3 x -6 6

7 Two waves with path difference For two waves with no initial phase difference ( ε apathdifferenceof Δ x wehave: ( ω [ ω α ] ( ω sin [ ω α ] E = E sin t k( x+δ x = E sin t+ E = E sin t kx = E t + α α = kδx The resulting wave is ε = = but Amplitude is a function of path difference kδx Δx E = E cos sin ωt k x+ Constructive interference: if Δ x<< λ, or Δx ± mλ then the resulting amplitude is ~ E Destructive interference: Δx ± (m+ λ then E PHYS 58 Spring SJSU Eradat Superposition 7

8 Exercise. Plot E, E, E + E, and ( E + E for the following two sinusoidal waves for < x< 5λ with λ = 5 nm: E = E sin( ωt ( kx+ ε and E = E sin( ωt ( kx+ ε a same frequency, E = E =, zero initial phase, both forward. b same frequency, E = E =, ε =, ε = π, both forward. c same frequency, E = E =, ε =, ε = π /, both forward. d same frequency, E = E =, ε =, ε = π, E forward, E backward. e same frequency, E = E =, ε =, ε =, both forward. f same frequency, E = E =, ε =, ε = π, both forward. g Compare the results of direct superposition with the formula derived in text for case a in slide 5 (Notice the difference between atan and atan functions in MATLAB PHYS 58 Spring SJSU Eradat Superposition 8

9 Phasors and complex number representation Each harmonic function is shown as a rotating vector (phasor projection of the phasor on the x axis is the instantaneous value of the function, length of the phasor is the maximum amplitude angle of the phasor with the positive x direction is the phase of the wave. y i ( t + E = E e ω α E(t= =E sin(ωt+ +α PHYS 58 Spring SJSU Eradat Superposition ωt α E x E(t=E cos(ωt+α 9

10 Example of superposition using phasors E( t = Ecos( ωt+ φ E( t = Ecos( ωt E p = E+ E a vector sum of E and E Magntude of E (from triangonomety E = E + E E cos( π φ p E = E + E + E cosφ p Using +cos φ=cos ( φ/ p E = E (+ cos φ = 4E cos ( φ/ p Amplitude of two indentical waves interfering with phase difference of φ is independent of time: Ep φ = E cos PHYS 58 Spring SJSU Eradat Superposition ωt

11 Superposition p of many waves Superposition of any number of coherent harmonic waves with a given frequency, ω and traveling in the same direction leads to a harmonic wave of that t same frequency. N E = E cos( α ± ωt = E cos( α ± ωt E i= i i n N N i= Ei EiE j cos( αi α j and tanα N i= j> i i= = + = α = ( kx + ε and α = ( kx + ε i i j j N i= E E i i sinα i cosα N N N j E Ei EiE j NEi i= j> i i= For coherent sources α = α and = + = i i ( For incoherent sources (random phases the second term is zero. ( ( ( = = i Flux density for N emmiters: E NE ; E NE incoherent coherent PHYS 58 Spring SJSU Eradat Superposition

12 Exercise. Write a MATLAB routine to calculate the amplitude and phase of N harmonic waves (cosine with same frequencies but varying initial phase and amplitudes. Assume the wavelength is 5 nm and V = c a The program should read the phase and amplitude of the waves from a file that has two columns and N rows. Test the program for the following waves E =, ε =, E =, ε = π / 4. Once made sure it is working, create a file with the folloing waves and plot their superposition from to 5λ. E =, ε =, E =, ε =, E =, ε =, E = 3, = 3, E5 =, 5 = 4, E6 =, 6 = 5, E7 =, 7 = 6 ε ε ε ε i π b Next run the program for N = and εi = ε+ π, where ε = and E i =. This time create the phases and amplitudes inside the routine and don't read from a file. PHYS 58 Spring SJSU Eradat Superposition

13 Addition of waves: different frequencies I Mathematics behind light modulation and light as a carrier of information. Two propagating waves are superimposed E = Ecos( kx ωt E = Ecos( kx ωt k > k and ω > ω with equal amplitudes and zero init al phases E = E + E = E [cos( k x ωt + cos( k x ω t] using cosα + cos β = cos ( α + β cos ( α β E = E cos ( ( cos ( ( k + k x ω + ω t k k x ω ω t Need to simplify this PHYS 58 Spring SJSU Eradat Superposition 3

14 Addition of waves: different frequencies II [ ω ] E = E cos kmx mt cos kx ωt Average angular frequency ω = ( ω+ ω / Average propagation number k = ( k + k / Modulation angular frequency ωm = ( ω ω Modulation propagation p number k = ( k k with the following definitions m / / Superimposed p wavefunction: E = E x tcos kx ωt (, [ ω ] Time-varying modulation amplitude E ( x, t = E cos k x t For large if then we will have a slowly varying ω ω ω ω >> ω m amplitude with a rapidly oscillating wave m m PHYS 58 Spring SJSU Eradat Superposition 4

15 Irradiance of two superimposed waves with different frequencies [ ω ] ( ω E ( x, t = 4Ecos kmx mt = E + cos kmx mt Beat frequency ω m = ω ω or oscillation frquency of the E ( x, t Amplitude,, oscilates at, the modulation freuency E ω m Irradiance, E, varies at ω, twice the modulation frequency m Two waves with different amplitudes produce beats with less contrast. PHYS 58 Spring SJSU Eradat Superposition 5

16 Beats 4 Irradiance ω m 3 Superposition of two waves (4E λ λ λ λ λ m Amplitude λ - - E Amplitude ω m λ Distance PHYS 58 Spring SJSU Eradat Superposition x -6 6

17 Group velocity In nondispersive media velocity of a wave is independent of its frequency. ω For a single frequency wave there is one velocity and that t is Vphase = k When a vave is composed of different frequency elements, the resulting disturbance will travel at a differnt velocity than phase velocity of its components. [ ω ] E = E cos kmx mt cos kx ωt ω Vphase = velocity of a constant phase point on the high frequency wave k ωm dω Vgroup = = velocity of a constant magnitude (amplitude or km dk ω modulation envelope Vg may be smaller, equal, or larger than v p To calculate the V and V we need the dispersion relation ω = ω( k p g PHYS 58 Spring SJSU Eradat Superposition 7

18 Dispersion relation ω vs. k or ω=ω(k Phase velocity for a given frequency is slope of a line on the dispersion curve that connects that point to the origin or ω/k. Group velocity for that frequency is the slope of the dispersion curve at that point or dω/dk. ω V ( ω V ( ω g ω V ( ω Normal dispersion: dω ω Vg < Vp or < dk k ω V ( ω V ( ω g k V ( ω g No dispersion: V g dω ω = Vp or = dk k k Normal dispersion: PHYS 58 Spring SJSU Eradat Superposition V g dω ω > Vp or > dk k k 8

19 Finite it waves Finite wave: any wave starts and ends in a certain time interval Any finite wave can be viewed as a really long pulse Any pulse is a result of superposition of numerous different frequency harmonic waves called Fourier components. Wave packet is a localized pulse that is composed of many waves that cancel each other everywhere else but at a certain interval in space. We need to study Fourier Analysis to understand actual waves, pulses, and wave packets. Width of a wave packet is proportional to the range of k m of the wave packet. Since each component of the wave packet has different phase velocity in the medium, through the relationship V p =ω/k, k of the components change in the dispersive media. As a result k m of the modulation disturbance changes and consequently group velocity changes. This results in change of the width of the wave packet. PHYS 58 Spring SJSU Eradat Superposition 9

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