Superposition of waves 1 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? 2 Linearity and superposition principle ∂2ψ (r,t) 1 ∂2ψ (r,t) The scaler 3D wave equation = is a linear ∂r 2 V 2 ∂t 2 differential equation (all derivatives apper in first power). So any n linear combination of its solutions ψ (r,t) = Ciψ i (r,t) is a solution. i=1 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=~1010 V/cm compared to sun light on earth ~10 V/cm. Electric field of the laser beam triggers nonlinear phenomena. 3 Superposition of two waves Two light rays with same frequency meet at point p traveled by x1 and x2 E1 = E01 sin[ωt − (kx1 + ε1)] = E01 sin[ωt +α1 ] E2 = E02 sin[ωt − (kx2 + ε 2 )] = E02 sin[ωt +α2 ] Where α1 = −(kx1 + ε1 ) and α2 = −(kx2 + ε 2 ) Magnitude of the composite wave is sum of the magnitudes at a point in space & time or: E = E1 + E2 = E0 sin (ωt +α ) where 2 2 2 E01 sinα1 + E02 sinα2 E0 = E01 + E02 + 2E01E02 cos(α2 −α1) and tanα = E01 cosα1 + E02 cosα2 The resulting wave has same frequency but different amplitude and phase. 2E01E02 cos(α2 −α1) is the interference term δ ≡ α2 −α1 is the phase difference. 4 Phase difference and interference 2π δ ≡ α2 −α1 = (kx1 + ε1) − (kx2 + ε 2 ) = (x1 − x2 ) + (ε1 −ε 2 ) = δ1 +δ2 λ Total phase difference between the two waves has two different origins. a) δ2 = (ε1 −ε 2 ) phase difference due to the initial phase of the waves. Waves with constant initial phase difference are said to be coherent. 2π b) δ1 = n(x1 − x2 ) = k0Λ is pahse difference due to the Optical Path λ0 Difference or OPD ≡ Λ = n(x1 − x2 ) Waves in-phase: δ ≡ α2 −α1 = 0,±2π ,±4π ,... then E0 is mximum Waves out of phase: δ ≡ α2 −α1 = ±π ,±3π ,... then E0 is minimum 2 Waves in-phase interfere constructively E = Emax = (E01 + E02 ) 2 Waves out of phase interfere destructively E = Emin = (E01 − E02 ) 2 If E01 = E02 then Emax = (2E01) and Emin = 0 Λ x − x = 1 2 is the number of waves in the medium 5 λ0 λ Addition of Two waves with same frequency Superposition of two waves 3 Superposition of two waves 2 2 1.5 1 1 0.5 e d e u d t u i l t 0 i l p 0 p m m A A -0.5 -1 -1 -2 -1.5 -3 -2 0 0.5 1 1.5 2 2.5 3 0 0.5 1 1.5 2 2.5 3 Distance -6 Distance -6 x 10 x 10 Superposition of two waves Superposition of two waves 2 1 0.8 1.5 0.6 1 0.4 0.5 0.2 e e d d u t u i l t i 0 p l 0 p m m A A -0.2 -0.5 -0.4 -1 -0.6 -1.5 -0.8 -1 -2 0 0.5 1 1.5 2 2.5 3 0 0.5 1 1.5 2 2.5 3 Distance -6 6 Distance -6 x 10 x 10 Two waves with path difference For two waves with no initial phase difference (ε1 = ε 2 = 0) but a path difference of ∆x we have: E1 = E01 sin (ωt − k(x + ∆x)) = E01 sin[ωt +α1 ] E2 = E02 sin (ωt − kx) = E02 sin[ωt +α2 ] α −α = k∆x 2 1 Amplitude is a function of path difference The resulting wave is k∆x ’ » ≈ ∆x ’ÿ E = 2E cos∆ ÷sin …ωt − k ∆ x + ÷Ÿ 0 « 2 ◊ « 2 ◊⁄ Constructive interference: if ∆x << λ, or ∆x/λ ≈ ±2m then the resulting amplitude is ~2E0 Destructive interference: ∆x ≈ ±mλ / 2 then E ≈ 0 7 Exercise 2 RV2-1) Plot E1, E2, E1+E2, and (E1+E2 ) for the following two sinusoidal waves for 0 < x < 5λ with λ = 500 nm: E1=E01 sin(ωt − (kx + ε1)) and E2 =E02 sin(ωt − (kx + ε 2 )) a) same frequency, E01=E02 =2, zero initial phase, both forward. b) same frequency, E01=E02 =2, ε1 = 0, ε 2 = π, both forward. c) same frequency, E01=E02 =2, ε1 = 0, ε 2 = π / 2, both forward. d) same frequency, E01=E02 =2, ε1 = 0, ε 2 = π, E1 forward, E2 backward. e) same frequency, E02 =2E01=2, ε1 = 0, ε 2 = 0, both forward. f) same frequency, E02 =2E01=2, ε1 = 0, ε 2 = π, both forward. g) Compare the results of direct superposition with the formula derived in text for case a (slide 4). (Notice the difference between a tan and a tan 2 functions in MATLAB 8 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 ) 1 E0 i(ωt+α ) α + t E = E e ω 0 ( n i s 0 α1 E = ) t ( ωt E x 9 E(t)=E0 cos(ωt+α1) Superposition using phasors E (t) = E cos(ωt +φ) E (t) = E cos(ωt) d1 d d 2 d d E = E + E a vector sum of E and E p 1 2 d 1 2 Magntude of Ep (from triangonomety) 2 2 2 2 Ep = E + E − 2E cos(π −φ) 2 2 2 2 Ep = E + E + 2E cosφ Using 1+cosφ=2cos2 (φ / 2) 2 2 2 2 Ep = 2E (1+ cosφ) = 4E cos (φ / 2) Amplitude of two wave interference ωt independent of time: φ E = 2E cos p 2 10 Superposition 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 same frequency. N E = E0i cos(αi ±ωt) = E0 cos(α ±ωt) i=1 N n N N E0i sinαi 2 2 i=1 E0 = E0i + 2 E0i E0 j cos(αi −α j ) and tanα = N i=1 j>i i=1 E0i cosαi i=1 N N N 2 2 2 For coherent sources αi = α j and E0 = E0i + 2 E0i E0 j = (NE0i ) i=1 j>i i=1 For incoherent sources (random phases) the second term is zero. 2 2 2 2 Flux density for N emmiters: E0 = NE01; E0 = NE0i ( )incoherent ( )coherent ( )11 Exercise RV2-2) 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 500 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 E1=1, ε1=0, E2 =1, ε2 =π /4. Once made sure it is working, create a file with the folloing waves and polot their superposition from 0 to 5λ. 0 E1=1, ε1=0, E2 =1, ε 2 =10, E3 =2, ε3 =20 , E4 =3, 0 0 0 0 ε 4 =30 , E5 =2, ε5 =40 , E6 =1, ε6 =50 , E7 =1, ε7 =60 i π b) Next run the program for N=100 and ε =ε + π, where ε = i 1 100 1 2 and Ei = 2. This time create the phases and amplitudes inside the routine and don't read from a file. 12 Addition of waves: different frequencies I Mathematics behind light modulation and light as a carrier of information. Two propagating waves are superimposed E1 = E01 cos(k1x −ω1t) E2 = E01 cos(k2 x −ω2t) k1 > k2 and ω1 > ω2 with equal amplitudes and zero inital phases E = E1 + E2 = E01[cos(k1x −ω1t) + cos(k2 x −ω2t)] 1 1 using cosα + cos β = 2cos (α + β )cos (α − β ) 2 2 1 1 E = 2E01 cos (k1 +k2 ) x − (ω1 +ω2 )t⁄×cos (k1 − k2 ) x − (ω1 −ω2 )t⁄ 2 2 Need to simplify this 13 Addition of waves: different frequencies II E = 2E01 cos[km x −ωmt]×cos kx −ωt⁄ with the following definitions 1 Average angular frequency ≡ ω = (ω1 +ω2 ) 2 1 Average propagation number ≡ k = (k1 +k2 ) 2 1 Modulation angular frequency ≡ ωm = (ω1 −ω2 ) 2 1 Modulation propagation number ≡ km = (k1 −k2 ) 2 Time-varying modulation amplitude ≡ E0 (x,t) = 2E01 cos[km x −ωmt] » ÿ Superimposed wavefunction: E = E0 (x,t)cos kx −ωt⁄ For large ω if ω1 ≈ ω2 then ω >> ωm we will have a slowly varying amplitude with a rapidly oscillating wave 14 Irradiance of two superimposed waves with different frequencies 2 2 2 2 E0 (x,t) = 4E01 cos [km x −ωmt] = 2E01 1+ cos (2km x − 2ωmt)⁄ 2 Beat frequency ≡ 2ωm = ω1 −ω2 or oscillation frquency of the E0 (x,t) Amplitude, E0, oscilates at ωm, the modulation freuency 2 Irradiance, E0 , varies at 2ωm, twice the modulation frequency Two waves with different amplitudes produce beats with less contast.