Fourier Approach to Wave Propagation

Transcription

1 Phys 531 Lecture October 005 Fourier Approach to Wave Propagation Last time, reviewed Fourier transform Write any function of space/time = sum of harmonic functions e i(k r ωt) Actual waves: harmonic functions restricted k = n ω /c Today, apply Fourier to wave propagtion Start to study diffraction 1 Outline: Diffraction Fourier approach Transfer function Fresnel approximation Gaussian example Note: we won t be following book very well - Hecht Ch. 10 takes different approach - Ch. 11: Fourier approach, based on Ch. 10 Next time, continue development

2 Diffraction Previously said ray optics fails - small feature sizes a - long propagation distances d Need d a /λ Otherwise see diffraction light spreads out Demo! 3 Want to understand diffraction and calculate effects Note: already have one way to understand: scattering picture Recall HW : a Plane wave incident on sphere diameter a 4

3 Ray optics: Transmitted light has shadow diameter a Propagates indefinitely Wrong! Scattering picture: Shadow due to forward scattered field In shadow, E tot = E inc + E scat 0 To sides, E scat fields cancel out 5 But forward scattering not perfectly forward at angle θ λ/a, E scat significant θ a At small angle, E scat from all atoms in phase 6

4 Similar to two slit interference a θ Get large peak when fields from slits in phase 7 Diffraction in scattering picture: E scat fields don t cancel perfectly for finite object General prediction: Diffraction angle θ λ/a Valid, but hard to calculate more precisely Come back to idea later 8

5 Fourier Treatment Use math Set up problem: Suppose monochromatic field, frequency ω propagating towards +z (perhaps at angle) Specify E(r, t) in plane z = 0 (= plane of slits, aperture) Ask: What is E(r, t) for z > 0? Don t worry about 3D objects like sphere Sphere disk 9 Monochromatic: write E(r, t) = E(r)e iωt just consider E(r) Field known at z = 0: Write E(x, y, z = 0) = A(x, y) Call A(x, y) = aperture function Usually look at diffraction from aperture A(x, y) = 0 for points outside aperture A(x, y) = E(x, y, 0) for points inside aperture (Stop using A for amplitude) 10

6 Example: Plane wave E inc = E 0 ei[k(z cos θ+x sin θ) ωt] travelling at angle θ to z-axis Incident on square aperture side a, centered at x = x 0, y = y 0 Then A(x, y) = E 0 e ikx sin θ ( x 0 else x 0, y y 0 < a/) Think of A(x, y) as initial condition want to solve for E(x, y, z) 11 Apply Fourier ideas First thought: E(x, y, z) = 1 (π) 3 E(k)e ik r d 3 k If we knew E(k), problem solved Do have A(x, y) = 1 (π) 3 E(k)e i(k xx+k y y) d 3 k Can we invert to get E(k) from A(x, y)? No: E(k) = E(x, y, z)e i(k xx+k y y+k z z) dx dy dz 1

7 Second thought: Have A(x, y) = 1 (π) with A(k x, k y ) = No problem getting A(k x, k y ) A(k x, k y )e i(k xx+k y y) dk x dk y A(x, y)e i(k xx+k y y) dx dy Can we get E(x, y, z) from A? Yes! 13 Develop with example: Suppose A(x, y) = E 0 e i(βx) harmonic function What function E(x, y, z) would give us this A? Already know answer: E(r) = E 0 e i(βx+k zz) for some kz Plane wave In this case, easy to guess form of solution 14

8 What is k z? Have k = k x + k y + k z = n ω /c ω, n given For our function k x = β and k y = 0, so k z = k β k z = k β Full solution is E(r) = E 0 e i(βx+z k β ) Question: Could I use k z = k β instead? 15 In general, if A(x, y) = E 0 e i(k xx+k y y), get solution E(r) = E 0 e i(k xx+k y y+κz) for κ k kx ky Solution to problem for particular form A(x, y) Important to understand this! Question: If A(x, y) = E 0, what is E(x, y, z)? 16

9 With Fourier transform, A(x, y) = sum of harmonic funcs So solution E(r) = sum of plane waves with κ = k kx ky If A(x, y) = 1 (π) A(k x, k y )e i(k xx+k y y) dkx dk y then E(r) = 1 (π) A(k x, k y )e i(k xx+k y y+κz) dkx dk y 17 Simple example: A(x, y) = E 0 cos(βx) One solution: A(x, y) = E ( 0 e iβx + e iβx ) = sum of harmonic funcs Then E(r) = E 0 (e i(βx+z k β ) + e i( βx+z k β ) = E 0 e iz k β cos(βx) 18

10 Another solution: Recall transform of e iβx is πδ(k x β) So So A(k x, k y ) = π E 0 [ δ(kx β) + δ(k x + β) ] δ(k y ) E(r) = 1 (π) A(k x, k y )e i(k xx+k y y+κz) dkx dk y = 1 { [ π E0 e i(βx+κz) + e i( βx+κz) ]} (π) for κ = = E 0 e iκz cos(βx) k β 19 Either method fine Note, solution is physically interesting: Two plane waves, angle θ = tan 1 (β/κ) θ Implement with glass plate, sinusoidal markings simple diffraction grating 0

11 Another example: Plane wave normally incident on square hole A(x, y) = Then A(k x, k y ) = = 1 ( x, y < a/) 0 (else) ( a/ = a sinc A(x, y)e i(k xx+k y y) dx dy a/ eik xx dx ( ) kx a ) ( a/ sinc a/ eik yy dy ( ) ky a ) 1 and E(r) = a (π) sinc ( ) kx a sinc e i ( k x x+k y y+z ( ) ky a k k x k y ) dk x dk y Can t do this integral analytically - square root in exponent is hard! Need to introduce some approximations First, study what we ll approximate

12 Transfer Function General result E(r) = 1 (π) A(k x, k y )e i(k xx+k y y+κz) dkx dk y Can write as E(r) = 1 (π) for H(k x, k y ) = e iz k k x k y A(k x, k y )H(k x, k y )e i(k xx+k y y) dk x dk y Call H = transfer function for free space 3 Note H depends on z = propagation distance More general: H d (k x, k y ) = e id k k x k y propagates field from z 0 to z 0 + d Call E(x, y, z 0 ) = input, E(x, y, z 0 + d) = ouput Linear system: output depends linearly on input Transfer function = linear coefficients but in Fourier space 4

13 H d = e id k k x +k y For k x + k y < k, have H = 1 k z is real But for k x + k y > k, have H = e d k x+k y k < 1 k z is imaginary! Plot magnitude and phase 5 6

14 Is it possible to have k x + k y > k? Yes: can make arbitrary apertures If feature size λ, will have A(k x, k y ) 0 for large k x, k y Example: square hole with a = 10 nm For large k x, k y, H decays with d E(r) decays with d Have seen before: evanescent wave 7 For aperture with small hole, field doesn t propagate away Can t fit wave through hole smaller than λ/π Limits imaging resolution of microscope: images of small features don t propagate But, can measure evanescent wave itself: called near field microscopy Place detector very close to surface resolution surface distance/π 8

15 Fresnel Approximation Note, large k x, k y large propagation angle θ kx + ky sin θ = k But usually interested in small θ paraxial Evanescent wave behavior irrelevant Suggests approximation k k x k y k k x + k y k so H d e ikd e id(k x +k y )/k 9 Called Fresnel approximation Gives diffracted field E(x, y, z) = e ikz (π) A(k x, k y )e i(k xx+k y y) e iz(k x+k y )/k dkx dk y Integrals more manageable Still hard to get analytic result but numerical integration is straightforward 30

16 Valid when next term in expansion is small Next term in Taylor series of d = d(k x + ky ) 8k 3 For propagation angle kx + ky θ, need kdθ 4 1 k k k x k y More physics of Fresnel approxmation next class For now: one example where analytic solution possible 31 Gaussian beam Suppose A(x, y) = E 0 e (x +y )/w 0 Gaussian function, width w 0 Make with glass filter: - transparent in center - smoothly becomes opaque at edge Turns out, this field produced naturally by laser practically important Calculate E(x, y, z) 3

17 Need transform A(k x, k y ) Transform of Gaussian e x /w0 is w 0 πe w0 k x /4 So A(k x, k y ) = E 0 πw0 e w 0 (k x+ky )/4 With Fresnel approximation E(r) = eikz (π) E 0πw 0 e w 0 (k x +k y )/4 e iz(k x +k y )/k e i(k xx+k y y) dkx dk y Define q = w 0 + iz/k 33 Then E(r) = eikz (π) E 0πw 0 e q (k x+k y )/4 e i(k x x+k y y) dk x dk y = e ikz E 0 πw0 1 π e q kx /4 e ik xx dkx 1 π e q ky /4 e ik yy dky Inverse transforms of Gaussians 34

18 So Solved! E(r) = E 0 e ikz πw 0 = E 0 e ikzw 0 q e (x +y )/q Field remains Gaussian But complicated since q is complex ( ) ( ) 1 q π e x /q 1 q π e y /q 35 Calculate E(r) Use 1 q = 1 w 0 + iz/k for = w 0 iz/k w z /k 1 w ( 1 i z ) kw0 w = w 0 + 4z k w 0 = q 4 w 0 36

19 Then E(r) = E 0 w4 0 q 4e (x +y )/w = E 0 w 0 w e (x +y )/w Irradiance remains Gaussian, but size expands w(z) = w 0 + λ z π w 0 λz πw 0 Divergence angle θ = λ/πw 0 λ/a feature size a 37 For large w 0, divergence is slow Light propagates uniformly Call solution Gaussian beam always Gaussian profile More on Gaussian beams later in course 38

20 Summary: Diffraction due to wave nature of light Can use Fourier analysis to calculate - E(x, y, 0) A(k x, k y ) - know k z = (k k x k y ) 1/ Transfer function: E(z) E(z + d) A d = H d A H d = e ik zd Fresnel approximation: expansion of H d Apply to Gaussian beam laser beam 39