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Topic 2: Scalar Diffraction V N I E R U S E I T H Y Modern Optics T O H F G E R D I N B U Topic 2: Scalar Diffraction Aim: Covers Scalar Diffraction theory to derive Rayleigh-Sommerfled diffraction. Take approximations to get Kirchhoff and Fresnel approx- imations. Contents: 1. Preliminary Theory. 2. General propagation between two planes. 3. Kirchhoff Diffraction 4. Fresnel Diffraction 5. Summary PTIC D O S G IE R L O P U P P A D E S C P I Scalar Diffraction -1- Autumn Term A S R Y TM H ENT of P V N I E R U S E I T H Y Modern Optics T O H F G E R D I N B U Scalar Wave Theory Light is really a vector electomagnetic wave with E and B field linked by Maxwell’s Equations. Full solution only possible in limited cases, so we have to make as- sumptions and approximations. Assume: Light field can be approximated by a complex scalar potential. (am- plitude) Valid for: λ Apertures and objects , (most optical systems). NOT Valid for: Very small apertures, Fibre Optics, Planar Wave Guides, Ignores Po- larisation. Also Assume: Scalar potential is a linear super-position of monochromatic compo- nents. So theory only valid for Linear Systems, (refractive index does not depend on wavelength). PTIC D O S G IE R L O P U P P A D E S C P I Scalar Diffraction -2- Autumn Term A S R Y TM H ENT of P V N I E R U S E I T H Y Modern Optics T O H F G E R D I N B U Scalar Potentials For light in free space the E and B field are linked by = ∇:E 0 = ∇:B 0 ∂E = ε ∇ ^ B µ 0 0 ∂t ∂B = ∇ ^ E ∂t which, for free space, results in the “Wave Equation” given by ∂2 2 1 E = ∇ E 0 c2 ∂t2 ; Assume light field represented by scalar potential Φ r t which MUST also obey the “Wave Equation”,so: ∂2Φ 2 1 = ∇ Φ 0 c2 ∂t2 Write the Component of scalar potential with angular frequency ω as ; = ω Φ r t u r exp ı t then substituting for Φ we get that, 2 2 ∇ + κ ] = [ u r 0 π=λ where κ = 2 or wave number. So that ur must obey Helmholtz Equation, (starting point for Scalar Wave Theory). PTIC D O S G IE R L O P U P P A D E S C P I Scalar Diffraction -3- Autumn Term A S R Y TM H ENT of P V N I E R U S E I T H Y Modern Optics T O H F G E R D I N B U Diffraction Between Planes y z y x x 0 z u(x,y;0) u(x,y;z) P P 0 1 In P0 the 2-D scalar potential is: ; = ; ux y;0 u0 x y where in plane P1 the scalar potential is: ; ux y;z where the planes are separated by distance z. ; ; Problem: Given u0 x y in plane P0 we want to calculate in u x y;z is any plane P1 separated from P0 by z. Looking for a 2-D Scalar Solution to the full wave equation. PTIC D O S G IE R L O P U P P A D E S C P I Scalar Diffraction -4- Autumn Term A S R Y TM H ENT of P V N I E R U S E I T H Y Modern Optics T O H F G E R D I N B U Fourier Transform Approach Take the 2-D Fourier Transform in the plane, wrt x ; y.soinplaneP1 we get ZZ ; = ; π + U u v;z u x y;z exp ı2 ux vy dxdy sincewehavethat ZZ ; = ; π + ux y;z U u v;z exp ı2 ux vy dudv ; then if we know U u v;z in any plane, then we can easily find ; ux y;z the required amplitude distribution. ; The amplitude ux y;z is a Linear Combination of term of ; π + U u v;z exp ı2 ux vy These terms are Orthogonal (from Fourier Theory), So each of these terms must individually obey the Helmholtz Equa- tion. Note: the term: ; π + U u v;z exp ı2 ux vy ; =κ; =κ in a Plane Wave with Amplitude U u v;z and Direction u v , π=λ where κ = 2 . PTIC D O S G IE R L O P U P P A D E S C P I Scalar Diffraction -5- Autumn Term A S R Y TM H ENT of P V N I E R U S E I T H Y Modern Optics T O H F G E R D I N B U Propagation of a Plane Wave From Helmholtz Equation we have that 2 2 ∇ + κ ] ; π + = [ U u v;z exp ı2 ux vy 0 Noting that the exp terms cancel, then we get that 2 ∂ ; U u v;z 2 1 2 2 π ; = + 4 u v U u v;z 0 ∂z2 λ2 so letting 2 1 2 2 γ = u v λ2 then we get the relation that ∂2 U 2 πγ = 8 ; +2 U 0 u v ∂z2 With the condition that ; = ; = f ; g U u v;0 U0 u v F u0 x y we get the solution that ; = ; πγ U u v;z U0 u v exp ı2 z This tells us how each component of the Fourier Transform propa- gates between plane P0 and plane P1,so: ZZ ; = ; πγ π + ux y;z U0 u v exp ı2 z exp ı2 ux vy dudv which is a general solution to the propagation problems valid for all z. PTIC D O S G IE R L O P U P P A D E S C P I Scalar Diffraction -6- Autumn Term A S R Y TM H ENT of P V N I E R U S E I T H Y Modern Optics T O H F G E R D I N B U Free Space Propagation Function Each Fourier Component (or Spatial Frequency) propagates as: ; = ; πγ U u v;z U0 u v exp ı2 z Define: Free Space Propagation Function ; = πγ H u v;z exp ı z so we can write: ; = ; ; U u v;z U0 u v H u v;z ; Look at the form of H u v;z . 2 2 2 1 + γ = u + v λ2 2 2 2 =λ γ Case 1: If u + v 1 then is REAL ; H u v Phase Shift so all spatial frequency components passed with Phase Shift. 2 2 2 > =λ γ Case 2: If u + v 1 them is IMAGINARY ; = πjγj H u v exp 2 z 2 2 2 ; + > =λ So Fourier Components of U0 u v with u v 1 decay with Negative Exponential. (evanescent wave) PTIC D O S G IE R L O P U P P A D E S C P I Scalar Diffraction -7- Autumn Term A S R Y TM H ENT of P V N I E R U S E I T H Y Modern Optics T O H F G E R D I N B U Frequency Limit for Propagation λ In plane P1 where z then the negative exponential decay will remove high frequency components. 2 2 2 ; = + > =λ U u v;z 0 u v 1 ; so Fourier Transform of ux y;z is of limited extent. =λ Maximum spatial frequency when u = 1 , this corresponds to a grat- ing with period λ. d θ λ > λ d sinθ = n for d d f λ No diffraction when d < . Information not transferred to plane P1. PTIC D O S G IE R L O P U P P A D E S C P I Scalar Diffraction -8- Autumn Term A S R Y TM H ENT of P V N I E R U S E I T H Y Modern Optics T O H F G E R D I N B U Convolution Relation We have that ; = ; ; U u v;z H u v;z U0 u v;z so by the Convolution Theorem,wehavethat ; = ; ; ux y;z h x y;z u0 x y;z ; ; With ux y;z the distribution in P1 due to u0 x y;z in P0. We then have that ZZ ; = πγ π + hx y;z exp ı2 z exp ı2 ux vy dudv 2 2 2 < =λ u +v 1 where r 1 2 2 + γ = u v λ2 “It-Can-Be-Shown” (with considerable difficulty), that κ 2π ∂ exp ı r ; = hx y;z λ2 ∂z κr where be have that π 2 2 2 2 2 + + κ = r = x y z and λ we therefore get that κ 1 1 z exp ı r ; = hx y;z ı λ κr r r which is known as “The Impulse Response Function for Free Space Propagation” PTIC D O S G IE R L O P U P P A D E S C P I Scalar Diffraction -9- Autumn Term A S R Y TM H ENT of P V N I E R U S E I T H Y Modern Optics T O H F G E R D I N B U Point Object y y x x r aδ( x,y ) 0 z P0 u(x,y;z) P1 In P0 we have a Delta Function, so: ; = δ ; u0 x y a x y So in plane P1 we have ; = ; ux y;z ah x y;z κ a 1 z exp ı r = ı λ κr r r where r is the distance from: ; ; 0 0;0 x y;z So the intensity in plane P1 is given by: b2 1 z 2 ; = + ix y;z 1 r2 κ2r2 r where the λ2 has been incorperated into the constant b.
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