Astronomy 203/403, Fall 1999 Physical optics 12. Lecture, 12 October 1999 12.1 Scalar diffraction Suppose you’ve designed and built an optical system for which you are assured by your ray-tracing program that the aberrations are zero, and you’re using it in a place where air turbulence does not distort the wavefronts of light. Then you will see perfect images, but you may be disappointed to discover that perfect images of point objects are not points. This has to do, of course, with the wave nature of light, and the breakdown of the ray approximation: diffraction. Diffraction is the only effect that blurs images with completely smooth, undistorted wavefronts (i.e. coherent light); when an optical system generates a spot diagram for a point object that is significantly smaller than the diffraction blur (or atmospheric-turbulence blur, whichever is larger), then the images it produces are dominated by diffraction, and are as perfect as is allowed by physics. In the following we will derive the important features of diffraction under two approximations frequently introduced in a first exposition of the subject: the scalar-field and far-field regimes. Consider one polarization component of an electric field E. Let us treat this component as a scalar quantity in the description of the propagation of electromagnetic waves. This component obeys the wave equation: 1 ∂ 2 ∇=2E E . (12.1) ct2 ∂ 2 It turns out that there is a spherical wave solution to the wave equation, η κω− Ertaf, = 0 eiraf t , (12.2) r κπλ= ωπν= η where 2 / is the wavenumber and 2 the angular frequency of the wave, and 0 is a constant. This is easily demonstrated: for the left side of Equation 12.1, we have − ωκκ 1 ∂ ∂ ηκe i td F e irie irI ∇=2E r2 Ert(,) =0 r2 − + r2 ∂r ∂r r2 dr G r2 r J H K (12.3) −itω η e κκ κ =−+−=−0 ejieκκκir ie ir22 re ir κ Ertaf,, r2 while on the left side there is 2 irκ 2 2 irκω− i t 1 ∂ ηωηe d − ω − e E ==0 e it 0 =−κ 2Ertaf, (12.4) ct2 ∂ 2 cr2 dr2 cr2 (Q.E.D.). Equation 12.2 is a description of a wave propagating from a point source (at which the wave’s amplitude is singular), for which the position of a given crest of the wave, with constant κωrt− , moves η toward increasing r (i.e. in the r direction), and for which the amplitude decreases as 1/r. The factor 0 , called the source strength, turns out to be related to the electric field at the source of the spherical wave, as 1999 University of Rochester 1 All rights reserved Astronomy 203/403, Fall 1999 we shall see. This solution is just what we need as a building block to use with a classic technique for the calculation of the propagation of light: Huygens’ principle. Consider a plane electromagnetic wave incident normally on a hole in an otherwise opaque screen. Some of the wave will pass through the hole and continue to propagate. What will be the electric field EF due to the light that got through, at some point F a great distance away from the hole? x da’ r’ r = r-r’ r F θ θ y y θ x z Figure 12.1: screen, aperture and observation point for calculation of diffraction of plane wave incident normally on the screen and travelling in the +z direction. The normal approach to this problem is to note that the hole can considered to have a constant electric field at some instant in time, and for each infinitesimal element of it to be an independent source of spherical ′ waves of the form we just derived, so that the contribution to EF by an area element da is ′′ ′ = EA(,)xyda it(κωr-) dEF r e , (12.5) =−′ where EA (x’,y’) is the source strength per unit area, and r rr is the vector distance from the area element to point F. The dissection of the aperture into independent spherical wave emitters and the addition of the results to describe further propagation is the essence of Huygens’ principle; our restriction to a single polarization component treated as a scalar simplifies the addition: we can integrate this expression over the aperture in a straightforward fashion to obtain EF . Each element of the aperture lies a different distance from point F, and the phase differences between the spherical waves arising from these pathlength differences will provide constructive or destructive interference. The integration of Equation 12.5 becomes quite straightforward if a simplifying approximation is made. With a glance at Figure 12.1, we can write expressions for the two relevant distances: 1999 University of Rochester 2 All rights reserved Astronomy 203/403, Fall 1999 rxyz=++222 22 (12.6) 2 xx′ + yy′ xy′ + ′ r =−afxx′ 2 +−bg yy′ += z2 r12 − + . r2 r2 If the distance to the observation point F is much greater than any dimension of the aperture (i.e. if r ≅any r’), then the terms with primed coordinates in Equation 12.6 are both small compared to unity, and the second much smaller than the first. Here and henceforth we will apply this far-field approximation, and employ the binomial theorem, Equation 7.9, to obtain a first-order approximation for the distance between an area element and point F: xx′ + yy′ F xx′ + yy′I r ≅−r 12≅−rG 1 J . (12.7) r22H r K Thus κκ− κ ′+ ′ eeeiirr = ixxyyrbg/ . (12.8) − κ ′+ ′ The use of a first-order approximation in an exponent makes the factor e ixxyyrbg/ first order in r’/r. (To see that this is true, consider for example the imaginary part of its first term, sinbgκκxx′≅′ / r xx / r , which is manifestly first order in x’/r.) When using Equation 12.7 in Equation 12.5, we thus must use the zeroth-order approximation for the denominator, r ≅ r , in order for the resulting expression to be first order: E (,)xyda′′ ′ κω − κ ′+ ′ dE = A eeir( -) t ixxyyrbg/ , (12.9) F r whence ir(κω-) t e −ixxyyrκ ′+ ′ / E =′′′E (,)xyebg da . (12.10) F r z A aperture Define κκθκ=≅ κκθκ =≅ xxxr/, yy yr / , (12.11) where the latter two approximate equalities apply if the θs are small angles (which we also assume henceforth), and then κω ir( -) t − κκ′+ ′ e ixe xy yj E κκ,(,)= E xye′′ dxdy ′′ . (12.12) Fxyejr zz A aperture ′′′ If furthermore we define a source strength per unit area EAbgxy, such that EE′′′bgbgxy,,= xy′′ inside, and AA (12.13) = 0 outside the aperture, 1999 University of Rochester 3 All rights reserved Astronomy 203/403, Fall 1999 then the aperture comes to be built into the source strength, and we have κω ∞ ∞ ir( -) t −′+′κκ κκ =′′′′′e ixejxy y EFxyej,(,)z z EA xye dxdy . (12.14) r −∞ −∞ Apart from the leading factor, the right-hand side of this expression is just the two-dimensional Fourier ′′′ transform of the source strength per unit area, EAbgxy, . We usually write the one-dimensional Fourier transform and its inverse transform as ∞ 1 − fx()= Fseaf ixs ds , π z 2 −∞ ∞ (12.15) 1 Fsaf= f af xeixs dx . π z 2 −∞ The corresponding two-dimensional transforms are ∞ ∞ 1 −+ fxy= Fsteixsytbg dsdt (,)π z z af , , 2 −∞ −∞ ∞ ∞ (12.16) 1 + Fst= f xyeixsytbg dxdy af,,π z z bg . 2 −∞ −∞ With these definitions it can be shown that ∞ ∞ ∞ ∞ 2 2 zzz fbg x,,, y dxdy= z Fa s tf dsdt (12.17) −∞ −∞ −∞ −∞ as we will presume has been done for you before, in an applied math class. This relation, called Rayleigh’s theorem, is useful in electricity and magnetism because there are many ways to express energy conservation by its use. That’s what we will do here, to discover the nature of the hitherto obscure source ′′′ strength per unit area EAbgxy, . First we group the terms of Equation 12.14 to look exactly like a Fourier transform, ∞ ∞ κω π ir( -) t ′′′−′+′κκ κκ = 1 2 exyEA(,) ixejxy y ′′ EFxyej, π z z edxdy , (12.18) 2 −∞ −∞ r and then we can apply Rayleigh’s theorem, to obtain ∞ ∞ ∞ ∞ ir(κω-) t 2 2 ∞ ∞ 2 2πexyE′′′(,) 4π 2 Eddκκ, κκ= A dx′′= dy E ′(,)x ′′ y dx ′′ dy . (12.19) zzz Fxyej x y zz2 z A −∞ −∞ −∞ −∞ r r −∞ −∞ Using Equation 12.11 we can change variables on the left hand side of 12.19, as follows: κ κ π 2 κκ==dx dy 4 ddxy dxdy , r r λ22r 1999 University of Rochester 4 All rights reserved Astronomy 203/403, Fall 1999 so that 2 ∞ ∞ 2 ∞ ∞ 44π 2 π 2 Edxdy=′′′′′E (x , y ) dx dy . (12.20) 22 zzz FA2 z λ r −∞ −∞ r −∞ −∞ Now cancel the common factors, move the factor of λ2 to the right-hand side, and multiply through by c /8π , and the terms look familiar: ∞ ∞ ∞ ∞ c 2 =′′′′′c λ 2 ππzzz EdxdyFAz E (xy , ) dxdy . (12.21) 88−∞ −∞ −∞ −∞ The left-hand side is clearly the power in electromagnetic radiation passing through a planar surface through point F. The right-hand side must therefore be the power passing through the aperture. We identify the source strength per unit area with the electric field in the plane of the aperture: ′′= λ ′′′−itω ExyNAbg,(,)E xye (12.22) so that ∞ ∞ ∞ ∞ c 22=′′c ππzzz EdxdyFNz Edxdy (12.23) 88−∞ −∞ −∞ −∞ simply expresses energy conservation. Now we combine Equations 12.22 and 12.14 to obtain κ ∞ ∞ ir −′+′κκ κκ =′′′′e ixejxy y EtFxyej,,λ z z ExyteN (,,) dxdy .