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Today in 218: by a circular or obstacle ‰ Circular-aperture diffraction and the Airy pattern ‰ Circular obstacles, and Poisson’s spot.

V773 Tau: AO off V773 Tau: AO on (and brightness Neptune orbit diameter, turned way up) seen from same distance

2 April 2004 Physics 218, Spring 2004 1 The circular aperture

Most experimental situations in (e.g. ) have circular , so the application of the Kirchhoff integral to diffraction from such apertures is of particular interest. We start with a plane wave incident normally on a circular hole with radius a in an otherwise opaque screen, and ask: what is the distribution of the intensity of on a screen a distance Ra >> away? The field in the aperture is constant, spatially: −itω ExytEeNN(,,)′′ = 0 , and the geometry is as follows:

2 April 2004 Physics 218, Spring 2004 2 xr′′= cosφ ′ x The circular aperture yr′′= sinφ ′ φ (continued) da′′′′= rdrdφ da’ Xq= cosΦ r’ Yq= sin Φ a X r y R

Φ Small angles: θ q X kq cos Φ kk≅= z,Z x rr kq sin Φ k ≅ Y y r

2 April 2004 Physics 218, Spring 2004 3 The circular aperture (continued)

Thus, ikr ∞∞ e −+ikx()xy′′ ky Ekkt(),,= Exyte (,,)′′ dxdy ′′ Fxyλr ∫∫ N −∞ −∞ a eikr = λ dr′′ r r ∫ 0 2π −it ⎛⎞ikr′ q ¥ dE′′′N0φφφ e exp⎜⎟− () cos cosΦ+ sin sin Φ ∫ ⎝⎠r 0 ω ikr()− t a 2 EeN0 ⎛⎞ikr′ q =−−Φω dr′′ r d ′exp⎜⎟ cos() ′ , rr∫∫ ⎝⎠ 00λ π

2 April 2004 Physics 218, Spring 2004φφ 4 The circular aperture (continued)

The aperture is symmetrical about the z axis, so we expect that the answer will be independent of the “screen” azimuthal coordinate Φ; without loss of generality, then, we can take Φ = 0. The integral over φ′ becomes 2π ⎛⎞ikr′ q I =−dφφ′′exp⎜⎟ cos . ∫ ⎝⎠r 0 Don’t try to integrate that directly; it’s a of the first kind, order zero: π 2π 1 iucos v ⎛⎞kr′ q JuJu00()−= () = e dv ⇒I =2.π J 0⎜⎟ 2 ∫ ⎝⎠r 0

2 April 2004 Physics 218, Spring 2004 5 Flashback: Bessel functions

The Bessel function of the first kind, of order m, can be represented by the integral −m 2π i + Ju()= eimvu()cos v dv. m 2π ∫ 0 Bessel functions of different order are related by the u d ⎡⎤mm m m uJ() u=⇔= uJ−−() u uJ u vJ () vdv . () du ⎣⎦mm11 m∫ m 0 Recurrence relations of are very useful when one has to integrate those special functions, as you’re about to see.

2 April 2004 Physics 218, Spring 2004 6 Flashback: Bessel functions (continued)

J (u) 1 0 Zeroes of the Bessel functions

J1 (u) J0 J1 J2 J (u) 0.5 2 2.405 0 0

5.520 3.832 5.136

0 8.654 7.016 8.417

11.792 10.174 11.620

0.5 14.931 13.324 14.796

10 50 510 u

2 April 2004 Physics 218, Spring 2004 7 The circular aperture (continued)

So the far field is ikr( −ω t) a 2πEeN0 ⎛⎞kqr′ EF () q, t= drrJ′′0 ⎜⎟ rr∫ ⎝⎠ λ 0 − 2 kaq/ r 2 Eeikr() t ω⎛⎞r = Nπ 0 ⎜⎟ vJ() v dv . rkq∫ 0 ⎝⎠ 0 Now use the recurrence relation,λ with m = 1: u uJ10() u= ∫ vJ v () dv ; 0 ikr()−ω t 2 2π EeN0 ⎛⎞r ⎛⎞⎛⎞kaq kaq EqtF (),.= ⎜⎟⎜⎟⎜⎟ J1 rkqrr⎝⎠⎝⎠⎝⎠ λ 2 April 2004 Physics 218, Spring 2004 8 The circular aperture (continued)

Rearrange the field in a somewhat more convenient form:

2 ikr( −ω t) π aEN0 e r ⎛⎞kaq EqtF (),2,= J1 ⎜⎟ rkaqr⎝⎠ ikr()λ − tω θ EAe 2Jka() or Et() ,= N0 1 , F rka λθ22 2 c ∗ cEθN0 A ⎡⎤2Jka1 () whence IkaFFF()==θθθ E()() , tE , t ⎢⎥ . 8 8 22r ⎣⎦ka πθ This leaves a minor problem: the expression is indeterminate at ka θ = 0. But the recurrence relation can help us again: πλ θ 2 April 2004 Physics 218, Spring 2004 9 The circular aperture (continued)

Take the recurrence relation at m = 1 and use the chain rule: d dJ1 uJ011() u=⎡⎣⎦ uJ() u ⎤=+ J u () u() u du du () dJ Ju() Ju=+1 () u 1 . 0 du u Note that JJ01(01) == and (00:) dJ 1 ()u dJJu() dJ 1=+11() 0 lim1 =+() 0 lim du duuu→→00 u du 1 dJ = 20,1 () or du () Ju dJ 1 lim1 ==1 () 0 . u→0 udu 2 2 April 2004 Physics 218, Spring 2004 10 The circular aperture (continued)

This resolves the indeterminacy: 222 22 cENN00 A⎡⎤1 cE A IF ()02==⎢⎥ , λπλ88πλ 22rr⎣⎦2 22 2 θ ⎡⎤2Jka1 () IkaIFF()= 0. ()⎢⎥Airy pattern ⎣⎦ka θ θ Because also has zeroes at finite values of θ , θ J1 θ ka IkaF ( ) has a set of concentric rings for which the intensity is zero (dark rings) The first of these lies at ka θ 1 = 3.832, or π 3.832 3.832 λ λ == =1.22 . First 1 ka2 a D dark ring

2 April 2004 Physics 218, Spring 2004 11 The Airy pattern

Linear scale Logarithmic scale

1 1

0.8 0.1

Ika( θ ) 0.6 I ()0 0.01 0.4 3 10 0.2

4 0 10 5 0 5 10 10 10 5 0 5 10

kaθ

2 April 2004 Physics 218, Spring 2004 12 The Airy pattern (continued)

Linear scale Logarithmic scale

2 April 2004 Physics 218, Spring 2004 13 The Airy pattern (continued)

A young triple-star system, T Tau, observed at λ = 2.2 µm with adaptive optics on the Palomar 200- inch . The brightest star has saturated the detector in the . Note the extensive nest of concentric dark rings around it. (Linear scale.)

2 April 2004 Physics 218, Spring 2004 14 The opaque circular obstacle

We can handle the case of diffraction by a circular obstacle quite easily, using the result just obtained. For the field, ikr( −ω t) ∞ 2πEeN0 ⎛⎞kqr′ EF () q, t= drrJ′′0 ⎜⎟ rr∫ ⎝⎠ λ a − ∞ ikr() tω ′ 2 EeNπ 0 ⎛⎞kqr = drrJ′′0 ⎜⎟ rr∫ ⎝⎠ 0 λ ikr()− t a 2 EeN0 ⎛⎞kqr′ − drrJ′′0 ⎜⎟ πrr∫ ω ⎝⎠ 0 2 ikr− t() ikz()− t EaeN0 λ r ⎛⎞kaq =−EeN012, J⎜⎟ ω r kaq⎝⎠ r 2 April 2004 Physics 218, Spring 2004ω 15 π

λ The opaque circular obstacle (continued) =−2coskr( 1 z r) and for the intensity, =−2 coskr () 1 cosθ c ∗ Iqt(), = EE θ2 2 FFFπ ⎛⎞ kq 8 ≅=2coskr ⎜⎟ 2cos 2 ⎜⎟ ⎡ ⎛⎞2 ⎝⎠22r ckar2 ⎢ ⎛⎞kaq =+EJπN0112⎜⎟⎜⎟ 82⎢ ⎜⎟r kaq⎝⎠ r ⎣ ⎝⎠ 2 ka r ⎛⎞kaq ik() r−−− z ik r ()z ⎤ −+2Jee1 ⎜⎟( ⎥ ) rkaq⎝⎠ r ⎦⎥ 2 ⎡⎤⎛⎞222 ckarkar2 ⎢⎥⎛⎞kaq2 ⎛⎞ kaq kq =+EJN0112⎜⎟⎜⎟ − 2cos. J 1 ⎜⎟ 82⎢⎥⎜⎟r kaq⎝⎠ r r kaq ⎝⎠ r 2 r ⎣⎦π ⎝⎠

2 April 2004 Physics 218, Spring 2004 16 The opaque circular obstacle (continued)

This result has the curious property of not being zero inside the shadow of the obstacle. In fact, there’s a sharp peak exactly in the center, with peak intensity ⎡ 2 ⎤ ckaka⎛⎞222 Iqt(),1=+− E2 ⎢ ⎜⎟ ⎥ . FN82π 0 ⎢ ⎜⎟rr⎥ ⎣ ⎝⎠ ⎦

And there are concentric bright and dark rings that also lie within the shadow, though generally it is much darker there than it is outside the shadow.

2 April 2004 Physics 218, Spring 2004 17 The opaque circular obstacle (continued)

Diffraction patterns at λ = 0.635µ m, seen 5 m away from 0.09375 inch, 0.15625 inch, and 0.1875 inch diameter spheres (Ioan Feier, Horst Friedsam and Merrick Penicka, Argonne National Laboratory).

2 April 2004 Physics 218, Spring 2004 18 Poisson’s spot

Not all effects named after famous physicists are meant to honor their namesakes. ‰ In 1818, the French Academy, led by neo-Newtonians like Laplace, Biot and Poisson, offered a prize for the best work on the theme of diffraction, expecting that the result would be a definitive refutation of the wave theory of light. ‰ Fresnel, supported by Ampère and Arago, offered a paper in which he developed the scalar theory of diffraction in much the same way we did, based on the wave theory. ‰ During Fresnel’s talk, Poisson pointed out that one of the consequences of Fresnel’s theory was the intensity peak in the center of circular shadows that we just found. 2 April 2004 Physics 218, Spring 2004 19 Poisson’s spot (continued)

‰ Poisson did this, of course, because he thought such a result was ridiculous; he meant it as a fatal objection to Fresnel’s theory. ‰ But right after the talk, Arago went into his lab, observed the intensity peak and concentric rings in the shadow directly, and proceeded to demonstrate it to the judges. ‰ Thus Fresnel was awarded the prize, the corpuscular theory of light stood refuted (until Einstein and Planck came Nick Nicola along), and the intensity peak has been (University of known ever since as Poisson’s spot. Melbourne) 2 April 2004 Physics 218, Spring 2004 20