PHY 6347 Spring 2018 Homework #10, Due Friday, April 6

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1 PHY 6347 Spring 28 Homework #, Due Friday, April 6. A plane wave ψ = ψ e ik x is incident from z < on an opaque screen that blocks the entire plane z = except for the opening 2 a < x < 2 a, 2 b < y < 2 b. (a) Compute the diffracted field ψ scatt from the scalar Kirchhoff-Neumann relation (.85) in the large-r regime. Asitsays, startingfromtheneumannversionof(.85),or(.83),withintegrandg N ( ψ/ n ) S in place of the Dirichlet ψ S ( G D / n ), the diffracted field is a/2 b/2 ψ diff = dx dy ˆÒ (ie ik x ) e ikr e ikˆr x. a/2 b/2 r The Neumann Green s function onthe boundary is twice e ik x x /4πÜ Ü. The Neumann boundary condition ψ/ n = ˆÒ ψ leads to the angular dependence ˆÞ i = ikcosθ. The area integral factors into x - and y -integrals that can be done separately, a/2 dx e ik(sinθ cosφ sinθcosφ)x = a sin( 2 ka x) a/2 2 ka, x b/2 dy e ik(sinθ sinφ sinθsinφ)y = b sin( 2 kb y) b/2 2 kb, y where x and y are the factors in parentheses in the exponentials, x = sinθcosφ sinθ cosφ, y = sinθsinφ sinθ sinφ. The diffracted wave is then ψ diff = ikabψ cosθ e ikr r sin( 2 ka x) ( 2 ka x) sin( 2 kb y) ( 2 kb y). (b) Compute the angular distribution of power dp/dω. The angular distribution of power is dp dω = r2 ψ 2 = k2 a 2 b 2 ψ 2 cos 2 θ sin( 2 ka x ) 2 sin( 2 kb y ) 2. ( 2 ka x) ( 2 kb y) You could also write these in terms of Cartesian x = a x and y = b y. Write sinx/x = S(x), the so-called sinc function; then, the outgoing power per solid angle is dp dω = k2 a 2 b 2 ψ 2 cos 2 θ S 2 ( 2 ka x)s 2 ( 2 kb y).

2 2 For long wavelength, or small ka, kb, S, For small λ, function S is confined to a small range about the incident direction. Within the first zero of the sin function, 2 ka sinθcosφ sinθ cosφ < π, sinθsinφ sinθ sinφ < λ a, As ka, kb become large S narrows to a δ-function. For what the intensity looks like, see the square diffraction link on the course web page. (c) Write an expression for the total power transmitted through the opening in the forward direction. Compute the transmission coefficient T in the limits ka, kb and ka, kb. Estimate the fraction of the power inside a microwave oven that leaks out through the openings in the screen (a microwave oven operates at f = 245MHz). The incoming power through the opening is P = Ê d 2 aˆþ Ë = ab ψ 2 cosθ ; the transmitted power in the forward direction is the integral over < θ < π 2, π/2 P = dφ and the transmission coefficient T = P/P is T = k2 ab cosθ sinθdθ dp dω ; dωs 2 ( 2 ka x)s 2 ( 2 kb y). The integral cannot be done in general in closed form, but is easy in the two limits. As ka, kb, S, and Ê dω =. and T = k2 abcosθ (ka,kb ). In the Dirichlet version this becomes k 2 ab/4πcosθ, which looks strange, the transmitted power becomes greater for oblique incidence. As ka and kb become large, the functions S 2 ( 2 ka x) and S 2 ( 2 ka y) become narrower and narrower, effectively becoming δ-functions in x and y, or θ θ, φ φ. In the integration over solid angle, it is fairly easy to change variables from θ and φ to x and y, since the Jacobian of the variable change is x θ y θ x φ y φ = sinθ cosθ sinθ cosθ.

3 3 When ka becomes large, any finite integration range in θ, φ is effectively infinite in x, y, and the projected transmission is T = k2 abcosθ = k2 ab cosθ dθdφ sinθ S 2 ( 2 ka x)s 2 ( 2 ka y) d x d y sinθ cosθ sinθ S 2 ( 2 ka x)s 2 ( 2 ka y) Believe it or not. = k2 ab sin 2 ( d 2 ka x) x ( 2 ka x) 2 =. kb = k2 ab ka d y sin 2 ( 2 kb y) ( 2 kb y) 2 This calculation tells you how it is that you can look through the holes in the screen of your microwave oven to see what s happening inside: visible light with λ = 5nm passes easily through holes of size mm (ka 2,, T ); while the microwave cooking power (f = 2.45GHz, λ = 2.2cm) mostly remains confined in the interior, with only a fraction k 2 a 2 /.4 (or k 2 a 2 /8 =.3, using the area of a circle of diameter a) leaking out. A microwave oven might operate at a power of kw. Spread over the area 25cm 2 of the door, the leakage per area is of order The published safety limit is 5mWcm 2. (kw).4 25cm 2 =.6mWcm 2.

4 4 2. This is another classic problem: An opaque screen occupies the x < half of the x-y plane. A plane wave ψ = Ô I e i(kz ωt) is incident on the screen from z <. (a) Show that in the usual scalar Kirchhoff approximation, for kz and for z x, y the diffracted field in the forward direction is ψ(x,y,z) = 2I π +i e i(kz ωt) e iξ 2 dξ, 2i ξ ξ = kx Ô 2kz. In the limit kr, and with cosθ = cosθ, the scalar Kirchhoff approximation (.79) and its Dirichlet and Neumann variations all give ψ diff = dx dy eikr 4π x > R 2ikψ, where ψ S = ψ = I /2. In Cartesian coordinates, the distance from source to observer is R = Ü Ü = (x x ) 2 +(y y ) 2 +z 2 /2 z + 2z [(x x ) 2 +(y y ) 2 ], where the last holds for x x, y y z. Thus, integrated over the illuminated portion of the x-y plane, ψ = ki/2 iz eikz dx e ik(x x) 2 /2z dy e ik(y y) 2 /2z. Change integration variables to ξ 2 = k(x x) 2 /2z and η = k(y y) 2 /2z; then ψ = I/2 iπ eikz dξ e iξ 2 dη e iη 2. ξ The answer is independent of y. Use Ê e iη 2 dη = Ô iπ = (+i) Ô π/2 [analytic continuation of the Gaussian integral; or see the two lines above (.32)]. The choice of sign of the square root follows because the real and imaginary parts of the integral Ê dη(cosη 2 +isinη 2 ) are both positive, and ψ = I /2 ikz +i e i Ô dξ e iξ 2. ξ You are of course worried about expanding for small x x when x is integrated to infinity, but the integral falls off exponentially when the exponent becomes larger than, and where the exponent is, k( x) 2 /z is of order, so x is of order Ô z/k, and x/z is of order / Ô kz.

5 (b) Show that the diffracted intensity I = ψ 2 can be written I = I(ξ) = 2 I C(ξ) S(ξ)+ 2 2, 5 where C(ξ) and S(ξ) are one representation of the Fresnel cosine and sine integrals in the form 2 ξ C(ξ) = dξ cos(ξ 2 2 ξ ), S(ξ) = dξ sin(ξ 2 ). π π What is I()? What are the behaviors of I(ξ) for large positive and negative values of ξ? Plot I(ξ). On consulting various sources, including Abramowitz and Stegun, Handbook of Mathematical Functions, and MathWorld, as well as Jackson, one finds that there are at least three forms of the Fresnel integrals. The perhaps standard form is defined as ξ C (ξ) = cos( π ξ 2 t2 )dt, S (ξ) = sin( π 2 t2 )dt, with two alternative versions, and C (ξ) = C 2 (ξ) = Ô ξ 2 ξ cost 2 dt, S π (ξ) = 2 ξ sint 2 dt π cost Ô t dt, S 2 (ξ) = Ô ξ sint Ô t dt, related by the obvious changes of variable C (ξ) = C ( Ô π 2 ξ) = C 2 ( π 2 ξ2 ) and S (ξ) = S ( Ô π 2 ξ) = S 2 ( π 2 ξ2 ). In his Problem. Jackson uses C, S ; so for the remainder of this problem, take C = C (ξ) and S = S (ξ). The integrands are even, so both the functions are odd, C( ξ) = C(ξ) and S( ξ) = S(ξ). For large argument both approach C(½) = S(½) = 2. Expressed using the Fresnel integrals, the phase integral is 2 e iξ 2 dξ 2 = cosξ 2 dξ 2 +i sinξ 2 dξ = π ξ π ξ π 2 +C(ξ) +i 2 +S(ξ), ξ and the diffracted amplitude and intensity are ψ = I /2 ikz +i e 2i As ξ, ξ ½, and ξ ½: [ 2 +C(ξ)]+i 2 +S(ξ), I = ψ 2 = 2 I 2 +C(ξ) S(ξ) 2. I [] I 4 + Ô ξ + ξ2 ; I π [ ] I + sinξ2 cosξ 2 Ô ; I [ ] I ξ ξ 2.

6 6 Fresnel Integrals: Diffracted Intensity:

7 7 (c) At the end of the eclipse of 2 August 27, an observer noticed that bands of light and shadow appeared to sweep over him with the return of the sun. What (in meters) is the scale of the width of these bands? The first few intensity maxima at ξ =.525, 2.938, and 3.862, with minima at ξ =.346, The separations are all different, with smaller contrast and closer spacing at larger ξ, but are all of order ξ =. Visible light has wavelengths between 4nm and 7nm, and the distance to the moon is 38km. This gives x = 2z λz k ξ = π = (55nm)(38km) π = 8m m as the scale of the modulation. A factor of 2 either way is also OK. This is consistent with the impressions of the observer.