MATH 262/CME 372: Applied Fourier Analysis and Winter 2016

Elements of Modern Signal Processing

Lecture 16 — February 25, 2016

Prof. Emmanuel Candes

Scribe: Carlos A. Sing-Long, Edited by E. Bates

1 Outline

Agenda: Fourier/wave optics

1. Fresnel-Kirchhoff theory

2. Rayleigh-Sommerfeld theory

3. Fresnel diffraction

4. Fraunhofer diffraction

5. Examples

Last Time: We began the study of diffraction with Maxwell’s equations with no sources, fromwhich we deduced that the electric field satisfies the wave equation. Considering a single componentand assuming the wave is monochromatic, we reduced the task of solving the wave equation tosolving the Helmholtz equation. Using the Green’s function, which is an outgoing spherical wavewith source at the origin, and Green’s theorem, we deduced that the wave field at any point canbe obtained from its values over any surface enclosing the point, provided we also know the valuesof its normal derivative on the surface.

2 Fresnel-Kirchhoff formulation of diffraction

Recall from last time the Green’s function to the Helmholtz equation, called the free-spaceGreen’s function:

G(x) =eik∥x∥

∥x∥,

where k = 2πλ is the wavenumber, and λ is the wavelength. For any u that is another solution to

the Helmholtz equation, we established the Fresnel-Kirchhoff integral formula:

u(x0) =1

4π

∫∂Ω

(u(x)

∂G

∂ν

∣∣∣∣x−x0

−G(x− x0)∂u

∂ν

∣∣∣∣x

)dS(x), (1)

where Ω ⊂ R3, and ν is the outward unit normal vector to ∂Ω.

1

We now use (1) to explain the diffraction of the wave field caused by an aperture. Consider areference frame with origin on the aperture, and with the z-axis orthogonal to the aperture plane.For R > 0, consider the domain

ΩR = x : ∥x∥ ≤ R, ⟨ez, x⟩ ≥ 0.

ΩR is the upper half of the ball of radius R centered at the aperture, and its boundary decomposesas ∂ΩR = ΣR ∪ ΓR (see Fig. 1). Here

ΣR = x : ∥x∥ ≤ R, ⟨ez, x⟩ = 0 and ΓR = x : ∥x∥ = R, ⟨ez, x⟩ > 0

are a disk and a hemisphere, respectively, of radius R and again centered at the origin. We assumethe aperture is infinitely thin so that it is a subset of ΣR.

x0

R

R

R

ez

x

(x,x0)

Figure 1: Description of the contour of integration, and the quantities used to deduce Fresnel-Kirchhoff’s diffraction formula (4). Here x is shown as a point on the aperture, which is a subsetof the aperture plane.

Applying (1) with Ω = ΩR, we obtain

u(x0) =1

4π

∫∂ΩR

(u(x)

∂G

∂ν

∣∣∣∣x−x0

−G(x− x0)∂u

∂ν

∣∣∣∣x

)dS(x)

=1

4π

∫ΣR

(u(x)

∂G

∂ν

∣∣∣∣x−x0

−G(x− x0)∂u

∂ν

∣∣∣∣x

)dS(x)

+1

4π

∫ΓR

(u(x)

∂G

∂ν

∣∣∣∣x−x0

−G(x− x0)∂u

∂ν

∣∣∣∣x

)dS(x).

We are interested in the interaction of the wave front exactly at the aperture, and so we wouldlike to remove the contribution from ΓR by focusing on the limiting case R → ∞. Indeed, forR ≫ 0, we have for x ∈ ΓR that ∥x−x0∥ ≈ ∥x∥, and ν(x) = x/∥x∥. Additionally, we can use theapproximation

∂G

∂ν

∣∣∣∣x−x0

≈ ∂G

∂ν

∣∣∣∣x

=

(ik − 1

∥x∥

)G(x),

2

and thus the contribution to (1) from ΓR can be approximated by

1

4π

∫ΓR

(u(x)

∂G

∂ν

∣∣∣∣x−x0

−G(x− x0)∂u

∂ν

∣∣∣∣x

)dS(x)

≈ 1

4π

∫Γ1

((ikR− 1)u(Rx)−R

∂u

∂ν

∣∣∣∣Rx

)eikR dS(x).

Unfortunately, it is not clear whether this last integral over Γ1 converges as R tends to infin-ity. Therefore, we impose an additional condition on the field, namely Sommerfeld’s radiationcondition

lim∥x∥→∞

∥x∥(iku(x)− ∂u

∂ν

∣∣∣∣x

)= 0.

Roughly speaking, this condition guarantees that we are only dealing with outgoing waves, thatis, there are no sources at infinity. Assuming the radiation condition holds, it is clear that thecontribution from ΓR vanishes as R → ∞. If we let Σ be the (infinite) aperture plane, we nowconclude

u(x0) =1

4π

∫Σ

(u(x)

∂G

∂ν

∣∣∣∣x−x0

−G(x− x0)∂u

∂ν

∣∣∣∣x

)dS(x).

Recall that in order to determine the wave field, we need to know both u and ∂u∂ν over the aperture

plane. The choice

u(x) = 0 and∂u

∂ν

∣∣∣∣x

= 0 for x ∈ Σ \A, (2)

yields Fresnel-Kirchhoff’s diffraction formula

u(x0) =1

4π

∫A

(u(x)

∂G

∂ν

∣∣∣∣x−x0

−G(x− x0)∂u

∂ν

∣∣∣∣x

)dS(x). (3)

Note that∂G

∂ν

∣∣∣∣x−x0

= −(ik − 1

∥x− x0∥

)G(x− x0) cos

(θ(x,x0)

),

where

cos(θ(x,x0)

)=

1

∥x− x0∥⟨ν(x), x− x0⟩

is the cosine of the angle between the z-axis and the line connecting x and x0.

When x0 is far away from the aperture, so that ∥x−x0∥ is much larger than the wavelength λ, wecan make the approximation

∂G

∂ν

∣∣∣∣x−x0

≈ −ikG(x− x0) cos(θ(x,x0)

),

to obtain the most general form of Fresnel-Kirchhoff’s diffraction formula:

u(x0) = − 1

4π

∫A

(iku(x) cos

(θ(x,x0)

)+

∂u

∂ν

∣∣∣∣x

)G(x− x0) dS(x). (4)

3

This formula enables us to predict diffraction effects, particularly when the wavelength is verysmall and observations are made away from the aperture. Otherwise, though, the predictions ofthis formula can be inaccurate. From an experimental point of view, Fresnel-Kirchhoff’s formuladoes not correctly explain Poisson’s spot1. And from a mathematical perspective, Sommerfeldnoticed that if u is analytic, then the choices made in (2) actually force u to be identically zero.

These inconsistencies do not mean the Fresnel-Kirchhoff formula has no predictive power. Instead,we should suspect that it may not be predictive in all regimes. A modification to the previousreasoning is required in order to obtain a consistent formulation of diffraction.

3 Rayleigh-Sommerfeld formulation of diffraction

In order to overcome the issues outlined in the previous section, we need to deduce an identity thatdoes not depend on the values of the normal derivative of u on the aperture plane. Note from (3)that we can accomplish this if we use a Green’s function that vanishes on the aperture plane. Inthis section we discuss the clever means by which Rayleigh and Sommerfeld did just that.

Let x0 be any point behind the aperture, which we will consider fixed, and let x0 be its reflectionwith respect to the aperture plane. Define

gx0(x) = G(x− x0)−G(x− x0) =eik∥x−x0∥

∥x− x0∥− eik∥x−x0∥

∥x− x0∥.

By linearity, g is easily seen to be a solution to the Helmholtz equation away from x0 and x0.Note that if x lies on the aperture plane, then ∥x − x0∥ = ∥x − x0∥, and so gx0(x) = 0 (seeFig. 2). We can repeat the same computations as in Lecture 15, using v(x) = gx0(x) and assumingSommerfeld’s radiation condition, to obtain

u(x0) =1

4π

∫Σu(x)

∂gx0

∂ν

∣∣∣∣x

dS(x).

Now the dependence on the normal derivative of the wave field on the aperture has disappeared!Consequently, we can write

u(x0) =1

4π

∫Au(x)

∂gx0

∂ν

∣∣∣∣x

dS(x).

Happily, this expression does not exhibit the inconsistencies of Fresnel-Kirchhoff’s formula, evenproviding a correct description of Poisson’s spot.

Note that for x on the aperture plane A, we have

∇gx0(x) =

(ik

∥x− x0∥− 1

∥x− x0∥2

)G(x− x0)(x− x0)

−(

ik

∥x− x0∥− 1

∥x− x0∥2

)G(x− x0)(x− x0)

= 2

(ik

∥x− x0∥− 1

∥x− x0∥2

)G(x− x0)(x− x0),

1R. L. Lucke, “Rayleigh-Sommerfeld diffraction and Poisson’s spot”, European Journal of Physics, 27 (2006), pg.193–204.

4

x0x0

x

A

Figure 2: Rayleigh-Sommerfeld’s placement of a reflected secondary source x0. Any x ∈ Σ is equidis-tance from the two sources, implying the modified Green’s function gx0 vanishes on the aperture plane.

and∂gx0

∂ν

∣∣∣∣x

= −2

(ik − 1

∥x− x0∥

)G(x− x0) cos

(θ(x,x0)

).

Since we are interested in the regime where the observation point x0 is far from the aperture, sothat ∥x − x0∥ is much larger than the wavelength λ, we can use the approximation (similar tobefore)

∂gx0

∂ν

∣∣∣∣x

≈ −2ikG(x− x0) cos(θ(x,x0)

).

We are led to

u(x0) =1

iλ

∫Au(x) cos

(θ(x,x0)

)G(x− x0) dS(x), (5)

known as the Rayleigh-Sommerfeld diffraction formula.

Here we see an analogy with the Huygens-Frensel principle. Namely, the Green’s function Grepresents an outgoing spherical wave with source at the origin. Therefore, the integral (5) is asuperposition of spherical waves with sources on the aperture. However, the phase and amplitude ofthese sources are not those of the wave field at the aperture. Instead we observe a phase shift of π/2from the factor of −i, and an amplitude attenuation given by the inclination factor cos

(θ(x,x0)

)(recall that the envelope u(x) is the intensity of the incoming wave). The absence of this factorwas the main shortcoming of Huygens and Frensel.

Note also that (5) is a convolution, as required for us to have constructed a proper Green’s function.Indeed, writing the impulse-response

1

iλcos(θ(x,x0)

)G(x− x0) =

1

iλ

(⟨−ez, x− x0⟩

∥x− x0∥

)G(x− x0) = h(x− x0),

we have

u(x0) =

∫Au(x)h(x− x0) dS(x).

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The linearity of this convolution operation corresponds to the principle of superposition, while thespatial invariance (with respect to translation) models homogeneity in the dielectric medium.

4 Fresnel diffraction

We now explore some regimes where we observe the wave field on a screen parallel to the apertureplane, but placed far from the aperture (relative to the wavelength). We use the coordinate systemof Fig. 3, denoting by x the orthogonal projection of x ∈ R3 onto the aperture plane Σ (i.e. thexy-axis). When x is itself in Σ, we simply have x = x. Furthermore, in the integral over Σ, thedifferential surface element reduces to dS(x) = dx.

x0

x0

z

x = x

ez

ex

eyA

apertureplane

screen

Figure 3: Coordinate system used to analyze the different diffraction regimes. The underscore indicatesthe component of a vector that lies on the aperture plane. If x is already in the aperture plane, thenx = x. When a point of interest x0 is far away from the aperture plane but well aligned with x,∥x− x0∥ is small relative to z = ⟨ez, x0⟩.

Let x be an element of the aperture A ⊂ Σ, and x0 a point of interest behind the aperture planewith vertical coordinate z = ⟨ez, x0⟩ > 0. Since the observation plane is parallel to the aperture,z is constant in the discussion that follows. Taylor expansion of

√1 + x2 gives

∥x− x0∥ =√

z2 + ∥x− x0∥2 = z

√1 +

(∥x− x0∥

z

)2

= z

[1 +

1

2

(∥x− x0∥

z

)2

+O

((∥x− x0∥

z

)4)]

, (6)

We will neglect higher order terms on the assumption that ∥x− x0∥ ≪ z, known as the paraxialapproximation (“near axis”). In doing so, we can make a zeroth order approximation of the

6

inclination factor

cos(θ(x,x0)

)=

z

∥x− x0∥= 1 +O

((∥x− x0∥

z

)2)

≈ 1.

We must be slightly more careful when approximating the phase, as the exponential function ismore sensitive to deviations. So we take the first order Fresnel approximation

eik∥x−x0∥ = eikz

[1+ 1

2

(∥x−x0∥

z

)2+O

((∥x−x0∥

z

)4)]

≈ eikzeik2z

∥x−x0∥2 .

These estimates give

G(x− x0) =eik∥x−x0∥

∥x− x0∥≈ 1

zeikze

ik2z

∥x−x0∥2 .

If we make the second assumption that our scalar diffraction theory holds, then the Rayleigh-Sommerfeld formula gives

u(x0) ≈1

iλzeikz

∫Au(x)e

ik2z

∥x−x0∥2 dx.

We see this formula once again yields a convolution of the wave field on the aperture with the kernel

h(x) =1

iλzeikze

ik2z

∥x∥2 .

By expanding the square, we deduce Fresnel’s diffraction integral

u(x0) =1

iλzeikze

ik2z

∥x0∥2∫Au(x)e

ik2z

∥x∥2e−ikz⟨x,x0⟩ dx.

This is a Fourier transform! It says the diffraction pattern will be the Fourier transform of thewave field on the aperture, after being modulated by a quadratic phase term and scaled by amultiplicative factor.

When is the diffraction integral accurate? The main source of errors will be the approximationused on the phase term. We have used the Taylor expression

√1 + ε = 1 +

ε

2− ε2

8+ · · · with ε =

(∥x− x0∥

z

)2

.

The Fresnel approximation proposed eikzε2/8 ≈ 1, which requires

kzε2

8=

kz

8

∥x− x0∥4

z4≪ π =⇒ 1

4λz∥x− x0∥2

(∥x− x0∥

z

)2

≪ 1. (7)

We introduce the characteristic size of the aperture a2, where a is taken such that

∥x− x0∥ ≤ a

7

for any x in the aperture and any paraxial point of interest x0 behind the aperture. Now considerthat for θ close to zero, we have sin θ ≈ θ. Hence

θ(x,x0) ≈ sin(θ(x,x0)

)=

∥x− x0∥∥x− x0∥

≈ ∥x− x0∥z

≤ a

z.

Then defining

θmax =a

zand Fresnel’s number

F =a2

λz,

we obtain that the condition (7) is equivalent to

Fθ2max

4≪ 1. (8)

In practice, the Fresnel number helps determine the regime in which diffraction occurs. Note thatin order to obtain Fresnel’s integral formula, the quadratic phase term needs to be relevant, thatis,

π ≪ k

2z∥x− x0∥2 =⇒ 1 ≪ a2

λz= F

We say there is Fresnel diffraction, or near-field diffraction, when F ≫ 1. On the other hand,we have argued that Fresnel’s diffraction integral will be accurate if (8) is satisfied. For instance,consider a circular aperture with a radius of 1cm, and light with wavelength λ = 0.5µm. Then wecan take

a2 = πcm2 =⇒ a =√πcm,

and in order to have a good approximation, we need

Fθ2max

4=

((π × 10−4)

(0.5× 10−6)

1

z

)((π × 10−4)

4

1

z2

)=

π2

200

1

z3≪ 1 =⇒ z ≫ 37cm,

For z = 37cm, we have

F =(π × 10−4)

(0.5× 10−6) (37× 10−2)≈ 15 and θmax ≈ 2.74.

In reality, Fresnel’s diffraction integral can be accurate even outside the regimes described. Inparticular, the higher order terms in (6) need not be small, but rather just not change the value ofthe integral significantly.

5 Fraunhofer diffraction

In Fresnel diffraction, we have a quadratic phase term modulating the wave field on the aperture.This quadratic term is relevant because F ≫ 1. However, the quadratic phase term can be ne-glected when F ≪ 1. In this Fraunhofer regime, we say Fraunhofer diffraction, or far-fielddiffraction, occurs. In this case, we obtain

u(x0) =1

iλzeikze

ik2z

∥x0∥2∫Au(x)e−

ikz⟨x,x0⟩ dx.

8

Up to modulation by a quadratic phase factor, this is simply the Fourier transform of the wavefield at the aperture. That is,

u(x0) =1

iλzeikze

ik2z

∥x0∥2 u

(k

zx0

).

We have assumed u = 0 on the part of the aperture plane away from the aperture iteself. Continuingour previous example, we require

F =π × 10−4

0.5× 10−6

1

z≪ 1 =⇒ z ≫ 628m.

While this distance seems excessive, we can actually observe the effects of Fraunhofer diffractionat much shorter distances by using lenses2. We will discuss this in the next lecture.

6 Examples

In this section we compute two Fraunhofer diffraction patterns, of which we saw pictures in Lecture15.

6.1 Rectangular aperture

Suppose we have a rectangular aperture aligned with the coordinate axis and with side lengths axand ay (see, for instance, Fig. 4a). If we assume the wave field has constant intensity u0 accrossthe aperture, we see that in the Fraunhofer regime

u(x0) =u0axayiλz

eikzeikz∥x0∥2 sinc

(kax2πz

⟨ex, x0⟩)sinc

(kay2πz

⟨ey, x0⟩).

On the screen we will observe the intensity (not phase) of the wave field:

I(x0) = |u(x0)|2 =(u0|A|4λz

)2

sinc

(kax2πz

⟨ex, x0⟩)2

sinc

(kay2πz

⟨ey, x0⟩)2

,

where |A| is the area of the aperture (see Fig. 4b). Amazingly, we observe dark lines separatingilluminated regions. These lines occur whenever

⟨ex, x0⟩ =2λz

axn or ⟨ey, x0⟩ =

2λz

ayn for an integer n.

6.2 Circular apertures

Suppose now we have a circular aperture of radius a (see Fig. 4c). We have that∫∥x∥≤1

dx =J1(∥x∥)∥x∥

,

2Another method is to have waves converging at the aperture.

9

(a) Aperture (b) Diffraction pattern

(c) Aperture (d) Diffraction pattern

Figure 4: Diffraction patterns for some apertures in the Fraunhofer regime. (a) Rectangular aperturewith ax = 1, ax = 2. (b) Diffraction pattern of the rectangular aperture. Note the rectangular patternand regions of dark lines. (c) Circular aperture with a = 1. (d) Diffraction pattern of the circularaperture. Note the circular pattern and the dark, concentric circles.

where J1 is the first Bessel function of the first kind. The pattern we observe in the Fraunhoferregime is (see Fig. 4d)

u(x0) =u0a

i2πeikze

ikz∥x0∥2

J1(akz ∥x0∥

)∥x0∥

,

and its intensity becomes

I(x0) = |u(x0)|2 =(u0a2π

)2 ∣∣J1 (akz ∥x0∥)∣∣2

∥x0∥2.

In this case we observe dark circles, occurring concentrically at radii

∥x0∥ =z

akj1,n where j1,n is the nth positive root of J1.

In particular, the first circular dark spot occurs at

z

akj1,1 = z

λ

2a

j1,1π

≈ 1.22λ

dz,

10

where d is the diameter of the circular aperture. The central bright region it encloses is called theAiry disk.

| z

|z

2z

akj1,2

2z

akj1,1

Figure 5: First Bessel function of the first kind, shown on the real axis. This corresponds to a radialcross section of the intensity function for Fraunhofer diffraction with a circular aperture. The widths ofthe first and second lobes are indicated.

6.3 Other apertures

Fig. 6 and Fig. 7 show diffraction patterns created by other shapes of apertures.

Figure 6: Numerical predictions of Fraunhofer diffraction from various rotated, regular apertures

11

Figure 7: Calculated Fraunhofer diffraction patterns for diamond and rectangular apertures. Note theeffect of rotation for the rectangular aperture by comparing the right two images to Fig. 4(a) and 4(b).

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