Elements of diffraction theory – part II

Svetoslav S. IvanovDepartment of Physics, St. Kliment Ohridski University of Sofia, 5 James Bourchier Blvd, 1164 Sofia, Bulgaria

(Dated: February 7, 2016)

We make a brief introduction to diffraction theory. We begin with the derivation of the Helmholtz-Kirchhoff integral theorem, which we use to derive the Fresnel-Kirchhoff diffraction formula. Next wederive formulas describing Fresnel’s diffraction and Fraunhofer’s diffraction and finally we calculatethe Fraunhofer diffraction pattern from rectangular and circular apertures.

Outline

1. Kirchhoff’s diffraction theory

2. Fresnel’s diffraction and Fraunhofer’s diffraction

3. Fraunhofer’s diffraction – rectangular and circular apertures

4. Fraunhofer’s diffraction – Diffraction gratings

5. Appendix

I. KIRCHHOFF’S DIFFRACTION THEORY

Let the point P0 from Fig. 1 a) be a light source emitting spherical waves. Consider a particular wave-front S havingradius r0. According to the Huygens-Fresnel (HF) principle each point of the front is a centre of a secondary sphericalwavelet and the wave-front S1 at a later instant is the envelope of these wavelets. Therefore the light disturbanceat a point P arises from the superposition of secondary waves that can be regarded as to originate from a surface,e.g. S, situated between P0 and P (Fig. 1 a). By disturbance we will mean a monochromatic scalar wave, e.g. onecomponent of the E field or the B field, say Ex (in absence of polarization coupling).

P

v

S

a) б)

FIG. 1: a) Illustration of the Huygens-Fresnel principle. P0 and P are the light source and the receiver, respectively, r0 is theradius of the wave front with surface S, S1 is the wave front at a later instant and χ is the inclination angle at some pointQ. b) Derivation of the Helmholtz-Kirchhoff integral (10). S and S′ denote the outer and the inner surfaces, respectively, v isvolume, n is unit vector normal to the surface, P is arbitrary internal point. This setting generalizes the setting from Fig. 1 a.

Kirchhoff realized that the Huygens-Fresnel principle can be put on a mathematical basis by using an already knownintegral theorem which expresses the disturbance at any point P in the field in terms of the values of the disturbanceand its derivatives at all points on an arbitrary closed surface S surrounding P (Fig. 1 b). Here we only briefly followKirchhoff’s steps following Ref. [1]:

1. we derive Helmholtz-Kirchhoff’s integral theorem (10);2. next we use this theorem to derive an approximate, yet very useful, Fresnel-Kirchhoff diffraction formula (15)

valid for small wavelengths (λ≪ r, s);3. we further approximate this formula for small aperture to get the Fresnel integral (16) and the Fraunhofer

integral (24);4. finally we derive analytical expressions describing Fraunhofer diffraction from rectangular (32) and circular (42)

apertures.These steps are described in separate sections, which can be read independently. You can directly go over to Sec. III.

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A. Derivation of Helmholtz-Kirchhoff’s integral theorem

Summary: This theorem gives the field disturbance U(P ) at the observer’s point P :

1. We use the wave equation (1) to describe the field E(x, y, z, t) at any point in space and time

2. We decompose E into spacial and temporal parts: U(x, y, z)e−iωt and write equation (3) for U(x, y, z)

3. We find U(P ), which depends on the values of U(x, y, z) on the wave front S (Eq. (10)).

In Fig. 1 a) we considered two points – a source (P0) and a receiver (P ), and, by the Huygens-Fresnel (HF)principle, we concluded that the disturbance at P was specified entirely by the disturbances at the points forming thesurface S. Thus the role of P0 was undertaken by S. Now consider Fig. 1 b), which generalizes the setting from Fig.1 a), where now we only have the necessary elements from Fig. 1 a) – S and P . In this subsection we will set the HFprinciple on a rigorous mathematical ground by showing that the disturbance at P is defined by and originates fromthe disturbance over S.Consider the vacuum wave equation for the E field (see Appendix A for derivation)

∇2E− 1

c2∂2E

∂t2= 0, (1)

where the field depends on position and time, i.e. E = E(x, y, z, t). Written down for any of the amplitudes Ex,

Ey and Ez, commonly denoted as V hereafter, this equation is identical, ∇2V − 1c2∂2V∂t2 = 0, as in vacuum these

amplitudes are independent (uncoupled). Now let us assume that our wave is monochromatic with frequency ω (thegeneralization to non-monochromatic waves is almost trivial). Then we can split V into a spacial and a temporarypart,

V (x, y, z, t) = U(x, y, z)e−iωt (2)

with the spatial part satisfying the Helmholtz equation for free propagation (see Appendix B):

(∇2 + k2)U = 0. (3)

In fact, U(x, y, z) defines the disturbance at position (x, y, z), while the time-dependent factor e−iωt is universal forall points in space and will therefore be omitted.Let us assume that U possesses continuous first- and second-order partial derivatives within and on the surface S.

If U ′ is another function that satisfies the same continuity requirements as U , we have by the divergence theorem (seeAppendix C) ∫

v

(U∇2U ′ − U ′∇2U)dv = −∫S

(U∂U ′

∂n− U ′ ∂U

∂n

)dS, (4)

where ∂/∂n denotes differentiation along the inward normal to S, denoted as n (cf. Fig. 1 b). If for U ′ we also have

(∇2 + k2)U ′ = 0, (5)

then apparently the left-hand side of Eq. (4) vanishes:∫S

(U∂U ′

∂n− U ′ ∂U

∂n

)dS = 0. (6)

Suppose we take U ′ = eiks/s, where s denotes the distance from P to the point (x, y, z). The rationale is that U ′

describes a spherical wavelet in accordance with the HF principle. We construct another integral like Eq. (6) wherethe integration is carried over the surface S′, which is a sphere centered at P with radius ε. We sum both integrals:∫

S

(U∂U ′

∂n− U ′ ∂U

∂n

)dS +

∫S′

(U∂U ′

∂n− U ′ ∂U

∂n

)dS = 0. (7)

Next we substitute U ′ to obtain (see Appendix B 3)∫S

(U∂

∂n

(eiks

s

)− eiks

s

∂U

∂n

)dS = −

∫S′

[Ueiks

s

(ik − 1

s

)− eiks

s

∂U

∂n

]dS′ =

−∫Ω

[Ueikε

ε

(ik − 1

ε

)− eikε

ε

∂U

∂n

]ε2dΩ′, (8)

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where dΩ denotes an element of the solid angle,∫ΩdΩ = 4π.

Since the integral over S is independent of ε, we replace the integral over S′ by its limiting value as ε→ 0; the firstand the third terms give no contribution in the limit (they become too small), and the only term, which survives is

−∫Ω

UeikεdΩ → −∫Ω

UdΩ = 4πU(P ). (9)

Hence we obtain the Helmholtz-Kirchhoff’s integral theorem

Helmholtz-Kirchhoff’s integral theorem

U(P ) =1

4π

∫S

(U∂

∂n

(eiks

s

)− eiks

s

∂U

∂n

)dS. (10)

Thus we have shown that the disturbance U(P ) is defined by the values of U and ∂U/∂n over S. In fact, this differsfrom our original claim that U(P ) is defined by the values of U over S only. It may, however, be shown from thetheory of Green’s functions that the value of U over S is sufficient to specify U at every internal point P . Still, formula(10) is sufficient to describe diffraction.

B. Derivation of Fresnel-Kirchhoff’s diffraction formula

Summary: This formula approximates Eq. (10) and gives the field disturbance U(P ) for the very typical situationfrom Fig. 2 a) for small wavelengths:

1. We specialize the integral theorem (10) for the situation from Fig. 2 a) and show that only the integral overA is not zero

2. We assume that the disturbance arising from P0 is a spherical wave: U (i) = A eikr

r , originating from P0

3. We assume small wavelength, λ≪ r, s, where r and s are distances, and arrive at Eq. (15).

Here we will specialize the theorem stated in Eq. (10) for the very typical situation illustrated in Fig. 2 a). Considera monochromatic wave from a point source P0, propagated through an opening in a plane screen, and let P be thepoint at which the disturbance U(P ) is to be determined.

a) б)

FIG. 2: a) Illustrating the derivation of the Fresnel-Kirchhoff diffraction formula. The surface S from Fig. 1 b) consists of A,B and C. Only the integral over A is non-zero, whereas the integrals over B and C have zero contribution. The assumption wemake is λ ≪ r, s, r, s being the distances. b) Definition of angles and directions. r and s point in the direction of increasing rand s, respectively, n is normal to the screen.

To find the disturbance U(P ) we take Kirchhoff’s integral (10) over a surface S formed by (see Fig. 2 a): theopening A, a portion B of the nonilluminated side of the screen and a portion C of a large sphere of radius R, centeredat P which, together with A and B, forms a closed surface S. We will show that only the integral over A is not zero.

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Kirchhoff’s integral (10) now gives

U(P ) =1

4π

[∫A+

∫B+

∫C

](U∂

∂n

(eiks

s

)− eiks

s

∂U

∂n

), (11)

where the symbol [. . .] represents the sum of the three integrals.Kirchhoff made the following approximation – he assumed that i) the disturbance U over A is the same as if the

screen is missing, and ii) the disturbance U on B is everywhere zero. This approximation is written in terms of thefollowing boundary conditions for U and ∂U

∂n :

on A : U = U (i) ∂U

∂n=∂U

∂n

(i)

, (12a)

on B : U = 0∂U

∂n= 0. (12b)

where (see Appendix A for U (i) and B3 for ∂U (i)/∂n)

U (i) =Aeikr

r,

∂U (i)

∂n=Aeikr

r

[ik − 1

r

]cos(n, r). (13)

Here U (i) represents a spherical wave originating from P0 and (n, r) is the angle between n and r (see Fig. 2 (bottom)).Because of Eq. (12b), we find that the integral over B is zero. The integral over C is zero, as well, as we are free to

choose the radius R so large that light from P0 has not yet reached the surface C at the time of interest. Thus thedisturbance on C is zero, so is the integral, too. Therefore from Eqs. (11) and (13) we have

U(P ) =1

4π

∫A

[U (i) ∂

∂n

(eiks

s

)− eiks

s

∂U

∂n

(i)]dS =

1

4π

∫A

[Aeikr

r

∂

∂n

(eiks

s

)− eiks

s

∂

∂n

(Aeikr

r

)]dS =

1

4π

∫A

[Aeikr

r

eiks

s

(ik − 1

s

)cos(n, s)− eiks

s

Aeikr

r

(ik − 1

r

)cos(n, r)

]dS. (14)

Now we assume that the wavelength is very small compared to the distances r and s. Thus we can approximateik − 1

r,s with ik and finally we arrive at the

Fresnel-Kirchhoff diffraction formula, valid for λ≪ r, s

U(P ) =iA

2λ

∫A

eik(r+s)

rs[cos(n, s)− cos(n, r)] dS. (15)

This formulation allows us to improve a bit our understanding of the Huygens-Fresnel principle: a light wave fallingon the aperture A propagates as if every element dS from A emitted a spherical wavelet, described by eiks/s. Theamplitude and the phase of this wavelet depend on the distance r between dS and P , since the wavelet is driven bythe spherical wave, emitted from P and described by eikr/r. The factor A naturally accounts for the intensity, andthe cosine factors correspond to Lambert’s law of surface brightness.

II. FRESNEL’S DIFFRACTION AND FRAUNHOFER’S DIFFRACTION

Summary: We derive Fresnel’s and Fraunhofer’s diffraction integrals.

1. We approximate Eq. (15) for a≪ r, s (small apertures) to derive Fresnel’s diffraction formula (16)

2. We make a stronger assumption that a2/λ ≪ r, s (even smaller apertures) to derive Fraunhofer’s diffractionformula (24)

Now that we have derived the Fresnel-Kirchhoff diffraction formula (15), we are ready to calculate diffraction.Though we can treat this problem numerically with ease, an analytic solution to this integral is highly desirablebut yet unknown in the general case for arbitrary apertures and positions of P0 and P . Therefore in this sectionwe will seek further approximation to the formula (15). Now we assume that the linear dimensions of the opening,

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although large compared to λ, are small compared to r and s. Combined with the previous assumption we now haveλ ≪ a ≪ r, s, where a is a characteristic size of the opening (e.g. radius, if it is circular, or the longer side forrectangular). In the following we apply this assumption to approximate each of the terms inside the integral formula(15).Due to the approximation a ≪ r, s, the factor cos(n, s)− cos(n, r) varies only slightly inside the aperture (cf. Fig.

3). We can then safely assume that cos(n, s) ≈ cos(n, s′) and cos(n, r) ≈ cos(n, r′), so that cos(n, s) − cos(n, r) ≈cos(n, s′)− cos(n, r′) is simply a constant. On the other hand, due to the approximation λ≪ a≪ r, s, as the elementdS explores the domain of integration A, r+s will change by a huge number of wavelengths, so that the factor eik(r+s)

will oscillate rapidly. Finally, since a ≪ r, s, the factor 1/rs may be replaced by 1/r′s′ (Fig. 3). As a result theintegral (15) reduces to

Fresnel’s diffraction formula, valid for λ≪ a≪ r, s

U(P ) ≈ iA

2λ

cos(n, s′)− cos(n, r′)

r′s′

∫Aeik(r+s)dS. (16)

This integral describes Fresnel’s diffraction. Unfortunately, only numeric treatment of this integral is possible.In the following, based on our assumption a ≪ r, s, we will approximate eik(r+s) too, thereby deriving a simplifiedintegral describing Fraunhofer diffraction.

Screen

0

P

Q

s`

s

r

z

y

x

r`O

P

FIG. 3: Diffraction at an aperture in a plane screen. We introduce Cartesian coordinates with origin O, where the xy-planeand the aperture plane coincide and positive z direction points into the half-space containing P . An arbitrary point Q withinthe aperture is specified by the coordinates (ξ, η).

We introduce a Cartesian reference system with origin O in the aperture and with the x- and y-axes in the planeof the aperture and choose the positive z direction to point in the half-space that contains the point P (Fig. 3). If(x0, y0, z0) and (x, y, z) are the coordinates of P0 and P , respectively, and (ξ, η) are the coordinates of an arbitrarypoint Q in the aperture, we have

r2 = |(ξ, η, 0)− (x0, y0, z0)|2 = (x0 − ξ)2 + (y0 − η)2 + z20 ,

s2 = |(ξ, η, 0)− (x, y, z)|2 = (x− ξ)2 + (y − η)2 + z2,

r′2 = |(0, 0, 0)− (x0, y0, z0)|2 = x20 + y20 + z20 ,

s′2 = |(0, 0, 0)− (x, y, z)|2 = x2 + y2 + z2.

Hence we find

r2 = r′2 − 2(x0ξ + y0η) + ξ2 + η2, (18a)

s2 = s′2 − 2(xξ + yη) + ξ2 + η2. (18b)

We can neglect the term ξ2 + η2 in the general case as a≪ r, s, while the terms x0ξ+ y0η and xξ+ yη depend on thepositions of P0 and P (the distances from the z axis) and can have considerable values. Therefore we write

r2 ≈ r′2 − 2(x0ξ + y0η), (19a)

s2 ≈ s′2 − 2(xξ + yη). (19b)

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Now we use that a≪ r, s and expand r and s in power series, where we use the formula√1 + x = 1 +

x

2+O(x2). (20)

We obtain

r ≈ r′ − x0ξ + y0η

r′, (21a)

s ≈ s′ − xξ + yη

s′, (21b)

where we have neglected the higher orders assuming that a2/λ≪ r, s. We substitute r and s into the integral (15):

U(P ) = C

∫Ae−ik(

x0ξ+y0η

r′ + xξ+yηs′ )dξdη, (22)

where

C =iA

2λ

cos(n, s′)− cos(n, r′)

r′s′eik(r

′+s′). (23)

Thus we derive the formula for

Fraunhofer’s diffraction, valid for λ≪ a≪ r, s and a2/λ≪ r, s

U(p, q) ≡ U(P ) = C

∫Ae−ik(pξ+qη)dξdη, (24)

wherep =

x0r′

+x

s′, q =

y0r′

+y

s′. (25)

The integral (24) describes Fraunhofer’s diffraction.Note that p and q depend on the directions of P0 and P measured from the origin O. In fact, p = q = 0 describes

the central direction, i.e. the situation when P0P contains O. Such P , fulfilling p = q = 0 for fixed P0, we will callthe centre of the diffraction pattern. We expect that the brightest spot is located at this centre.Let us denote with I0 the intensity at the centre, I0 = |U(0, 0)|2. From Eq. (24) we have

U(0, 0) = C

∫Adξdη = CD, (26)

where D is the area of the aperture. Hence we find a simple physical meaning for C:

C =

√I0D

. (27)

A. Fresnel’s number F

The integral (24) describes Fraunhofer’s diffraction and was obtained by neglecting quadratic and higher orderterms in ξ and η in Eq. (21), under the assumption a2/λ≪ r, s. When this assumption does not hold, we have to usethe more complicated formula (16) describing the more general case of Fresnel’s diffraction. To distinguish betweenboth types of diffraction we introduce Fresnel’s number

F =a2

λs. (28)

Fresnel’s diffraction and Fraunhofer’s diffraction are observed for F ≈ 1 and F ≪ 1, respectively.

B. Remark – Fourier transform

The Fraunhofer integral (24) can be rewritten in the way

U(p, q) ≡ U(P ) = C

∫∞M(ξ, η)e−ik(pξ+qη)dξdη, (29)

where the integration is over an infinite plane in the aperture plane and M(ξ, η) describes the aperture: M(ξ, η) = 1if (ξ, η) ∈ A and 0 otherwise. Hence the diffraction image U(p, q) is the Fourier transform of the aperture’s shapeM(ξ, η). Thus one can encode a secret message, which can be decoded with a laser beam. :)

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III. FRAUNHOFER’S DIFFRACTION – SIMPLE EXAMPLES

In the following section we will study the Fraunhofer diffraction by rectangular and circular openings.

a) б)

FIG. 4: Rectangular aperture (a) and circular aperture (b).

A. Rectangular aperture

Consider a rectangular opening with sides 2a and 2b and centre O, as shown in Fig. 4 a). Now that we have derivedthe Fraunhofer integral (24) it is straightforward to obtain the diffraction pattern:

U(p, q) = C

∫Ae−ik(pξ+qη)dξdη = C

∫ a

−a

∫ b

−be−ik(pξ+qη)dξdη = C

∫ a

−ae−ikpξdξ

∫ b

−be−ikqηdη. (30)

Note that ∫ a

−ae−ikpξdξ =

e−ikpa − eikpa

−ikp= 2

sin kpa

kp. (31)

We obtain

U(p, q) = 4ab Csin kpa

kpa

sin kqb

kqb. (32)

Hence for the intensity we get

I(p, q) ∝ |U(p, q)|2 = I0

(sin kpa

kpa

)2(sin kqb

kqb

)2

, (33)

where I0 = 4ab C is the intensity at the centre of the pattern, where p = q = 0 (cf. Eq. (27)).Let us investigate the function y = (sinx/x)2, the base of our result, shown in Fig. 5 a) (top). The principal

maximum is for x = 0 and y = 0 for x = ±π,±2π, . . .. The secondary maxima occur when x − tanx = 0. As xincreases, the solution to this equation asymptotically approaches xm = (2m + 1)π/2, where m is integer. Can youfigure out why? Thus we see that the intensity I(p, q), shown in Fig. 5 a) (bottom), is zero along two sets of linesparallel to the sides of the rectangle, given by

kpa = ±mπ, kqb = ±nπ (m,n = 1, 2, 3, . . .) (34)

Homework: Find the diffraction from a single slit, which is effectively an infinite linear source (as opposed to apoint source), e.g. a luminous wire, the light from which is diffracted by an infinitely long narrow slit, parallel to thesource. Let both be parallel to the y-axis. Hint: for coherent (incoherent) source you must integrate the disturbance(32) (the intensity (33)). Since q = y/s′ + y0/r

′, where y0/r′ is fixed and is defined by the position of a particular

element from the wire, the integral is over q from −∞ to ∞.

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4 2 2 4

0.2

0.4

0.6

0.8

1.0

4 2 2 4

0.2

0.4

0.6

0.8

1.0

a) )

FIG. 5: Top left: The function y = (sinx/x)2. The zeroes occur for x = ±π,±2π, . . ., while the secondary maxima asymp-totically reach the values xm = (2m + 1)π/2 for integer m. Bottom left: Simulated Fraunhofer diffraction pattern from arectangular aperture. The picture is courtesy of Hecht, Eugene, Optics, 2nd Ed, Addison Wesley, 1987. Top right: The

function y =[2J1(x)

x

]2. Bottom right: Simulated Fraunhofer diffraction from a circular aperture.

B. Circular aperture

Now we study the Fraunhofer diffraction from a circular opening, shown in Fig. 4 b). Again we need to solve theintegral (24), where now we use polar coordinates (ρ, θ) to specify a point in the aperture:

ρ cos θ = ξ, ρ sin θ = η. (35)

Before we calculate the integral, let us first write the expression −ik(pξ + qη) in the new coordinates and do a bit ofsimplification:

−ik(pξ + qη) = −ikρ(p cos θ + q sin θ) = −ikρ√p2 + q2

(p√

p2 + q2cos θ +

q√p2 + q2

sin θ

). (36)

Let us introduce a new variable

w =√p2 + q2. (37)

Because we have −1 ≤ p/w ≤ 1, −1 ≤ q/w ≤ 1 and (p/w)2 + (q/w)2 = 1, we can set p/w = cosψ and q/w = sinψ.From now on we will specify the observational point P by the polar variables (w,ψ) rather than (p, q). We find

−ik(pξ + qη) = −ikρw (cosψ cos θ + sinψ sin θ) = −ikρw cos(θ − ψ). (38)

Note that w is the sine of the angle which the direction (p, q) makes with the central direction p = q = 0; w representsthe sine of the angle between a point P and the centre of the diffraction pattern (see Appendix D for proof; kind ofobvious for x0 = y0 = 0). The integral now becomes

U(w,ψ) = C

∫ a

0

∫ 2π

0

e−ikρw cos(θ−ψ)ρdρdθ, (39)

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where we have used the Jacobian relating Cartesian to polar coordinates: dξdη = ρdρdθ. Note that∫ 2π

0

ei x cosαdα = 2πJ0(x), (40)

where J0(x) belongs to the family of the Bessel functions Jn(x). Thus we obtain (we drop the ψ from U , as U hasbecome independent of ψ)

U(w) = 2πC

∫ a

0

J0(kρw)ρdρ, (41)

which finally yields (here I used Wolfram Mathematica)

U(w) = Cπa22J1(kaw)

kaw. (42)

For the intensity we obtain

I(w) ∝ |U(w)|2 = I0

(2J1(kaw)

kaw

)2

. (43)

The function y =[2J1(x)x

]2is plotted in Fig. 5 b (top). Note that the central maximum is very strong compared to

the rest and comprises almost 98% of all diffracted light. Therefore the net effect of the diffraction through a circularaperture is to spread a point source of light, such as a star, into a larger circular shape, as shown in Fig. b 5 (bottom).In this sense the first zero, located at kaw ≈ 1.22π, is a good indicator and fundamentally defines the resolution ofan optical imaging system. Hence we obtain the limiting angle under which two stars appear as a single spot on thediffraction screen,

wmin = 1.22λ

d, d = 2a. (44)

C. Extending or contracting the aperture

Consider aperture A, which corresponds to diffraction image U(p, q) given by the integral (24):

U(p, q) = C

∫Ae−ik(pξ+qη)dξdη. (45)

Now let us extend or contract the aperture in one direction by factor µ, say in ξ direction. We can easily obtain thediffraction image from the new aperture A′. For illustrative purposes we consider rectangular aperture, though theprocedure for any aperture is absolutely the same. The original diffraction image is given by (Eq. (30))

U(p, q) = C

∫ a

−a

∫ b

−be−ik(pξ+qη)dξdη (46)

and the new image is given by

U ′(p, q) = C

∫ µa

−µa

∫ b

−be−ik(pξ+qη)dξdη = µC

∫ µa

−µa

∫ b

−be−ik(µp

ξµ+qη)d

(ξ

µ

)dη = µC

∫ a

−a

∫ b

−be−ik(µpξ

′+qη)dξ′dη. (47)

Thus we obtain

U ′(p, q) = µU(µp, q), (48)

which means that the image is contracted in p-direction by a factor of µ and the intensity is increased by µ2. See Fig.6 for comparison of diffraction at different apertures.

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FIG. 6: Top: Various apertures – square, rectangular, circular and elliptic. Bottom: The corresponding diffraction images.Extending the aperture in one dimension results in contracting the diffraction image in the same dimension by the same factor.

D. Multiple diffraction openings

Consider Fraunhofer diffraction from a screen containing a large number of identical openings. Does the resultremind you of the interference principle? Can you derive diffraction from a grating?

IV. FRAUNHOFER’S DIFFRACTION – DIFFRACTION GRATINGS

p

0.2

0.4

0.6

0.8

1.0

FIG. 7: Left: Diffraction grating. Plane wave falls at angle θ1, while the viewer’s angle θ2 is variable. Right: Intensity vs p forN=5, s = λ/2π and d = 4s = 2λ/π.

The diffraction image from a grating is obtained with the integral (24)

U(p, q) = C

∫Ae−ik(pξ+qη)dξdη, (49)

with p = sin θ2 − sin θ1 (see Fig. 7). The aperture A is a collection of N slits and the grating extends from 0 to a.Because the diffraction problem from a grating is one-dimensional, we perform the integration over η right away. We

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obtain

U(p) = C ′N∑n=1

∫ nd

nd−se−ikpξdξ =

C ′

−ikp

N∑n=1

(e−ikpnd − e−ikp(nd−s)

)=

=iC ′

kp

(1− eikps

)e−ikpd

N−1∑n=0

(e−ikpd)n =iC ′

kpe−ikpd

(1− eikps

) 1− e−ikpdN

1− e−ikpd. (50)

We note that 1− e−ix = 2ie−ix/2 sinx/2, yielding

U(p) = sC ′eikp/2(d−s−d(N−1)) sin(kps/2)

kps/2

sin(kpdN/2)

sin(kpd/2). (51)

The intensity is

I(p) ∝ |U(p)|2 = I0

(sin(kps/2)

kps/2

)2(sin(kpdN/2)

sin(kpd/2)

)2

(52)

and is shown in Fig. 7. Here I0 = I(0). Show that I0 ∝ N2.

From limx→0 sinx/x = 1 we find that(

sin(kpdN/2)N sin(kpd/2)

)2attains its maximal value 1 for kpd/2 = mπ, i.e. for p = mλ/d.

The envelop’s zeros occur for kps/2 = mπ, i.e. for p = mλ/s.

A. Spectral resolution of a grating

0.2

0.4

0.6

0.8

1.0 space to first minimum

FIG. 8: The limit of resolution is determined by the Rayleigh criterion as applied to the diffraction maxima, i.e., two wavelengthsλr and λb are just resolved when the maximum of one lies at the first minimum of the other.

Consider two wavelengths λr and λb, shown in Fig. 8. They can be resolved if the maximum of λb lies at the firstminimum of λr. The maximums are located at mλr/d and mλb/d, respectively, where m = 1 in the figure, and theseparation is m(λr − λb)/d. The distance to the first minimum is λr/Nd. Therefore λr and λb can be resolved if

m(λr − λb)

d≤ λrNd

. (53)

The resolving power is defined as R = λλr−λb

, so we have

R = mN. (54)

12

APPENDIX A: WAVE EQUATION IN VACUUM

Here we derive the wave equation for the E field in a region with no charges (ρ = 0) and no currents (j = 0). Weuse Maxwell’s equations:

∇ ·E = 0, (A1a)

∇ ·B = 0, (A1b)

∇×E = −∂B∂t, (A1c)

∇×B =1

c2∂E

∂t. (A1d)

We operate with ∇× on both sides of Eq. (A1c) to get

∇× (∇×E) = −∇×(∂B

∂t

)= − ∂

∂t(∇×B) , (A2)

where we have exchanged the time and the spacial derivatives. Note that

∇× (∇×E) = ∇(∇ ·E)−∇ · ∇E = −∇2E, (A3)

where we have used Eq. (A1a). Using the above identity and Eq. (A1d) we obtain

−∇2E = − ∂

∂t

(1

c2∂E

∂t

)≡ − 1

c2∂2E

∂t2. (A4)

Hereby we arrive at the wave equation for vacuum (Eq. (1)).

APPENDIX B: SOLUTION TO THE WAVE EQUATION

1. The wave equation

We consider a solution to Eq. (1), which represents a spherical wave, as we deal with such waves (cf. Fig. 1).

Spherical waves originate from a point and thus have central symmetry, i.e. V only depends on r = |r| =√x2 + y2 + z2

and t. Therefore, for convenience, we switch to spherical coordinates, where a spherical solution has the formV = V (r, t). It is explicitly given as

V (r, t) =A

re−i(ωt−kr). (B1)

Though simple, we will not consider the mathematical derivation of this solution, but rather we will exercise ourphysical intuition to understand it. We know that the phase of a spherical wave is constant over a sphere, and thisis what the exponential factor accounts for. We also know that the intensity of the wave I(r, t) (note that I ∝ |V |2)must decrease as 1/r2 with r. Hence the factor 1/r. A defines intensity. A non-monochromatic analogue is simply asum (or integral) over the frequency range ω with amplitude A(ω).We can separate Eq. (B1) into a position-dependent and a time-dependent factor, i.e.

V (r, t) = U(r)e−iωt, where U(r) =A

reikr, (B2)

which is Eq. (2) in spherical coordinates, when central symmetry is imposed.

2. The Helmholtz equation

The Helmholtz equation (3) is derived by plugging V (r, t) from Eq. (2) (or Eq. (B2) if we want to impose sphericalsymmetry) into the wave equation (1). We only need to calculate the time derivative:

∂2

∂t2V = −ω2V. (B3)

13

Thus we get, using Eq. (B2),

∇2V − 1

c2∂2V

∂t2≡ ∇2(Ue−iωt) +

ω2

c2(Ue−iωt) ≡ (∇2U + k2U)e−iωt = 0, (B4)

which is exactly Eq. (3) for k = ω/c.

3. Directional derivatives

For the calculation of the directional derivative ∂∂n

(eiks

s

)in Eq. (11), we have replaced ∂/∂n with ∂/∂s over S′,

since at each point from S′ the vector n points along the direction of increasing s (with s being the distance from P ).Thus we have

∂

∂n

(eiks

s

)=

∂

∂s

(eiks

s

)=eiks

s

(ik − 1

s

). (B5)

Alternatively, we could have used the definition of a directional derivative

∂

∂n= n · ∇. (B6)

It is very helpful in the more general case of Eq. (13), where the derivative is worked out as follows:

∂U (i)

∂n= n · ∇

(Aeikr

r

)= n ·

(r∂

∂r

Aeikr

r

)= (n · r)Ae

ikr

r

[ik − 1

r

]=Aeikr

r

[ik − 1

r

]cos(n, r), (B7)

where n · r = cos(n, r), r = rr is a unit vector parallel to r.

APPENDIX C: PROOF OF EQ. (4)

We will prove that ∫v

(U∇2V − V∇2U)dv = −∫S

(U∂V

∂n− V

∂U

∂n

)dS. (C1)

First note that

∇ · (aA) = a∇ ·A+ (∇a) · (∇A), (C2)

so that

a∇ ·A = ∇ · (aA)− (∇a) · (∇A). (C3)

Therefore we have

U∇2V = U∇ · (∇V ) = ∇ · (U∇V )− (∇U) · (∇V ) (C4)

Now from Eq. (C4) we obtain∫v

(U∇2V − V∇2U)dv =

∫v

(∇ · (U∇V )− (∇U) · (∇V )−∇ · (V∇U) + (∇V ) · (∇U))dv =∫v

(∇ · (U∇V )−∇ · (V∇U))dv =

∫v

∇ · (U∇V − V∇U)dv. (C5)

By using the divergence theorem ∫v

∇ ·Adv =

∫S

dS ·A =

∫s

n ·AdS, (C6)

where dS = ndS, with n being a normal vector to dS, we continue Eq. (C5):∫v

(U∇2V − V∇2U)dv =

∫S

n · (U∇V − V∇U)dS =

∫s

(Un · ∇V − V n · ∇U)dS =

∫S

(U∂V

∂n− V

∂U

∂n

)dS, (C7)

where we have used the definition for directional derivative (B6). Thus we have proved Eq. (4).

14

APPENDIX D: POLAR COORDINATES (w,ψ) IN THE DIFFRACTION PATTERN

Here we prove that w is approximately the sine of the angle which the direction (p, q) makes with the the centraldirection p = q = 0, as we claimed in Sec. III B. Note that the vector corresponding to the direction (p, q) is −s′,whereas the vector corresponding to the central direction (p = 0, q = 0) is r′. The sine of the angle between the twovectors is given by the length of their cross product

sin(−s′, r′) = sin(s′, r′) =

∣∣∣∣s′s′ × r′

r′

∣∣∣∣ . (D1)

Recall that s′ = −(x, y, z) and r′ = −(x0, y0, z0) (cf. Fig. 2 (bottom) and Eq. (17)). We neglect (x/z)2, (y/z)2,(x0/z0)

2, (y0/z0)2 and thereby obtain

sin(s′, r′) ≈ (x0z − xz0)2 + (y0z − yz0)

2

z2z20. (D2)

On the other hand,

w =√p2 + q2 =

√(x0r′

+x

s′

)2+(y0r′

+y

s′

)2, (D3)

where r′ and s′ are given by Eq. (17). Again we neglect (x/z)2, (y/z)2, (x0/z0)2, (y0/z0)

2 and thereby obtain

w ≈ (x0z − xz0)2 + (y0z − yz0)

2

z2z20. (D4)

Hence we prove that w ≈ sin(−s′, r′), so that w gives the angular distance between a point P and the centre of thediffraction pattern.

APPENDIX E: LIST OF ASSUMPTIONS

Here is a (not comprehensive) list of the assumptions we made concerning the derivation of:The Helmholtz-Kirchhoff theorem (10):

• Monochromatic wave: In Sec. IA we derived the theorem for monochromatic waves. It can be generalizedrelatively easy [1].

The Fresnel-Kirchhoff’s diffraction formula (15):

• The contour of integration S: In Secs. I A and IB we assumed that i) in the aperture A the disturbance U isthe same as if there is no screen at all; ii) in the nonilluminated region B, U and its derivative are zero.

• Short wavelength: In Sec. IA we assumed that the wavelength λ is much shorter than the distances r and s,i.e. λ≪ r, s.

The Fresnel diffraction formula (16):

• Small aperture size: In Secs. IA, I B and II we assumed that the linear dimensions of the opening, althoughlarge compared to λ, are small compared to r and s, i.e. a≪ r, s, where a is a characteristic size of the aperture(radius for circular, diagonal for rectangle, etc.).

The Fraunhofer diffraction formula (24):

• Even smaller aperture size: In Secs. I A, I B, II for the derivation of the Fraunhofer diffraction formula (24) weassumed that a2/λ≪ r, s, which is a stronger condition than a≪ r, s.

[1] M. Born and E. Wolf, Principles of Optics, 7th edition (1999).