Scalar Diffraction Optics

1Introduction¶

In this chapter we will study how light propagates as a wave. In the study of the double-slit experiment we concluded from the interference pattern observed on a screen that light is a wave. To demonstrate more convincingly that light is indeed a wave, we require a detailed quantitative model of the propagation of light, which gives experimentally verifiable predictions.

But a precise description of the propagation of light is not only important for fundamental science, it also has many practical applications. For example, if a sample must be analysed by illuminating it and measuring the scattered light, the fact that the detected light has not only been affected by the sample, but by both the sample and propagation has to be taken into account. Another example is lithography. If a pattern has to be printed onto a substrate using a mask that is illuminated and there is a certain distance between the mask and the photoresist, the light which reaches the resist does not have the exact shape of the mask due to propagation effects. Thus, the mask needs to be designed to compensate for these effects.

From electromagnetic theory, we know that in homogeneous matter (i.e. the permittivity is constant), every component $U$ of a time-harmonic electromagnetic field satisfies the scalar Helmholtz Eq. (1):

where $k=\omega\sqrt{\epsilon \mu_0}$ is the wave number of the light in matter with permittivity $\epsilon$ and refractive index $n=\sqrt{\epsilon/\epsilon_0}$.

When the refractive index is not constant, Maxwell’s equations are no longer equivalent to the wave equation for the individual electromagnetic field components and there is then coupling between the components due to the curl operators in Maxwell’s equation. When the variation of the refractive index is slow on the scale of the wavelength, the scalar wave equation may still be a good approximation, but for structures that vary on the scale of the wavelength (i.e. on the scale of ten microns or less), the scalar wave equation is not sufficiently accurate.

2Propagation of light through a homogenous medium¶

We will describe two equivalent methods to compute the propagation of the field through homogeneous matter, namely the angular spectrum method and the Rayleigh-Sommerfeld diffraction formula. Our goal is to derive the field in some point $(x,y,z)$ with $z>0$, given the field in the plane $z=0$, as is illustrated in Figure 2. Although both methods in the end describe the same, they give physical insight into different aspects of propagation.

2.1Angular Spectrum Method¶

The inverse Fourier transform implies:

The Fourier transform has important properties that are essential for this analysis. By defining $k_x=2\pi \xi$, $k_y=2\pi \eta$, Eq. (3) can be written as

The variables in the Fourier plane: $(\xi,\eta)$ and $(k_x, k_y)$ are called spatial frequencies.

Eq. (4) says that we can write $U_0(x,y)=U(x,y,z=0)$ as an integral (a sum) of plane waves^[1] with wave vector $\mathbf{k}=(k_x,k_y, k_z)^T$, each with its own weight (i.e. complex amplitude) ${\cal F}(U_0)\left(\frac{k_x}{2\pi},\frac{k_y}{2\pi}\right)$. We know how each plane wave with complex amplitude ${\cal F}(U_0)\left(\frac{k_x}{2\pi},\frac{k_y}{2\pi}\right)$ and wave vector $\mathbf{k}=(k_x,k_y,k_z)^T$ propagates over a distance $z>0$

Therefore, the field $U(x,y,z)$ in the plane $z$ (for some $z>0$) is given by

where

with $\lambda$ the wavelength of the light as measured in the material (hence, $\lambda=\lambda_0/n$, with $\lambda_0$ the wavelength in vacuum). The sign in front of the square root in Eq. (7) could in principle be chosen negative: one would then also obtain a solution of the Helmholtz equation. The choice of the sign of $k_z$ is determined by the direction in which the light propagates, which in turn depends on the location of the sources and on the convention chosen for the time dependance. We have to choose here the + sign in front of the square root because the sources are in $z<0$ and the time dependence of time-harmonic fields is (as always in this book) given by $e^{-i\omega t}$ with $\omega >0$.

Eq. (6) can be written alternatively as

where now $k_z$ has to be considered as a function of $(\xi, \eta)$:

Note that one can interpret this as a diagonalisation of the propagation operator.

We can observe something interesting: if $k_x^2+k_y^2 > \left(\frac{2\pi }{\lambda}\right)^2$, then $k_z$ becomes imaginary, and $\exp(i k_z z)$ decays exponentially for increasing $z$:

These exponentially decaying waves are called evanescent in the positive $z$-direction. We have met evanescent waves already in the context of total internal reflection discussed in the context of total internal reflection. The physical consequences of evanescent waves in the angular spectrum decomposition are important for understanding diffraction limits.

The waves for which $k_z$ is real have constant amplitude: only their phase changes due to propagation. These waves therefore are called propagating waves.

Remark. In homogeneous space, the scalar Helmholtz equation for every electric field component is equivalent to Maxwell’s equations and hence we may propagate each component $E_x$, $E_y$ and $E_z$ individually using the angular spectrum method. If the data in the plane $z=0$ of these field components are physically consistent, the electric field thus obtained will automatically satisfy the condition that the electric field is free of divergence, i.e.

everywhere in $z>0$. This is equivalent to the statement that the electric vectors of the plane waves in the angular spectrum are perpendicular to their wave vectors. Alternatively, one can propagate only the $E_x$- and $E_y$-components and afterwards determine $E_z$ from the condition that Eq. (11) must be satisfied.

2.2Rayleigh-Sommerfeld Diffraction Integral¶

Another method to propagate a wave field is by using the Rayleigh-Sommerfeld integral. A very good approximation of this integral states that each point in the plane $z=0$ emits spherical waves with amplitude proportional to the field in the plane $z=0$. To find the field in a point $(x, y,z)$, we have to add the contributions from all these point sources together. This corresponds to the Huygens-Fresnel principle postulated earlier in the Spatial Coherence section of the Interference chapter. Because a more rigorous derivation starting from the Helmholtz equation^[2] would be rather lengthy, we will just give the final result:

where we wrote again $U_0(x,y)=U(x,y,0)$ and where

Remarks.

1. The formula Eq. (12) is not completely rigorous because a term that is a factor $1/(kr)$ smaller (and which in practice therefore is very much smaller) has been omitted.

2. In Eq. (12) there is an additional factor $z/r$ that has been omitted in the standard time-harmonic spherical wave formulation. This factor means that the amplitudes of the spherical waves in the Rayleigh-Sommerfeld diffraction integral depend on direction (although their wave fronts are spherical), the amplitudes being largest in the forward direction.

3. The angular spectrum method amounts to a multiplication by $\exp(i z k_z)$ in Fourier space, while the Rayleigh-Sommerfeld integral is a convolution with respect to $(x,y)$. It is a property of the Fourier transform that a multiplication in Fourier space corresponds to a convolution in real space and vice versa. Indeed a mathematical result called Weyl’s identity implies that the rigorous Rayleigh-Sommerfeld formulation and the plane wave expansion (i.e. the angular spectrum method) give identical results.

3Intuition for the Spatial Fourier Transform in Optics¶

Since spatial Fourier transformations play an important role in our discussion of the propagation of light, it is important to understand them not just mathematically, but also intuitively.

What happens when an object is illuminated and the reflected or transmitted light is detected at some distance from the object? Let us look at transmission for example. When the object is much larger than the wavelength, a transmission function $\tau(x,y)$ is often defined and the field transmitted by the object is then assumed to be simply the product of the incident field and the function $\tau(x,y)$. For example, for a hole in a metallic screen with diameter large compared to the wavelength, the transmission function would be 1 inside the hole and 0 outside. However, if the object has features of the size of the order of the wavelength, this simple model of multiplying by a transmission function breaks down and the transmitted field must instead be determined by solving Maxwell’s equations. This is not easy, but some software packages can do it.

Now suppose that the transmitted electric field has been obtained in a plane $z=0$ very close to the object (a distance within a fraction of a wavelength). This field is called the transmitted near field and it may have been obtained by simply multiplying the incident field with a transmission function $\tau(x,y)$ or by solving Maxwell’s equations. This transmitted near field is a kind of footprint of the object. But it should be clear that, although it is quite common in optics to speak in terms of “imaging an object”, strictly speaking we do not image an object as such, but we image the transmitted or reflected near fields which are a kind of copy of the object. After the transmitted near field has been obtained, we apply the angular spectrum method to propagate the individual plane waves through homogeneous matter (e.g. air) from the object to the detector plane or to an optical element like a lens.

Let $U_0(x,y)=U(x,y,0)$ be a component of the transmitted near field. The first step is to Fourier transform it, by which the field component is decomposed in plane waves. To each plane wave, characterized by the wave numbers $k_x$ and $k_y$, the Fourier transform assigns a complex amplitude $\mathcal{F}(U_0)\left(\frac{k_x}{2\pi},\frac{k_y}{2\pi}\right)$, the magnitude of which indicates how important the role is which this particular wave plays in the formation of the near field. So what can be said about the object field $U_0(x,y)$, by looking at the magnitude of its spatial Fourier transform $|\mathcal{F}(U_0)\left(\frac{k_x}{2\pi},\frac{k_y}{2\pi}\right)|$?

Suppose $U_0(x,y)$ has sharp features, i.e. there are regions where $U_0(x,y)$ varies rapidly as a function of $x$ and $y$. To describe these features as a combination of plane waves, these waves must also vary rapidly as a function of $x$ and $y$, which means that the length of their wave vectors $\sqrt{k_x^2+k_y^2}$ must be large. Thus, the sharper the features that $U_0(x,y)$ has, the larger we can expect $|\mathcal{F}(U_0)\left(\frac{k_x}{2\pi},\frac{k_y}{2\pi}\right)|$ to be for large $\sqrt{k_x^2+k_y^2}$, i.e. high spatial frequencies can be expected to have large amplitude. Similarly, the slowly varying, broad features of $U_0(x,y)$ are described by slowly fluctuating waves, i.e. by $\mathcal{F}(U_0)\left(\frac{k_x}{2\pi},\frac{k_y}{2\pi}\right)$ for small $\sqrt{k_x^2+k_y^2}$, i.e. for low spatial frequencies. This is illustrated in Figure 4.

To investigate these concepts further we choose a certain field, take its Fourier transform, remove the higher spatial frequencies and then invert the Fourier transform. We then expect that the resulting field has lost its sharp features and only retains its broad features, i.e. the image is blurred. Conversely, if we remove the lower spatial frequencies but retain the higher, then the result will only show its sharp features, i.e. its contours. These effects are shown in Figure 6. Recall that when $k_x^2+k_y^2 > \left(\frac{2\pi}{\lambda}\right)^2$, the plane wave decays exponentially as the field propagates. Because by propagation through homogeneous space, the information contained in the high spatial frequencies corresponding to evanescent waves is lost (only exponentially small amplitudes of the evanescent waves remain), perfect imaging is impossible, no matter how well-designed an optical system is.

It is this fact that motivates near-field microscopy, which tries to detect these evanescent waves by scanning close to the sample, thus obtaining subwavelength resolution.

So we have seen how we can guess properties of some object field $U_0(x,y)$ given the amplitude of its spatial Fourier transform $|\mathcal{F}(U_0)\left(\frac{k_x}{2\pi},\frac{k_y}{2\pi}\right)|$. But what about the phase of $\mathcal{F}(U_0)\left(\frac{k_x}{2\pi},\frac{k_y}{2\pi}\right)$? Although one cannot really guess properties of $U_0(x,y)$ by looking at the phase of $\mathcal{F}(U_0)\left(\frac{k_x}{2\pi},\frac{k_y}{2\pi}\right)$ the same way as we can by looking at its amplitude, it is in fact the phase that plays a larger role in defining $U_0(x,y)$. This is illustrated in Figure 8: if the amplitude information of $\mathcal{F}(U_0)\left(\frac{k_x}{2\pi},\frac{k_y}{2\pi}\right)$ is removed, features of the original $U_0(x,y)$ may still be retrieved. However, if we only know the amplitude $|\mathcal{F}(U_0)(k_x,k_y)|$ but not the phase, then the original object is completely lost. Thus, the phase of a field $\mathcal{F}(U_0)$ is very important, arguably often more important than its amplitude. However, we cannot measure the phase of a field directly, only its intensity $I=|\mathcal{F}(U_0)|^2$ from which we can calculate the amplitude $|\mathcal{F}(U_0)|$. It is this fact that makes phase retrieval an entire field of study on its own: how can we find the phase of a field, given that we can only perform intensity measurements? This question is related to a new field of optics called “lensless imaging”, where amplitudes and phases are retrieved from intensity measurements and the image is reconstructed computationally. Interesting as this topic may be, we will not treat it in these notes and refer instead to master courses in optics ^[3].

Remark. The importance of the phase for the field can also be seen by looking at the plane wave expansion Eq. (6). We have seen that the field in a plane $z=\text{constant}$ can be obtained by propagating the plane waves by multiplying their amplitudes by the phase factors $\exp(i z k_z)$, which depends on the propagation distance $z$. If one leaves the evanescent waves out of consideration (since after some distance they hardly contribute to the field anyway), it follows that only the phases of the plane waves change upon propagation, while their amplitudes (the moduli of their complex amplitudes) do not change. Yet, depending on the propagation distance $z$, widely differing light patterns are obtained (see e.g. Figure 12).

Another aspect of the Fourier transform is the uncertainty principle. It states that many waves of different frequencies have to be added to get a function that is confined to a small space^[4]. Stated differently, if $U(x,y)$ is confined to a very small region, then $\mathcal{F}(U)(k_x,k_y)$ must be very spread out. This can also be illustrated by the scaling property of the Fourier transform:

which simply states that the more $h(x)$ is squeezed by increasing $a$, the more its Fourier transform $\mathcal{F}(h)$ spreads out. This principle is illustrated in Figure 9. The uncertainty principle is familiar from quantum physics where it is stated that a particle cannot have both a definite momentum and a definite position. In fact, this is just one particular manifestation of the uncertainty principle just described. A quantum state $|\psi\rangle$ can be described in the position basis $\psi_{x}(x)$ as well as in the momentum basis $\psi_p(p)$. The basis transformation that links these two expressions is the Fourier transform

Hence, the two are obviously subject to the uncertainty principle! In fact, any two quantum observables which are related by a Fourier transform (also called conjugate variables), such as position and momentum, obey this uncertainty relation. The uncertainty relation states:

Since after propagation over a distance $z$, the evanescent waves do not contribute to the Fourier transform of the field, it follows that this Fourier transform has maximum width $\Delta k_x = k$. By the uncertainty principle it follows that after propagation, the minimum width of the field is $\Delta x, \Delta y \approx 2\pi/k=\lambda$.

This poses a fundamental limit to resolution given by the wavelength of the light.

4Fresnel and Fraunhofer Approximations¶

The Fresnel and Fraunhofer approximation are two approximations of the Rayleigh-Sommerfeld integral Eq. (12). The approximations are accurate provided the propagation distance $z$ is sufficiently large. In the Fraunhofer approximation, $z$ has to be quite large, i.e. larger than for the Fresnel approximation, which is already accurate for typical distances occurring in optical systems. Putting it differently: in order of most accurate to least accurate (i.e. only valid for large propagation distances), the diffraction integrals would rank as:

4.1Fresnel Approximation¶

We assume that $z$ in Eq. (12) is so large that in the denominator we can approximate $r$ by $z$:

The reason why we can not apply the same approximation for $r$ in the exponent, is that in the exponent $r$ is multiplied by $k= 2\pi /\lambda$, which is a very large number at optical frequencies, so any error introduced by approximating $r$ would be drastically magnified by multiplying by $k$ which can easily lead to a completely different value of $\exp(ikr)=\cos(kr)+i\sin(kr)$. To approximate $r$ in $\exp(ikr)$ we must be more careful and apply a Taylor expansion. Recall that

We know that the Taylor expansion around $s=0$ implies:

Since we assume that $z$ is large, $\frac{(x-x')^2+(y-y')^2}{z^2}$ is small, so we have

With this approximation, we arrive at the Fresnel diffraction integral, which can be written in the following equivalent forms:

Especially the last expression is interesting, because it shows that

Note that this propagator depends on the distance of propagation $z$.

Remark. By Fourier transforming Eq. (20), one can get the plane wave amplitudes of the Fresnel approximation. It turns out that these amplitudes are equal to ${\cal F}(U_0)$ multiplied by a phase factor which is obtained from a paraxial approximation of the exact phase factor $\exp(i z k_z)$, i.e. $k_z$ is approximated by a quadratic function of $k_x, k_y$. Therefore the Fresnel approximation is also called the paraxial approximation. In fact, it can be shown that the Fresnel diffraction integral is a solution of the paraxial wave equation and conversely, that every solution of the paraxial wave equation can be written as a Fresnel diffraction integral^[5].

4.2Fraunhofer Approximation¶

To obtain the Fraunhofer approximation, we will make one further approximation to $r$ in $\exp(ikr)$ in addition to the Fresnel approximation:

Hence we have omitted the quadratic terms $x'^2+y'^2$, and in comparison with the Fresnel diffraction integral, we omit the factor $\exp\left[ik(x'^2+y'^2)/(2z)\right]$ to obtain the Fraunhofer diffraction integral:

This leads to the following important observation:

Note that to obtain the Fraunhofer field in $(x,y,z)$, the Fourier transform has to be evaluated at spatial frequencies $x/(\lambda z)$ and $y/(\lambda z)$. These spatial frequencies scale with $1/z$ and the overall field $U(x,y,z)$ is proportional to $1/z$. This implies that when $z$ is increased the field pattern spreads out without changing its shape while its amplitude goes down so that the total energy is preserved. Stated differently, apart from the factor $1/z$ in front of the integral, the Fraunhofer field depends on the angles $x/z$ and $y/z$. Therefore for large propagation distances $z$ for which the Fraunhofer approximation is accurate, the field becomes broader (it diverges) when the propagation distance increases.

Remarks.

1. The Fresnel approximation is, like the Fraunhofer approximation, a Fourier transform of the field in the starting plane ($z=0$), evaluated in spatial frequencies which depend on the point of observation:

However, in contrast to the Fraunhofer approximation, the Fresnel approximation depends additionally in a different way on the propagation distance $z$, namely through the exponent of the Fresnel propagator in the integrand. Therefore the Fresnel approximation gives quite diverse patterns depending on the propagation distance $z$, as is shown in Figure 12.

2. In the Rayleigh-Sommerfeld diffraction integral, the field is written as a superposition of spherical waves. In the Fresnel approximation the spherical waves are approximated by parabolic waves. Finally, in the Fraunhofer approximation the propagation distance is so large that the parabolic waves can be approximated by plane waves.

3. Let $f_{a,b}(x,y)=f(x-a,y-b)$ be the function obtained from $f$ by translation. From the general property of the Fourier transform:

it follows that when the field $U_0$ is translated, the intensity in the Fraunhofer far field is not changed. In contrast, due to the additional quadratic phase factor in the integrand of the Fresnel integral, the intensity of the Fresnel field in general changes when $U_0$ is translated.

4. Suppose that $U_0$ is the field immediately behind an aperture ${\cal A}$ with diameter $D$ in an opaque screen. It can then be shown that points $(x,y,z)$ of observation, for which the Fresnel and Fraunhofer diffraction integrals are sufficiently accurate, satisfy:

The Fresnel number is defined by

When $N_F < 0.1$ the Fraunhofer approximation is accurate, while for $N_F>0.1$ it is better to use the Fresnel approximation (see Figure 12). Suppose that $D= 1~\text{mm}$ and the wavelength is that of green light: $\lambda=550~\text{nm}$, then Fraunhofer’s approximation is accurate if $z> 10~\text{m}$.

5. The points of observation where the Fraunhofer approximation can be used must in any case satisfy:

When $x/z>1$, the spatial frequency $k_x = \frac{2 \pi x}{z \lambda} > k$ associated with this point corresponds to an evanescent wave. An evanescent wave obviously cannot contribute to the Fraunhofer far field because it exponentially decreases with distance $z$. In practice the Fresnel and Fraunhofer approximations are used only when $x/z$ and $y/z$ are smaller than 0.3.

6. In any expression for an optical field, one may always omit factors of constant phase, i.e. an overall phase which does not depend on position. If one is only interested in the field in certain planes $z=\text{constant}$, then a factor like $\exp(ikz)$ may also be omitted. Further, in some cases also a position dependent phase factor in front of the Fresnel and Fraunhofer diffraction integrals is omitted, namely when only the intensity is of interest. In exercises it is usually mentioned that this factor may be omitted: if this is not stated, it should be retained in the formulae.

4.3Examples of Fresnel and Fraunhofer approximations¶

Fresnel approximation of the field of two point sources.

Consider two mutual coherent time-harmonic point sources in $\mathbf{r}_s^+=(a/2,0,0)$ and $\mathbf{r}_s^-=(-a/2,0,0)$. The fields in $\mathbf{r}=(x,y,z)$ emitted are according to the time-harmonic field equation (see the Interference chapter) proportional to

We apply the Fresnel approximation for large $z$:

Hence,

where in the denominator we replaced $|\mathbf{r}-\mathbf{r}_s^\pm|$ by $z$. Note that the Fraunhofer approximation amounts to $e^{ i k a^2/(8z)} \approx 1$ while the phase factor $e^{i k \frac{x^2+y^2}{2z} }$ remains. The intensity on a screen $z=\text{constant}$ of the sum of the two fields for the case that the sources have equal strength and emit in phase is:

It is seen that the intensity results from the interference of two plane waves: $\exp[\pm i k ax/(\lambda z)]$ and is given by a cosine function (see Figure 10). Note that for two point sources, the intensity pattern is the same in the Fresnel and the Fraunhofer approximation. However, this is special for two point sources: when more than two point sources are considered, the Fresnel and Fraunhofer patterns are different. The intensity pattern is independent of $y$, and vanishes on lines

and has maxima on lines

for integer $m$.

Fraunhofer approximation of a rectangular aperture in a screen.

Let the screen be $z=0$ and the aperture be given by $-a/2 < x < a/2$, $-b/2 < y < b/2$. The transmission function $\tau(x,y)$ is:

where

and similarly for $1_{[-b/2, b/2]}(y)$. Let the aperture be illuminated by a perpendicular incident plane wave with unit amplitude. Then the field immediately behind the screen is:

We have

where $\text{ sinc}(u)=\sin(u)/u$. Hence,

The Fraunhofer far field at large distance $z$ from a rectangular aperture in mask is obtained by substituting Eq. (40) into Eq. (23).

Remarks.

1. The first zero along the $x$-direction from the center $x=0$ occurs for

The distance between the first two zeros along the $x$-axis is $2\lambda z/a$ and is thus larger when the width along the $x$-direction of the aperture is smaller.

2. The inequalities Eq. (29) imply that when $a< \lambda$, the far field pattern does not have any zeros as function of $x$. When $a$ is further decreased it becomes more and more difficult to deduce the width $a$ from the Fraunhofer intensity. This is an illustration of the fact that information about features that are than the wavelength cannot propagate to the far field.

3. As illustrated in Figure 11, the Fraunhofer diffraction pattern as function of diffraction angle is narrowest in the direction in which the aperture is widest.

Fresnel approximation of a rectangular aperture in a mask

The integral in the Fresnel approximation for the field of a rectangular aperture in a mask can be computed analytically and leads to functions that are actually called “Fresnel integrals” which can be studied using the Cornu spirals. We will not go deeper in this matter but simply show the results of the simulations in Figure 12. The distance to the mask increases ($N_F$ decreases), from very close to the mask at the bottom right, to further from the mask at the bottom left, to rather far from the mask in the upper right, to Fraunhofer distance in the upper left figures. Note the change in scale along the axis in the figures and the decrease of intensity with propagation distance. It is seen that the pattern changes and broadens drastically with distance from what is more or less a copy of the aperture, to a patterns that is equal to the Fourier transform of the aperture. Once the Fraunhofer approximation is accurate, a further increase of distance only results in a widening of the pattern and a decrease of overall amplitude without change of shape. In contrast, in the region where the Fresnel approximation is accurate, the shape of the pattern is seen to change a lot with distance to the mask.

Fraunhofer approximation of a periodic array of slits

We can now predict what is the diffraction pattern of an array of slits of finite width. It follows from the Fraunhofer pattern of a single rectangular aperture that, if the sides parallel to the $y$-direction are very long, the Fraunhofer diffraction pattern as function of angle in the $y$-direction is very narrow. In Figure 11b the Fraunhofer diffraction pattern of a rectangular aperture is shown, of which the width in the $y$-direction is 10 times that in the $x$-direction. The diffraction pattern is then strongly concentrated along the $x$-axis. If we only consider the Fraunhofer pattern for $y/z=0$ while still considering it as a function of $x/z$, it suffices to compute the Fourier transform only with respect to $x$. The problem then becomes a diffraction problem for a one-dimensional slit.

We consider now an array of such slits of which the long sides are all parallel to the $y$-axis and we neglect from now on the $y$-variable. Suppose $W_{\text{slit}}(x)$ is the function that describes the transmission of a single slit. Then the transmission function of $M$ equidistant slits is

where $p$ is the distance of neighbouring slits, i.e. $p$ is the period of the row. If the illumination is by a perpendicular incident plane wave with unit amplitude, the transmitted near field $U_0(x)$ is simply $\tau(x)$. Then

where for a slit with width $a$ we have according to Eq. (39):

There must obviously hold $a< p$. Using

we get

The intensity of the Fraunhofer far ield is:

where $\theta=x/z$ is the diffraction angle. The factor

is, due to the factor $M$ under the sinus in the numerator, a fast oscillating function of $\theta$ while $|{\cal F}(W_{slit})(\theta/\lambda)|^2$ is a slowly varying envelope. This is a manifestation of the property of the Fourier transform that small details of a structure (e.g. the size of a single slit) cause large scale features of the far field pattern, whereas large scale properties such as the length $M p$ of the total structure, cause quickly changing features. This is illustrated in Figure 13.

The diffraction amplitude is maximum for angles where both the denominator and numerator of Eq. (48) vanish:

These directions are called diffraction orders and since

which follows by applying l’H^{o}pital’s rule, the intensity of the m$^{th}$ order is

hence it is proportional to the square of the number of periods and to the squared modulus of the envelope in $\theta_m$.

Between two adjacent diffraction orders, Eq. (48) has $M-1$ zeros and $M-2$ secondary maxima (see Figure 13). The angular width of a diffraction order is half the angular distance to the nearest zeros next to the order, i.e.

If there are more slits the intensity peaks into which the energy is diffracted are narrower and the peaks are higher.

As explained above, there holds in the Fraunhofer far field: $\theta= x/z<1$. This sets a limit to the number of diffracted orders:

Hence, the larger the ratio of the period and the wavelength, the more diffraction orders.

The property eq:diff:diffraction-order that the diffraction orders $m\neq 0$ depend on wavelength is used to separate wavelengths. Grating spectrometers use periodic structures such as an array of slits to very accurately separate and measure wavelengths. The $m$th diffraction order of two wavelengths $\lambda_1$ and $\lambda_2>\lambda_1$ are just separated if

which implies with $\Delta \lambda=\lambda_2-\lambda_1$ and $\lambda=\lambda_1$ that

It follows that the resolution is higher when there are more slits and for larger diffraction order. However, the disadvantage of using higer diffraction orders is that often their intensity is less. For a grating with 1000 periods one can obtain a resolution of $\Delta \lambda/\lambda =10^{-3}$ in the first order.

It should be remarked that a grating is obtained for any periodic variation of the refractive index. If the proper transmission function for the unit cell of the grating is substituted for $W_{slit}$, the formulae above also give the Fraunhofer far field of such more general diffraction gratings. By changing the unit cell, the envelope of the diffraction pattern can be changed and a certain order can be given more intensity. In Figure 14 a so-called blazed grating is shown which is used in reflection and which has a strong first diffracted order for a certain angle of incidence.

Remark. A periodic row of slits is an example of a diffraction grating. A grating is a periodic structure, i.e. the refractive index is a periodic function of position. Structures can be periodic in one, two and three directions. A crystal acts as a three-dimensional grating whose period is the period of the crystal, which typically is a few Angstrom. Electromagnetic waves with wavelength less than one Angstrom are called x-rays. When a beam of x-rays illuminates a crystal, a detector in the far field measures the Fraunhofer diffraction pattern given by the intensity of the Fourier transform of the refracted near field. These diffraction orders of crystals for x-rays where discovered by Von Laue and are used to study the atomic structure of crystals.

5Fraunhofer Diffraction Revisited¶

Fraunhofer diffraction patterns can qualitatively be explained by considering directions in which destructive and constructive interferences occur. Consider two mutually coherent point sources $S_1$, $S_2$ on the $x$-axis as shown in Figure 15. We assume that these point sources are in phase. On a screen at large distance $z$ an interference pattern is observed. If the distance $z$ of the screen is very large, the spherical wave fronts emitted by the point sources are almost plane at the screen and the field is the Fraunhofer far field of the two point sources. In point $P$ on the screen at a distance $x$ above the $z$-axis the optical path differences of the waves emitted by the two sources is approximately given by $S_2Q=a \theta$, where $\theta=x/z$ is assumed small. Hence constructive interference occurs for angles $\theta$ such that $S_2Q=m\lambda$ for some integer $m$, i.e. when

Destructive interference occurs when the path length difference satisfies $S_2Q=\lambda/2 + m \lambda$ for some integer $m$, hence for angles

If the point sources have the same strength, their fields perfectly cancel for these angles.

Now consider a slit as shown in Figure 16 which is illuminated by a perpendicular incident plane wave. By the Huygens-Fresnel principle, the field at a screen far from the slit is the sum of the fields of point sources in the aperture, with strengths proportional to the field in the slit at the position of the point sources. When the slit is illuminated by a plane wave at perpendicular incidence, all point sources are in phase and have equal strength. Divide the slit in two equal halves as shown in Figure 16. The point sources in the slit can be arranged into pairs, of which one point source is in the upper half of the slit and the other is at the equivalent position (at distance $a/2$ from the other point source) in the lower half of the slit. Let $\theta$ be an angle for which the two point sources of a pair cancel each other i.e.

since the distance between the point sources is $a/2$. By translating the pair of point sources through the slits, it follows that both half slits perfectly cancel each other for these angles. In this way we have found the angles $\theta = m\lambda/a$ with $m$ odd for which destructive interference occurs. Destructive interference for $m$ even can be derived by further subdivisions of the aperture.

In general it is easier to find the angles for which the far field vanishes than to find (local) maxima of the field. An exception is the case of a diffraction grating. It follows from Figure 17 that there will be constructive interference between adjacent periods, and hence for all periods, for angles for which the distance $SQ$ in Figure 17 is a multiple of the wavelength, i.e. for

which corresponds to the direction of the diffraction orders. For other angles the phases of the fields of the different periods differ widely and therefore the fields almost cancel at these angles when there are many periods. This explains that for a diffraction grating of many periods, the far field intensity is highly concentrated in particular directions given by the orders Eq. (59) which depend only on the ratio of the wavelength and the period.

6Fourier Optics¶

In this section we apply diffraction theory to a lens. We consider in particular the focusing of a parallel beam and the imaging of an object.

6.1Focusing of a Parallel Beam¶

A lens induces a local phase change to an incident field in proportion to the local thickness of the lens. Let a plane wave which propagates parallel to the optical axis be incident on a positive lens. In Gaussian geometrical optics the incident rays are all focused into the image focal point. According to the Principle of Fermat, all rays have traveled the same optical distance when they intersect in the image focal point where they constructively interfere and cause an intensity maximum. The wavefronts are in the focal region pats of spheres with center the focal point and cut off by the cone with the focal point as top and opening angle $2 a/f$, as shown in Figure 18. Behind the focal point, there is a second cone with again spherical wavefronts, but there the light is propagating away from the focal point. According to Gaussian geometrical optics it is in image space completely dark outside of the two cones in Figure 18. However, as we will show, in diffraction optics this is not true.

We assume that the lens is thin and choose as origin of the coordinate system the center of the thin lens with the positive $z$-axis along the optical axis. Let $f_i$ be the $z$-coordinate of the image focal point according to Gaussian geometrical optics. Then $(0,0,f_i)$ is the image focal point. Let $(x,y,z)$ be a point between the lens and this focal point. According to geometrical optics the field in $(x,y,z)$ is proportional to

where we have included the time-dependence. Indeed the surfaces of constant phase:

are spheres with center the focal point. For increasing time, these spheres converge to the focal point, while the amplitude increases as the distance to the focal point decreases, so that energy is preserved.

Remark. For a point $(x,y,z)$ to the right of the focal point, i.e. for $z>f_i$, the spherical wave fronts propagate away from the focal point and therefore there $-ik$ should be replaced by $+ik$ in the exponent in Eq. (60).

The exit pupil of the lens is in the plane $z=0$ where, according to Eq. (60), the field is

where the time dependence has been omitted and

i.e. $1_{⊙_a}(x,y)=1$ for $(x,y)$ in the exit pupil of the lens and $=0$ otherwise. If $a/f_i$ is sufficiently small, we may replace the distance $\sqrt{x^2 +y^2 + f_i^2}$ between a point in the exit pupil and the image focal point in the denominator of Eq. (62) by $f_i$. This is not allowed in the exponent, however, because of the multiplication by the large wave number $k$. In the exponent we therefore use instead the first two terms of the Taylor series Eq. (18):

which is valid for $a/f_i$ sufficiently small. Then Eq. (62) becomes: