# Aperture Diffraction

That's why I researched and wrote my Selecting the Sharpest Aperture piece years ago. Diffraction-Limited MTF. F(diffraction) = 1,800 / f/stop. Diffraction is worst at small f/numbers, like f/22, and not an issue at large f/numbers like f/4. The resolution at which the graph falls to zero is calculated as 1,800 / f/stop.

## Diffraction by Apertures

1. There is a limit to how high you can make the effective f-number. If the aperture becomes too small, then aperture diffraction will begin to affect the image sharpness. Particularly with small CCD detectors, one has to experiment to find out the optimum f-number for sharpness.
2. Circular Aperture Diffraction When light from a point source passes through a small circular aperture, it does not produce a bright dot as an image, but rather a diffuse circular disc known as Airy's discsurrounded by much fainter concentric circular rings.

Plane waves passing through finite openings emerge propagating in all directions by a process called diffraction. Antennas that radiate or receive plane waves within finite apertures are aperture antennas. Examples include the parabolic reflector antennas used for radio astronomy, radar, and receiving satellite television signals, as well as the lenses and finite apertures employed in cameras, microscopes, telescopes, and many optical communications systems. As in the case of dipole antennas; we assume reciprocity and knowledge of the source fields or equivalent currents.

Since we have already derived expressions for fields radiated by arbitrary current distributions, one approach to finding aperture-radiated fields is to determine current distributions equivalent to the given aperture fields. Then these equivalent currents can be replaced by a continuous array of Hertzian dipoles for which we know the radiated far fields.

Consider a uniform current sheet (overline{underline{mathrm J}} ) [A m-1] occupying the x-z plane, as illustrated in Figure 11.1.1. Maxwell’s equations are then satisfied by:

[underline{mathrm{overline E}}=hat{z} mathrm{underline E}_{0} mathrm{e}^{-mathrm{jky}} qquad underline{mathrm{overline H}}=hat{x}left(underline{mathrm{E}}_{mathrm{o}} / eta_{mathrm{o}}right) mathrm{e}^{-mathrm{jky}} qquad(text { for } mathrm{y}>0)]

[overline{mathrm{underline E}}=hat{z} mathrm{underline E}_{mathrm{o}} mathrm{e}^{+mathrm{j} mathrm{ky}} qquad overline{mathrm{underline H}}=-hat{x}left(mathrm{underline E}_{mathrm{o}} / eta_{mathrm{o}}right) mathrm{e}^{+mathrm{j} mathrm{ky}} qquad(text { for } mathrm{y}<0)]

The electric field ( mathrm{hat{z} underline{E}_{0} equiv overline{underline E}_{0}}) must satisfy the boundary condition (2.6.11) that:

[overline{mathrm{underline J}}_{mathrm{S}}=hat{y} timesleft[overline{mathrm{underline H}}left(mathrm{y}=0_{+}right)-overline{mathrm{underline H}}left(mathrm{y}=0_{-}right)right]=hat{y} timesleft[hat{x} frac{mathrm{underline E}_{0}}{eta_{mathrm{o}}}+hat{x} frac{mathrm{underline E}_{0}}{eta_{mathrm{o}}}right]]

[overline{mathrm{underline J}}_{mathrm{S}}=-hat{z} 2 frac{mathrm{underline E}_{0}}{ underline{eta_{mathrm{o}}}}=-2 frac{overline{mathrm{underline E}}_{0}}{eta_{mathrm{o}}}left[mathrm{Am}^{-1}right]]

Therefore we can consider any plane wave emerging from an aperture as emanating from an equivalent current sheet ( underline{mathrm{overline J}}_{mathrm{underline S}}) given by (11.1.4) provided we neglect radiation from the charges and currents induced at the aperture edges. They can generally be neglected if the aperture is large compared to a wavelength and if we remain close to the y axis in the direction of propagation, because then the aperture area dominates the observable radiating area. This approximation (11.1.4) for a finite aperture is valid even if the strength of the plane wave varies across the aperture slowly relative to a wavelength.

The equivalent current sheet (11.1.4) radiates according to (11.1.5) [from (10.2.8)], where we represent the current sheet by an equivalent array of Hertzian dipoles of length dz and current ( mathrm{overline{underline I}=underline{overline J}_{s} d x}):

The far fields radiated by the z-polarized current sheet ( mathrm{overline{mathrm{underline J}}_S (x,z)}) in the aperture A are then:

[begin{align}overline{mathrm{underline E}}_{mathrm{ff}}(theta, phi) & cong hat{theta} frac{mathrm{j} eta_{mathrm{o}}}{2 lambda mathrm{r}} sin theta int_{mathrm{A}} underline{mathrm{J}}_{mathrm{Z}}(mathrm{x}, mathrm{z}) mathrm{e}^{-mathrm{j} mathrm{kr}(mathrm{x}, mathrm{z})} mathrm{d} mathrm{x} mathrm{dz} nonumber & cong-hat{theta} frac{mathrm{j}}{lambda mathrm{r}} sin theta int_{mathrm{A}} mathrm{underline E}_{mathrm{o}}(mathrm{x}, mathrm{y}) mathrm{e}^{-mathrm{j} mathrm{kr}(mathrm{x}, mathrm{z})} mathrm{dx} mathrm{dz}end{align}]

To simplify the integral we can assume all rays are parallel by using the Fraunhofer far-field approximation:

where we define position within the aperture ( overline{mathrm{r}}^{prime} equiv mathrm{x} hat{x}+mathrm{z} hat{z}), and the distance ( mathrm{r_{0}=left(x^{2}+y^{2}+z^{2}right)^{0.5}}). The Fraunhofer approximation is generally used when ( mathrm{r}_{mathrm{o}}>2 mathrm{L}^{2} / lambda). Then:

[overline{mathrm{underline E}}_{mathrm{ff}}(theta, phi) cong-hat{theta} frac{mathrm{j}}{lambda mathrm{r}} mathrm{e}^{-mathrm{j} mathrm{kr}_{mathrm{o}}} sin theta int_{mathrm{A}} mathrm{underline E}_{mathrm{oz}}(mathrm{x}, mathrm{z}) mathrm{e}^{+j k hat{r} bullet mathrm {overline r}^prime} mathrm{dx} mathrm{dz}]

Those points in space too close to the aperture for the Fraunhofer approximation to apply lie in the Fresnel region where r < ~ 2L2/λ, as shown in (11.1.4). If we restrict ourselves to angles close to the y axis we can define the angles ( alpha_{mathrm x}) and ( alpha_{mathrm z}) from the y axis in the x and z directions, respectively, as illustrated in Figure 11.1.1, so that:

[mathrm{hat{r} bullet bar{r}^{prime} cong x sin alpha_{x}+z sin alpha_{z} cong x alpha_{x}+z alpha_{z}}]

Therefore, close to the y axis (11.1.8) can be approximated55 as:

[overline{mathrm{underline E}}_{mathrm{ff}}left(alpha_{mathrm{x}}, alpha_{mathrm{z}}right) cong-hat{theta} frac{mathrm{j}}{lambda mathrm{r}} mathrm{e}^{-mathrm{j} mathrm{kr}_{mathrm{o}}} int_{mathrm{A}} mathrm{underline E}_{mathrm{oz}}(mathrm{x}, mathrm{z}) mathrm{e}^{+mathrm{j} 2 pileft(mathrm{x} alpha_{mathrm{x}}+mathrm{z} alpha_{mathrm{z}}right) / lambda} mathrm{dx} mathrm{dz}]

which is the Fourier transform of the aperture field distribution (mathrm{underline{E}_{o z}(x, z)} ), times a factor that depends on distance r and wavelength λ. Unlike the usual Fourier transform for converting signals between the time and frequency domains, this reversible transform in (11.1.10) is between the aperture spatial domain and the far-field angular domain.

55 In the Huygen’s approximation a factor of (1 + cos(alpha))/2 is added to improve the accuracy, but this has little impact near the y axis. In this expression α is the angle from the direction of propagation (y axis) in any direction.

For reference, the Fourier transform for signals is:

[mathrm{underline{S}(f)=int_{-infty}^{infty} s(t) e^{-j 2 pi f t} d t}]

[mathrm{s(t)=int_{-infty}^{infty} underline{S}(f) e^{j 2 pi f t} d f}]

The Fourier transform (11.1.11) has exactly the same form as the integral of (11.1.10) if we replace the aperture coordinates x and z with their wavelength-normalized equivalents x/λ and z/λ, analogous to time t; α is analogous to frequency f.

Assume the aperture of Figure 11.1.1 is z-polarized, has dimensions (mathrm{L}_{mathrm{x}} times mathrm{L}_{mathrm{z}}), and is uniformly illuminated with amplitude (mathrm{underline E}_{0}). Then its far fields can be computed using (11.1.10):

[underline{mathrm{overline E}}_{mathrm{ff}}left(alpha_{mathrm{x}}, alpha_{mathrm{z}}right) cong-hat{theta} frac{mathrm{j}}{lambda mathrm{r}} mathrm{e}^{-mathrm{j} mathrm{kr}_{mathrm{o}}} int_{-mathrm{L}_{mathrm{z}} / 2}^{+mathrm{L}_{mathrm{z}} / 2} mathrm{e}^{+mathrm{j} 2 pi alpha_{mathrm{z}} mathrm{z} / lambda} int_{-mathrm{L}_{mathrm{x}} / 2}^{+mathrm{L}_{mathrm{x}} / 2} underline{mathrm{E}}_{mathrm{o}}(mathrm{x}, mathrm{z}) mathrm{e}^{+mathrm{j} 2 pi alpha_{mathrm{x}} mathrm{x} / lambda} mathrm{dx} mathrm{dz}]

The inner integral yields:

[begin{align}int_{-mathrm{L}_{mathrm{x}} / 2}^{+mathrm{L}_{mathrm{x}} / 2} mathrm{underline E}_{mathrm{o}}(mathrm{x}, mathrm{z}) mathrm{e}^{+mathrm{j} 2 pi alpha_{mathrm{x}} mathrm{x} / lambda} mathrm{dx} &=mathrm{underline E}_{mathrm o} frac{lambda}{mathrm{j} 2 pi alpha_{mathrm{x}}}left[mathrm{e}^{+mathrm{j} pi alpha_{mathrm{x}} mathrm{L}_{mathrm{x}} / lambda}-mathrm{e}^{-mathrm{j} pi alpha_{mathrm{x}} mathrm{L}_{mathrm{x}} / lambda}right] nonumber &=mathrm{underline E}_{mathrm{o}} frac{sin left(pi alpha_{mathrm{x}} mathrm{L}_{mathrm{x}} / lambdaright)}{pi alpha_{mathrm{x}} / lambda}end{align}]

The outer integral yields a similar result, so the far field is:

### Aperture Diffraction Photography

[overline{mathrm{underline E}}_{mathrm{ff}}(theta, phi) cong-hat{theta} frac{mathrm{j}}{lambda mathrm{r}} mathrm{underline E}_{mathrm{oz}} mathrm{e}^{-mathrm{j} mathrm{kr}_{mathrm{o}}} bullet mathrm{L}_{mathrm{x}} mathrm{L}_{mathrm{z}} frac{sin left(pi alpha_{mathrm{x}} mathrm{L}_{mathrm{x}} / lambdaright)}{pi alpha_{mathrm{x}} mathrm{L}_{mathrm{x}} / lambda} frac{sin left(pi alpha_{mathrm{z}} mathrm{L}_{mathrm{z}} / lambdaright)}{pi alpha_{mathrm{Z}} mathrm{L}_{mathrm{z}} / lambda}]

The total power Pt radiated through the aperture is simply (mathrm{A}left mathrm{underline E}_{mathrm{o}}right ^{2} / 2 eta_{mathrm{o}} ), where A = LxLz, so the antenna gain (mathrm{G}left(alpha_{mathrm{X}}, alpha_{mathrm{Z}}right)) given by (10.3.1) is:

[begin{align}
mathrm{G}left(alpha_{mathrm{x}}, alpha_{mathrm{z}}right) & cong frac{left mathrm{E}_{mathrm{ff}}left(alpha_{mathrm{x}}, alpha_{mathrm{z}}right)right ^{2} / 2 eta_{mathrm{o}}}{mathrm{P}_{mathrm{t}} / 4 pi mathrm{r}^{2}}
& cong mathrm{A} frac{4 pi}{lambda^{2}}left(frac{sin ^{2}left(pi alpha_{mathrm{x}} mathrm{L}_{mathrm{x}} / lambdaright)}{left(pi alpha_{mathrm{x}} mathrm{L}_{mathrm{x}} / lambdaright)^{2}}right)left(frac{sin ^{2}left(pi alpha_{mathrm{z}} mathrm{L}_{mathrm{z}} / lambdaright)}{left(pi alpha_{mathrm{z}} mathrm{L}_{mathrm{z}} / lambdaright)^{2}}right)
end{align}]

The function (sin x)/x appears so often in electrical engineering that it has its own symbol ‘sinc(x)’. Note that sinc(0) = 1 since (mathrm{sin (x) cong x-left(x^{3} / 6right) text { for } x<<1} ). This gain pattern is plotted in Figure 11.1.2. The first nulls occur when (pi alpha_{1} mathrm{L}_{mathrm{i}} / lambda=pi ) (i = x or z), and therefore (alpha_{text {null }}=lambda / mathrm{L} ) where a narrower beamwidth α corresponds to a wider aperture L. The on-axis gain is:

Equation (11.1.18) applies to any uniformly illuminated aperture antenna, and such antennas have on-axis effective areas A(θ,(phi)) that approach their physical areas A, and have peak gains (mathrm{G}_{mathrm{o}}=4 pi mathrm{A} / lambda^{2} ). The antenna pattern of Figure 11.1.2 vaguely resembles that of circular apertures as well, and the same nominal angle to first null, λ/L, roughly applies to all. Such diffraction pattterns largely explain the limiting angular resolution of telescopes, cameras, animal eyes, and photolithographic equipment used for fabricating integrated circuits.

The coupling between two facing aperture antennas having effective areas A1 and A2 is:

[mathrm{P_{r_{2}}=frac{P_{t_{1}}}{4 pi r^{2}} G_{1} A_{2}=frac{P_{t_{1}}}{lambda^{2} r^{2}} A_{1} A_{2}=P_{t_{1}}left(frac{lambda}{4 pi r}right)^{2} G_{1} G_{2}}]

where ( mathrm{P_{r_{2}}}) and ( mathrm{P_{t_{1}}}) are the power received by antenna 2 and the total power transmitted by antenna 1, respectively. For (11.1.19) to be valid, (r^{2} lambda^{2}>>A_{1} A_{2} ); if A1 = A2 = D2 , then we require (mathrm{r}>>mathrm{D}^{2} / lambda) for validity. Otherwise (11.1.19) could predict that more power would be received than was transmitted.

Example (PageIndex{A})

What is the angle between the first nulls of the diffraction pattern for a visible laser (λ = 0.5 microns) illuminating a 1-mm square aperture (about the size of a human iris)? What is the approximate diffraction-limited angular resolution of the human visual system? How does this compare to the maximum angular diameters of Venus, Jupiter, and the moon (~1, ~1, and ~30 arc minutes in diameter, respectively)?

Solution

### Aperture Diffraction Formula

The first null occurs at

### Aperture Diffraction Aberration

( phi=sin ^{-1}(lambda / mathrm{L}) cong 5 times 10^{-7} / 10^{-3}=5 times 10^{-4} text {radians }=0.029^{circ} cong 1.7 text{arc}) minutes.

This is 70 percent larger than the planets Venus and Jupiter at their points of closest approach to Earth, and ~6 percent of the lunar diameter. Cleverly designed neuronal connections in the human visual system improve on this for linear features, as can a dark-adapted iris, which has a larger diameter.

Example (PageIndex{B})

### Fresnel Diffraction Circular Aperture

A cell-phone dipole antenna radiates one watt toward a uniformly illuminated square aperture antenna of area A = one square meter. If Pr = 10-9 watts are required by the receiver for satisfactory link performance, how far apart r can these two terminals be? Does this depend on the shape of the aperture antenna if A remains constant?

Solution

[ mathrm{P_{r}=A P_{t} G_{t} / 4 pi r^{2} Rightarrow r=left(A P_{t} G_{t} / 4 pi P_{r}right)^{0.5}=left(1 times 1 times 1.5 / 4 pi 10^{-9}right)^{0.5} cong 10.9 mathrm{km}}]

The on axis gain Gt of a uniformly illuminated constant-phase aperture antenna is given by (11.1.18). The denominator depends only on the power transmitted through the aperture A, not on its shape. The numerator depends only on the on-axis far field Eff, given by (11.1.10), which again is independent of shape because the phase term in the integral is unity over the entire aperture. Since the on-axis gain is independent of aperture shape, so is the effective area A since (mathrm{A}(theta, phi)=mathrm{G}(theta, phi) lambda^{2} / 4 pi ). The on-axis effective area of a uniformly illuminated aperture approximates its physical area.