Circular Aperture and Resolution

Section 48.6 Circular Aperture and Resolution

Diffraction through a circular aperture spreads the wave over a larger bright circular area surrounded by alternating dark and bright rings. The circular aperture may be just a circular hole in a opaque object. The circular aperture may also be a circular lens through which light could pass such as in the viewing tube of a telescope or microscope. Because of the spreading of light, diffraction limits images that can be resolved in instruments that use lens.

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Consider a circular aperture of diameter \(D\) and we wish to study the diffraction pattern observed on a screen a distance \(L\) away as shown in Figure 48.22. The diffraction through a circular aperture gives rise to a central bright spot which is surrounded by a dark ring which itself is surrounded by a bright ring and so forth as illustrated in Figure 48.22. The circular rings in the diffraction through a circular aperture are also called Airy rings since the intensity of the diffraction function turns out to be a special function called the Airy function.

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Figure 48.22. Diffraction from a circular aperture.
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The radius \(R\) of the central bright spot on the screen is not necessarily equal to the diameter of the aperture but depends on the wavelength \(\lambda\) and the distance \(L\) also.

\begin{equation} R \approx 1.22 \frac{L\lambda}{D}.\tag{48.27} \end{equation}

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To obtain the mathematical expression for the intensity of the diffraction pattern we proceed in the same way as we have done in the case of diffraction through a slit. Specifically, we seek the superposition of the secondary wavelets of the wavefront emerging from the circular slit. The integration is now done over the domain of the slit. We will not present the derivation here and refer the student to more advanced textbooks in optics. We do however wish to discuss some important aspects of the result.

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Sometimes it is more convenient to write Eq. (48.27) for the central bright spot in terms of the angle \(\theta\) subtended at the slit. As the distance to the screen is far greater than the radius of the bright spot, we can use small angle approximation and write

\begin{equation} \theta \approx \tan\theta = \frac{R}{L} = 1.22 \frac{\lambda}{D}.\tag{48.28} \end{equation}

Light from a point object passing through a circular aperture will spread out in accordance with Eq. (48.27) producing a bright circle on the screen instead of a point along with larger circular rings around the circle. As a result, central bright cicles corresponding to two close by sources, such as two stars, may overlap making it difficult, or even impossible, to distinguish them.

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As in other instances, we use Raleigh criterion to set resolvability of two images. Specifically, two images on a screen will be considered just resolvable if the center of one circle is at the edge of the other circle as illustrated in Figure 48.23.

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Figure 48.23. Raleigh criterion of resolution.
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In terms of angular separation of the centers of the two images, the Raleigh criterion states that the angle of separation of the centers of the images must greater than a minimum angle given by the first dark ring of the diffraction pattern of one of the objects.

\begin{equation} \text{Angular separation,}\ \ \theta > \theta_R = \frac{1.22 \lambda}{D}\ \ \ (\text{Raleigh Criterion})\tag{48.29} \end{equation}

Figure 48.24 shows two stars whose directions in the sky have angular separation \(\theta\text{.}\) Suppose we project their images by a lens of diameter \(D\text{.}\) Each will form diffraction pattern at the focal plane of the lens. If stars are too close, central disks of the images will overlap. Therefore, unless, their separation angle \(\theta\) is larger than the angle of airy disk, \(\theta_R\text{,}\) the images will blur into each other.

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Figure 48.24. Angular separation of two images. For the stars to be resolvable the angle \(\theta\) must be larger than \(\theta_R\) given by \(1.22\lambda/D\text{.}\)
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Since loss of resolution due to diffraction cannot be eliminated by grinding a better lens or adding additional optical elements we say that the resolution is diffraction-limited. An image that is diffraction-limited can be improved by changing the aperture or observing in a different part of the electromagnetic spectrum, e.g., by using a smaller \(\lambda\) will make \(\theta_R\) smaller.

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Example 48.25. Radius of Central Bright Disk in Diffraction through a Lens.

A converging lens of diameter \(8\, \text{cm}\) and focal length \(20\, \text{cm}\) is used to focus the image of a star. Find the radius of the central bright spot by using \(550\, \text{nm}\) for the wavelength of light from the star. Ignore the effect of spherical aberration.

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Answer.

\(1.68\,\mu\text{m}\text{.}\)

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Solution.

Let \(D\) be the diameter of the lens and \(f\) its focal length. Then, treating the lens as a circular aperture the passage of the beam through the lens will cause diffraction and the central bright spot will have the following radius at the focal plane,

\begin{equation*} R = 1.22\dfrac{f\:\lambda}{D} = 1.22\dfrac{20\:\text{cm}\times 550\:\text{nm}}{8\:\text{cm}} = 1.68\:\mu\text{m}. \end{equation*}

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Example 48.26. Applying Raleigh Criterion on Diffraction-limited Image from a Circular Lens.

A circular converging lens of diameter \(100\, \text{mm}\) and focal length \(50\, \text{cm}\) is to be used to project the images of two far away point sources of wavelength \(632.8\, \text{nm}\text{.}\) The image distance can be taken to be equal to the focal length and the image assumed to be on the focal plane of the lens. Looking from the lens’s center, what should be the minimum angular separation of the two objects so as to satisfy the Raleigh criterion?

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Answer.

\(7.71\times 10^{-6}\text{ rad}, \text{ or, } 4.42\times 10^{-4}\text{ deg}\)

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Solution.

The angular separation \(\theta\) must be larger than the angle for the Raleigh criterion. Therefore,

\begin{equation*} \theta_\text{min}= 1.22\times \frac{632.8\text{ nm}}{100\text{ mm}} = 7.71\times 10^{-6}\text{ rad} = 4.42\times 10^{-4}\text{ deg}. \end{equation*}

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Example 48.27. Effect of Diffraction on Resolution of an Image of Moon in Yerkes Observatory.

What is the minimum distance between two points on the Moon that the \(40\text{-in}\) refracting telescope at Yerkes observatory in Wisconsin can resolve if the resolution was limited due to the diffraction only? Use \(540\, \text{nm}\) as the wavelength of light and a distance of \(4\, \times 10^8\,\text{m}\) from the Earth to the Moon. Note: the resolution in reality for this case is much worse than just being diffraction-limited.

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Answer.

\(260\:\text{m}\)

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Solution.

In order for the two images to be resolvable, the angle \(\theta\) in Figure 48.28 following figure must be greater than the half width of the central max for image \(I_1\text{.}\)

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Figure 48.28. Figure for Example 48.27 solution.
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Therefore, we get the condition

\begin{equation*} \theta = 1.22\:\dfrac{\lambda}{D} = 1.22\times \dfrac{540 \:\text{nm}}{40\times 2.54\:\text{cm}} = 6.48\times10^{-7}\:\text{rad}. \end{equation*}

Using this angle with the distance \(x\) to the moon we find the separation \(y\) on the moon to be

\begin{equation*} y \approx x\: \theta = 4 \times 10^8\:\text{m}\times 6.48\times10^{-7}\:\text{rad} = 260\:\text{m}. \end{equation*}

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Example 48.29. Distance Two Objects can be Moved and Still Resolvable by Eye If Diffraction-Limited.

Two lamps producing light of frequency \(589\, \text{nm}\) are fixed \(1\, \text{meter}\) apart on a wooden plank. How far the plank can be moved from the eye so that the eye can still resolve them if the resolution is affected solely by the diffraction of light in the eye? Assume light enters the eye through a pupil of diameter \(4.5\, \text{mm}\text{.}\) Note: the resolution in reality for this case is much worse than just being diffraction-limited.

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Answer.

\(6.3\ \text{km}\text{.}\)

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Solution.

Refer to Figure 48.28 of the last problem. We need to find \(x\) here.

\begin{equation*} x = \dfrac{y}{\theta},\ \ \text{and}\ \ \theta = 1.22\:\dfrac{\lambda}{D}. \end{equation*}

Therefore,

\begin{equation*} x = \dfrac{y\: D}{1.22\:\lambda} = 6.3\:\text{km}. \end{equation*}

This is too far for the eye to see. The main problem will be the low intensity of light reaching the eye.

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