Section 51.6 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.
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 51.6.1. We find that 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.
Sometimes it is more convenient to write this equation 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
Light from a point object passing through a circular aperture will spread out in accordance with Eq. (51.6.1) 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.
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 51.6.2.
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.
Figure 51.6.3 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.
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.
Checkpoint 51.6.4. Radius of Central Bright Disk in Diffraction through a Lens.
A converging lens of diameter 8 cm and focal length 20 cm is used to focus the image of a star. Find the radius of the central bright spot by using 550 nm for the wavelength of light from the star. Ignore the effect of spherical aberration.
1.68 \(\mu\)m.
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,
Checkpoint 51.6.5. Applying Raleigh Criterion on Diffraction-limited Image from a Circular Lens.
A circular converging lens of diameter 100 mm and focal length 50 cm is to be used to project the images of two far away point sources of wavelength 632.8 nm. 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?
Use \(\theta_R\text{.}\)
\(7.71\times 10^{-6}\text{ rad}, \text{ or, } 4.42\times 10^{-4}\text{ deg}\)
The angular separation \(\theta\) must be larger than the angle for the Raleigh criterion. Therefore,
Checkpoint 51.6.6. 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-in refracting telescope at Yerkes observatory in Wisconsin can resolve if the resolution was limited due to the diffraction only? Use 540 nm as the wavelength of light and a distance of 4 \(\times\) \(10^8\) m from the Earth to the Moon. Note: the resolution in reality for this case is much worse than just being diffraction-limited.
Use Raleigh criterion.
\(260\:\text{m}\)
In order for the two images to be resolvable, the angle \(\theta\) in Figure 51.6.7 following figure must be greater than the half width of the central max for image \(I_1\text{.}\)
Therefore, we get the condition
Using this angle with the distance \(x\) to the moon we find the separation \(y\) on the moon to be
Checkpoint 51.6.8. Distance Two Objects can be Moved and Still Resolvable by Eye If Diffraction-Limited.
Two lamps producing light of frequency 589 nm are fixed 1-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 mm. Note: the resolution in reality for this case is much worse than just being diffraction-limited.
Use Raleigh criterion.
6.3 km.
Refer to Figure 51.6.7 of the last problem. We need to find \(x\) here. \[ x = \dfrac{y}{\theta},\ \ \text{and}\ \ \theta = 1.22\:\dfrac{\lambda}{D}. \] Therefore, \[ x = \dfrac{y\: D}{1.22\:\lambda} = 6.3\:\text{km}. \] This is too far for the eye to see. The main problem will be the low intensity of light reaching the eye.