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# 4.6: Circular Apertures and Resolution

• • Contributed by OpenStax
• General Physics at OpenStax CNX

Learning Objectives

By the end of this section, you will be able to:

• Describe the diffraction limit on resolution
• Describe the diffraction limit on beam propagation

Light diffracts as it moves through space, bending around obstacles, interfering constructively and destructively. This can be used as a spectroscopic tool—a diffraction grating disperses light according to wavelength, for example, and is used to produce spectra—but diffraction also limits the detail we can obtain in images.

Figure $$\PageIndex{1a}$$ shows the effect of passing light through a small circular aperture. Instead of a bright spot with sharp edges, we obtain a spot with a fuzzy edge surrounded by circles of light. This pattern is caused by diffraction, similar to that produced by a single slit. Light from different parts of the circular aperture interferes constructively and destructively. The effect is most noticeable when the aperture is small, but the effect is there for large apertures as well. Figure $$\PageIndex{1}$$: (a) Monochromatic light passed through a small circular aperture produces this diffraction pattern. (b) Two point-light sources that are close to one another produce overlapping images because of diffraction. (c) If the sources are closer together, they cannot be distinguished or resolved.

How does diffraction affect the detail that can be observed when light passes through an aperture? Figure $$\PageIndex{1b}$$ shows the diffraction pattern produced by two point-light sources that are close to one another. The pattern is similar to that for a single point source, and it is still possible to tell that there are two light sources rather than one. If they are closer together, as in Figure $$\PageIndex{1c}$$, we cannot distinguish them, thus limiting the detail or resolution we can obtain. This limit is an inescapable consequence of the wave nature of light.

Diffraction limits the resolution in many situations. The acuity of our vision is limited because light passes through the pupil, which is the circular aperture of the eye. Be aware that the diffraction-like spreading of light is due to the limited diameter of a light beam, not the interaction with an aperture. Thus, light passing through a lens with a diameter D shows this effect and spreads, blurring the image, just as light passing through an aperture of diameter D does. Thus, diffraction limits the resolution of any system having a lens or mirror. Telescopes are also limited by diffraction, because of the finite diameter D of the primary mirror.

Just what is the limit? To answer that question, consider the diffraction pattern for a circular aperture, which has a central maximum that is wider and brighter than the maxima surrounding it (similar to a slit) (Figure $$\PageIndex{1a}$$). It can be shown that, for a circular aperture of diameter D, the first minimum in the diffraction pattern occurs at $$\theta = 1.22 \lambda/D$$ (providing the aperture is large compared with the wavelength of light, which is the case for most optical instruments). The accepted criterion for determining the diffraction limit to resolution based on this angle is known as the Rayleigh criterion, which was developed by Lord Rayleigh in the nineteenth century.

Rayleigh Criterion

The diffraction limit to resolution states that two images are just resolvable when the center of the diffraction pattern of one is directly over the first minimum of the diffraction pattern of the other (Figure $$\PageIndex{1b}$$).

The first minimum is at an angle of $$\theta = 1.22 \lambda/D$$, so that two point objects are just resolvable if they are separated by the angle

$\theta = 1.22 \lambda/D \label{Rayleigh}$

where λ is the wavelength of light (or other electromagnetic radiation) and D is the diameter of the aperture, lens, mirror, etc., with which the two objects are observed. In this expression, θ has units of radians. This angle is also commonly known as the diffraction limit. Figure $$\PageIndex{2}$$: (a) Graph of intensity of the diffraction pattern for a circular aperture. Note that, similar to a single slit, the central maximum is wider and brighter than those to the sides. (b) Two point objects produce overlapping diffraction patterns. Shown here is the Rayleigh criterion for being just resolvable. The central maximum of one pattern lies on the first minimum of the other.

All attempts to observe the size and shape of objects are limited by the wavelength of the probe. Even the small wavelength of light prohibits exact precision. When extremely small wavelength probes are used, as with an electron microscope, the system is disturbed, still limiting our knowledge. Heisenberg’s uncertainty principle asserts that this limit is fundamental and inescapable, as we shall see in the chapter on quantum mechanics.

Example $$\PageIndex{1}$$: Calculating Diffraction Limits of the Hubble Space Telescope

The primary mirror of the orbiting Hubble Space Telescope has a diameter of 2.40 m. Being in orbit, this telescope avoids the degrading effects of atmospheric distortion on its resolution. (a) What is the angle between two just-resolvable point light sources (perhaps two stars)? Assume an average light wavelength of 550 nm. (b) If these two stars are at a distance of 2 million light-years, which is the distance of the Andromeda Galaxy, how close together can they be and still be resolved? (A light-year, or ly, is the distance light travels in 1 year.)

Strategy

The Rayleigh criterion stated in Equation \ref{Rayleigh}, $$\theta = 1.22 \lambda/D$$, gives the smallest possible angle θ between point sources, or the best obtainable resolution. Once this angle is known, we can calculate the distance between the stars, since we are given how far away they are.

Solution

1. The Rayleigh criterion for the minimum resolvable angle is $\theta = 1.22 \dfrac{\lambda}{D}.$ Entering known values gives $\theta = 1.22\dfrac{550 \times 10^{-9} m}{2.40 \,m} = 2.80 \times 10^{-7} rad.$
2. The distance s between two objects a distance r away and separated by an angle θ is $$s = r\theta$$. Substituting known values gives$s = (2.0 \times 10^6 \,ly)(2.80 \times 10^{-7} \,rad) = 0.56 \,ly.$

Significance

The angle found in part (a) is extraordinarily small (less than 1/50,000 of a degree), because the primary mirror is so large compared with the wavelength of light. As noticed, diffraction effects are most noticeable when light interacts with objects having sizes on the order of the wavelength of light. However, the effect is still there, and there is a diffraction limit to what is observable. The actual resolution of the Hubble Telescope is not quite as good as that found here. As with all instruments, there are other effects, such as nonuniformities in mirrors or aberrations in lenses that further limit resolution. However, Figure $$\PageIndex{3}$$ gives an indication of the extent of the detail observable with the Hubble because of its size and quality, and especially because it is above Earth’s atmosphere. Figure $$\PageIndex{3}$$: These two photographs of the M82 Galaxy give an idea of the observable detail using (a) a ground-based telescope and (b) the Hubble Space Telescope. (credit a: modification of work by “Ricnun”/Wikimedia Commons)

The answer in part (b) indicates that two stars separated by about half a light-year can be resolved. The average distance between stars in a galaxy is on the order of five light-years in the outer parts and about one light-year near the galactic center. Therefore, the Hubble can resolve most of the individual stars in Andromeda Galaxy, even though it lies at such a huge distance that its light takes 2 million years to reach us. Figure $$\PageIndex{4}$$ shows another mirror used to observe radio waves from outer space. Figure $$\PageIndex{4}$$: A 305-m-diameter paraboloid at Arecibo in Puerto Rico is lined with reflective material, making it into a radio telescope. It is the largest curved focusing dish in the world. Although D for Arecibo is much larger than for the Hubble Telescope, it detects radiation of a much longer wavelength and its diffraction limit is significantly poorer than Hubble’s. The Arecibo telescope is still very useful, because important information is carried by radio waves that is not carried by visible light. (credit: Jeff Hitchcock)

Exercise $$\PageIndex{1}$$

What is the angular resolution of the Arecibo telescope shown in Figure $$\PageIndex{4}$$ when operated at 21-cm wavelength? How does it compare to the resolution of the Hubble Telescope?

Solution

$$8.4 \times 10^{-4} \,rad$$, 3000 times broader than the Hubble Telescope

Diffraction is not only a problem for optical instruments but also for the electromagnetic radiation itself. Any beam of light having a finite diameter D and a wavelength λ exhibits diffraction spreading. The beam spreads out with an angle θ given by Equation \ref{Rayleigh}, $$\theta = 1.22 \lambda/D$$. Take, for example, a laser beam made of rays as parallel as possible (angles between rays as close to θ = 0° as possible) instead spreads out at an angle $$\theta = 1.22 \lambda/D$$, where D is the diameter of the beam and λ is its wavelength. This spreading is impossible to observe for a flashlight because its beam is not very parallel to start with. However, for long-distance transmission of laser beams or microwave signals, diffraction spreading can be significant (Figure $$\PageIndex{5}$$). To avoid this, we can increase D. This is done for laser light sent to the moon to measure its distance from Earth. The laser beam is expanded through a telescope to make D much larger and θ smaller. Figure $$\PageIndex{5}$$: The beam produced by this microwave transmission antenna spreads out at a minimum angle $$\theta = 1.22 \lambda/D$$ due to diffraction. It is impossible to produce a near-parallel beam because the beam has a limited diameter.

In most biology laboratories, resolution is an issue when the use of the microscope is introduced. The smaller the distance x by which two objects can be separated and still be seen as distinct, the greater the resolution. The resolving power of a lens is defined as that distance x. An expression for resolving power is obtained from the Rayleigh criterion. Figure $$\PageIndex{6a}$$ shows two point objects separated by a distance x. According to the Rayleigh criterion, resolution is possible when the minimum angular separation is

$\theta = 1.22 \dfrac{\lambda}{D} = \dfrac{x}{d},$

where d is the distance between the specimen and the objective lens, and we have used the small angle approximation (i.e., we have assumed that x is much smaller than d), so that $$tan \,\theta \approx sin \,\theta$$. Therefore, the resolving power is

$x= 1.22 \dfrac{\lambda d}{D}. \nonumber$

Another way to look at this is by the concept of numerical aperture (NA), which is a measure of the maximum acceptance angle at which a lens will take light and still contain it within the lens. Figure $$\PageIndex{1b}$$ shows a lens and an object at point P. The NA here is a measure of the ability of the lens to gather light and resolve fine detail. The angle subtended by the lens at its focus is defined to be $$\theta = 2\alpha$$. From the figure and again using the small angle approximation, we can write

$sin \,\alpha = \dfrac{D/2}{d} = \dfrac{D}{2d}.$

The NA for a lens is $$NA = n \,sin \,\alpha$$, where n is the index of refraction of the medium between the objective lens and the object at point P. From this definition for NA, we can see that

$x = 1.22 \dfrac{\lambda d}{D} = 1.22 \dfrac{\lambda}{2 \,sin \,\alpha} = 0.61 \dfrac{\lambda n}{NA}.$

In a microscope, NA is important because it relates to the resolving power of a lens. A lens with a large NA is able to resolve finer details. Lenses with larger NA are also able to collect more light and so give a brighter image. Another way to describe this situation is that the larger the NA, the larger the cone of light that can be brought into the lens, so more of the diffraction modes are collected. Thus the microscope has more information to form a clear image, and its resolving power is higher. Figure $$\PageIndex{6}$$: (a) Two points separated by a distance x and positioned a distance d away from the objective. (b) Terms and symbols used in discussion of resolving power for a lens and an object at point P (credit a: modification of work by “Infopro”/Wikimedia Commons).

One of the consequences of diffraction is that the focal point of a beam has a finite width and intensity distribution. Imagine focusing when only considering geometric optics, as in Figure $$\PageIndex{7a}$$. The focal point is regarded as an infinitely small point with a huge intensity and the capacity to incinerate most samples, irrespective of the NA of the objective lens—an unphysical oversimplification. For wave optics, due to diffraction, we take into account the phenomenon in which the focal point spreads to become a focal spot (Figure $$\PageIndex{7b}$$) with the size of the spot decreasing with increasing NA. Consequently, the intensity in the focal spot increases with increasing NA. The higher the NA, the greater the chances of photodegrading the specimen. However, the spot never becomes a true point. Figure $$\PageIndex{7}$$:(a) In geometric optics, the focus is modelled as a point, but it is not physically possible to produce such a point because it implies infinite intensity. (b) In wave optics, the focus is an extended region.

In a different type of microscope, molecules within a specimen are made to emit light through a mechanism called fluorescence. By controlling the molecules emitting light, it has become possible to construct images with resolution much finer than the Rayleigh criterion, thus circumventing the diffraction limit. The development of super-resolved fluorescence microscopy led to the 2014 Nobel Prize in Chemistry.

Optical Resolution Simulation

In this Optical Resolution Model, two diffraction patterns for light through two circular apertures are shown side by side in this simulation by Fu-Kwun Hwang. Watch the patterns merge as you decrease the aperture diameters.