The Rayleigh Criterion – Douglas College Physics 1207 (2023)

Chapter 12 – Wave Optics

Learning Objectives

  • Discuss the Rayleigh criterion.
  • Calculate the limits of resolution for microscopes and telescopes

Light diffracts as it moves through space, bending around obstacles, interfering constructively and destructively. While this can be used as a spectroscopic tool—a diffraction grating disperses light according to wavelength, for example, and is used to produce spectra—diffraction also limits the detail we can obtain in images. The figure below shows the effect of passing light through a small circular aperture. Instead of a bright spot with sharp edges, a spot with a fuzzy edge surrounded by circles of light is obtained. 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, too.

How does diffraction affect the detail that can be observed when light passes through an aperture? Figure 1(b) 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 just barely possible to tell that there are two light sources rather than one. If they were closer together, as in Figure 1(c), we could not distinguish them, thus limiting the detail or resolution we can obtain. This limit is an inescapable consequence of the wave nature of light.

There are many situations in which diffraction limits the resolution. The acuity of our vision is limited because light passes through the pupil, the circular aperture of our 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 Ddoes. So 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 their primary mirror.

Take-Home Experiment: Resolution of the Eye

Draw two lines on a white sheet of paper (several mm apart). How far away can you be and still distinguish the two lines? What does this tell you about the size of the eye’s pupil? Can you be quantitative? (The size of an adult’s pupil is discussed in another chapter.)

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) [see Figure 2(a)]. It can be shown that, for a circular aperture of diameter D, the first minimum in the diffraction pattern occurs atθ = 1.22λ/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 was developed by Lord Rayleigh in the 19th century. The Rayleigh criterion for the diffraction limit to resolution states that two images are just resolvable when the centre of the diffraction pattern of one is directly over the first minimum of the diffraction pattern of the other. See Figure 2(b). The first minimum is at an angle of θ= 1.22λ/ D, so that two point objects are just resolvable if they are separated by the angle

θ= 1.22λ/ D

where 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, θhas units of radians.

Connections: Limits to Knowledge

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 as with an electron microscope are used, the system is disturbed, still limiting our knowledge, much as making an electrical measurement alters a circuit. Heisenberg’s uncertainty principle asserts that this limit is fundamental and inescapable, as we shall see in quantum mechanics.

Example 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 the 2 million light year 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 the equationθ = 1.22λ/D gives the smallest possible angle θbetween point sources, or the best obtainable resolution. Once this angle is found, the distance between stars can be calculated, since we are given how far away they are.

Solution for (a)

The Rayleigh criterion for the minimum resolvable angle is

θ = 1.22λ/D

Entering known values gives

The Rayleigh Criterion – Douglas College Physics 1207 (3)

Solution for (b)

The distance s between two objects a distance r away and separated by an angle θ is s = rθ, by definition of an angle in radians.

Substituting known values gives

The Rayleigh Criterion – Douglas College Physics 1207 (4)

Discussion

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 non-uniformities in mirrors or aberrations in lenses that further limit resolution. However, Figure 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 the Earth’s atmosphere.

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 5 light years in the outer parts and about 1 light year near the galactic centre. 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 for its light to reach us. Figure 4 shows another mirror used to observe radio waves from outer space.

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 wavelengthλ exhibits diffraction spreading. The beam spreads out with an angleθ given by the equationθ = 1.22λ/D. Take, for example, a laser beam made of rays as parallel as possible (angles between rays as close toθ = 0oas possible) instead spreads out at an angleθ = 1.22λ/D, whereD is the diameter of the beam andλis 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 (see Figure 5). To avoid this, we can increase D. This is done for laser light sent to the Moon to measure its distance from the Earth. The laser beam is expanded through a telescope to makeD much larger andθ smaller.

In most biology laboratories, resolution is presented when the use of the microscope is introduced. The ability of a lens to produce sharp images of two closely spaced point objects is called resolution. 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. In Figure 6(a) we have two point objects separated by a distancex. According to the Rayleigh criterion, resolution is possible when the minimum angular separation is

θ = 1.22 λ/D = 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 thatx is much smaller than d), so that tanθ ≈ sinθ≈ θ when is in radians.

Therefore, the resolving power is

x = 1.22 λ d / D

A fun interactive showing cell sizes can be found at https://www.cellsalive.com/howbig.htm

The Rayleigh Criterion – Douglas College Physics 1207 (8)

As I write this in March 2020, Covid-19 is a major concern as we do not yet have a vaccine.Coronavirus are about 82-94 nm in diameter, excluding the spikes. [1]

Another way to look at this is by re-examining the concept of Numerical Aperture (NA) discussed in the previous chapter that deal with the lenses in a microscope, but did not talk about the wave light nature of light. Chapter Microscopes. There,NA is a measure of the maximum acceptance angle at which the fibre will take light and still contain it within the fibre.Figure 6(b) shows a lens and an object at point P. TheNA 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 θ = 2 α. From the figure and again using the small angle approximation, we can write

θ = x / d = D / d but sin θ = θ = D/d for small angles.

but by definition α = θ / 2 or θ = 2α, so

sinθ / 2 = θ / 2 = x / 2d = D / 2d gives

sinα =D / 2d or 2 sinα =D / d

The NA for a lens is NA = n sinα, wheren is the index of refraction of the medium between the objective lens and the object at point P.

From this definition forNA, we can see from substituting the equation above, that

x = 1.22 λ d / D = (1.22λ) / (2sinα ) = 0.61λ (n / NA)

In a microscope,NA is important because it relates to the resolving power of a lens. A lens with a largeNA will be able to resolve finer details. Lenses with largerNA will also be able to collect more light and so give a brighter image. Another way to describe this situation is that the larger theNA, the larger the cone of light that can be brought into the lens, and so more of the diffraction modes will be collected. Thus the microscope has more information to form a clear image, and so its resolving power will be higher.

One of the consequences of diffraction is that the focal point of a beam has a finite width and intensity distribution. Consider focusing when only considering geometric optics, shown in Figure 7(a). The focal point is infinitely small with a huge intensity and the capacity to incinerate most samples irrespective of theNA of the objective lens. For wave optics, due to diffraction, the focal point spreads to become a focal spot (see Figure 7(b)) with the size of the spot decreasing with increasing NA. Consequently, the intensity in the focal spot increases with increasing NA. The higher theNA, the greater the chances of photodegrading the specimen. However, the spot never becomes a true point.

If you want to learn more about this and other topics, visit The Microscope Organization’s website at https://microscope-microscope.org/microscope-info/numerical-aperture/

  • Diffraction limits resolution.
  • For a circular aperture, lens, or mirror, the Rayleigh criterion states that two images are just resolvable when the centre of the diffraction pattern of one is directly over the first minimum of the diffraction pattern of the other.
  • This occurs for two point objects separated by the angle θ = 1.22 λ / D, where λis the wavelength of light (or other electromagnetic radiation) andD is the diameter of the aperture, lens, mirror, etc. This equation also gives the angular spreading of a source of light having a diameter D.

Conceptual Questions

1: A beam of light always spreads out. Why can a beam not be created with parallel rays to prevent spreading? Why can lenses, mirrors, or apertures not be used to correct the spreading?

Problems & Exercises

1: The 300-m-diameter Arecibo radio telescope pictured in Figure 4 detects radio waves with a 4.00 cm average wavelength.

(a) What is the angle between two just-resolvable point sources for this telescope?

(b) How close together could these point sources be at the 2 million light year distance of the Andromeda galaxy?

2: What is the smallest detail that could be observed on the Moon when it is 384,000 km away from the Hubble Telescope? Assume the angular resolution found for the Hubble Telescope is the same as explained in Example 1 — the Hubble Telescope has a mirror diameter of 2.40 metres and the wavelength of the light is 550.0 nm.

3: Diffraction spreading for a flashlight is insignificant compared with other limitations in its optics, such as spherical aberrations in its mirror. To show this, calculate the minimum angular spreading of a flashlight beam that is originally 5.00 cm in diameter with an average wavelength of 600 nm.

4: (a) What is the minimum angular spread of a 633-nm wavelength He-Ne laser beam that is originally 1.00 mm in diameter?

(b) If this laser is aimed at a mountain cliff 15.0 km away, how big will the illuminated spot be?

(c) How big a spot would be illuminated on the Moon, neglecting atmospheric effects? (This might be done to hit a corner reflector to measure the round-trip time and, hence, distance.) Explicitly show how you follow the steps in Chapter Problem-Solving Strategies for Wave Optics.

5: A telescope can be used to enlarge the diameter of a laser beam and limit diffraction spreading. The laser beam is sent through the telescope in opposite the normal direction and can then be projected onto a satellite or the Moon.

(a) If this is done with the Mount Wilson telescope, producing a 2.54-m-diameter beam of 633-nm light, what is the minimum angular spread of the beam?

(b) Neglecting atmospheric effects, what is the size of the spot this beam would make on the Moon, assuming a lunar distance 3.84 x 108 m?

6: The limit to the eye’s acuity is actually related to diffraction by the pupil.

(a) What is the angle between two just-resolvable points of light for a 3.00-mm-diameter pupil, assuming an average wavelength of 550 nm?

(b) Take your result to be the practical limit for the eye. What is the greatest possible distance a car can be from you if you can resolve its two headlights, given they are 1.30 m apart?

(c) What is the distance between two just-resolvable points held at an arm’s length (0.800 m) from your eye?

(d) How does your answer to (c) compare to details you normally observe in everyday circumstances?

7:a) What is the minimum diameter mirror on a telescope that would allow you to see details as small as 5.00 km on the Moon some 384,000 km away? Assume an average wavelength of 550 nm for the light received. Find the angle in radians you need first.

b)What is the minimum diameter mirror on a telescope that would allow you to see details as small as 5.00 metres? This is what you would need to see evidence of the Lunar Landing.

8: You are told not to shoot until you see the whites of their eyes. If the eyes are separated by 6.5 cm and the diameter of your pupil is 5.0 mm, at what distance can you resolve the two eyes using light of wavelength 555 nm?

9: (a) The planet Pluto and its Moon Charon are separated by 19,600 km. Neglecting atmospheric effects, should the 5.08-m-diameter Mount Palomar telescope be able to resolve these bodies when they are 4.50 x 109 km from Earth? Assume an average wavelength of 550 nm.

(b) In actuality, it is just barely possible to discern that Pluto and Charon are separate bodies using an Earth-based telescope. What are the reasons for this?

10: When dots are placed on a page from a laser printer, they must be close enough so that you do not see the individual dots of ink. To do this, the separation of the dots must be less than Raleigh’s criterion. Take the pupil of the eye to be 3.0 mm and the distance from the paper to the eye of 35 cm; find the minimum separation of two dots such that they cannot be resolved. How many dots per inch (dpi) does this correspond to?

11: Unreasonable Results

An amateur astronomer wants to build a telescope with a diffraction limit that will allow him to see if there are people on the moons of Jupiter.

(a) What diameter mirror is needed to be able to see 1.00 m detail on a Jovian Moon at a distance of 7.50 x 108 km from Earth? The wavelength of light averages 600 nm.

(b) What is unreasonable about this result?

(c) Which assumptions are unreasonable or inconsistent?

12: Construct Your Own Problem

Consider diffraction limits for an electromagnetic wave interacting with a circular object. Construct a problem in which you calculate the limit of angular resolution with a device, using this circular object (such as a lens, mirror, or antenna) to make observations. Also calculate the limit to spatial resolution (such as the size of features observable on the Moon) for observations at a specific distance from the device. Among the things to be considered are the wavelength of electromagnetic radiation used, the size of the circular object, and the distance to the system or phenomenon being observed.

Microscopes

13: A basic simple magnification microscope is being used in light with a wavelength of 555 nm. It has an objective lens with a diameter of 5.00 cm. This is the aperture. The specimen is 15.0 cm away from the objective lens. What is the resolution in meters? In micrometers which are also called microns? The width of a human hair is 20 microns. Will this microscope be able to resolve human hair?

14: Covid-19 virus has a size of 88 nm. An optical compound microscope such as that found in most first year college labs has an objective lens with a diameter of 2.0 cm and the virus is on a slide that is being viewed from a distance of 0.50 cm. If it is being viewed in light with a wavelength of 555 nm. a) what is the resolution in a) metres b) nanometres of this microscope? Can you see the virus using this microscope? Yes or no?

15: What wavelength of electromagnetic radiation do you need to view the a Covid-19 virus particle with a size of 82 nanometres? Assume the diameter of the objective lens is 2.0 mm and it is being viewed from 3.0 cm away.

Glossary

Rayleigh criterion
two images are just resolvable when the centre of the diffraction pattern of one is directly over the first minimum of the diffraction pattern of the other

Solutions

Problems & Exercises

1: (a) 1.63 x 10 -4 rads (b) 326 ly

2: angle = 2.80 x 10-7 radians, distance = 3.84 x 108m so s=x=107 metres

3:1.46 x 10-5 rad

5: (a) 3.04 x 10-7 rad (b) Diameter of 235 m

6: angle =2.24 x10-4 radians = x/distance so distance = 5.81 x 103 m = 5.81 km c) s= 1.79 x10-4 m = 0.179 mm. Take a ruler and hold it away from you. Most of you can easily see the 1 mm separation so 0.2 mm makes sense.

7:a) angle = 1.30 x 10-5 radians D = 5.15 cm or about 2 inches. This is about the size of Galileo’s telescope.

b)angle = 1.30 x 10-8radians D = 51.3 m. I often get asked by the general public why they cannot see the Lunar Landing Site with their telescope which is usually about 2 inches in diameter. This is why.

8: 480 m

9: (a) Yes. Should easily be able to discern. (b) The fact that it is just barely possible to discern that these are separate bodies indicates the severity of atmospheric aberrations. We do not have perfect skies.

10: distance = 7.9 x 10-5 m Now to a ratio 1 inch=2.54 cm. 2.54 x10-2 m / 7.9 x 10-5 m = 321 dots per inch. Common. https://computer.howstuffworks.com/printer-ink3.htm states that common resolutions are 300-600 dpi.

13: Microscopes: x=1.22 λ (d/D) = 2.01×10-6 m = 2.03 microns which is much less than the 20 micron hair so it will be able to resolve it.

14: 1.7 x10-7 m = 170 nm No, the virus cannot be seen as it is smaller than the resolution of this microscope.

15: 4.5 x10-9 m = 4.5 nm much smaller than visible light. This is ultraviolet electromagnetic radiation.

  1. Neuman BW, Adair BD, Yoshioka C, Quispe JD, Orca G, Kuhn P, et al. (August 2006). "Supramolecular architecture of severe acute respiratory syndrome coronavirus revealed by electron cryomicroscopy". Journal of Virology. 80 (16): 7918–28. doi:10.1128/JVI.00645-06. PMC 1563832. PMID 16873249. Particle diameters ranged from 50 to 150 nm, excluding the spikes, with mean particle diameters of 82 to 94 nm.

FAQs

What is Rayleigh criterion in physics? ›

“Rayleigh criterion for the diffraction limit to resolution states that two images are just resolvable when the centre of the diffraction pattern of one is directly over the first minimum of the diffraction pattern of the other.”

What is the Rayleigh criterion used to determine? ›

The Rayleigh criterion specifies the minimum separation between two light sources that may be resolved into distinct objects. where I(0) is the peak irradiance at the centre of the diffraction pattern, D=2a is the diameter of the aperture, k is the wave number and J1(u) is the first order Bessel function.

What is meant by Rayleigh criterion of resolution? ›

The Rayleigh criterion for 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.

What is the Rayleigh formula? ›

The Rayleigh equation describing the angular distribution of resultant scattering is:[2]iθ/I0={(n′/n)−1}2(NV2/λ4r2)(1+cos2θ)where iθ is the intensity of light scattered at angle θ, I0 is the initial light source intensity, n′ is the refractive index of the particles, n is the refractive index of the suspension medium, ...

Who discovered Rayleigh criterion? ›

To characterize the resolution of a microscope, it is often practical to following the theory of “Rayleigh Criterion” which was invented by John William Strutt, 3rd Baron Rayleigh in 1896.

What is Rayleigh's criterion for spectral resolution? ›

The Rayleigh criterion requires that the maximum of one signal must be at a wavelength that corresponds to a minimum of the other. Deconvolution of the spectrum may be necessary to determine if two signals are considered resolved according to this criterion.

What is Rayleigh criterion of resolving power of two spectral lines? ›

According to Rayleigh's criterion, two nearby images are said to be resolved if they are separated by at least a certain minimum distance so that the position of diffraction central maximum of the first image coincides with the first diffracted minimum of the second image in any order of diffraction and vice versa.

Where does 1.22 come from in Rayleigh criterion? ›

Where does the 1.22 in Rayleigh's criterion come from? It comes from the combination of simple wave theory, the shape of the telescope's aperture and Lord Rayleigh's definition of “resolvable”.

What do you mean by Rayleigh criterion for resolution derive an expression for resolving power for diffraction grating? ›

According to Rayleigh`s criterion, two wavelengths will be just resolved if principal maxima of λ + dλ in the direction θn+dθn falls on the first minima of λ in the direction θn +dθn. We know, in grating equation for maxima is (a+b) sinθ = nλ And minima is N (a+b) sinθ = mλ (m = nN ±1 ) At p2.

What is the formula for resolution? ›

In order to increase the resolution (d=λ/2 NA), the specimen must be viewed using either shorter wavelength (λ) light or through an imaging medium with a relatively high refractive index or with optical components which have a high NA (or, indeed, a combination of all of these factors).

What do you understand by the resolving power of an optical instrument explain the Rayleigh? ›

The ability of an optical instrument to form distinctly separate images of two objects very close is called its resolving power. Rayleigh Criterion of Resolution: It is defined as the inverse of the distance or angular separation between two objects which can be just resolved when viewed through the optical instrument.

What is the SI unit of resolving power? ›

Resolving power doesn't have any SI unit. This is because the resolving power is the ratio of a mean wavelength of a pair of spectral lines and the difference of wavelength between them. Since both the quantities have the same unit, the resolving power has no unit.

Where does 1.22 come from in Rayleigh criterion? ›

Where does the 1.22 in Rayleigh's criterion come from? It comes from the combination of simple wave theory, the shape of the telescope's aperture and Lord Rayleigh's definition of “resolvable”.

What is Abbe criterion? ›

The Abbe diffraction limit determines the spot size to which a light beam can be focused. With current technology, this limits optical microscopy–based techniques using visible light—such as micro-Raman spectroscopy—to supermicron particles.

What is Rayleigh criterion of resolving power of two spectral lines? ›

According to Rayleigh's criterion, two nearby images are said to be resolved if they are separated by at least a certain minimum distance so that the position of diffraction central maximum of the first image coincides with the first diffracted minimum of the second image in any order of diffraction and vice versa.

What is grating in physics? ›

A diffraction grating is an optical element that divides(disperses) light composed of lots of different wavelengths(e.g., white light) into light components by wavelength. The simplest type of grating is one with a large number of evenly spaced parallel slits.

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