diffraction and interference of light experiment

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Diffraction and Interference (Light)

• is the bending or spreading of a wave around an obstacle or through an opening.
• is most apparent when the size of the obstacle or opening ( a ) and the wavelength of the wave ( λ ) are of the same order ( a  ~ λ ).
• Umbra : a region of total shadow; the source of the wave is completely obscured.
• Penumbra : a region of partial shadow; the source of the wave is partially obscured.
• Sound waves are millions of times longer than light waves.
• Diffraction is more easily observed for sound than light in everyday experience.
• You can hear around corners, but you can't see around them.
• Light sources cast noticeable shadows.
• When waves occupy the same place at the same time they interfere or superpose .
• This is known as the principle of linear superposition .
• It was first performed by the English scientist Thomas Young at the end of the 18th century and is often called Young's double slit experiment .
• Description
• Monochromatic light is incident on an opaque barrier with two parallel, narrow openings (often called slits ).
• Light diffracts through each opening.
• Interference can be produced for other shaped openings, so in general an interference pattern is produced.

• Monochromatic light is incident upon a single opening or obstacle.
• Light diffracts through the opening or around the obstacle.
• An interference pattern arises from the interference of light waves from extreme ends of the opening or opposite sides of the obstacle.

• Monochromatic light is incident upon a series of equally small, equally spaced, equally shaped openings or obstacles.
• Light diffracts through the openings or around the obstacles.
• An interference pattern arises where the bright regions are distinct and widely separated.
• grooves in phonograph records
• scratches in a car windshield caused by rough wiper blades
• an optical device consisting of a series of tightly spaced, narrow grooves etched onto a plastic, glass, or metal surface.
• a device for illustrating multiple source interference in the classroom
• a device for producing a spectrum

• Mathematically, there is a nonzero value of the intensity for every value of frequency or wavelength.
• Imagine a smooth curve on a graph where frequency or wavelength is on the x-axis and intensity is on the y-axis.
• This is an informal way of saying that the characteristics of the spectrum are affected by the temperature of the source.
• Mathematically, there is a finite set of wavelengths or frequencies, each with its own intensity.
• Imagine a data table with frequency or wavelength in the first column and intensity in the second column that has a limited number of unique entries.
• Since different elements have different energy level structures, discrete spectra are often described as the "fingerprints" of the elements.
• This is why discrete spectra are often said to be made up of spectral lines .

17.2 Applications of Diffraction, Interference, and Coherence

Section learning objectives.

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

• Explain behaviors of waves, including reflection, refraction, diffraction, interference, and coherence, and describe applications based on these behaviors
• Perform calculations related to applications based on wave properties of light

Teacher Support

The learning objectives in this section will help your students master the following standards:

• (D) investigate behaviors of waves, including reflection, refraction, diffraction, interference, resonance, and the Doppler effect and
• (F) describe the role of wave characteristics and behaviors in medical and industrial applications .

Section Key Terms

Wave-Based Applications of Light

[BL] Define the terms laser, diffraction grating, and resolution .

[OL] Remind students of the meaning of coherent light in terms of wave properties. Ask students to name examples of lasers and diffraction gratings. Suggest common ones they miss, such as lasers used in surgery, as pointers, for reading CDs, and diffraction gratings on the surface of CDs, iridescent minerals, backs of beetles, and in spectroscopes.

[AL] Explain how, for very short wavelengths ( λ < 10 −12  m λ < 10 −12  m ), the limit of resolution is related to Heisenberg’s uncertainty principle.

Misconception Alert

Perfect resolution is impossible. There will always be some blurring of images, no matter what the size of the aperture or the wavelength of light used to make an image.

In 1917, Albert Einstein was thinking about photons and excited atoms. He considered an atom excited by a certain amount of energy and what would happen if that atom were hit by a photon with the same amount of energy. He suggested that the atom would emit a photon with that amount of energy, and it would be accompanied by the original photon. The exciting part is that you would have two photons with the same energy and they would be in phase. Those photons could go on to hit other excited atoms, and soon you would have a stream of in-phase photons. Such a light stream is said to be coherent . Some four decades later, Einstein’s idea found application in a process called, light amplification by stimulated emission of radiation . Take the first letters of all the words (except by and “of”) and write them in order. You get the word laser (see Figure 17.2 (a)), which is the name of the device that produces such a beam of light.

Laser beams are directional, very intense, and narrow (only about 0.5 mm in diameter). These properties lead to a number of applications in industry and medicine. The following are just a few examples:

• This chapter began with a picture of a compact disc (see Figure 17.1 ). Those audio and data-storage devices began replacing cassette tapes during the 1990s. CDs are read by interpreting variations in reflections of a laser beam from the surface.
• Some barcode scanners use a laser beam.
• Lasers are used in industry to cut steel and other metals.
• Lasers are bounced off reflectors that astronauts left on the Moon. The time it takes for the light to make the round trip can be used to make precise calculations of the Earth-Moon distance.
• Laser beams are used to produce holograms. The name hologram means entire picture (from the Greek holo- , as in holistic ), because the image is three-dimensional. A viewer can move around the image and see it from different perspectives. Holograms take advantage of the wave properties of light, as opposed to traditional photography which is based on geometric optics. A holographic image is produced by constructive and destructive interference of a split laser beam.
• One of the advantages of using a laser as a surgical tool is that it is accompanied by very little bleeding.
• Laser eye surgery has improved the vision of many people, without the need for corrective lenses. A laser beam is used to change the shape of the lens of the eye, thus changing its focal length.

Virtual Physics

This animation allows you to examine the workings of a laser. First view the picture of a real laser. Change the energy of the incoming photons, and see if you can match it to an excitation level that will produce pairs of coherent photons. Change the excitation level and try to match it to the incoming photon energy.

In the animation there is only one excited atom. Is that the case for a real laser? Explain.

• No, a laser would have two excited atoms.
• No, a laser would have several million excited atoms.
• Yes, a laser would have only one excited atom.
• No, a laser would have on the order of 10 23 excited atoms.

An interesting thing happens if you pass light through a large number of evenly-spaced parallel slits. Such an arrangement of slits is called a diffraction grating . An interference pattern is created that is very similar to the one formed by double-slit diffraction (see Figure 17.8 and Figure 17.9 ). A diffraction grating can be manufactured by scratching glass with a sharp tool to form a number of precisely positioned parallel lines, which act like slits. Diffraction gratings work both for transmission of light, as in Figure 17.13 , and for reflection of light, as on the butterfly wings or the Australian opal shown in Figure 17.14 , or the CD pictured in the opening illustration of this chapter. In addition to their use as novelty items, diffraction gratings are commonly used for spectroscopic dispersion and analysis of light. What makes them particularly useful is the fact that they form a sharper pattern than do double slits. That is, their bright regions are narrower and brighter, while their dark regions are darker. Figure 17.15 shows idealized graphs demonstrating the sharper pattern. Natural diffraction gratings occur in the feathers of certain birds. Tiny, fingerlike structures in regular patterns act as reflection gratings, producing constructive interference that gives the feathers colors not solely due to their pigmentation. The effect is called iridescence .

Diffraction Grating

• A CD (compact disc) or DVD
• A measuring tape
• Sunlight near a white wall

Instructions

• Hold the CD in direct sunlight near the wall, and move it around until a circular rainbow pattern appears on the wall.
• Measure the distance from the CD to the wall and the distance from the center of the circular pattern to a color in the rainbow. Use those two distances to calculate tan θ tan θ . Find sin θ sin θ .
• Look up the wavelength of the color you chose. That is λ λ .
• Solve d sin θ = m λ d sin θ = m λ for d .

How do you know what number to use for m ?

• Count the rainbow rings preceding the chosen color.
• Calculate mfrom the frequency of the light of the chosen color.
• Calculate m from the wavelength of the light of the chosen color.
• The value of m is fixed for every color.

Fun In Physics

Can you see the grooves on a CD or DVD (see Figure 17.16 )? You may think you can because you know they are there, but they are extremely narrow—1,600 in a millimeter. Because the width of the grooves is similar to wavelengths of visible light, they form a diffraction grating. That is why you see rainbows on a CD. The colors are attractive, but they are incidental to the functions of storing and retrieving audio and other data.

The grooves are actually one continuous groove that spirals outward from the center. Data are recorded in the grooves as binary code (zeroes and ones) in small pits. Information in the pits is detected by a laser that tracks along the groove. It gets even more complicated: The speed of rotation must be varied as the laser tracks toward the circumference so that the linear speed along the groove remains constant. There is also an error correction mechanism to prevent the laser beam from getting off track. A diffraction grating is used to create the first two maxima on either side of the track. If those maxima are not the same distance from the track, an error is indicated and then corrected.

The pits are reflective because they have been coated with a thin layer of aluminum. That allows the laser beam to be reflected back and directed toward a photodiode detector. The signal can then be processed and converted to the audio we hear.

The longest wavelength of visible light is about 780 nm . How does that compare to the distance between CD grooves?

• The grooves are about 3 times the longest wavelength of visible light.
• The grooves are about 2 times the longest wavelength of visible light.
• The grooves are about 2 times the shortest wavelength of visible light.
• The grooves are about 3 times the shortest wavelength of visible light.

Links To Physics

Biology: dic microscopy.

If you were completely transparent, it would be hard to recognize you from your photograph. The same problem arises when using a traditional microscope to view or photograph small transparent objects such as cells and microbes. Microscopes using differential interference contrast (DIC) solve the problem by making it possible to view microscopic objects with enhanced contrast, as shown in Figure 17.17 .

A DIC microscope separates a polarized light source into two beams polarized at right angles to each other and coherent with each other, that is, in phase. After passing through the sample, the beams are recombined and realigned so they have the same plane of polarization. They then create an interference pattern caused by the differences in their optical path and the refractive indices of the parts of the sample they passed through. The result is an image with contrast and shadowing that could not be observed with traditional optics.

Where are diffraction gratings used? Diffraction gratings are key components of monochromators —devices that separate the various wavelengths of incoming light and allow a beam with only a specific wavelength to pass through. Monochromators are used, for example, in optical imaging of particular wavelengths from biological or medical samples. A diffraction grating can be chosen to specifically analyze a wavelength of light emitted by molecules in diseased cells in a biopsy sample, or to help excite strategic molecules in the sample with a selected frequency of light. Another important use is in optical fiber technologies where fibers are designed to provide optimum performance at specific wavelengths. A range of diffraction gratings is available for selecting specific wavelengths for such use.

Diffraction gratings are used in spectroscopes to separate a light source into its component wavelengths. When a material is heated to incandescence, it gives off wavelengths of light characteristic of the chemical makeup of the material. A pure substance will produce a spectrum that is unique, thus allowing identification of the substance. Spectroscopes are also used to measure wavelengths both shorter and longer than visible light. Such instruments have become especially useful to astronomers and chemists. Figure 17.18 shows a diagram of a spectroscope.

Light diffracts as it moves through space, bending around obstacles and interfering constructively and destructively. While diffraction allows light to be used as a spectroscopic tool, it also limits the detail we can obtain in images.

• The bands produced by diffraction gratings are dimmer but sharper than the bands produced by two slits.
• The bands produced by diffraction gratings are brighter, though less sharp, than the bands produced by two slits.
• The bands produced by diffraction gratings are brighter and sharper than the bands produced by two slits.
• The bands produced by diffraction gratings are dimmer and less sharp, but more widely dispersed, than the bands produced by two slits.

Figure 17.19 (a) 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 17.19 (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 are closer together, as in Figure 17.19 (c), you cannot distinguish them, thus limiting the detail, or resolution , you can obtain. That limit is an inescapable consequence of the wave nature of light.

There are many situations in which diffraction limits the resolution. The acuity of vision is limited because light passes through the pupil, 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 of D shows the diffraction effect and spreads, blurring the image, just as light passing through an aperture of diameter D does. 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.

Calculations Involving Diffraction Gratings and Resolution

[BL] Review problem-solving techniques: Identify the knowns and unknowns. Convert measurements of the same property to the same units of measurements. Choose the equation and rearrange it, if necessary, to solve for the unknown.

[OL] Review the meaning of arcsine in particular and of inverse trigonometric functions in general. Explain the radian as a unit of measure for angles, and relate it to degrees.

Early in the chapter, it was mentioned that when light passes from one medium to another, its speed and wavelength change, but its frequency remains constant. The equation

shows how to the wavelength in a given medium, λ n λ n , is related to the wavelength in a vacuum, λ λ , and the refractive index, n , of the medium. The equation is useful for calculating the change in wavelength of a monochromatic laser beam in various media. The analysis of a diffraction grating is very similar to that for a double slit. As you know from the discussion of double slits in Young’s double-slit experiment, light is diffracted by, and spreads out after passing through, each slit. Rays travel at an angle θ θ relative to the incident direction. Each ray travels a different distance to a common point on a screen far away. The rays start in phase, and they can be in or out of phase when they reach a screen, depending on the difference in the path lengths traveled. Each ray travels a distance that differs by d sin θ d sin θ from that of its neighbor, where d is the distance between slits. If d sin θ d sin θ equals an integral number of wavelengths, the rays all arrive in phase, and constructive interference (a maximum) is obtained. Thus, the condition necessary to obtain constructive interference for a diffraction grating is

where d is the distance between slits in the grating, λ λ is the wavelength of the light, and m is the order of the maximum. Note that this is exactly the same equation as for two slits separated by d . However, the slits are usually closer in diffraction gratings than in double slits, producing fewer maxima at larger angles.

Watch Physics

This video explains the geometry behind the diffraction pattern produced by a diffraction grating.

• It is the same as the equation for destructive interference for a double-slit diffraction pattern.
• It is the same as the equation for constructive interference for a double-slit diffraction pattern.
• It is the same as the equation for constructive interference for a single-slit diffraction pattern.
• It is the same as the equation for destructive interference for a single-slit diffraction pattern.

Just what is the resolution limit of an aperture or lens? To answer that question, consider the diffraction pattern for a circular aperture, which, similar to the diffraction pattern of light passing through a slit, has a central maximum that is wider and brighter than the maxima surrounding it (see Figure 17.19 (a)). It can be shown that, for a circular aperture of diameter D , the first minimum in the diffraction pattern occurs at θ = 1.22 λ / D θ = 1.22 λ / D , provided that 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 diffraction 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 center of the diffraction pattern of one is directly over the first minimum of the diffraction pattern of the other. See Figure 17.20 (b). The first minimum is at an angle of θ = 1.22 λ / D θ = 1.22 λ / D , so that two point objects are just resolvable if they are separated by the angle

where λ λ is the wavelength of the light (or other electromagnetic radiation) and D is the diameter of the aperture, lens, mirror, etc., with which the two objects are observed. In the expression above, θ θ has units of radians.

• A sheet of white paper
• A black pen or pencil
• Draw two lines several mm apart on a white sheet of paper.
• Move away from the sheet as it is held upright, and measure the distance at which you can just distinguish (resolve) the lines as separate.
• Use θ = 1.22 λ D θ = 1.22 λ D to calculate D the diameter of your pupil. Use the distance between the lines and the maximum distance at which they were resolved to calculate θ θ . Use the average wavelength for visible light as the value for λ λ .
• Compare your answer to the average pupil diameter of 3 mm.
• The limit for resolution is when the minimum of the pattern for one of the lines is directly over the first minimum of the pattern for the other line.
• The limit for resolution is when the maximum of the pattern for one of the lines is directly over the first minimum of the pattern for the other line.
• The limit for resolution is when the maximum of the pattern for one of the lines is directly over the second minimum of the pattern for the other line.
• The limit for resolution is when the minimum of the pattern for one of the lines is directly over the second maximum of the pattern for the other line.

Worked Example

Change of wavelength.

A monochromatic laser beam of green light with a wavelength of 550 nm passes from air to water. The refractive index of water is 1.33. What will be the wavelength of the light after it enters the water?

You can assume that the refractive index of air is the same as that of light in a vacuum because they are so close. You then have all the information you need to solve for λ n λ n .

The refractive index of air is 1.0003, so the approximation holds for three significant figures. You would not see the light change color, however. Color is determined by frequency, not wavelength.

A diffraction grating has 2000 lines per centimeter. At what angle will the first-order maximum form for green light with a wavelength of 520 nm?

You are given enough information to calculate d , and you are given the values of λ λ and m . You will have to find the arcsin of a number to find θ θ .

First find d .

Rearrange the equation for constructive interference conditions for a diffraction grating, and substitute the known values.

d sin θ = m λ θ = sin − 1 m λ d = sin − 1 ( ( 1 ) ( 520 ) 5,000 ) = 5.97 d sin θ = m λ θ = sin − 1 m λ d = sin − 1 ( ( 1 ) ( 520 ) 5,000 ) = 5.97

This angle seems reasonable for the first maximum. Recall that the meaning of sin ‒1 (or arcsin) is the angle with a sine that is (the unknown) . Remember that the value of sin θ sin θ will not be greater than 1 for any value of θ θ .

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

The diameter of the beam is the same as if it were coming through an aperture of that size, so D = 1.00 mm. You are given λ λ , and you must solve for θ θ .

The conversion factor for radians to degrees is 1.000 radian = 57.3°. The spread is very small and would not be noticeable over short distances. The angle represents the angular separation of the central maximum and the first minimum.

Practice Problems

A beam of yellow light has a wavelength of 600 nm in a vacuum and a wavelength of 397 nm in Plexiglas. What is the refractive index of Plexiglas?

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

• 1.83 × 10 –4 rad
• 2.24 × 10 –4 rad

Check Your Understanding

Use these questions to assess student achievement of the section’s learning objectives. If students are struggling with a specific objective, these questions will help identify which and direct students to the relevant content.

• The pattern is colorful.
• The pattern is faded.
• The pattern is sharper.
• The pattern is curved.
• Light is always polarized.
• Light is always reflected.
• Light is always refracted.
• Light is always diffracted.
• The pattern formed by a diffraction grating has broader and brighter bands.
• The pattern formed by a diffraction grating has broader and duller bands.
• The pattern formed by a diffraction grating has narrower and duller bands.
• The pattern formed by a diffraction grating has narrower and brighter bands.
• The slits in a diffraction grating are broader, with space between them that is greater than the separation of the two slits in two-slit diffraction.
• The slits in a diffraction grating are broader, with space between them that is the same as the separation of the two slits in two-slit diffraction.
• The slits in a diffraction grating are narrower, with space between them that is the same as the separation of the two slits in two-slit diffraction.
• The slits in a diffraction grating are narrower, with space between them that is greater than the separation of the two slits in two-slit diffraction.

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• Authors: Paul Peter Urone, Roger Hinrichs
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• Book title: Physics
• Publication date: Mar 26, 2020
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• Understanding Interference and Diffraction in Optics
• Physics concepts

Welcome to our article on Interference and Diffraction , two fundamental phenomena in the field of Optics. These concepts are crucial in understanding the behavior of light and its interactions with matter, making them essential for anyone interested in the world of physics. Interference and diffraction have been studied for centuries, with scientists and philosophers alike trying to unravel their mysteries. From the ancient Greeks to modern physicists, the fascination with these phenomena has only grown as we continue to uncover new insights and applications. In this article, we will dive deep into the world of Optics and explore the concepts of interference and diffraction . We will discuss their origins, principles, and practical applications, providing you with a comprehensive understanding of these fascinating phenomena. So, whether you're a student, a researcher, or simply someone with a curious mind, join us on this journey as we unravel the complexities of interference and diffraction in optics. To start off, we will define interference and diffraction and explain how they are related to each other.

Diffraction

These include the Young's double-slit formula, the single-slit diffraction formula, and the grating equation. We will also provide step-by-step examples to help you better understand how these formulas work. Next, we will discuss some common applications of interference and diffraction in everyday life. These include technologies such as holograms, fiber optics, and anti-reflection coatings. We will also touch upon their use in fields such as astronomy and microscopy. For those interested in conducting their own experiments, we will provide a list of materials and instructions for a simple double-slit interference experiment.

Diving into Diffraction

Hands-on experiment.

To conduct this experiment, you will need the following materials: a laser pointer, a piece of cardboard, a ruler, and two razor blades. First, cut two slits about 1mm wide and 1cm apart on the cardboard using the razor blades. Place the cardboard with the slits in front of the laser pointer, making sure that the laser beam passes through both slits. Next, place a white screen a few feet away from the cardboard.

You should see a pattern of light and dark bands on the screen. This is known as an interference pattern. By changing the distance between the slits or the distance between the cardboard and the screen, you can observe how the interference pattern changes. This simple experiment allows us to visualize how light waves interfere with each other, creating patterns that demonstrate the principles of interference and diffraction.

Real-World Applications

Latest research and advancements, exploring interference.

Interference occurs when two or more waves overlap and interact with each other, resulting in either constructive or destructive interference. This concept has numerous applications, from creating colorful patterns in soap bubbles to producing high-resolution images with holograms. By understanding the principles of interference, we can gain insight into the nature of light and use it to our advantage in various fields of science and technology. To fully comprehend interference, we must first understand the wave nature of light. Light is an electromagnetic wave that travels through space at a constant speed.

When two light waves meet, they superimpose on each other, creating a new wave with an amplitude that is the sum of the amplitudes of the individual waves. This process is known as interference. One of the most common examples of interference is the colorful patterns we see in soap bubbles. The thin film of soap creates a surface that reflects light waves, and when two waves reflect off this surface and overlap, they interfere with each other. Depending on the thickness of the soap film, certain colors will be enhanced or canceled out, resulting in the iridescent hues we see. Another application of interference is in the production of holograms.

Holograms are three-dimensional images created by using laser light to interfere with each other. The resulting pattern, known as an interference pattern, is recorded on a photographic plate and can be viewed by shining a laser on it. This technology has revolutionized imaging and is used in security features on credit cards and passports. In conclusion, the concept of interference is essential in understanding optics and has numerous real-world applications. By exploring this phenomenon, we can gain a deeper understanding of light and use it in various fields of science and technology.

Calculating with Formulas

• Path Difference: The path difference between two waves can be calculated using the formula: d = nλ , where d is the path difference, n is the number of wavelengths, and λ is the wavelength of the light.
• Constructive Interference: This occurs when two waves with a path difference of d = mλ , where m is an integer, reinforce each other and result in a brighter light. The formula for constructive interference is: I = I 1 + I 2 , where I is the intensity of the resulting light and I 1 and I 2 are the intensities of the individual waves.
• Destructive Interference: This occurs when two waves with a path difference of d = (m + 1/2)λ , where m is an integer, cancel each other out and result in a darker light. The formula for destructive interference is: I = I 1 - I 2 , where I is the intensity of the resulting light and I 1 and I 2 are the intensities of the individual waves.
• Diffracted Angle: The angle at which diffraction occurs can be calculated using the formula: sinθ = nλ/d , where θ is the angle of diffraction, n is the number of wavelengths, and d is the width of the diffracting object.
• Diffracted Intensity: The intensity of the diffracted light can be calculated using the formula: I = I 0 (sinθ/θ) 2 , where I 0 is the intensity of the incident light.

These calculations are crucial for various applications, including designing optical devices and conducting experiments. In conclusion, interference and diffraction are fascinating concepts that play a crucial role in our understanding of light and sound. From simple everyday examples to complex technological applications, these phenomena are all around us. By understanding the basic principles and formulas, you can gain a deeper appreciation for the world of optics. We hope this article has been informative and has sparked your interest in further exploring these concepts.

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17.1 Understanding Diffraction and Interference

Learning objectives.

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

• Explain wave behavior of light, including diffraction and interference, including the role of constructive and destructive interference in Young’s single-slit and double-slit experiments
• Perform calculations involving diffraction and interference, in particular the wavelength of light using data from a two-slit interference pattern

Diffraction and Interference

We know that visible light is the type of electromagnetic wave to which our eyes responds. As we have seen previously, light obeys the equation

where c = 3.00 × 10 8 c = 3.00 × 10 8 m/s is the speed of light in vacuum, f is the frequency of the electromagnetic wave in Hz (or s –1 ), and λ λ is its wavelength in m. The range of visible wavelengths is approximately 380 to 750 nm. As is true for all waves, light travels in straight lines and acts like a ray when it interacts with objects several times as large as its wavelength. However, when it interacts with smaller objects, it displays its wave characteristics prominently. Interference is the identifying behavior of a wave.

In Figure 17.2 , both the ray and wave characteristics of light can be seen. The laser beam emitted by the observatory represents ray behavior, as it travels in a straight line. Passing a pure, one-wavelength beam through vertical slits with a width close to the wavelength of the beam reveals the wave character of light. Here we see the beam spreading out horizontally into a pattern of bright and dark regions that are caused by systematic constructive and destructive interference. As it is characteristic of wave behavior, interference is observed for water waves, sound waves, and light waves.

That interference is a characteristic of energy propagation by waves is demonstrated more convincingly by water waves. Figure 17.3 shows water waves passing through gaps between some rocks. You can easily see that the gaps are similar in width to the wavelength of the waves and that this causes an interference pattern as the waves pass beyond the gaps. A cross-section across the waves in the foreground would show the crests and troughs characteristic of an interference pattern.

Light has wave characteristics in various media as well as in a vacuum. When light goes from a vacuum to some medium, such as water, its speed and wavelength change, but its frequency, f , remains the same. The speed of light in a medium is v = c / n v = c / n , where n is its index of refraction. If you divide both sides of the equation c = f λ c = f λ by n , you get c / n = v = f λ / n c / n = v = f λ / n . Therefore, v = f λ n v = f λ n , where λ n λ n is the wavelength in a medium, and

where λ λ is the wavelength in vacuum and n is the medium’s index of refraction. It follows that the wavelength of light is smaller in any medium than it is in vacuum. In water, for example, which has n = 1.333, the range of visible wavelengths is (380 nm)/1.333 to (760 nm)/1.333, or λ n = λ n = 285–570 nm. Although wavelengths change while traveling from one medium to another, colors do not, since colors are associated with frequency.

The Dutch scientist Christiaan Huygens (1629–1695) developed a useful technique for determining in detail how and where waves propagate. He used wavefronts , which are the points on a wave’s surface that share the same, constant phase (such as all the points that make up the crest of a water wave). Huygens’s principle states, “Every point on a wavefront is a source of wavelets that spread out in the forward direction at the same speed as the wave itself. The new wavefront is a line tangent to all of the wavelets.”

Figure 17.4 shows how Huygens’s principle is applied. A wavefront is the long edge that moves; for example, the crest or the trough. Each point on the wavefront emits a semicircular wave that moves at the propagation speed v . These are drawn later at a time, t , so that they have moved a distance s = v t s = v t . The new wavefront is a line tangent to the wavelets and is where the wave is located at time t . Huygens’s principle works for all types of waves, including water waves, sound waves, and light waves. It will be useful not only in describing how light waves propagate, but also in how they interfere.

What happens when a wave passes through an opening, such as light shining through an open door into a dark room? For light, you expect to see a sharp shadow of the doorway on the floor of the room, and you expect no light to bend around corners into other parts of the room. When sound passes through a door, you hear it everywhere in the room and, thus, you understand that sound spreads out when passing through such an opening. What is the difference between the behavior of sound waves and light waves in this case? The answer is that the wavelengths that make up the light are very short, so that the light acts like a ray. Sound has wavelengths on the order of the size of the door, and so it bends around corners.

If light passes through smaller openings, often called slits, you can use Huygens’s principle to show that light bends as sound does (see Figure 17.5 ). The bending of a wave around the edges of an opening or an obstacle is called diffraction . Diffraction is a wave characteristic that occurs for all types of waves. If diffraction is observed for a phenomenon, it is evidence that the phenomenon is produced by waves. Thus, the horizontal diffraction of the laser beam after it passes through slits in Figure 17.2 is evidence that light has the properties of a wave.

Once again, water waves present a familiar example of a wave phenomenon that is easy to observe and understand, as shown in Figure 17.6 .

Watch Physics

Single-slit interference.

This video works through the math needed to predict diffraction patterns that are caused by single-slit interference.

Which values of m denote the location of destructive interference in a single-slit diffraction pattern?

• whole integers, excluding zero
• whole integers
• real numbers excluding zero
• real numbers

The fact that Huygens’s principle worked was not considered enough evidence to prove that light is a wave. People were also reluctant to accept light’s wave nature because it contradicted the ideas of Isaac Newton, who was still held in high esteem. The acceptance of the wave character of light came after 1801, when the English physicist and physician Thomas Young (1773–1829) did his now-classic double-slit experiment (see Figure 17.7 ).

When light passes through narrow slits, it is diffracted into semicircular waves, as shown in Figure 17.8 (a). Pure constructive interference occurs where the waves line up crest to crest or trough to trough. Pure destructive interference occurs where they line up crest to trough. The light must fall on a screen and be scattered into our eyes for the pattern to be visible. An analogous pattern for water waves is shown in Figure 17.8 (b). Note that regions of constructive and destructive interference move out from the slits at well-defined angles to the original beam. Those angles depend on wavelength and the distance between the slits, as you will see below.

Virtual Physics

Wave interference.

This simulation demonstrates most of the wave phenomena discussed in this section. First, observe interference between two sources of electromagnetic radiation without adding slits. See how water waves, sound, and light all show interference patterns. Stay with light waves and use only one source. Create diffraction patterns with one slit and then with two. You may have to adjust slit width to see the pattern.

Visually compare the slit width to the wavelength. When do you get the best-defined diffraction pattern?

• when the slit width is larger than the wavelength
• when the slit width is smaller than the wavelength
• when the slit width is comparable to the wavelength
• when the slit width is infinite

Calculations Involving Diffraction and Interference

The fact that the wavelength of light of one color, or monochromatic light, can be calculated from its two-slit diffraction pattern in Young’s experiments supports the conclusion that light has wave properties. To understand the basis of such calculations, consider how two waves travel from the slits to the screen. Each slit is a different distance from a given point on the screen. Thus different numbers of wavelengths fit into each path. Waves start out from the slits in phase (crest to crest), but they will end up out of phase (crest to trough) at the screen if the paths differ in length by half a wavelength, interfering destructively. If the paths differ by a whole wavelength, then the waves arrive in phase (crest to crest) at the screen, interfering constructively. More generally, if the paths taken by the two waves differ by any half-integral number of wavelengths ( 1 2 λ ,   3 2 λ ,   5 2 λ ,  etc .) ( 1 2 λ ,   3 2 λ ,   5 2 λ ,  etc .) , then destructive interference occurs. Similarly, if the paths taken by the two waves differ by any integral number of wavelengths ( λ ,   2 λ ,   3 λ ,  etc .) ( λ ,   2 λ ,   3 λ ,  etc .) , then constructive interference occurs.

Figure 17.10 shows how to determine the path-length difference for waves traveling from two slits to a common point on a screen. If the screen is a large distance away compared with the distance between the slits, then the angle θ θ between the path and a line from the slits perpendicular to the screen (see the figure) is nearly the same for each path. That approximation and simple trigonometry show the length difference, Δ L Δ L , to be d sin θ d sin θ , where d is the distance between the slits,

To obtain constructive interference for a double slit, the path-length difference must be an integral multiple of the wavelength, or

Similarly, to obtain destructive interference for a double slit, the path-length difference must be a half-integral multiple of the wavelength, or

The number m is the order of the interference. For example, m = 4 is fourth-order interference.

Figure 17.11 shows how the intensity of the bands of constructive interference decreases with increasing angle.

Light passing through a single slit forms a diffraction pattern somewhat different from that formed by double slits. Figure 17.12 shows a single-slit diffraction pattern. Note that the central maximum is larger than those on either side, and that the intensity decreases rapidly on either side.

The analysis of single-slit diffraction is illustrated in Figure 17.13 . Assuming the screen is very far away compared with the size of the slit, rays heading toward a common destination are nearly parallel. That approximation allows a series of trigonometric operations that result in the equations for the minima produced by destructive interference.

When rays travel straight ahead, they remain in phase and a central maximum is obtained. However, when rays travel at an angle θ θ relative to the original direction of the beam, each ray travels a different distance to the screen, and they can arrive in or out of phase. Thus, a ray from the center travels a distance λ / 2 λ / 2 farther than the ray from the top edge of the slit, they arrive out of phase, and they interfere destructively. Similarly, for every ray between the top and the center of the slit, there is a ray between the center and the bottom of the slit that travels a distance λ / 2 λ / 2 farther to the common point on the screen, and so interferes destructively. Symmetrically, there will be another minimum at the same angle below the direct ray.

Below we summarize the equations needed for the calculations to follow.

The speed of light in a vacuum, c , the wavelength of the light, λ λ , and its frequency, f , are related as follows.

The wavelength of light in a medium, λ n λ n , compared to its wavelength in a vacuum, λ λ , is given by

To calculate the positions of constructive interference for a double slit, the path-length difference must be an integral multiple, m , of the wavelength. λ λ

where d is the distance between the slits and θ θ is the angle between a line from the slits to the maximum and a line perpendicular to the barrier in which the slits are located. To calculate the positions of destructive interference for a double slit, the path-length difference must be a half-integral multiple of the wavelength:

For a single-slit diffraction pattern, the width of the slit, D , the distance of the first ( m = 1) destructive interference minimum, y , the distance from the slit to the screen, L , and the wavelength, λ λ , are given by

Also, for single-slit diffraction,

where θ θ is the angle between a line from the slit to the minimum and a line perpendicular to the screen, and m is the order of the minimum.

Worked Example

Two-slit interference.

Suppose you pass light from a He-Ne laser through two slits separated by 0.0100 mm, and you find that the third bright line on a screen is formed at an angle of 10.95º relative to the incident beam. What is the wavelength of the light?

The third bright line is due to third-order constructive interference, which means that m = 3. You are given d = 0.0100 mm and θ θ = 10.95º. The wavelength can thus be found using the equation d sin θ = m λ d sin θ = m λ for constructive interference.

The equation is d sin θ = m λ d sin θ = m λ . Solving for the wavelength, λ λ , gives

Substituting known values yields

To three digits, 633 nm is the wavelength of light emitted by the common He-Ne laser. Not by coincidence, this red color is similar to that emitted by neon lights. More important, however, is the fact that interference patterns can be used to measure wavelength. Young did that for visible wavelengths. His analytical technique is still widely used to measure electromagnetic spectra. For a given order, the angle for constructive interference increases with λ λ , so spectra (measurements of intensity versus wavelength) can be obtained.

Single-Slit Diffraction

Visible light of wavelength 550 nm falls on a single slit and produces its second diffraction minimum at an angle of 45.0° relative to the incident direction of the light. What is the width of the slit?

From the given information, and assuming the screen is far away from the slit, you can use the equation D sin θ = m λ D sin θ = m λ to find D .

Quantities given are λ λ = 550 nm, m = 2, and θ 2 θ 2 = 45.0°. Solving the equation D sin θ = m λ D sin θ = m λ for D and substituting known values gives

You see that the slit is narrow (it is only a few times greater than the wavelength of light). That is consistent with the fact that light must interact with an object comparable in size to its wavelength in order to exhibit significant wave effects, such as this single-slit diffraction pattern.

Practice Problems

What is the width of a single slit through which 610-nm orange light passes to form a first diffraction minimum at an angle of 30.0°?

Check Your Understanding

• The wavelength first decreases and then increases.
• The wavelength first increases and then decreases.
• The wavelength increases.
• The wavelength decreases.
• This is a diffraction effect. Your whole body acts as the origin for a new wavefront.
• This is a diffraction effect. Every point on the edge of your shadow acts as the origin for a new wavefront.
• This is a refraction effect. Your whole body acts as the origin for a new wavefront.
• This is a refraction effect. Every point on the edge of your shadow acts as the origin for a new wavefront.

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The observation of interference effects definitively indicates the presence of overlapping waves. Thomas Young postulated that light is a wave and is subject to the superposition principle; his great experimental achievement was to demonstrate the constructive and destructive interference of light (c. 1801). In a modern version of Young’s experiment, differing in its essentials only in the source of light, a laser equally illuminates two parallel slits in an otherwise opaque surface. The light passing through the two slits is observed on a distant screen. When the widths of the slits are significantly greater than the wavelength of the light, the rules of geometrical optics hold—the light casts two shadows, and there are two illuminated regions on the screen. However, as the slits are narrowed in width, the light diffracts into the geometrical shadow, and the light waves overlap on the screen. (Diffraction is itself caused by the wave nature of light, being another example of an interference effect—it is discussed in more detail below.)

The superposition principle determines the resulting intensity pattern on the illuminated screen. Constructive interference occurs whenever the difference in paths from the two slits to a point on the screen equals an integral number of wavelengths (0, λ, 2λ,…). This path difference guarantees that crests from the two waves arrive simultaneously. Destructive interference arises from path differences that equal a half-integral number of wavelengths (λ/2, 3λ/2,…). Young used geometrical arguments to show that the superposition of the two waves results in a series of equally spaced bands, or fringes, of high intensity, corresponding to regions of constructive interference, separated by dark regions of complete destructive interference.

An important parameter in the double-slit geometry is the ratio of the wavelength of the light λ to the spacing of the slits d . If λ/ d is much smaller than 1, the spacing between consecutive interference fringes will be small, and the interference effects may not be observable. Using narrowly separated slits, Young was able to separate the interference fringes. In this way he determined the wavelengths of the colours of visible light. The very short wavelengths of visible light explain why interference effects are observed only in special circumstances—the spacing between the sources of the interfering light waves must be very small to separate regions of constructive and destructive interference.

Observing interference effects is challenging because of two other difficulties. Most light sources emit a continuous range of wavelengths, which result in many overlapping interference patterns, each with a different fringe spacing. The multiple interference patterns wash out the most pronounced interference effects, such as the regions of complete darkness. Second, for an interference pattern to be observable over any extended period of time, the two sources of light must be coherent with respect to each other. This means that the light sources must maintain a constant phase relationship. For example, two harmonic waves of the same frequency always have a fixed phase relationship at every point in space, being either in phase, out of phase, or in some intermediate relationship. However, most light sources do not emit true harmonic waves; instead, they emit waves that undergo random phase changes millions of times per second. Such light is called incoherent . Interference still occurs when light waves from two incoherent sources overlap in space, but the interference pattern fluctuates randomly as the phases of the waves shift randomly. Detectors of light, including the eye, cannot register the quickly shifting interference patterns, and only a time-averaged intensity is observed. Laser light is approximately monochromatic (consisting of a single wavelength) and is highly coherent; it is thus an ideal source for revealing interference effects.

After 1802, Young’s measurements of the wavelengths of visible light could be combined with the relatively crude determinations of the speed of light available at the time in order to calculate the approximate frequencies of light. For example, the frequency of green light is about 6 × 10 14 Hz ( hertz , or cycles per second). This frequency is many orders of magnitude larger than the frequencies of common mechanical waves. For comparison, humans can hear sound waves with frequencies up to about 2 × 10 4 Hz. Exactly what was oscillating at such a high rate remained a mystery for another 60 years.

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Diffraction and Interference of Light

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The Scalar Theory of Everything (STOE) model of single photon diffraction is a model with photons being directed by plenum forces along their trajectory. Previous papers explored the screen projections of a diffraction pattern from a first mask through second mask slits. The images on a screen by an edge of the second mask also demonstrate correlation of observed pattern and simulation pattern. The STOE is consistent and all other models of diffraction are inconsistent with these observations.

INTRODUCTION The goal of this lab is to demonstrate the experimental validity of the theoretical predictions made by equations describing the diffraction pattern of light through a double slit. Objectives: 1. To examine how light waves through a double slit interfere constructively and destructively. 2. To evaluate the differences and similarities of the interference patterns for different slit widths under red light. Background:

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CHAPTER OUTLINE 37.1 Conditions for Interference 37.2 Young's Double-Slit Experiment 37.3 Light Waves in Interference 37.4 Intensity Distribution of the Double-Slit Interference Pattern 37.5 Change of Phase Due to Refl ection 37.6 Interference in Thin Films 37.7 The Michelson Interferometer ANSWERS TO QUESTIONS Q37.1 (a) Two waves interfere constructively if their path difference is zero, or an integral multiple of the wavelength, according to δ λ = m , with m = 0 1 2 3 , , , ,. .. . (b) Two waves interfere destructively if their path difference is a half wavelength, or an odd multiple of λ 2 , described by δ λ = + ⎛ ⎝ ⎞ ⎠ m 1 2 , with m = 0 1 2 3 , , , ,. .. . Q37.2 The light from the fl ashlights consists of many different wavelengths (that's why it's white) with random time differences between the light waves. There is no coherence between the two sources. The light from the two fl ashlights does not maintain a constant phase relationship over time. These three equivalent statements mean no possibility of an interference pattern. *Q37.3 (i) The angles in the interference pattern are controlled by λ/d, which we estimate in each case: (a) 0.45 µm/400 µm ≈ 1.1 × 10 −3 (b) 0.7 µm/400 µm ≈ 1.6 × 10 −3 (c) and (d) 0.7 µm/800 µm ≈ 0.9 × 10 −3. The ranking is b > a > c = d. (ii) Now we consider Lλ/d: (a) 4 m (0.45 µm/400 µm) ≈ 4.4 mm (b) 4 m (0.7 µm/400 µm) ≈ 7 mm (c) 4 m(0.7 µm/800 µm) ≈ 3 mm (d) 8 m(0.7 µm/800 µm) ≈ 7 mm. The ranking is b = d > a > c. *Q37.4 Yes. A single beam of laser light going into the slits divides up into several fuzzy-edged beams diverging from the point halfway between the slits. *Q37.5 Answer (c). Underwater, the wavelength of the light decreases according to λ λ water air water = n. Since the angles between positions of light and dark bands are proportional to λ, the underwater fringe separations decrease. Q37.6 Every color produces its own pattern, with a spacing between the maxima that is characteristic of the wavelength. With white light, the central maximum is white. The fi rst side maximum is a full spectrum with violet on the inner edge and red on the outer edge on each side. Each side maximum farther out is in principle a full spectrum, but they overlap one another and are hard to distinguish. Using monochromatic light can eliminate this problem. *Q37.7 With two fi ne slits separated by a distance d slightly less than λ, the equation d sin θ = 0 has the usual solution θ = 0. But d sin θ = 1λ has no solution. There is no fi rst side maximum. d sin θ = (1/2)λ has a solution. Minima fl ank the central maximum on each side. Answer (b). 353

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