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Electro-Magnetic Waves

Electromagnetic waves

7-26-99

At this point in the course we'll move into optics. This might seem like a separate topic from electricity and magnetism, but optics is really a sub-topic of electricity and magnetism. This is because optics deals with the behavior of light, and light is one example of an electromagnetic wave.

Light and other electromagnetic waves

Light is not the only example of an electromagnetic wave. Other electromagnetic waves include the microwaves you use to heat up leftovers for dinner, and the radio waves that are broadcast from radio stations. An electromagnetic wave can be created by accelerating charges; moving charges back and forth will produce oscillating electric and magnetic fields, and these travel at the speed of light. It would really be more accurate to call the speed "the speed of an electromagnetic wave", because light is just one example of an electromagnetic wave.

speed of light in vacuum: c = 3.00 x 10⁸ m/s

As we'll go into later in the course when we get to relativity, c is the ultimate speed limit in the universe. Nothing can travel faster than light in a vacuum.

There is a wonderful connection between c, the speed of light in a vacuum, and the constants that appeared in the electricity and magnetism equations, the permittivity of free space and the permeability of free space. James Clerk Maxwell, who showed that all of electricity and magnetism could be boiled down to four basic equations, also worked out that:



This clearly shows the link between optics, electricity, and magnetism.

Creating an electromagnetic wave

We've already learned how moving charges (currents) produce magnetic fields. A constant current produces a constant magnetic field, while a changing current produces a changing field. We can go the other way, and use a magnetic field to produce a current, as long as the magnetic field is changing. This is what induced emf is all about. A steadily-changing magnetic field can induce a constant voltage, while an oscillating magnetic field can induce an oscillating voltage.

Focus on these two facts:

1. an oscillating electric field generates an oscillating magnetic field
2. an oscillating magnetic field generates an oscillating electric field

Those two points are key to understanding electromagnetic waves.

An electromagnetic wave (such as a radio wave) propagates outwards from the source (an antenna, perhaps) at the speed of light. What this means in practice is that the source has created oscillating electric and magnetic fields, perpendicular to each other, that travel away from the source. The E and B fields, along with being perpendicular to each other, are perpendicular to the direction the wave travels, meaning that an electromagnetic wave is a transverse wave. The energy of the wave is stored in the electric and magnetic fields.

Properties of electromagnetic waves

Something interesting about light, and electromagnetic waves in general, is that no medium is required for the wave to travel through. Other waves, such as sound waves, can not travel through a vacuum. An electromagnetic wave is perfectly happy to do that.

An electromagnetic wave, although it carries no mass, does carry energy. It also has momentum, and can exert pressure (known as radiation pressure). The reason tails of comets point away from the Sun is the radiation pressure exerted on the tail by the light (and other forms of radiation) from the Sun.

The energy carried by an electromagnetic wave is proportional to the frequency of the wave. The wavelength and frequency of the wave are connected via the speed of light:



Electromagnetic waves are split into different categories based on their frequency (or, equivalently, on their wavelength). In other words, we split up the electromagnetic spectrum based on frequency. Visible light, for example, ranges from violet to red. Violet light has a wavelength of 400 nm, and a frequency of 7.5 x 10¹⁴ Hz. Red light has a wavelength of 700 nm, and a frequency of 4.3 x 10¹⁴ Hz. Any electromagnetic wave with a frequency (or wavelength) between those extremes can be seen by humans.

Visible light makes up a very small part of the full electromagnetic spectrum. Electromagnetic waves that are of higher energy than visible light (higher frequency, shorter wavelength) include ultraviolet light, X-rays, and gamma rays. Lower energy waves (lower frequency, longer wavelength) include infrared light, microwaves, and radio and television waves.

Energy in an electromagnetic wave

The energy in an electromagnetic wave is tied up in the electric and magnetic fields. In general, the energy per unit volume in an electric field is given by:



In a magnetic field, the energy per unit volume is:



An electromagnetic wave has both electric and magnetic fields, so the total energy density associated with an electromagnetic wave is:



It turns out that for an electromagnetic wave, the energy associated with the electric field is equal to the energy associated with the magnetic field, so the energy density can be written in terms of just one or the other:



This also implies that in an electromagnetic wave, E = cB.

A more common way to handle the energy is to look at how much energy is carried by the wave from one place to another. A good measure of this is the intensity of the wave, which is the power that passes perpendicularly through an area divided by the area. The intensity, S, and the energy density are related by a factor of c:



Generally, it's most useful to use the average power, or average intensity, of the wave. To find the average values, you have to use some average for the electric field E and the magnetic field B. The root mean square averages are used; the relationship between the peak and rms values is:

Reflection

The reflection and refraction of light

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Rays and wave fronts

Light is a very complex phenomenon, but in many situations its behavior can be understood with a simple model based on rays and wave fronts. A ray is a thin beam of light that travels in a straight line. A wave front is the line (not necessarily straight) or surface connecting all the light that left a source at the same time. For a source like the Sun, rays radiate out in all directions; the wave fronts are spheres centered on the Sun. If the source is a long way away, the wave fronts can be treated as parallel lines.

Rays and wave fronts can generally be used to represent light when the light is interacting with objects that are much larger than the wavelength of light, which is about 500 nm. In particular, we'll use rays and wave fronts to analyze how light interacts with mirrors and lenses.

The law of reflection

Objects can be seen by the light they emit, or, more often, by the light they reflect. Reflected light obeys the law of reflection, that the angle of reflection equals the angle of incidence.



For objects such as mirrors, with surfaces so smooth that any hills or valleys on the surface are smaller than the wavelength of light, the law of reflection applies on a large scale. All the light travelling in one direction and reflecting from the mirror is reflected in one direction; reflection from such objects is known as specular reflection.

Most objects exhibit diffuse reflection, with light being reflected in all directions. All objects obey the law of reflection on a microscopic level, but if the irregularities on the surface of an object are larger than the wavelength of light, which is usually the case, the light reflects off in all directions.

Plane mirrors

A plane mirror is simply a mirror with a flat surface; all of us use plane mirrors every day, so we've got plenty of experience with them. Images produced by plane mirrors have a number of properties, including:

1. the image produced is upright
2. the image is the same size as the object (i.e., the magnification is m = 1)
3. the image is the same distance from the mirror as the object appears to be (i.e., the image distance = the object distance)
4. the image is a virtual image, as opposed to a real image, because the light rays do not actually pass through the image. This also implies that an image could not be focused on a screen placed at the location where the image is.

A little geometry

Dealing with light in terms of rays is known as geometrical optics, for good reason: there is a lot of geometry involved. It's relatively straight-forward geometry, all based on similar triangles, but we should review that for a plane mirror.

Consider an object placed a certain distance in front of a mirror, as shown in the diagram. To figure out where the image of this object is located, a ray diagram can be used. In a ray diagram, rays of light are drawn from the object to the mirror, along with the rays that reflect off the mirror. The image will be found where the reflected rays intersect. Note that the reflected rays obey the law of reflection. What you notice is that the reflected rays diverge from the mirror; they must be extended back to find the place where they intersect, and that's where the image is.



Analyzing this a little further, it's easy to see that the height of the image is the same as the height of the object. Using the similar triangles ABC and EDC, it can also be seen that the distance from the object to the mirror is the same as the distance from the image to the mirror.

Spherical mirrors

Light reflecting off a flat mirror is one thing, but what happens when light reflects off a curved surface? We'll take a look at what happens when light reflects from a spherical mirror, because it turns out that, using reasonable approximations, this analysis is fairly straight-forward. The image you see is located either where the reflected light converges, or where the reflected light appears to diverge from.

A spherical mirror is simply a piece cut out of a reflective sphere. It has a center of curvature, C, which corresponds to the center of the sphere it was cut from; a radius of curvature, R, which corresponds to the radius of the sphere; and a focal point (the point where parallel light rays are focused to) which is located half the distance from the mirror to the center of curvature. The focal length, f, is therefore:

focal length of a spherical mirror : f = R / 2

This is actually an approximation. Parabolic mirrors are really the only mirrors that focus parallel rays to a single focal point, but as long as the rays don't get too far from the principal axis then the equation above applies for spherical mirrors. The diagram shows the principal axis, focal point (F), and center of curvature for both a concave and convex spherical mirror.



Spherical mirrors are either concave (converging) mirrors or convex (diverging) mirrors, depending on which side of the spherical surface is reflective. If the inside surface is reflective, the mirror is concave; if the outside is reflective, it's a convex mirror. Concave mirrors can form either real or virtual images, depending on where the object is relative to the focal point. A convex mirror can only form virtual images. A real image is an image that the light rays from the object actually pass through; a virtual image is formed because the light rays can be extended back to meet at the image position, but they don't actually go through the image position.

Ray diagrams

To determine where the image is, it is very helpful to draw a ray diagram. The image will be located at the place where the rays intersect. You could just draw random rays from the object to the mirror and follow the reflected rays, but there are three rays in particular that are very easy to draw.

Only two rays are necessary to locate the image on a ray diagram, but it's useful to add the third as a check. The first is the parallel ray; it is drawn from the tip of the object parallel to the principal axis. It then reflects off the mirror and either passes through the focal point, or can be extended back to pass through the focal point.

The second ray is the chief ray. This is drawn from the tip of the object to the mirror through the center of curvature. This ray will hit the mirror at a 90° angle, reflecting back the way it came. The chief and parallel rays meet at the tip of the image.

The third ray, the focal ray, is a mirror image of the parallel ray. The focal ray is drawn from the tip of the object through (or towards) the focal point, reflecting off the mirror parallel to the principal axis. All three rays should meet at the same point.



A ray diagram for a concave mirror is shown above. This shows a few different things. For this object, located beyond the center of curvature from the mirror, the image lies between the focal point (F) and the center of curvature. The image is inverted compared to the object, and it is also a real image, because the light rays actually pass through the point where the image is located.

With a concave mirror, any object beyond C will always have an image that is real, inverted compared to the object, and between F and C. You can always trade the object and image places (that just reverses all the arrows on the ray diagram), so any object placed between F and C will have an image that is real, inverted, and beyond C. What happens when the object is between F and the mirror? You should draw the ray diagram to convince yourself that the image will be behind the mirror, making it a virtual image, and it will be upright compared to the object.

A ray diagram for a convex mirror

What happens with a convex mirror? In this case the ray diagram looks like this:



As the ray diagram shows, the image for a convex mirror is virtual, and upright compared to the object. A convex mirror is the kind of mirror used for security in stores, and is also the kind of mirror used on the passenger side of many cars ("Objects in mirror are closer than they appear."). A convex mirror will reflect a set of parallel rays in all directions; conversely, it will also take light from all directions and reflect it in one direction, which is exactly how it's used in stores and cars.

The mirror equation

Drawing a ray diagram is a great way to get a rough idea of how big the image of an object is, and where the image is located. We can also calculate these things precisely, using something known as the mirror equation. The textbook does a nice job of deriving this equation in section 25.6, using the geometry of similar triangles.



Magnification

In most cases the height of the image differs from the height of the object, meaning that the mirror has done some magnifying (or reducing). The magnification, m, is defined as the ratio of the image height to the object height, which is closely related to the ratio of the image distance to the object distance:



A magnification of 1 (plus or minus) means that the image is the same size as the object. If m has a magnitude greater than 1 the image is larger than the object, and an m with a magnitude less than 1 means the image is smaller than the object. If the magnification is positive, the image is upright compared to the object; if m is negative, the image is inverted compared to the object.

Sign conventions

What does a positive or negative image height or image distance mean? To figure out what the signs mean, take the side of the mirror where the object is to be the positive side. Any distances measured on that side are positive. Distances measured on the other side are negative.

f, the focal length, is positive for a concave mirror, and negative for a convex mirror.

When the image distance is positive, the image is on the same side of the mirror as the object, and it is real and inverted. When the image distance is negative, the image is behind the mirror, so the image is virtual and upright.

A negative m means that the image is inverted. Positive means an upright image.

Steps for analyzing mirror problems

There are basically three steps to follow to analyze any mirror problem, which generally means determining where the image of an object is located, and determining what kind of image it is (real or virtual, upright or inverted).

• Step 1 - Draw a ray diagram. The more careful you are in constructing this, the better idea you'll have of where the image is.
  
• Step 2 - Apply the mirror equation to determine the image distance. (Or to find the object distance, or the focal length, depending on what is given.)
  
• Step 3 - Make sure steps 1 and 2 are consistent with each other.

An example

A Star Wars action figure, 8.0 cm tall, is placed 23.0 cm in front of a concave mirror with a focal length of 10.0 cm. Where is the image? How tall is the image? What are the characteristics of the image?

The first step is to draw the ray diagram, which should tell you that the image is real, inverted, smaller than the object, and between the focal point and the center of curvature. The location of the image can be found from the mirror equation:



which can be rearranged to:



The image distance is positive, meaning that it is on the same side of the mirror as the object. This agrees with the ray diagram. Note that we don't need to worry about converting distances to meters; just make sure everything has the same units, and whatever unit goes into the equation is what comes out.

Calculating the magnification gives:



Solving for the image height gives:



The negative sign for the magnification, and the image height, tells us that the image is inverted compared to the object.

To summarize, the image is real, inverted, 6.2 cm high, and 17.7 cm in front of the mirror.

Example 2 - a convex mirror

The same Star Wars action figure, 8.0 cm tall, is placed 6.0 cm in front of a convex mirror with a focal length of -12.0 cm. Where is the image in this case, and what are the image characteristics?

Again, the first step is to draw a ray diagram. This should tell you that the image is located behind the mirror; that it is an upright, virtual image; that it is a little smaller than the object; and that the image is between the mirror and the focal point.

The second step is to confirm all those observations. The mirror equation, rearranged as in the first example, gives:



Solving for the magnification gives:



This gives an image height of 0.667 x 8 = 5.3 cm.

All of these results are consistent with the conclusions drawn from the ray diagram. The image is 5.3 cm high, virtual, upright compared to the object, and 4.0 cm behind the mirror.

Refraction

When we talk about the speed of light, we're usually talking about the speed of light in a vacuum, which is 3.00 x 10⁸ m/s. When light travels through something else, such as glass, diamond, or plastic, it travels at a different speed. The speed of light in a given material is related to a quantity called the index of refraction, n, which is defined as the ratio of the speed of light in vacuum to the speed of light in the medium:

index of refraction : n = c / v

When light travels from one medium to another, the speed changes, as does the wavelength. The index of refraction can also be stated in terms of wavelength:



Although the speed changes and wavelength changes, the frequency of the light will be constant. The frequency, wavelength, and speed are related by:



The change in speed that occurs when light passes from one medium to another is responsible for the bending of light, or refraction, that takes place at an interface. If light is travelling from medium 1 into medium 2, and angles are measured from the normal to the interface, the angle of transmission of the light into the second medium is related to the angle of incidence by Snell's law :

Lenses

Total internal reflection, and Lenses

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Total internal reflection



When light crosses an interface into a medium with a higher index of refraction, the light bends towards the normal. Conversely, light traveling across an interface from higher n to lower n will bend away from the normal. This has an interesting implication: at some angle, known as the critical angle, light travelling from a medium with higher n to a medium with lower n will be refracted at 90°; in other words, refracted along the interface. If the light hits the interface at any angle larger than this critical angle, it will not pass through to the second medium at all. Instead, all of it will be reflected back into the first medium, a process known as total internal reflection.

The critical angle can be found from Snell's law, putting in an angle of 90° for the angle of the refracted ray. This gives:



For any angle of incidence larger than the critical angle, Snell's law will not be able to be solved for the angle of refraction, because it will show that the refracted angle has a sine larger than 1, which is not possible. In that case all the light is totally reflected off the interface, obeying the law of reflection.

Optical fibers are based entirely on this principle of total internal reflection. An optical fiber is a flexible strand of glass. A fiber optic cable is usually made up of many of these strands, each carrying a signal made up of pulses of laser light. The light travels along the optical fiber, reflecting off the walls of the fiber. With a straight or smoothly bending fiber, the light will hit the wall at an angle higher than the critical angle and will all be reflected back into the fiber. Even though the light undergoes a large number of reflections when traveling along a fiber, no light is lost.

Depth perception

Because light is refracted at interfaces, objects you see across an interface appear to be shifted relative to where they really are. If you look straight down at an object at the bottom of a glass of water, for example, it looks closer to you than it really is. Looking perpendicular to an interface, the apparent depth is related to the actual depth by:



An example

A beam of light travels from water into a piece of diamond in the shape of a triangle, as shown in the diagram. Step-by-step, follow the beam until it emerges from the piece of diamond.



(a) How fast is the light traveling inside the piece of diamond?

The speed can be calculated from the index of refraction:



(b) What is , the angle between the normal and the beam of light inside the diamond at the water-diamond interface?

A diagram helps for this. In fact, let's look at the complete diagram of the whole path, and use this for the rest of the questions.



The angle we need can be found from Snell's law:



(c) The beam travels up to the air-diamond interface. What is , the angle between the normal and the beam of light inside the diamond at the air-diamond interface?

This is found using a bit of geometry. All you need to know is that the sum of the three angles inside a triangle is 180°. If is 24.9°, this means that the third angle in that triangle must be 25.1°. So:



(d) What is the critical angle for the diamond-air interface?



(e) What happens to the light at the diamond-air interface?

Because the angle of incidence (64.9°) is larger than the critical angle, the light is totally reflected internally.

(f) The light is reflected off the interface, obeying the law of reflection. It then strikes the diamond-water interface. What happens to it here?

Again, the place to start is by determining the angle of incidence, . A little geometry shows that:



The critical angle at this interface is :



Because the angle of incidence is less than the critical angle, the beam will escape from the piece of diamond here. The angle of refraction can be found from Snell's law:



Lenses

There are many similarities between lenses and mirrors. The mirror equation, relating focal length and the image and object distances for mirrors, is the same as the lens equation used for lenses.There are also some differences, however; the most important being that with a mirror, light is reflected, while with a lens an image is formed by light that is refracted by, and transmitted through, the lens. Also, lenses have two focal points, one on each side of the lens.

The surfaces of lenses, like spherical mirrors, can be treated as pieces cut from spheres. A lens is double sided, however, and the two sides may or may not have the same curvature. A general rule of thumb is that when the lens is thickest in the center, it is a converging lens, and can form real or virtual images. When the lens is thickest at the outside, it is a diverging lens, and it can only form virtual images.

Consider first a converging lens, the most common type being a double convex lens. As with mirrors, a ray diagram should be drawn to get an idea of where the image is and what the image characteristics are. Drawing a ray diagram for a lens is very similar to drawing one for a mirror. The parallel ray is drawn the same way, parallel to the optic axis, and through (or extended back through) the focal point. The chief difference in the ray diagram is with the chief ray. That is drawn from the tip of the object straight through the center of the lens. Wherever the two rays meet is where the image is. The third ray, which can be used as a check, is drawn from the tip of the object through the focal point that is on the same side of the lens as the object. That ray travels through the lens, and is refracted so it travels parallel to the optic axis on the other side of the lens.



The two sides of the lens are referred to as the object side (the side where the object is) and the image side. For lenses, a positive image distance means that the image is real and is on the image side. A negative image distance means the image is on the same side of the lens as the object; this must be a virtual image.

Using that sign convention gives a lens equation identical to the spherical mirror equation:



The other signs work the same way as for mirrors. The focal length, f, is positive for a converging lens, and negative for a diverging lens.

The magnification factor is also given by the same equation:



Useful devices like microscopes and telescopes rely on at least two lenses (or mirrors). A microscope, for example, is a compound lens system with two converging lenses. One important thing to note is that with two lenses (and you can extend the argument for more than two), the magnification factor m for the two lens system is the product of the two individual magnification factors.

It works this way. The first lens takes an object and creates an image at a particular point, with a certain magnification factor (say, 3 times as large). The second lens uses the image created by the first lens as the object, and creates a final image, introducing a second magnification factor (say, a factor of seven). The magnification factor of the final image compared to the original object is the product of the two magnification factors (3 x 7 = 21, in this case).

Ray diagram for a diverging lens

Consider now the ray diagram for a diverging lens. Diverging lenses come in a few different shapes, but all diverging lens are fatter on the edge than they are in the center. A good example of a diverging lens is a bi-concave lens, as shown in the diagram. The object in this case is beyond the focal point, and, as usual, the place where the refracted rays appear to diverge from is the location of the image. A diverging lens always gives a virtual image, because the refracted rays have to be extended back to meet.



Note that a diverging lens will refract parallel rays so that they diverge from each other, while a converging lens refracts parallel rays toward each other.

An example

We can use the ray diagram above to do an example. If the focal length of the diverging lens is -12.0 cm (f is always negative for a diverging lens), and the object is 22.0 cm from the lens and 5.0 cm tall, where is the image and how tall is it?

Working out the image distance using the lens equation gives:



This can be rearranged to:



The negative sign signifies that the image is virtual, and on the same side of the lens as the object. This is consistent with the ray diagram.

The magnification of the lens for this object distance is:



So the image has a height of 5 x 0.35 = 1.75 cm.

Sign convention

The sign convention for lenses is similar to that for mirrors. Again, take the side of the lens where the object is to be the positive side. Because a lens transmits light rather than reflecting it like a mirror does, the other side of the lens is the positive side for images. In other words, if the image is on the far side of the lens as the object, the image distance is positive and the image is real. If the image and object are on the same side of the lens, the image distance is negative and the image is virtual.

For converging mirrors, the focal length is positive. Similarly, a converging lens always has a positive f, and a diverging lens has a negative f.

The signs associated with magnification also work the same way for lenses and mirrors. A positive magnification corresponds to an upright image, while a negative magnification corresponds to an inverted image. As usual, upright and inverted are taken relative to the orientation of the object.

Note that in certain cases involving more than one lens the object distance can be negative. This occurs when the image from the first lens lies on the far side of the second lens; that image is the object for the second lens, and is called a virtual object.

Interference

Interference

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The wave nature of light

When we discussed the reflection and refraction of light, light was interacting with mirrors and lenses. These objects are much larger than the wavelength of light, so the analysis can be done using geometrical optics, a simple model that uses rays and wave fronts. In this chapter we'll need a more sophisticated model, physical optics, which treats light as a wave. The wave properties of light are important in understanding how light interacts with objects such as narrow openings or thin films that are about the size of the wavelength of light.

Because physical optics deals with light as a wave, it is helpful to have a quick review of waves. The principle of linear superposition is particularly important.

Linear superposition

When two or more waves come together, they will interfere with each other. This interference may be constructive or destructive. If you take two waves and bring them together, they will add wherever a peak from one matches a peak from the other. That's constructive interference. Wherever a peak from one wave matches a trough in another wave, however, they will cancel each other out (or partially cancel, if the amplitudes are different); that's destructive interference.

The most interesting cases of interference usually involve identical waves, with the same amplitude and wavelength, coming together. Consider the case of just two waves, although we can generalize to more than two. If these two waves come from the same source, or from sources that are emitting waves in phase, then the waves will interfere constructively at a certain point if the distance traveled by one wave is the same as, or differs by an integral number of wavelengths from, the path length traveled by the second wave. For the waves to interfere destructively, the path lengths must differ by an integral number of wavelengths plus half a wavelength.



Young's double slit

Light, because of its wave properties, will show constructive and destructive interference. This was first shown in 1801 by Thomas Young, who sent sunlight through two narrow slits and showed that an interference pattern could be seen on a screen placed behind the two slits. The interference pattern was a set of alternating bright and dark lines, corresponding to where the light from one slit was alternately constructively and destructively interfering with the light from the second slit.

You might think it would be easier to simply set up two light sources and look at their interference pattern, but the phase relationship between the waves is critically important, and two sources tend to have a randomly varying phase relationship. With a single source shining on two slits, the relative phase of the light emitted from the two slits is kept constant.

This makes use of Huygen's principle, the idea that each point on a wave can be considered to be a source of secondary waves. Applying this to the two slits, each slit acts as a source of light of the same wavelength, with the light from the two slits interfering constructively or destructively to produce an interference pattern of bright and dark lines.



This pattern of bright and dark lines is known as a fringe pattern, and is easy to see on a screen. The bright fringe in the middle is caused by light from the two slits traveling the same distance to the screen; this is known as the zero-order fringe.The dark fringes on either side of the zero-order fringe are caused by light from one slit traveling half a wavelength further than light from the other slit. These are followed by the first-order fringes (one on each side of the zero-order fringe), caused by light from one slit traveling a wavelength further than light from the other slit, and so on.



The diagram above shows the geometry for the fringe pattern. For two slits separated by a distance d, and emitting light at a particular wavelength, light will constructively interfere at certain angles. These angles are found by applying the condition for constructive interference, which in this case becomes:



The angles at which dark fringes occur can be found be applying the condition for destructive interference:



If the interference pattern was being viewed on a screen a distance L from the slits, the wavelength can be found from the equation:



where y is the distance from the center of the interference pattern to the mth bright line in the pattern. That applies as long as the angle is small (i.e., y must be small compared to L).

Dispersion

Although we talk about an index of refraction for a particular material, that is really an average value. The index of refraction actually depends on the frequency of light (or, equivalently, the wavelength). For visible light, light of different colors means light of different wavelength. Red light has a wavelength of about 700 nm, while violet, at the other end of the visible spectrum, has a wavelength of about 400 nm.

This doesn't mean that all violet light is at 400 nm. There are different shades of violet, so violet light actually covers a range of wavelengths near 400 nm. Likewise, all the different shades of red light cover a range near 700 nm.

Because the refractive index depends on the wavelength, light of different colors (i.e., wavelengths) travels at different speeds in a particular material, so they will be refracted through slightly different angles inside the material. This is called dispersion, because light is dispersed into colors by the material.

When you see a rainbow in the sky, you're seeing something produced by dispersion and internal reflection of light in water droplets in the atmosphere. Light from the sun enters a spherical raindrop, and the different colors are refracted at different angles, reflected off the back of the drop, and then bent again when they emerge from the drop. The different colors, which were all combined in white light, are now dispersed and travel in slightly different directions. You see red light coming from water droplets higher in the sky than violet light. The other colors are found between these, making a rainbow.

Rainbows are usually seen as half circles. If you were in a plane or on a very tall building or mountain, however, you could see a complete circle. In double rainbows the second, dimmer, band, which is higher in the sky than the first, comes from light reflected twice inside a raindrop. This reverses the order of the colors in the second band.

Diffraction

Diffraction; thin-film interference

8-3-99

Diffraction

We discussed diffraction in PY105 when we talked about sound waves; diffraction is the bending of waves that occurs when a wave passes through a single narrow opening. The analysis of the resulting diffraction pattern from a single slit is similar to what we did for the double slit. With the double slit, each slit acted as an emitter of waves, and these waves interfered with each other. For the single slit, each part of the slit can be thought of as an emitter of waves, and all these waves interfere to produce the interference pattern we call the diffraction pattern.

After we do the analysis, we'll find that the equation that gives the angles at which fringes appear for a single slit is very similar to the one for the double slit, one obvious difference being that the slit width (W) is used in place of d, the distance between slits. A big difference between the single and double slits, however, is that the equation that gives the bright fringes for the double slit gives dark fringes for the single slit.

To see why this is, consider the diagram below, showing light going away from the slit in one particular direction.



In the diagram above, let's say that the light leaving the edge of the slit (ray 1) arrives at the screen half a wavelength out of phase with the light leaving the middle of the slit (ray 5). These two rays would interfere destructively, as would rays 2 and 6, 3 and 7, and 4 and 8. In other words, the light from one half of the opening cancels out the light from the other half. The rays are half a wavelength out of phase because of the extra path length traveled by one ray; in this case that extra distance is :



The factors of 2 cancel, leaving:



The argument can be extended to show that :



The bright fringes fall between the dark ones, with the central bright fringe being twice as wide, and considerably brighter, than the rest.

Diffraction effects with a double slit

Note that diffraction can be observed in a double-slit interference pattern. Essentially, this is because each slit emits a diffraction pattern, and the diffraction patterns interfere with each other. The shape of the diffraction pattern is determined by the width (W) of the slits, while the shape of the interference pattern is determined by d, the distance between the slits. If W is much larger than d, the pattern will be dominated by interference effects; if W and d are about the same size the two effects will contribute equally to the fringe pattern. Generally what you see is a fringe pattern that has missing interference fringes; these fall at places where dark fringes occur in the diffraction pattern.

Diffraction gratings

We've talked about what happens when light encounters a single slit (diffraction) and what happens when light hits a double slit (interference); what happens when light encounters an entire array of identical, equally-spaced slits? Such an array is known as a diffraction grating. The name is a bit misleading, because the structure in the pattern observed is dominated by interference effects.

With a double slit, the interference pattern is made up of wide peaks where constructive interference takes place. As more slits are added, the peaks in the pattern become sharper and narrower. With a large number of slits, the peaks are very sharp. The positions of the peaks, which come from the constructive interference between light coming from each slit, are found at the same angles as the peaks for the double slit; only the sharpness is affected.



Why is the pattern much sharper? In the double slit, between each peak of constructive interference is a single location where destructive interference takes place. Between the central peak (m = 0) and the next one (m = 1), there is a place where one wave travels 1/2 a wavelength further than the other, and that's where destructive interference takes place. For three slits, however, there are two places where destructive interference takes place. One is located at the point where the path lengths differ by 1/3 of a wavelength, while the other is at the place where the path lengths differ by 2/3 of a wavelength. For 4 slits, there are three places, for 5 slits there are four places, etc. Completely constructive interference, however, takes place only when the path lengths differ by an integral number of wavelengths. For a diffraction grating, then, with a large number of slits, the pattern is sharp because of all the destructive interference taking place between the bright peaks where constructive interference takes place.

Diffraction gratings, like prisms, disperse white light into individual colors. If the grating spacing (d, the distance between slits) is known and careful measurements are made of the angles at which light of a particular color occurs in the interference pattern, the wavelength of the light can be calculated.

Thin-film interference

Interference between light waves is the reason that thin films, such as soap bubbles, show colorful patterns. This is known as thin-film interference, because it is the interference of light waves reflecting off the top surface of a film with the waves reflecting from the bottom surface. To obtain a nice colored pattern, the thickness of the film has to be similar to the wavelength of light.

An important consideration in determining whether these waves interfere constructively or destructively is the fact that whenever light reflects off a surface of higher index of refraction, the wave is inverted. Peaks become troughs, and troughs become peaks. This is referred to as a 180° phase shift in the wave, but the easiest way to think of it is as an effective shift in the wave by half a wavelength.

Summarizing this, reflected waves experience a 180° phase shift (half a wavelength) when reflecting from a higher-n medium (n2 > n1), and no phase shift when reflecting from a medium of lower index of refraction (n2 < n1).

For completely constructive interference to occur, the two reflected waves must be shifted by an integer multiple of wavelengths relative to one another. This relative shift includes any phase shifts introduced by reflections off a higher-n medium, as well as the extra distance traveled by the wave that goes down and back through the film.

Note that one has to be very careful in dealing with the wavelength, because the wavelength depends on the index of refraction. Generally, in dealing with thin-film interference the key wavelength is the wavelength in the film itself. If the film has an index of refraction n, this wavelength is related to the wavelength in vacuum by:



A step-by step approach

Many people have trouble with thin-film interference problems. As usual, applying a systematic, step-by-step approach is best. The overall goal is to figure out the shift of the wave reflecting from one surface of the film relative to the wave that reflects off the other surface. Depending on the situation, this shift is set equal to the condition for constructive interference, or the condition for destructive interference.



Note that typical thin-film interference problems involve "normally-incident" light. The light rays are not drawn perpendicular to the interfaces on the diagram to make it easy to distinguish between the incident and reflected rays. In the discussion below it is assumed that the incident and reflected rays are perpendicular to the interfaces.

A good method for analyzing a thin-film problem involves these steps:

Step 1. Write down , the shift for the wave reflecting off the top surface of the film.



Step 2. Write down , the shift for the wave reflecting off the film's bottom surface.

One contribution to this shift comes from the extra distance travelled. If the film thickness is t, this wave goes down and back through the film, so its path length is longer by 2t. The other contribution to this shift can be either 0 or , depending on what happens when it reflects (this reflection occurs at point b on the diagram).



Step 3. Calculate the relative shift by subtracting the individual shifts.



Step 4. Set the relative shift equal to the condition for constructive interference, or the condition for destructive interference, depending on the situation. If a certain film looks red in reflected light, for instance, that means we have constructive interference for red light. If the film is dark, the light must be interfering destructively.



Step 5. Rearrange the equation (if necessary) to get all factors of on one side.

Step 6. Remember that the wavelength in your equation is the wavelength in the film itself. Since the film is medium 2 in the diagram above, we can label it . The wavelength in the film is related to the wavelength in vacuum by:



Step 7. Solve. Your equation should give you a relationship between t, the film thickness, and either the wavelength in vacuum or the wavelength in the film.

Example - a film of oil on water

Working through an example is a good way to see how the step-by-step approach is applied. In this case, white light in air shines on an oil film that floats on water. When looking straight down at the film, the reflected light is red, with a wavelength of 636 nm. What is the minimum possible thickness of the film?



Step 1. Because oil has a higher index of refraction than air, the wave reflecting off the top surface of the film is shifted by half a wavelength.



Step 2. Because water has a lower index of refraction than oil, the wave reflecting off the bottom surface of the film does not have a half-wavelength shift, but it does travel the extra distance of 2t.



Step 3. The relative shift is thus:



Step 4. Now, is this constructive interference or destructive interference? Because the film looks red, there is constructive interference taking place for the red light.



Step 5. Moving all factors of the wavelength to the right side of the equation gives:



Note that this looks like an equation for destructive interference! It isn't, because we used the condition for constructive interference in step 4. It looks like a destructive interference equation only because one reflected wave experienced a shift.

Step 6. The wavelength in the equation above is the wavelength in the thin film. Writing the equation so this is obvious can be done in a couple of different ways:



Step 7. The equation can now be solved. In this situation, we are asked to find the minimum thickness of the film. This means choosing the minimum value of m, which in this case is m = 0. The question specified the wavelength of red light in vacuum, so:



This is not the only thickness that gives completely constructive interference for this wavelength. Others can be found by using m = 1, m = 2, etc. in the equation in step 6.

If 106 nm gives constructive interference for red light, what about the other colors? They are not completely cancelled out, because 106 nm is not the right thickness to give completely destructive interference for any wavelength in the visible spectrum. The other colors do not reflect as intensely as red light, so the film looks red.

Why is it the wavelength in the film itself that matters?

The light reflecting off the top surface of the film does not pass through the film at all, so how can it be the wavelength in the film that is important in thin-film interference? A diagram can help clarify this. The diagram looks a little complicated at first glance, but it really is straightforward once you understand what it shows.



Figure A shows a wave incident on a thin film. Each half wavelength has been numbered, so we can keep track of it. Note that the thickness of the film is exactly half the wavelength of the wave when it is in the film.

Figure B shows the situation two periods later, after two complete wavelengths have encountered the film. Part of the wave is reflected off the top surface of the film; note that this reflected wave is flipped by 180°, so peaks are now troughs and troughs are now peaks. This is because the wave is reflecting off a higher-n medium.

Another part of the wave reflects off the bottom surface of the film. This does not flip the wave, because the reflection is from a lower-n medium. When this wave re-emerges into the first medium, it destructively interferes with the wave that reflects off the top surface. This occurs because the film thickness is exactly half the wavelength of the wave in the film. Because a half wavelength fits in the film, the peaks of one reflected wave line up precisely with the troughs of the other (and vice versa), so the waves cancel. Destructive interference would also occur with the film thickness being equal to 1 wavelength of the wave in the film, or 1.5 wavelengths, 2 wavelengths, etc.

If the thickness was 1/4, 3/4, 5/4, etc. the wavelength in the film, constructive interference occurs. This is only true when one of the reflected waves experiences a half wavelength shift (because of the relative sizes of the refractive indices). If neither wave, or both waves, experiences a shift of , there would be constructive interference whenever the film thickness was 0.5, 1, 1.5, 2, etc. wavelengths, and destructive interference if the film was 1/4, 3/4, 5/4, etc. of the wavelength in the film.

One final philosophical note, to really make your head spin if it isn't already. In the diagram above we drew the two reflected waves and saw how they cancelled out. This means none of the wave energy is reflected back into the first medium. Where does it go? It must all be transmitted into the third medium (that's the whole point of a non-reflective coating, to transmit as much light as possible through a lens). So, even though we did the analysis by drawing the waves reflecting back, in some sense they really don't reflect back at all, because all the light ends up in medium 3.

Non-reflective coatings

Destructive interference is exploited in making non-reflective coatings for lenses. The coating material generally has an index of refraction less than that of glass, so both reflected waves have a shift. A film thickness of 1/4 the wavelength in the film results in destructive interference (this is derived below)



For non-reflective coatings in a case like this, where the index of refraction of the coating is between the other two indices of refraction, the minimum film thickness can be found by applying the step-by-step approach:

Polarized Light

Polarization; and The Human Eye

8-4-99

Polarization

To talk about the polarization of an electromagnetic wave, it's easiest to look at polarized light. Just remember that whatever applies to light generally applies to other forms of electromagnetic waves, too. So, what is meant by polarized light? It's light in which there's a preferred direction for the electric and magnetic field vectors in the wave. In unpolarized light, there is no preferred direction: the waves come in with electric and magnetic field vectors in random directions. In linearly polarized light, the electric field vectors are all along one line (and so are the magnetic field vectors, because they're perpendicular to the electric field vectors). Most light sources emit unpolarized light, but there are several ways light can be polarized.

One way to polarize light is by reflection. Light reflecting off a surface will tend to be polarized, with the direction of polarization (the way the electric field vectors point) being parallel to the plane of the interface.

Another way to polarize light is by selectively absorbing light with electric field vectors pointing in a particular direction. Certain materials, known as dichroic materials, do this, absorbing light polarized one way but not absorbing light polarized perpendicular to that direction. If the material is thick enough to absorb all the light polarized in one direction, the light emerging from the material will be linearly polarized. Polarizers (such as the lenses of polarizing sunglasses) are made from this kind of material.

If unpolarized light passes through a polarizer, the intensity of the transmitted light will be 1/2 of what it was coming in. If linearly polarized light passes through a polarizer, the intensity of the light transmitted is given by Malus' law:



A third way to polarize light is by scattering. Light scattering off atoms and molecules in the atmosphere is unpolarized if the light keeps traveling in the same direction, is linearly polarized if at scatters in a direction perpendicular to the way it was traveling, and somewhere between linearly polarized and unpolarized if it scatters of at another angle.

There are plenty of materials that affect the polarization of light. Certain materials (such as calcite) exhibit a property known as birefringence. A crystal of birefringent material affects light polarized in a particular direction differently from light polarized at 90 degrees to that direction; it refracts light polarized one way at a different angle than it refracts light polarized the other way. Looking through a birefringent crystal at something, you'd see a double image.

Liquid crystal displays, such as those in digital watches and calculators, also exploit the properties of polarized light.

Polarization by reflection

One way to polarize light is by reflection. If a beam of light strikes an interface so that there is a 90° angle between the reflected and refracted beams, the reflected beam will be linearly polarized. The direction of polarization (the way the electric field vectors point)is parallel to the plane of the interface.

The special angle of incidence that satisfies this condition, where the reflected and refracted beams are perpendicular to each other, is known as the Brewster angle. The Brewster angle, the angle of incidence required to produce a linearly-polarized reflected beam, is given by:



This expression can be derived using Snell's law, and the law of reflection. The diagram below shows some of the geometry involved.



Using Snell's law:



The scattering of light in the atmosphere

The way light scatters off molecules in the atmosphere explains why the sky is blue and why the sun looks red at sunrise and sunset. In a nutshell, it's because the molecules scatter light at the blue end of the visible spectrum much more than light at the red end of the visible spectrum. This is because the scattering of light (i.e., the probability that light will interact with molecules when it passes through the atmosphere) is inversely proportional to the wavelength to the fourth power.



Violet light, with a wavelength of about 400 nm, is almost 10 times as likely to be scattered than red light, which has a wavelength of about 700 nm. At noon, when the Sun is high in the sky, light from the Sun passes through a relatively thin layer of atmosphere, so only a small fraction of the light will be scattered. The Sun looks yellow-white because all the colors are represented almost equally. At sunrise or sunset, on the other hand, light from the Sun has to pass through much more atmosphere to reach our eyes. Along the way, most of the light towards the blue end of the spectrum is scattered in other directions, but much less of the light towards the red end of the spectrum is scattered, making the Sun appear to be orange or red.

So why is the sky blue? Again, let's look at it when the Sun is high in the sky. Some of the light from the Sun traveling towards other parts of the Earth is scattered towards us by the molecules in the atmosphere. Most of this scattered light is light from the blue end of the spectrum, so the sky appears blue.

Why can't this same argument be applied to clouds? Why do they look white, and not blue? It's because of the size of the water droplets in clouds. The droplets are much larger than the molecules in the atmosphere, and they scatter light of all colors equally. This makes them look white.



Optics of the eye

The human eye is a wonderful instrument, relying on refraction and lenses to form images. There are many similarities between the human eye and a camera, including:

• a diaphragm to control the amount of light that gets through to the lens. This is the shutter in a camera, and the pupil, at the center of the iris, in the human eye.
  
• a lens to focus the light and create an image. The image is real and inverted.
  
• a method of sensing the image. In a camera, film is used to record the image; in the eye, the image is focused on the retina, and a system of rods and cones is the front end of an image-processing system that converts the image to electrical impulses and sends the information along the optic nerve to the brain.

The way the eye focuses light is interesting, because most of the refraction that takes place is not done by the lens itself, but by the aqueous humor, a liquid on top of the lens. Light is refracted when it comes into the eye by this liquid, refracted a little more by the lens, and then a bit more by the vitreous humor, the jelly-like substance that fills the space between the lens and the retina.

The lens is critical in forming a sharp image, however; this is one of the most amazing features of the human eye, that it can adjust so quickly when focusing objects at different distances. This process of adjustment is known as accommodation.

Consider the lens equation:



With a camera, the lens has a fixed focal length. If the object distance is changed, the image distance (the distance between the lens and the film) is adjusted by moving the lens. This can't be done with the human eye: the image distance, the distance between the lens and the retina, is fixed. If the object distance is changed (i.e., the eye is trying to focus objects that are at different distances), then the focal length of the eye is adjusted to create a sharp image. This is done by changing the shape of the lens; a muscle known as the ciliary muscle does this job.

Nearsightedness

A person who is nearsighted can only create sharp images of close objects. Objects that are further away look fuzzy because the eye brings them in to focus at a point in front of the retina. To correct for this, a diverging lens is placed in front of the eye, diverging the light rays just enough so that when the rays are converged by the eye they converge on the retina, creating a focused image.

Farsightedness

A farsighted person can only create clear images of objects that are far away. Close objects are brought to a focus behind the retina, which is why they look fuzzy. To correct for this, a converging lens is placed in front of the eye, allowing images to be brought into sharp focus at the retina.

Diopters

If you go to an optometrist to get glasses or contact lenses, you will get a prescription specified in units of diopters. This is a measure of the refractive power of the lens needed, which means it's a measure of the focal length of the lens. The two are, in fact, inversely related:

refractive power in diopters = 1 / focal length in meters

A diopter has units of 1 / m .

If the lenses you get are specified as 5.0 diopters, it means they have a focal length of 0.2 m, meaning that they are converging lenses that bring parallel rays of light to a focus 0.2 m beyond the lens. Similarly, lenses of -2.0 diopters correspond to a focal length of -0.5 m; these would be diverging lenses with a focal point 0.5 m from the lens.

Lens aberrations

Lenses can distort the image of an object for a number of reasons. One kind of distortion is known as spherical aberration; this occurs because parallel rays of light are not all focused to a single point by a spherical lens. The further a ray is from the principal axis, the more it misses the focal point.

A second kind of distortion is chromatic aberration. This occurs because a lens will have a slightly different index of refraction for light of different wavelengths (i.e., light of different colors). In other words, chromatic aberration is caused by dispersion. Parallel rays of red light, therefore, would be brought to a different focal point than parallel rays of light of another color, leading to a blurry image.

To minimize chromatic aberration, many high-quality lenses are made up of two lenses made from different materials. One lens will be converging, and the other diverging; the compound lens will still be converging overall, generally. The chromatic aberration introduced by one lens is corrected for in the second lens, bringing parallel rays of any color light to about the same focal point.

MicroScopes & Telescopes

Optical instruments

8-5-99

Making things look bigger

When you use an optical instrument, whether it be something very simple like a magnifying glass, or more complicated like a telescope or microscope, you're usually trying to make things look bigger so you can more easily see fine details. One thing to remember about this is that if you want to make things look bigger, you're always going to use converging mirrors or lenses. Diverging mirrors or lenses always give smaller images.

When using a converging lens, it's helpful to remember these rules of thumb. If the object is very far away, the image will be tiny and very close to the focal point. As the object moves towards the lens, the image moves out from the focal point, growing as it does so. The object and image are exactly the same size when the object is at 2F, twice the focal distance from the lens. Moving the object from 2F towards F, the image keeps moving out away from the lens, and growing, until it goes to infinity when the object is at F, the focal point. Moving the object still closer to the lens, the image steadily comes in towards the lens from minus infinity, and gets smaller the closer the object is to the lens.

Note that similar rules of thumb apply for a converging mirror, too.

Multiple lenses

Many useful devices, such as microscopes and telescopes, use more than one lens to form images. To analyze any system with more than one lens, work in steps. Each lens takes an object and creates an image. The original object is the object for the first lens, and that creates an image. That image is the object for the second lens, and so on. We won't use more than two lenses, and we can do a couple of examples to see how you analyze problems like this.

A microscope

A basic microscope is made up of two converging lenses. One reason for using two lenses rather than just one is that it's easier to get higher magnification. If you want an overall magnification of 35, for instance, you can use one lens to magnify by a factor of 5, and the second by a factor of 7. This is generally easier to do than to get magnification by a factor of 35 out of a single lens.

A microscope arrangement is shown below, along with the ray diagram showing how the first lens creates a real image. This image is the object for the second lens, and the image created by the second lens is the one you'd see when you looked through the microscope.



Note that the final image is virtual, and is inverted compared to the original object. This is true for many types of microscopes and telescopes, that the image produced is inverted compared to the object.

An example using the microscope

Let's use the ray diagram for the microscope and work out a numerical example. The parameters we need to specify are:



To work out the image distance for the image formed by the objective lens, use the lens equation, rearranged to:



The magnification of the image in the objective lens is:



So the height of the image is -1.8 x 1.0 = -1.8 mm.

This image is the object for the second lens, and the object distance has to be calculated:



The image, virtual in this case, is located at a distance of:



The magnification for the eyepiece is:



So the height of the final image is -1.8 mm x 3.85 = -6.9 mm.

The overall magnification of the two lens system is:



This is equal to the final height divided by the height of the object, as it should be. Note that, applying the sign conventions, the final image is virtual, and inverted compared to the object. This is consistent with the ray diagram.

Telescopes

A telescope needs at least two lenses. This is because you use a telescope to look at an object very far away, so the first lens creates a small image close to its focal point. The telescope is designed so the real, inverted image created by the first lens is just a little closer to the second lens than its focal length. As with the magnifying glass, this gives a magnified virtual image. This final image is also inverted compared to the original object. With astronomical telescopes, this doesn't really matter, but if you're looking at something on the Earth you generally want an upright image. This can be obtained with a third lens.

Note that the overall effect of the telescope is to magnify, which means the absolute value of the magnification must be larger than 1. The first lens (the objective) has a magnification smaller than one, so the second lens (the eyepiece) must magnify by a larger factor than the first lens reduces by. To a good approximation, the overall magnification is equal to the ratio of the focal lengths. With o standing for objective and e for eyepiece, the magnification is given by:

m = - f_o / f_e, with the minus sign meaning that the image is inverted.

Resolving power

The resolving power of an optical instrument, such as your eye, or a telescope, is its ability to separate far-away objects that are close together into individual images, as opposed to a single merged image. If you look at two stars in the sky, for example, you can tell they are two stars if they're separated by a large enough angle. Some stars, however, are so close together that they look like one star. You can only see that they are two stars by looking at them through a telescope. So, why does the telescope resolve the stars into separate objects while your eye can not? It's all because of diffraction.

If you look at a far-away object, the image of the object will form a diffraction pattern on your retina. For two far-away objects separated by a small angle, the diffraction patterns will overlap. You are able to resolve the two objects as long as the central peaks in the two diffraction patterns don't overlap. The limit is when one central peak falls at the position of the first dark fringe for the second diffraction pattern. This is known as the Rayleigh criterion. Once the two central peaks start to overlap, in other words, the two objects look like one.

The size of the central peak in the diffraction pattern depends on the size of the aperture (the opening you look through). For your eye, this is your pupil. A telescope, or even a camera, has a much larger aperture, and therefore more resolving power. The minimum angular separation is given by:



The factor of 1.22 applies to circular apertures like your pupil, a telescope, or a camera lens.

The closer you are to two objects, the greater the angular separation between them. Up close, then, two objects are easily resolved. As you get further from the objects, however, they will eventually merge to become one.

X-ray diffraction

Things that look a lot like diffraction gratings, orderly arrays of equally-spaced objects, are found in nature; these are crystals. Many solid materials (salt, diamond, graphite, metals, etc.) have a crystal structure, in which the atoms are arranged in a repeating, orderly, 3-dimensional pattern. This is a lot like a diffraction grating, only a three-dimensional grating. Atoms in a typical solid are separated by an angstrom or a few angstroms; . This is much smaller than the wavelength of visible light, but x-rays have wavelengths of about this size. X-rays interact with crystals, then, in a way very similar to the way light interacts with a grating.

X-ray diffraction is a very powerful tool used to study crystal structure. By examining the x-ray diffraction pattern, the type of crystal structure (i.e., the pattern in which the atoms are arranged) can be identified, and the spacing between atoms can be determined.

The two diagrams below can help to understand how x-ray diffraction works. Each represents atoms arranged in a particular crystal structure.



You can think of the diffraction pattern like this. When x-rays come in at a particular angle, they reflect off the different planes of atoms as if they were plane mirrors. However, for a particular set of planes, the reflected waves interfere with each other. A reflected x-ray signal is only observed if the conditions are right for constructive interference. If d is the distance between planes, reflected x-rays are only observed under these conditions:



That's known as Bragg's law. The important thing to notice is that the angles at which you see reflected x-rays are related to the spacing between planes of atoms. By measuring the angles at which you see reflected x-rays, you can deduce the spacing between planes and determine the structure of the crystal.