Lesson Explainer: The Properties of Laser Light Physics

In this explainer, we will learn how to describe the properties of laser light and use technical terms to refer to these properties.

We will see that light that comes from lasers is not like light that comes from other sources such as a light bulb or the Sun. We will name and describe the properties of laser light that make it so unusual.

Before we consider light that comes from a laser, let’s consider the familiar case of light emitted from an incandescent light bulb. If we could see individual photons given out by the light bulb, we would see many photons of many different wavelengths. This is shown in the following diagram.

In this diagram, we have labeled some of the wavelengths given out by the light bulb. We see that a broad range of waves are emitted from the light bulb. For example, the wave labeled “A” has a much longer wavelength than the wave labeled “C”. In fact, some of the beams that come from the light bulb do not even have a discernible wavelength. For example, wave “B” does not have a measurable wavelength.

Furthermore, we see that all of the photons are traveling in different directions. This means that the light from the light bulb varies in wavelength, frequency, and direction.

Let’s now compare this to light given out by a laser. In the following diagram, let’s say that this rectangle is our laser that gives out red light. There are a few things that we can see that tell us why laser light is special.

For example, we see that all of the photons given out by the laser are the same wavelength, and the amplitude, or height, of each photon is the same. Each photon given out by the laser is also “in phase,” which means that all of the peaks and troughs of each photon line up with all other photons in the beam. Light that behaves in this way is called “coherent” light. Note that we also see that all of the photons from the laser are moving in the same direction.

Definition: Coherent Light

Light that is made up of photons that are all the same wavelength and are in phase is called “coherent light.”

We see that the light from the light bulb that we considered previously does not behave in the same way as the light that came from the laser. As a result, we call the light from the light bulb “incoherent” light.

Coherence is a hallmark of laser light.

Example 1: Identifying Waveforms That Make Up Incoherent Light

The diagram represents the resultant waveform of the waves emitted from an incandescent light source. Which of the following diagrams most correctly represents a group of the waves emitted from the incandescent light source?

Answer

In this question, we are given a single waveform and are asked to identify which of the given groups of waves could combine to give this waveform.

We want to work out which of the given sets of waves, I or II, would combine to give the red waveform at the top of this example.

We see that the resultant waveform that we are trying to form does not have a clearly discernable wavelength. We also see that the amplitude, or height, of the wave varies from peak to peak. It looks like a very irregular wave.

Let’s begin by thinking about what would happen if we were to combine the group of waves given in I. These waves all have the same amplitude and wavelength and are all in phase. Therefore, if we combined them, we would expect the resultant waveform to also have the same wavelength and a consistent amplitude throughout. The following diagram shows how two in-phase waves of equal amplitude combine to give a resultant wave with a larger amplitude.

Note that the resultant wave would not have the same amplitude as each of the individual waves here. For example, in this diagram, we see the example of two waves which are in phase and both have amplitude 𝐴. The bottom wave here is the resultant wave that is obtained from combining the two waves above.

We see that the resultant wave is in phase with the two component waves and has the same wavelength. However, the amplitude of the resultant wave is the sum of the amplitudes of the component waves. This is because all of the peaks and troughs will combine to produce a wave with higher peaks and lower troughs. Here, since each component wave has amplitude 𝐴, the resultant wave has amplitude 𝐴+𝐴=2𝐴. If, for example, we had added three in-phase waves with amplitude 𝐴, then the resultant wave would have amplitude 𝐴+𝐴+𝐴=3𝐴.

The waves in I would produce a beam of light that is close to coherent, like laser light.

However, the waves in II each have a different wavelength and a different amplitude. The resultant waveform from combining these waves will therefore be irregular.

We can see this because each wave has a different wavelength, and so the different amplitudes add up in different ways at each point. This means that the waveform that results from group II will have an irregular amplitude.

Note that each wave in II has a consistent amplitude and wavelength. However, since each one is different to every other one, these will not combine in a consistent way.

This means that group II will combine to give a waveform that looks like the waveform we are trying to reproduce in this question, so our answer here is II.

A good way to compare coherent light with incoherent light is to make a graph of the intensity of light, 𝐼, against the wavelength of the light, 𝜆. A graph of this is shown here, with two different light sources represented by the different colored curves.

Let’s first look at the orange curve that represents the light given off by an incandescent light bulb. We see that this curve is quite broad, which tells us that the light bulb gives out lots of different wavelengths of light. This is typical because the light bulb produces incoherent light.

The red curve on this graph represents the light given out by the laser. We see that this curve is much narrower, and the peak intensity is much higher. This tells us that the range of wavelengths that the laser produces is much smaller than the incoherent light from the light bulb. This is because the laser produces coherent light, so all of the photons emitted by the laser are the same wavelength.

If the laser was producing perfectly coherent light, then its wavelength-versus-intensity curve would just be a vertical line. This is because the laser would be producing only light of exactly one wavelength. However, in reality, lasers do not produce perfectly coherent light and so there is a small amount of spreading of the wavelength-versus-intensity curve.

If a perfect laser could emit light of one wavelength, we would say that the light produced is “monochromatic.” However, while a real laser produces a very small range of wavelengths, it is not perfectly monochromatic.

Definition: Monochromatic Light

Light that is made up of only one wavelength, or a very small range of wavelengths, of photons is called “monochromatic.”

Despite the fact that real lasers do not produce exactly one wavelength of light, we still call laser light monochromatic. We do this while acknowledging that, in this case, we are using monochromatic to mean “light with a very small range of wavelengths.”

This means that laser light is both coherent and monochromatic. This is different from light that comes from other sources, such as light bulbs or the Sun. These sources produce light that is incoherent and has a very wide range of wavelengths.

Example 2: Identifying Intensity against Wavelength Curves of Coherent and Incoherent Light

The graph shows how the output intensity of two light sources varies with the wavelength of the light that they emit. Both light sources emit most strongly at one peak wavelength, with the output decreasing as the wavelength varies from the peak wavelength.

  1. Which color curve represents the light emitted by an incoherent light source?
  2. Which color curve represents a more monochromatic light source?

Answer

Part 1

In this question, we are given a graph that plots the intensity of the light (in watts per square metre) versus the wavelength of the light (in nanometres). This graph essentially tells us how much light is emitted at each wavelength by each source.

We see that both of these curves peak at the same wavelength. However, the blue curve represents a light source that emits a much wider range of wavelengths than the red curve. The spread of wavelengths for the blue curve is larger than the spread of wavelengths for the red curve.

There are two ways in which the two curves are different. Firstly, the base of the blue curve is much wider than the red curve. That is, the range between the maximum wavelength and the minimum wavelength emitted is larger for the blue spectrum than the red spectrum. Additionally, a crucial difference is that the blue curve has a near-constant gradient except for very close to the peak wavelength, while the red curve has a gradient which changes rapidly both at its base and at its peak. This tells us that the light source that corresponds to the blue spectrum emits a much larger range of wavelengths than the source corresponding to the red curve. These differences are shown in the following diagram.

For the first part of this question, we want to identify which of these curves corresponds to light emitted by an incoherent light source.

We have seen that in order to have a coherent light source, the light that it emits must be made up of photons that all have the same wavelength. Incoherent light, therefore, is light which is made up of photons that have a large range of wavelengths.

In terms of the graph we have in this question, an incoherent light source will have a wider curve with a gradient that changes slowly (away from the peak wavelength). This is because it will give out a large range of wavelengths. This means that the blue curve represents the light emitted by an incoherent light source, because it is much wider than the red curve. This curve represents light that exists over a broad range of wavelengths, which is incoherent light.

Part 2

The second part of this question asks us to identify which of the curves on the graph represents light that is emitted by a monochromatic light source.

We can recall that monochromatic light is light that is made up of only one wavelength or is made up of a very small range of wavelengths. This means that, for the graph in this question, a monochromatic light source will be very narrow, since the light exists only over a small number of wavelengths.

From the graph, we can therefore see that the monochromatic light source is represented by the red curve.

Example 3: Comparing Component Waves for Different Intensity against Wavelength Curves

The graph shows how the output power of three laser light sources varies with the wavelength of the light that they emit. The laser light sources emit most strongly at one peak wavelength, with the output decreasing as the wavelength varies from the peak wavelength. Diagrams I and II represent groups of waves emitted from two of the light sources. Diagram II represents the waves emitted by the light source that produced the blue-colored spectral distribution curve on the graph. What is the color of the curve that would correspond to the waves represented by diagram I?

Answer

In this example, we have three curves that represent the spectra (range of wavelengths) of light emitted by three different laser light sources. We see that the width at the base of each of these curves is the same, which tells us that the range of wavelengths emitted by each laser is the same. We can also see that the wavelength of the peak intensity is the same for each laser.

However, the gradient of each curve changes differently depending on the height of the curve. For example, the red curve has a peak intensity that is much larger than the peak intensity of the green curve. Since the width of each curve is the same but the height of the red curve is much larger, the gradient of the red curve must change much quicker than the gradient of the green curve.

This means that light corresponding to the red spectrum is much brighter for colors corresponding to wavelengths close to the peak wavelength than for colors corresponding to the maximum and minimum wavelengths. However, for light corresponding to the green spectrum, the difference in brightness of colors corresponding to wavelengths close to the peak wavelength and those colors corresponding to the maximum and minimum wavelengths is much less than for the red spectrum. The red spectrum is closer to representing monochromatic light than the green spectrum.

We are then given the component waves that make up light from each of the three light sources. We are told that the waves in diagram II, the second set of waves, are emitted from the light source represented by the blue curve. We can see that the waves in diagram II vary very slightly in wavelength. For example, the wavelength of the black wave at the bottom is slightly longer than that of the red wave at the top. The initial displacements of the waves are also different from each other, meaning that the peaks do not ever all line up perfectly with one another. We can see this in the diagram below, where we have boxed one peak of each wave and we can see that they do not line up perfectly.

This means that the intensity of the resultant wave from these waves will not be as large as it would be if all of the waves had the same wavelength and were in phase. We know that when these waves combine, they have an intensity given by the blue spectra.

Let’s now consider diagram I. These waves have wavelengths that are very different from one another, and the initial displacements of the waves are very different too. This is shown in the following diagram.

This means that when we combine these waves, the peak intensity will be smaller than the peak intensity of the waves from diagram I. This is because the wavelengths in diagram I are so different, so none of these wavelengths can dominate the resultant wave.

This tells us that the curve that represents the light given in diagram I must have a lower intensity than the blue curve described in diagram II. This means that it is the green curve that represents the light waves in diagram I.

Let’s note that in order to produce the light represented by the red curve in this graph, we would need to have a group of waves that are all the same wavelength (or very close to the same wavelength). These waves would also need to be in phase, meaning that the amplitude of the resultant wave will be much larger than the amplitude of each individual component wave.

We can see another property of laser light if we consider what happens to the emitted light if it continues to travel through space. Let’s again consider both our light bulb and a laser source, so that we can compare the differences we see.

Firstly, let’s look at a section of the light coming from the light bulb. If we look at the light that leaves the light bulb toward the right, we can see that even though all of the photons are traveling to the right, they are all moving in slightly different directions. This means that the light will spread out a lot as it travels away from the light bulb.

However, we see a difference if we look at the light coming from the laser. The photons here are all traveling in the exact same direction, which is parallel to the laser. This means that the beam of light will not spread out. Instead, it remains confined to a very thin beam of light.

When light behaves in this way, we say that the light is “collimated.” The width of a beam of perfectly collimated light does not change as the beam travels in space.

Definition: Collimated Light

A beam of light is “collimated” if the width of the beam remains constant as the beam travels through space.

Light from our light bulb is not collimated because each photon is moving in its own individual direction, and so light spreads out. Laser light is collimated because all of the photons it emits are moving in the same direction from the very beginning. Laser photons are effectively emitted parallel to one another and so do not cause the beam to widen.

A property of light that is related to collimation is the “attenuation” of light. Attenuation of light is the loss of intensity of the light as the light beam travels away from its source.

Definition: Attenuated Light

A beam of light “attenuates” if the intensity of light decreases as the light travels. We say that such a beam of light is attenuated.

Let’s first consider a laser light source. Let’s say that the intensity of the light when it leaves the laser is 𝐼, and the intensity of the light is 𝐼 at a distance of 1 m. We can understand the intensity of the light as the number of photons that are in a given area. The higher the density of photons, the higher the intensity of the light. We would say that the light attenuates if the density of photons, and hence the intensity of the beam, decreases as the beam travels.

In the following diagram, we can see how each photon in the laser beam is emitted parallel to the beam. We have seen how this means that the laser does not spread out as it travels; it remains a collimated beam.

Since the laser beam is tightly collimated, there are the same number of photons at each point along the beam. In particular, this means that at the start of the beam and 1 m along the beam, the intensity of the light must be the same. This means that, for the laser, we can say that 𝐼=𝐼. We can say that the laser beam does not experience any attenuation.

Let’s now consider an incandescent light bulb. Let’s again label the intensity of the light immediately after it leaves the bulb as 𝐼 and label the intensity of the light at a distance of 1 m as 𝐼.

In the following diagram, we show a light bulb and a section of the light it emits. We can see that the photons emitted from the light bulb are all travelling in different directions. This means that the photons spread out as they travel away from the light bulb, resulting in a beam that is not collimated. The intensities at 0 m and 1 m are labeled on this diagram as 𝐼 and 𝐼 respectively.

Closer to the light bulb, the photons are closer together, and as they spread out further away from the bulb, the same number of photons take up a larger area in space. This means that as the beam travels away from the bulb, the intensity of the beam decreases. In other words, 𝐼<𝐼.

Example 4: Exploring Collimation and Attenuation of Different Light Sources

The diagram represents the beams emitted by a red laser light source and an incandescent white light source.

  1. Which light source produces a more collimated beam?
  2. For the red laser light source, how does the light intensity 𝐼 at the emitting surface of the beam compare to the light intensity 𝐼 of the beam at a distance 𝑑 from the emitting surface?
  3. For the incandescent white light source, how does the light intensity 𝐼 at the emitting surface of the beam compare to the light intensity 𝐼 of the beam at a distance 𝑑 from the emitting surface?
  4. For the red laser light source, how does the light intensity 𝐼 of the beam at a distance 𝑑 from the emitting surface compare to the light intensity 𝐼 of the beam at a distance 𝑑 from the emitting surface?
  5. For the incandescent white light source, how does the light intensity 𝐼 of the beam at a distance 𝑑 from the emitting surface compare to the light intensity 𝐼 of the beam at a distance 𝑑 from the emitting surface?

Answer

Part 1

The first part of this question asks us to identify which light source pictured above produces the most collimated beam. Let’s recall that a collimated beam of light is one that does not spread out as it travels, so it remains in a beam of constant width.

From the diagram, we can see that the incandescent white light spreads out as it travels. This is because the photons produced by the light source are all traveling in different directions as they leave the light source, and hence they do not stay confined to a narrow beam.

However, we can see that the red laser produces light that does stay in a narrow beam as it travels. This is because all of the photons produced by the laser are traveling in the same direction, which is parallel to the laser, as they leave the laser source. This means that the red laser light source produces the most collimated beam here.

Part 2

For this part of the question, we are asked to compare the intensity of the light at the emitting surface of the red laser to the intensity of the light after it has traveled a distance 𝑑, as shown in the diagram.

The important thing to remember here is that the laser produces a collimated beam, so the photons in the beam do not spread out and remain in a narrow beam. This means that the intensity of the beam, which we can think of as the density of photons in an area of space, does not change as the beam travels.

Since the photons remain in a narrow and constant-width beam, the density of the photons is the same at the emitting surface and after the beam has traveled the distance 𝑑. This means that, using the notation in the question, we can say that 𝐼=𝐼.

Part 3

This part of the example also asks us to compare the intensity of the light at the emitting surface to the intensity of the light after it has traveled a distance 𝑑, but for the incandescent white light instead of the red laser source.

While the laser produces a collimated beam, the incandescent light produces a beam that spreads out as it travels away from the source. Since the photons are all traveling in different directions when they leave the incandescent light, the width of the beam increases as it travels away from the emitting surface. This means that the density of photons decreases along the beam.

As a result, the light intensity of the beam decreases as the beam travels and spreads out. This means that, for the incandescent white light, 𝐼>𝐼.

Part 4

This part of the question again asks us to consider the intensity of the laser beam, but this time at distances 𝑑 and 𝑑. At these distances, the intensities of the beam are 𝐼 and 𝐼 respectively.

Once again, we can see that because the laser produces a collimated beam, the density of photons does not change along the laser beam. In particular, at a distance 𝑑, the width of the beam is the same as it was at a distance 𝑑.

This means that the light intensity of the laser beam does not change, and hence 𝐼=𝐼.

Part 5

For the incandescent white light, the beam of light continues to widen as the beam travels. The beam is therefore wider at a distance of 𝑑 than it is at a distance of 𝑑, but the number of photons in the beam does not change as it travels.

This tells us that the density of photons decreases as the beam travels from 𝑑 to 𝑑. As a result, the light intensity of the beam continues to decrease as the beam travels, and we know that 𝐼<𝐼. We can equivalently say that 𝐼>𝐼.

We will now discuss how we are able to see laser light at all.

In order for our eyes to see any light, we need photons to leave the light source and hit our eyes. For an incoherent light source, this happens regularly. When photons leave, for example, a light bulb, they are traveling in many different directions, and so by chance many of these photons end up entering our eyes. This lets us see much of the light that is emitted by the bulb.

However, due to the properties of laser light, the process is not so simple for lasers. All of the photons that make up laser light leave the laser traveling in the same direction that is parallel to the laser. This means that the photons will only enter our eyes if we are looking at the laser head-on. Please note that you should never try this; looking at a laser head-on can permanently damage your eyes because the light has a very high intensity.

If, however, we are looking at the laser from a side-on angle, no photons from the beam will enter our eyes. This means that from a side-on view, the laser will be completely invisible to us.

The only reason we can sometimes see a laser beam is because of a process called “scattering.”

To understand this process, let’s consider two different lasers. The first laser is the same red laser we have been thinking about before. The second laser gives out green light instead of red light. These lasers are shown in the following diagram. We are viewing these two lasers side-on. This means that the photons in the laser beams are moving perpendicular to the direction that we are looking from.

In the case shown here, the red laser is seen to be a steady beam of red light. However, for the green laser, we only see a few points of green light.

The difference in the appearance of the two laser beams comes from the amount of scattering that each beam experiences.

Let’s consider just the red laser beam for a moment. We have seen that a laser produces collimated light, so all of the light is moving from left to right in the picture here. However, if this were completely true, then we wouldn’t be able to see the laser beam at all because no photons from the beam would reach our eyes.

The reason we can see the laser beam is because some photons get scattered toward our eyes. That is, some photons bounce off of an object in the way of the beam, such as an air molecule or water vapor. An example of this is sketched below.

Here, we see a sketch of a photon with a red wavelength traveling through the air from left to right. The photon interacts with a molecule of water vapor. This causes the photon to change direction and, in this case, redirects the photon into our eye. This allows us to see the laser beam.

For the case of the red laser, we can see the whole beam of light. This means that a lot of scattering is happening and lots of photons are redirected toward our eyes. This is shown in the diagram below, where we see that many red-wavelength photons are scattered toward our eye.

This could be caused by shining the laser through a very humid place. This would mean that there is a lot of water vapor in the air, so there are lots of molecules to scatter the laser beam.

If we now look at the green laser, we cannot see the whole beam. Instead, we only see a few dots of green light. This tells us that, in comparison to the red beam, there are relatively few scattering events happening.

In the following diagram, we show the green laser and see that only a few green-wavelength photons are scattered toward our eye. This means that we cannot see much of the beam, but only a few point-like spots of light.

Only at a few spots along the path of the green beam are photons scattered toward our eye, allowing us to see the green light. At other points along the path of the green laser, the light either is not scattered at all or is scattered in a direction that does not reach our eye. So, overall, very little scattering is happening along the path of the green laser.

Example 5: Explaining the Appearance of a Laser Beam

The diagram represents the beam emitted by a green laser light source. The beam is viewed perpendicularly to its length. Which of the following most correctly explains why the green beam appears to consist of many point-like particles?

  1. The point-like particles seen are individual electrons ionized from the air by the laser beam.
  2. The point-like particles seen are individual photons in the beam, and the beam has such low intensity that there is significant space between individual photons.
  3. Most of the light emitted travels parallel to the beam and only a few rays are scattered enough to travel perpendicularly to the direction of the beam.
  4. The beam produces an interference pattern of light and dark regions.
  5. The beam produces a longitudinal wave.

Answer

In this question, we have a diagram of a laser beam and we are asked to identify which one of the given statements explains the appearance of the laser. We should notice that we do not see a solid beam of light here. Instead, we see a series of green dots along the path of the laser.

Let’s look at each of the given statements in turn and see if they provide a good explanation for the appearance of the laser.

Statement A suggests that the point-like particles that we see are individual electrons which have been ionized from the air by the laser beam. This would mean that we are seeing individual electrons with our eyes. Since electrons are subatomic particles, they are much too small to be able to be seen by our eyes. This means that this cannot be the explanation for the appearance of the laser beam here.

Statement B says the the points of light we see are the individual photons of the beam, and these photons are very spread out because the beam has a low intensity. We would expect photons to travel parallel to the laser, and hence they would not reach our eyes on that path. This means that this cannot explain why we see the beam as we do.

Statement C suggests that most of the photons travel parallel to the beam, but some photons are scattered enough that they reach our eyes. This is a good explanation for why we see the beam as we do. Most photons travel parallel to the beam and hence we do not see them. However a few photons scatter off something in the path of the beam, such as an air molecule, and redivert into our eyes so that we can see them. This explains why we only see the beam as a few point-like particles of light. This is the most likely explanation for the appearance of the beam we have seen so far, but let’s check the remaining options too.

Statement D says that the beam produces an interference pattern of light and dark regions. Interference patterns cannot be seen as a beam of light travels; they are only visible once light hits a target. This means that this explanation cannot be the reason why we see the light as we do.

Statement E suggests that the appearance of the beam is because the laser beam produces a longitudinal wave. A wave being “longitudinal” tells us about the direction it is traveling in, but not about how that wave will look. This statement therefore cannot explain why we see the laser as we do.

This means that, having considered all of the statements given, we have decided that statement C is the correct explanation for the appearance of the beam.

Note that it is possible for a laser to be completely invisible to us but still experience a lot of scattering. This can happen if the laser produces a wavelength of light that is outside of the visible spectrum, such as mircowaves or X-rays.

If this is the case, we cannot judge the amount of scattering that the beam experiences by how much of the light we can see. The beam may experience a lot of scattering toward our eye, but we cannot see any of the light because the wavelength of the light is too long or too short to be visible to our eye.

Key Points

  • Lasers emit coherent light, so all photons given out by a laser have the same wavelength and the same waveform.
  • Laser beams are collimated, so they do not spread out as they travel through space and instead stay in a narrow beam.
  • Since laser beams are collimated, they experience less attenuation (loss of intensity) over distance compared with incoherent light sources. An incoherent light source spreads out as it travels, and hence the light attenuates with distance.
  • Lasers scatter by different amounts as they travel, which affects how much of a laser beam we can see. Generally, the more visible a laser beam is, the more scattering it has undergone.

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