Chapter 27 Lecture Notes

For all of these lecture notes files, I’m going to be using the associated chapter resource files on the 10164 website as a general outline. We will start our coverage of Chapter 27 with a simple definition of interference.

Interference is a phenomenon that is easiest to study when we have two sources of waves with the same wavelength. There are two kinds of interference we will study: constructive and destructive. Constructive interference (CI) occurs when an observer at point P in the diagram below sees waves coming in from two different sources. And those waves arrive at P in such a way that their maxima are coincident, so the waves add together, resulting in the same wave in terms of wavelength but twice the amplitude.

Notice that source 1 is at a distance of 2.25 wavelengths (or waves) from point P, and source 2 is at a distance of 3.25 waves from point P. The difference between these two distances is called the “path difference”. If two waves have a path difference of 0, 1, 2, 3, … or any integer number of waves, then the observer at P will see constructive interference.

In the case of destructive interference (DI), waves arrive from two different sources at point P in such a way that the maximum from one wave is coincident with the minimum of the other wave. Remember that these two waves (if we are talking about light) are simply electric and magnetic fields. So if we have two electric fields at point P simultaneously, and one points up with a certain magnitude and the other points down with the same magnitude, then the fields will perfectly cancel out.

Notice here that source 1 is at a new distance of 2.75 waves from point P while source 2 is at the same original distance of 3.25 waves from point P. The path difference between the waves this time is 0.5 waves. If two waves have a path difference of 0.5, 1.5, 2.5, … or any (integer + 0.5) number of waves, then the observer at P will see destructive interference.

In these two examples, the location of source 1 has moved by a distance of 0.5 waves. We sometimes call this a “phase shift”. Two waves are “in phase” if they can both originate from the same place (or experience constructive interference if they originate from two different places). The phase angle of a wave is a numerical way of stating this. If wave 1 is in phase with wave 2, then the phase angle between them is 0°. If wave 1 is exactly out of phase (DI occurs), then the phase angle is 180°.

When we talk about examples of intererence, we assume that the wave sources are “coherent,” meaning that they remain in a constant phase in relation to one another. Incandescent light sources are not coherent. The light they emit changes phase randomly, but laser light is coherent, so it is the most useful for demonstrating interference.

At this point, it would be a good idea for you to read, study and practice sections 27.1 and 27.2 in your textbook so you can read through the book’s definitions and see our first example, which is double-slit interference. We start with this because it is in some ways the easiest kind of interference to understand.

Once you have finished with those two sections in your textbook, return to these notes, and we will do a worksheet.

—In double-slit interference, we illuminate two slits with a single source of coherent light. As the light passes through each slit, it spreads out. This is a wave phenomenon known as “diffraction” and is covered in your book (for sound waves) back in Chapter 17.3. It might be useful for you to read that short section of the book, but it is not required (no homework or exam questions will deal explicitly with this phenomenon).

But it is important for you to know that diffraction is something that WAVES do, not particles. If I fire a series of small bullets through a narrow opening (without hitting either edge of the opening), they will pass straight through the opening. But if send a wave through a narrow opening, the wave will spread out (diffract). We’ll come back to this shortly.

—

Now, let’s cover the specific case of two-slit interference. The two slits each act as sources of light. The light passing through the slits is projected onto a screen at some distance L from the slits, which are separated by a distance d. Some of the light goes straight through the slit and doesn’t get diffracted at all, and it hits the screen opposite the slits. Some light diffracts with some angle θ from its original path.

This figure from example 1 in chapter 27.2 shows you how the variables d, L and θ are defined.

There is an additional variable here, y, which is the distance from the center of the pattern (the point on the screen directly opposite the center point between the slits) and where the light ray we are studying hits the screen.

Notice the right triangle formed by the dashed line, the screen and the top blue line. For that triangle, we can relate y, L and θ as follows: tan θ = y/L.

Usually, L is much larger than either y or d, in which case the triangle gets very skinny, so we can also say sin θ ≈ y/L in this case.

Next, we will look at path differences between the two sources.

From figure 27.6 (just above example 1 in Chapter 27.2), we can see the two slits acting as sources, and we following light rays from these two sources traveling in almost perfectly parallel directions to converge on a single point on the distant screen.

Part (b) of this diagram is crucial. Here, we can see that there is a path difference between our two sources of light. Your book calls it ∆ℓ, but I usually call it δ. Notice from the diagram that the red line in part (b) makes a right angle with the bottom blue line. So there is a right triangle with a hypotenuse of d.

If this is a very skinny triangle (which we will always assume), then we can say:

tan θ = sin θ = ∆ℓ/d or δ/d.

And if sin θ = δ/d, then we can say δ = d sin θ.

The angle θ in our little right triangle is the SAME angle θ as in our larger triangle from the previous page (you can see θ used to label both angles in part (b) of the diagram above.

Therefore, we can say the path difference can be written two different ways: δ = d (sin θ) = d (y/L) or just dy/L.

A final point on this diagram: See the numbers on the right edge? Those numbers by each bright spot are called “order numbers” (m). They give us a way to refer to which part of the pattern we are looking at.

So our equation for two-slit interference can be written asδ = d sin θ or δ = dy/L

Our five variables are:

d = distance between two slitsy = distance from center of pattern to where ray hits screenL = distance between slits and wallθ = angle between axis of symmetry and path of light rayδ = path difference between the two light rays

What’s happening on the screen?

At the center of the pattern, right along the axis of symmetry, we should have a bright spot (or a bright “fringe”) caused by constructive interference. That’s marked by the order number m=0 on the right-hand edge of figure 27.6, which I reproduced on the previous page. The path difference between our two rays here is exactly zero, so we see constructive interference.

If we move a little bit above or below the center of the pattern, we will arrive at a dark spot in the pattern in between the order numbers m=0 and m=1 on the right-hand edge of figure 27.6. That’s the first dark fringe, and it is a point where the path difference between our two light ray sources is exactly 0.5 wavelengths, so we see destructive interference.

Want to know the exact y-coordinate of this first dark fringe? Easy enough, just set our path difference, δ, equal to one-half of a wavelength. So δ = 0.5L.

0.5λ = dy/L

If we now solve for y, we get y = 0.5λL/d.

Moving further from the center of the pattern, we arrive at a bright spot (marked by a m=1 on the right-hand edge of figure 27.6). That’s the 1st maximum (we often call the center of the pattern the zeroth maximum for consistency). This is the place where the path difference between the two sources is 1 wave.

And so on… Your first worksheet (27.1) will get you some practice using this simple equation. Once you have finished it or at least seriously attempted it and gotten stuck, please proceed to the next page to see my detailed solution.

If you have any questions about the way I solved worksheet 27.1, please post to the “Chapter 27 worksheets” discussion forum on the Physics 10164 course shell. I will be checking this forum often to answer questions.

There is similarly a forum for questions about the “Chapter 27 Homework” for the same purpose, and I plan to have two forums for each of the six remaining homework assignments this semester, shared by all three classes, on the same course shell.

You may have noticed the wording with “order number.” That’s a shorthand. We say maxima occur when the path difference is equal to mλ where m = 0, 1, 2, … The value of m is the order number for bright fringes. For dark fringes, you can say the path difference is (m + 0.5)λ and m = 0 is the first fringe.

—Now for a SLIGHT (optional) detour before we move on because there is something deeply disturbing actually happening here that I think you will find interesting.

We have been assuming all along here that the light passing through the two-slit apparatus is an electromagnetic wave. The whole concept of interference depends on the fact that we are dealing with a wave phenomenon. But it turns out light is not quite so simple. It has properties of both waves and particles. We know this from some “Modern Physics” experiments (like the photoelectric effect) we will study later that light can also be thought of as tiny little energy-carrying particles called “photons.”

These particles also diffract and can experience interference effects, so we think of photons as wave packets, like below. That is, they have a wavelength associated with them but they are localized. Each one carries an energy that corresponds with its wavelength or frequency.

If you shine laser light on the wall for, say, 5 seconds, then the wall will absorb a certain amount of energy. If you tell me the wavelength of the light, I can tell you how many photons hit the wall because I know how much energy is carried by each individual photon.

Usually, the number is incredibly high, but in theory, we can lower the intensity of the light so low that essentially only one photon per second is passing through our apparatus. But if only one photon is passing through the apparatus, then there should be no opportunity for interference. You can only get interference if there are TWO sources of light.

And yet, as the photons pass through the apparatus, one at a time, if we put some kind of detector on the wall, the image that will build up over time is the same interference pattern we would see if the laser were at full intensity! How is this possible?! Does this mean the same photon is passing through both slits somehow?

We could test this by putting some kind of gate in front of each that would beep when a photon passes through it. That way, we would know which path each photon takes. If you do that, the pattern on the wall collapses into two dots directly opposite the two slits. In other words, the act of detecting the photon causes a change in its behavior.

You may think this is a product of the interaction between the gate and the photon, some kind of physical effect that disrupts the wave nature of the light. But what if we only put a gate over the top slit S1 and leave slit S2 alone? The exact same effect happens!

In other words, the light traveling through slit S2 CHANGES ITS BEHAVIOR even though we didn’t physically interact with it in any way. The only thing that happened is that we acquired knowledge (from the gate on slit S1, whether it beeps or not) about which slit the photon was choosing to pass through.

Take the gate over S1 away, and the diffraction pattern resumes.

The wave-particle duality thing isn’t just limited to light, you know. All matter has wave/particle duality, but it is easiest to detect in very small things. For example, electrons also act as particles and wave packets. If you fire a beam of electrons through a narrow opening, it will diffract like a wave. If you fire it at a two-slit apparatus, you will get an interference pattern just like for light.

And if you block slit S1, the interference pattern will vanish, just like for light. So this behavior isn’t unique to light.

What is happening here? How is it possible for a photon or an electron to change its nature just because we acquire knowledge about it in a way that doesn’t interact with the particle at all?

That’s a sticky question at the heart of quantum mechanics that has been debated for decades. If you are interested in proceeding further down this rabbit-hole to learn about the Copenhagen Interpretation or the Many Worlds hypothesis, both of which try to explain this bizarre behavior, I encourage you to follow a link at the end of the 10164 page, called “Quantum Physics and Reality” which leads to a book excerpt that summarizes the whole issue better than I am able to.

http://personal.tcu.edu/dingram/edu/pine3.html

Here ends the optional part of the lecture notes. Now I will return to your regularly scheduled discussion of interference as though there is absolutely nothing weird going on at all and everything fits very nicely with very simple equations.

—For more two slit interference problems, here are some links to videos from old exams:

https://www.youtube.com/watch?v=mBKqvoqc-6whttps://www.youtube.com/watch?v=GYXByNFjLVIhttps://www.youtube.com/watch?v=tNFCGJge5nQhttps://www.youtube.com/watch?v=hoB-yioqQQc

—We are going to veer away from our discussion of light passing through slits to discuss a different kind of interference called “thin film interference” before returning to slits. Don’t blame me, I’m just following the order laid out in your textbook!

This would be a good time for you to Read, Study and Practice chapter 27.3. Note that in this chapter, we will not study the last 1/3 of the section where it starts talking about air wedges. Read the section, then, and work through exam 3. Try some of the suggested problems from this section in the Chapter 27 Resources file (the assignment is found on your WileyPLUS page with full solutions).

Once you have done this, proceed to the next page of the notes.

You have probably noticed thin film interference if you’ve ever seen a colorful reflection on concrete or asphalt after it rains. Often, the oil or gasoline that accumulates in small amounts on the roadway gets mixed in with the water. Since oil is less dense than water, it rests on the top surface of the water as a thin film, often no more than tens of molecules thick. Light reflecting off the film at different angles travels different distances within the film, so we see different colors.

The physics of thin films involves interference again. So we model it simply as two sources. In the diagram below, a thin film of gasoline rests on a surface of water, and light is incident on the thin film at “nearly normal”.

That means the light is coming in perpendicular to the film, but we show it at an angle so we can see what’s going on more easily. Upon reaching the surface of the film, the incident light splits into two sources.

Source 1 reflects off the boundary between the thin film and the air.

Source 2 transmits through the boundary, reflects off the boundary between the film and the water, then transmits through the air-film boundary again. All of this while traveling an extra distance of 2t (where t is the thickness of the film, the diagram uses ℓ, but we will use t) within the film.

Some of the light transmits through the film/water boundary, but it doesn’t participate in the interference we need to study, so we won’t bother with it.

Before we go further with thin films specifically, we need to have a more general discussion about interference. There are three things that can cause a phase difference between two sources. The first one we have already covered, and that is a difference in distance between the sources and the point of interest where the waves meet. We have already covered that in two-slit interference and seen what can cause these “path differences” (measured in meters or waves) or “phase differences” (measured in waves or sometimes angles).

The second thing that can cause a phase difference is reflection. Much of the theoretical and mechanical basis for the rules of reflection are covered back in chapter 17 if you are interested. All that’s important for you to know is the following rules about reflection.

Assume light is traveling through a medium of index nA and reflects off a boundary of material of index nB.

1) If light transmits through the boundary, there is no phase shift under any circumstances.

2) If light reflects off the boundary and nB is larger, then the light experiences a phase shift of half a wave (0.5 λ).

3) If light reflects off the boundary and nB is smaller, then the light experiences no phase shift.

In the diagram I showed you on the previous page, source 1, which is traveling through air and reflects off the gasoline film, experiences a phase shift of half a wave (n for the gasoline is higher than n for air, so we say source 1 is “reflecting off a higher n”). Source 2 is traveling through the gasoline and reflects off the water. Since the water has a lower n than the gasoline, source 2 experiences no phase shift upon reflection. We say source 2 is “reflecting off a lower n”.

—

The final thing that can cause a phase difference between two sources occurs when one of the sources travels through a substance with a different n.

Here is a sketch of two waves. The wave on top (wave A) travels through empty space with no change in phase or wavelength. The wave on bottom (wave B) travels through a rectangular region filled with a substance of higher n.

When light travels through a higher index, the speed decreases by a factor n (we saw this in Chapter 26), but the frequency remains the same. The wavelength ALSO decreases by the same factor n. It has to in order to keep the wave equation valid.

So wave A and wave B were initially in synch at first with no phase difference, but because wave B went through that rectangle, when it emerged on the other side, waves A and B were no longer in synch.

If we measure from one side of the rectangle to the other, wave A has about 3.8 waves. But wave B has about 6.0 waves in the rectangle, so when wave B emerges, there is now a phase difference between A and B of 2.2 waves (which is the same as 0.2 waves).

The number of waves in the box for B is equal to the length of the box divided by the wavelength inside the box. We will use this concept for thin films. Now let’s return to the example.

First, let’s consider reflection. As I said previously, the phase shift for wave 1 due to reflection is 0.5 waves. The phase shift due to reflection for wave 2 is zero.

Next, we have to consider that wave 2 travels an extra distance compared to wave 1, and it travels that extra distance inside the film of the gasoline. So what is the resulting phase shift?

We can find this by just counting the number of waves in the film. The distance in the film is 2t (back and forth through a distance t), and the wavelength in the film is the base wavelength in air (λ0) divided by the index (n) of the film.

In other words, the path difference for wave 2 in the film is:

2t/λn where λn is the wavelength in the film, which is λ0/n. Of course, λ0 is the wavelength of light in air.

So we can rewrite that path difference [2t/(λ0/n)] as 2nt/λ0, where we are using n for the gasoline as opposed to the other n’s in the problem.

Let’s put it all together.

The phase shift for wave 1 is 0.5 waves (for reflection).

The phase shift for wave 2 is 0 (for reflection) + 2nt/λ0 (for the extra distance in the film).

Interference depends on the DIFFERENCE between those two phase shifts, so δ2 - δ1 = 2nt/λ0 - 0.5.

I’ll admit, finding the phase difference between our two sources is a LOT more complicated than the 2-slit case, when it was just d sin θ. But now that we have found it, the rest is not so bad.

Let’s say this problem asks “What colors of light are brightly reflected by this thin film?” Then that clues us in to use the rules for constructive interference. So we would take our phase difference and set it equal to the condition of constructive interference like this:

δ2 - δ1 = 2nt/λ0 - 0.5 = 0, 1, 2, 3, … waves

So if we know n and t for the film, there are actually several possible values for λ0 that result in constructive interference.

What if instead the problem says “What colors of light do not reflect brightly?” That’s asking for destructive interference, so we would write:

δ2 - δ1 = 2nt/λ0 - 0.5 = 0.5, 1.5, 2.5, … waves

Again, several possible solutions. How do we pick which solution to use? Maybe the problem asks for the smallest possible wavelength. Or it asks for answers that are in the visible light range (from 400 - 700 nm). You’ll see examples like this.

A good warm-up problem to look at before tackling a couple of worksheets would be problem #68 from your WileyPLUS chapter assignment.

If you are ready, you’ve read through 27.3, looked at the example, done some sample problems from WileyPLUS that are recommended in the resources file, then maybe you are ready to try our first thin film worksheet, 27.2. Once you have finished it or at least seriously attempted it and gotten stuck, please proceed to the next page to see my detailed solution.

If you have any questions about worksheet 27.2, please post to the appropriate “Chapter 27 worksheets” forum on d2l.tcu.edu, and I will respond there (or send me an email).

—Do you understand why, in the worksheet, I assigned a reflection phase shift of 0.5 waves to BOTH sources?

Do you understand why I set the phase difference equal to 0.5 and then again 1.5?

Do you see why I chose 102 nm as my final answer instead of 307 nm?

Do you see why I stopped solving problems after I did the 0.5 and 1.5 cases?

If you can’t answer these questions confidently, then you should ask me about them and/or go back and do a little bit more studying of the chapter.

Next, I’d like you to try another thin film problem, which is worksheet 27.3. Once you have finished it or at least seriously attempted it and gotten stuck, please proceed to the next page to see my detailed solution.

If you have any questions about worksheet 27.3, please post to the appropriate “Chapter 27 worksheets” forum on d2l.tcu.edu, and I will respond there (or send me an email).

—Do you understand in this case why I chose the reflection phase shift of 0.5 for wave A and 0 for wave B?

Do you understand why I set the phase difference equal to 0, 1, 2, 3, … ?

Do you understand why I stopped after solving my phase difference equation for the 4th time?

If you feel you need more practice with thin film problems, there are some videos online from old exams to help you:

https://www.youtube.com/watch?v=PbWroM53rzohttps://www.youtube.com/watch?v=8l8pkEsBfDQhttps://www.youtube.com/watch?v=LYSkzeieKV8https://www.youtube.com/watch?v=qjPbY1YGWSk

In addition, you can find these problems on some more recent exams as well. Looks also for examples of thin-film transmission! Same principle but a little tricky at first.

—Next, we will move on to single-slit interference.

At this point in the sequence, it would be a good idea for you to read, study and practice section 27.5. The section starts off just talking about diffraction (continuing the discussion that started back in 17.3 I referred to earlier) but quickly moves into the issue of single-slit interference.

Study figures 27.18 and 27.19 closely as well as the surrounding discussion. Concept Simulation 27.1 allows you to adjust the variables (such as wavelength, slit width, distance to screen) to see how the diffraction pattern changes, which is very helpful. I also recommend you work through example 6 and some of the practice problems for this section indicated in your Chapter 27 Resources.

When you are done with this, please proceed to the brief discussion of the topic on the next page and the associated worksheet.

We will start with figure 27.19. From the book’s discussion of diffraction, I hope you have by now learned the details of Huygens’ Principle, which says that a plane wave (like light) incident on an opening behaves at the opening as though there are multiple sources all along the slit. Light spreads out from these sources rather than just continuing in its original direction only.

Figure 27.19 imagines 5 different light sources. Pay close attention to what happens with sources 1 and 3. We can treat these two sources just like we dealt with two slit interference.In two slit interference, we found that when d sin θ = 0.5 waves or (λ/2) meters, we found a minimum in the pattern due to destructive interference.

In single slit interference, the book uses W for the width of the slit, but I traditionally use a. Since a is in use on the formula sheet, that is the letter I will use to refer to the slit width here.

Instead of d, our two sources (1 and 3) are separated by a/2. So the first minimum of the pattern will occur when we satisfy the equation (a/2) sin θ = 0.5 waves = (λ/2) meters. If there is a minimum there for sources 1 and 3, then there will also be a minimum for the pair of sources just below (call them 1.1 and 3.1), and so on. Every pair of sources gives the same result.

So for the location of the first minimum, we have:(a/2) sin θ = 0.5 waves = (λ/2) meters

Or, to put it another way, when a sin θ = λ.

Remember also that y and L are also present in this type of problem, as shown in the diagram above and to the left. And it is still the case that when L is very large compared to y (which it usually is), then θ is fairly small and we can say for the first minimum in the pattern:

a sin θ = a tan θ = ay/L = λ

We are generally ONLY going to deal with the location of the first minimum. That’s because the main application of single slit interference is understanding how angular resolution works with our eyesight and instruments, a topic for the next section. The resulting single-slit pattern on the wall looks like the diagram on the right.

We are interested in the width of that central bright fringe, which is just equal to 2y, where y marks the displacement between the center of the pattern and the first dark fringe.

Time to practice using the equations above in worksheet 27.4.

Once you have finished it or at least seriously attempted it and gotten stuck, please proceed to the next page to see my detailed solution.

If you have any questions about worksheet 27.4, please post to the appropriate “Chapter 27 worksheets” forum on d2l.tcu.edu, and I will respond there (or send me an email).

—Now we will move along to the next (closely related) section of the book, 27.6, which deals with the issue of resolving power. In this section, we are going to be talking about circular apertures (like the pupil, or a telescope) rather than slit apertures like we did for single slit and double slit interference. The difference is shown in these two diagrams from the book:

On the left is the diffraction/interference pattern of light for a vertical slit. On the right is the pattern for a circular aperture. There is still a central maximum, but the location of the first minimum is a bit different:

For a single slit of width a, we have a sin θ = λ.

For a circular aperture of diameter a, we have a sin θ = 1.22 λ.

Don’t worry about where the 1.22 comes from. It’s complicated.

Ok, for a circular aperture, we have a similar equation for the location of the first minimum as for single slit interference. So how does this apply to resolving power?

Imagine the light from two objects passing through the same circular aperture. Each of them will create an interference pattern as shown above. Our ability to resolve the two objects as separate objects is just fine as long as those two central maxima do not overlap.

Right at the limit where the two central maxima would overlap, that’s the minimum angular separation (θmin above) between two objects (like car headlights) that we can resolve.

For objects very far away separated by some distance, we can make a skinny triangle like this:

This is similar to the skinny triangle we make for the location of the first minimum in the diffraction pattern. And, like that skinny triangle, we can say:

sin θmin ≈ tan θmin = (separation of objects)/(distance to objects)

Recall our equation for the first minimum location is:

a sin θmin = 1.22 λ, which can be written sin θmin = 1.22 λ/a

Looking at the symmetry in the diagram on the previous page, notice in part a, the same angle θmin is in both sides of the circular aperture, which suggests we can make the equation:

1.22 λ/a = (separation of objects)/(distance to objects).

This is the same equation that appears on your formula sheet. I understand the derivation here is a little complex conceptually, but we are just going to focus on applying this simple formula. There are four unknowns, you’ll get three, and you’ll solve for the fourth.

Our goal in this chapter, as you have seen, is not to derive any of the formulas but instead to learn how to apply them in different situations. With that said, I hope you will now read, study, and practice chapter 27.6 and work through example 7 (the book uses D for the aperture diameter, whereas I am sticking with a).

When you are ready, try worksheet 27.5. Once you have finished it or at least seriously attempted it and gotten stuck, please proceed to the next page to see my detailed solution.

If you have any questions about worksheet 27.5, please post to the appropriate “Chapter 27 worksheets” forum on d2l.tcu.edu, and I will respond there (or send me an email).

—You can also practice with problems 34, 35, 39 and 63 from the Chapter 26 assignment for this topic.

I have one video problem solution from an old exam on this topic, found at:

https://www.youtube.com/watch?v=GObY8vvbHKs

You can also find some other problems to practice with on recent exams, though they aren’t as common as other optics problems (because I find them to be trivial), so I’ll point out a couple to save you from searching:

Spring 2019, Exam 4A, #2Spring 2016, Exam 4B, #4

—Now it is time to move on to the final topic for this chapter: diffraction gratings.

I recommend you read, study and practice with the material in chapter 27.7 and use the Chapter 27 resources to guide you to the appropriate problems before moving on in these notes. My notes are always a briefer summary of what you can find in the book, so it is better to use the book first to get a basic understanding, then read these notes to see how I hit the main points a second time.

When you have done this, please move on to the next page where I will begin covering this concept and introduce the last worksheet of this chapter.

A diffraction grating is an example of multi-slit interference. The details of how the maxima are formed are discussed in the chapter, although they gloss over some of the complex details. It boils down to a pretty simple equation that defines the location of bright fringes (unlike the single slit case where we were looking for the first dark fringe):

Again, m is the order number. So this equation will tell you where you can expect a maximum (a bright fringe) to occur. With diffraction gratings, they can either be transmission (light shines through lots of tiny parallel slits) or reflection (light bounces off lots of narrowly spaced grooves, like the surface of a CD or DVD, where you will see a rainbow-like reflection due to this effect).

The reason for the rainbow is the wavelength dependence in the equation above. Notice how the angular position of a maximum depends on wavelength, so the longer the wavelength, the bigger the angle.

The angle O here is measured with respect to the normal to the grating as you see in this figure from the textbook showing the angular position of the first and second maxima.

The result of this wavelength dependence is the rainbow you see in reflected light. The first order spectrum is closest to the normal, with the short wavelength (blue) end showing the smallest angular position. For second and third order maxima, you can see from the diagram that they may overlap.

You may also realize that there is a limit as to how many orders you can see. At some point, the order number gets to be so high that it is impossible to solve for the angular position.

I encourage you to do example 9 in your textbook section 27.7 if you haven’t already, then work on some of the relevant problems from the WileyPLUS assignment, like 43, 44, 45, 46, 48, 49 and 50. You will learn some of the language associated with diffraction gratings.

For example, instead of giving you the separation between two slits (d), the problem will often give you the slit density (or line density), like 2500 lines per cm. And you have to invert that to get the separation between the slits. After some practice, you will see that these are fairly basic problems, where you are given (sometimes in a roundabout way) three of the four unknowns, and you have to solve for the 4th.

When you are ready, try worksheet 27.6. Once you have finished it or at least seriously attempted it and gotten stuck, please proceed to the next page to see my detailed solution.

If you have any questions about worksheet 27.5, please post to the appropriate “Chapter 27 worksheets” forum on d2l.tcu.edu, and I will respond there (or send me an email).

—I have three video solutions regarding diffraction gratings from old exams on my web page, if you need further practice:

https://www.youtube.com/watch?v=3Zl5b2QtJKohttps://www.youtube.com/watch?v=aVkYBqDbUO4https://www.youtube.com/watch?v=l8NZznPNa_c

You can also find these problems on some recent old exams without much difficulty, I think.

—This concludes my lecture notes for Chapter 27. It is now time for you to get started on the Chapter 27 homework.

As always, I hope you will solve or seriously attempt each problem before asking for help. I will be checking the “Chapter 27 homework” forum on our course shell occasionally and contributing helpful comments in response to any questions you may have about the homework.

Your Chapter 27 homework is due on Tue Apr 21 at 11:59pm. I know this is a bit of a short turnaround after your exam, but I promise the next two chapters will have a bit less material that I think is easier to work through, so the next two weeks leading up to exam 4 should be comparatively relaxed!