Electromagnetic Waves Notes: Part 2Level 4: LensesMirrors work by using the law of reflection to warp light. Lenses use Snell’s law and refraction to bend light.

There are different types of lenses. We are going to focus on the two most common: converging lenses and diverging lenses (see the picture). Converging lenses (which are also sometimes called convex lenses) are fatter at the principle axis and thinner at the edges. Diverging lenses are the exact opposite- thinnest at the principle axis.

Converging and diverging lenses don’t need to look exactly like the ones in the picture on the right. They can also look like this:

You should know the names and be able to recognize these guys:

They will still follow the general rule: converging are thicker in the middle. Diverging are thinner in the middle.

Converging LensLet’s take a look at how a converging lens works. You already understand the mechanism behind this- Snell’s Law. Light moves from the air (n=1) into glass or plastic, where the index of the refraction is much higher. This means the light slows down and is bent.

Now that the light is in the glass, it continues to move in a straight line until it hits the opposite side, where it once more encounters the air. Again, it bends. Use one of the rays in the picture on the left to walk yourself through this process. Make sure you stop at each change in material and make sure you understand why the light bends that direction.

Let’s say we send a bunch of parallel rays toward converging lens- what will happen? This may be shocking- but the light will appear to converge, or come together.

Ever used a magnifying glass to burn a leaf? That was a converging lens. You were taking light from a source that was an infinite (or close enough for, you know, math) to enter the lens as parallel and watching it converge on one hot little point. That point was the focal point.

Converging Lenses and Ray DiagramsGuess what? That mantra I taught you before- Flat to focal that can still be used, but we modify it a little. For lenses, we will say : Flat to focal, through the center.

In real life, when a light ray encounters a lens, it bends twice as it passes through the medium (see above if this is confusing). To make things easier, we are actually going to just draw the ray going to the center of the lens and then bending.

Flat to Focal- Start at the object and trace a ray that moves parallel to the principle axis and passes through the lens. As we discussed above, the light will bend to pass through the focal point.

That’s one line done. Next, we do through the center. This one is easy. You start at the top of the object, and go straight through the center of the lens.

Where the two lines meet- that is where the image is created.

Let’s be clear here- if the object was a candle, the light would leave the candle pass through the lens and converge at the image on the right. This would be a real image, because it is formed by actual light rays converging- not back rays, like we saw with convex mirrors (or plane mirrors, for that matter). The upside down image you drew in the ray diagram (on the right) is not what your eye would see if you were to look through the lens from the opposite side of the lens- it is what you would see if you were to put a piece of paper there. Please ask me to demonstrate this for you, if I haven’t already. It will make the difference between a real and virtual image amazingly clear, very quickly.

Lenses and Virtual ImagesWhat happens if you put the object between the focal point and a convex lens?

Try drawing the ray diagram before moving on. Flat to focal- through the middle.

You should have drawn something like this:

The two rays diverge. What does this mean for the image. Well, image you had an observer on the other side of the lens. To that person, it would look as though the rays of light were coming from a large object.

This is a virtual image. This isn’t something projected onto paper- this is what you would actually see if you were on that side of the lens.

Did you catch that difference? A real image is where light rays actually converge. If you put a surface there you will see a projection of the object. A virtual image happens when back rays cause your brain to think the object actually is larger. This isn’t really projected onto anywhere- it is what you perceive.

Figure 1-Real Image

Figure 2-Virtual Image

Whether you get a real image or a virtual image for a convex lens depends on where the object is placed. Look through the following different places and make sure you are familiar with them. Think about how you could do a ray diagram to show you these.

Lenses and the Mirror Lens EquationAs you might have guessed from the name, you can use the Mirror Lens Equation on lenses, too. This is nice, because it lets you back your ray diagrams up with some math, which is prettier and more fun. It is important to note that for converging (convex) lenses, we say the focal point is positive. For diverging lenses (which we will get to in a moment) we say the focal is negative.

Practice Problem: Drawing Ray Diagrams and Calculating StuffsA glass 15 cm high is placed 40 cm in front of a double convex lens of radius 20 cm. Describe the image fully and drawthe diagram. And by fully describe I of course mean position, size, inverted, real….?

Solution

Practice: More Practice with Convex LensesA piece of chocolate candy is placed 10 cm from a converging lens of focal length 15cm. If the chocolate is 4.3 cm high,describe the image fully and draw the diagram.

Solution

Diverging LensesWhen light passes through a diverging lens, it diverges or spreads apart. This means the light rays never converge for a divergent lens. Is it possible to form a real image with a diverting lens?

Nope. In order to form a real image, you need rays of light to converge. That isn’t possible with a diverging lens, so divergent lenses only make virtual images.

If you were on the right side of this lens, think about what your brain would see. To your mind (which doesn’t know light can bend), it seems as through all these rays are coming from some point source on the other side of the lens (see picture below). In other words, it looks as through the light is coming from a

focal point on the opposite side. Because of this, we say that a divergent lens as a negative focal point.

Divergent Lenses and Ray DiagramsFor divergent lenses can still follow are lens mantra: flat to focal, through the middle.

Practice: Ray Diagrams for Divergent Lenses

Solution

It isn’t going to matter if the object is inside the focal point or not. That’s the sweet thing about divergent mirrors. They will always produce an image that is:

virtual Located on the same side as the object (which means you will need to be on the opposite side of the lens to see

it) Upright reduced in size

In this picture, you see two lenses. Which one is the convex lens? Which one is the divergent lens?

If you pulled the lens on the left away from the face, would you still see an enlarged eye? What if you pulled the lens on the right away from the face? Would you still see a small face?

Using the Mirror Lens Equation with Divergent LensesAs we discussed before, you can still use the Mirror Lens Equation on these. You just need to remember that the divergent lenses have a negative focal point. Here is a nifty little cheat sheet. You don’t want to memorize it (we have a lot to cram into that brain of yours in the next month or so), but you can use it to get a sense of how the math should work out/what signs to use.

Practice Problem: Divergent LensesWhere must an object 20 cm high be placed if a diverging/double concave lens of focal length 22 cm is to form a virtual

image 18 cm from the lens on the same side as the object? Find the position of the object and draw the ray diagram.

Solution

Multiple Optical SystemsEvery now and then the AP test gets extra sadistic and combines multiple lenses or has a lens and a mirror. This shouldn’t scare you- just deal with the first lens/mirror first, find out where the image is. Then treat that image as the object for the next mirror/lens.

AP Practice Problem: Multiple Optical Systems1986B6. An object is placed 3 centimeters to the left of a convex (converging) lens of focal length f = 2 cm, asshown below.

a. Sketch a ray diagram on the figure above to construct the image. It may be helpful to use a straightedge.b. Determine the ratio of image size to object size.The converging lens is removed and a concave (diverging) lens of focal length f = –3 centimeters is placed as shownbelow.

c. Sketch a ray diagram on the figure above to construct the image.d. Calculate the distance of this image from the lens.e. State whether the image is real or virtual.The two lenses and the object are then placed as shown below.

f. Construct a complete ray diagram to show the final position of the image produced by the two-lens system.

Solution

Level 5: DiffractionIn quantum, we spent a lot of time talking about the particle nature of light. We kept referencing the fact that scientist strongly believed that light was a wave, but we didn’t talk too much about why. in this section, we are going to look at a couple of the key principles and experiments that led scientists to believe light is wave (which, remember, it sometimes kinda sorta is).

Huygens’s PrincipleVideo Lecture: Huygen’s Principle: Twu Light 1: Huygen’s Principle 1: Light Wave and Huygen’s Principle Very short and to-the-point lecture on the theory that started it all. Our first steps inot exploring the wave nature of light.

Double SlitRefresher: Double Slit. We watched this segment in a video during quantum mechanics. This should help you remember what happens in the double slit experiment.

Video Lecture: Twu Light 2: Young’s Double Slit This is a great video that will introduce you to most of what you are going to need to know about the double slit experiment. Watch it first, then we will go into it in more detail next.

Let’s talk about how to do those calculations again. I know Ms. Twu covered it, but sometimes it’s nice to see it written out. If we pick a position y that is a particular distance above the midpoint between the differences between the two slits. We want to figure out if this is a spot is bright with constructive interference or dark with interference.

First of all, we are curious about whether light, when it hits the screen, is in phase (the peaks add constructively) or out out of phase (the peaks) (see the picture below). Imagine one wave of light leaving from the top slit. We’ll call that Wave A. Imagine another wave of light (Wave B) leaving the lower slit. They will go in all sorts of directions once the leave, but let’s follow the two that will end up at position y.

Let’s say Wave A hits the wall when it is at a peak, and Wave B hits the wall when it is at its peak. These two would constructively interfere and you would get a bright patch. They would be in phase (like the picture on the right). Okay, now imagine that Wave B hit when it was lagging exactly one wavelength behind. Would they still constructively interfere?

y

L

d

r1

r2

d sin

You betcha! The peak would still be at the peak, so there would be constructive interference.

What if they met and they were half a wavelength off? They would be out of phase. This means they would destructively interfere and the screen would be dark.

Let’s put some math to that reasoning. We need to figure out if the waves are in phase or out of phase.

from Cliff Notes

Look at this equation. This equation lets us know what positions will have bright patches. Putting m=0 into the equation gives you a y of zero and that makes sense- it means the first bright patch is straight across from the double slit. Putting +1 in will allow you to calculate the distance y above the double slit that the next bright patch appears. Putting -1 in for m will tell you the distance below the double slit that the bright patch on the opposite side appears.

How would we calculate the dark patches? For that, we can use:

Same idea here. On the previous page, you can see the m values for the different light and dark patches.

from Princeton Review

Practice ProblemRead through this problem very carefully. The apparatus they describe is just a double slit set up. This will walk you through what a typical double slit problem is like.

Solution

Double Slit Diffraction PatternsYou should be able to produce the intensity graph of for a double slit below. Notice the light is the most intense at m=0

A few things you should keep in mind:

The spacing between the bright maximums are fairly even. The wider the distance between the slits becomes, the closer the bright maximums will become.

AP-like Practice Problem

Single Slit ExperimentThe crazy thing about the double slit experiment is that you don’t actually need a double slit to get the interference pattern. If you were to send light through a single slit, you would actually see the wave interfere with itself and create a very similar pattern.

If we wanted to calculate the position of the minimums (places of destructive interference) we would use the equation:

y=mλLa

where λ is the wavelength of the light, m is the number of the minimum (see picture below), L is the distance to the screen and a is the width of the slit.

Keep in mind that this equation can only be used to find the dark patches- not the light patches.

We can also use this equation:

d sinθ=mλ

where d is the width of the slit, θ is the angle to whatever position on the screen we are interested in (from the maximum) and λ is the wavelength in question.

Supplemental: Twu Light 4: Single Slit Diffraction There is actually quite a bit of math involved in this and a lot of great, logical reasoning. I normally would go through it all with you, but we are getting short on time before the test. If you have the time, it is a very good idea to watch this (especially if you are going into the sciences). If you don’t however, you can go with what I give you here. Sorry! I’m totally ripping

off your education for once. Also, the AP Test hasn’t had you use this quantitatively on FR yet.

There are surprisingly few example problems out there for this in AP Physics B land, so watch Ms. Twu walk you through a great sample problem. Before she solves it, pause and make sure you can use the equations above to find the answer (you can use either equation to find part a- one of them requires a little trig, however).

Video Example: Ms. Twu Light 5: Single Slit Diffraction Problem

Comparing Double and Single Slit Experiment Results“While the patterns look similar, the single slit diffraction pattern of a given slit width is more gently varying than the two slit pattern of similar slit width, and the intensity of the maxima outside the central maximum is much less.” (People’s Education AP Prep)

Figure 3: Diffraction Pattern for Single Slit

Diffraction GratingsWhen we look at the double slit experiment, we see a bunch of fairly spread out maximums. If we wanted to get something much more precise, we could actually a ton of slits. A barrier with lots of very finely, evenly spaced slits is called a diffraction grating. This allows us to get a much more precise picture of the interference patterns.

A diffraction grating is used in a particular device called a spectrometer. A spectrometer allows you to accurately measures the wavelength of an unknown light source. You can send a mystery wave of light (such as one from a star, or a particular black body) and you should be able to measure the wavelength of light. What sort of equations does this require? Well, that’s the sweet part. You already know these. A diffraction grating will actually produce maximums at the same points as a double slit- the maximums will just be sharper and more defined.

Level 6: Thin Films and Polarization

Introduction to Thin FilmsThis section is going to be about the type of constructive and destructive interference we often encounter when we look at thin films. Thin films are just a very thin layer of a material covering another material with a completely different index of refraction. That was really confusing- they are easier to recognize in a picture. Here are some pictures:

The thin film is just a layer of material with a different index of refraction. That’s what they look like in physics textbooks. Here’s what they look like in real life:

or

Figure 4: The girl on the right is wearing a thin film on her glasses. The girl on the left didn't take physics.

We are going to explore this strange and completely useful phenomenon in this section.

Light’s Behavior at a BoundaryBefore we can fully understand this idea, we are going to need to review a little about how waves act when they encounter boundaries. These pictures from The Physics Classroom do a great job reminding us of how waves reflect and transmit when they encounter more dense (and less dense) mediums.

Light acts the same way when it encounters a new medium. Imagine a light wave comes along and, though it is not strictly an accurate picture, let’s image the light bobbing up and down up and down. When it encounters a new, more dense medium (one with a higher index of refraction, two things will happen at the same time: first, some of the wave will be transmitted. This transmitted part will continue its normal pattern- up, down, up *new material* down, up… The other part of the wave will reflected- and here is where things get interesting. If the material it’s entering is more dense, the pattern will suddenly shift- the light wave will be inverted –up, down, up *new material* up down. Did you catch that, because it is pretty important…If the light wave hits a material that is less dense, the reflected wave won’t change the pattern: up, down, up *new material* down, up.

Is the material the wave is entering more or less dense? What do you think?

Thin FilmsThis inversion of the wave (or lack thereof) has some pretty interesting consequences when it comes to thin films. For example’s sake, let’s look at how antiglare reflective coating on glasses work. I have a link for a great animation below. Please open the animation and refer to it as I walk you through this. It is very helpful to have something to look at. If you need to, have group member keep it open on their screen while the other reads off this section.

Thin Film Animation: This is the animation. Ta. Da.

Okay, on the left, you see a light wave hitting the material. Notice, as it enters a more dense medium, its wavelength decreases slightly (as we would expect when we enter a new material where the speed of light would decrease. Just to the right, we see the reflected wave. Notice it is now flipped and out of phase with the original wave. The wave on the far left shows what happens once the wave is in the material. Now, it is hitting the next boundary. When it hits this next boundary, it does the same thing all over again- part of it is transmitted (which is not shown in the animation, sadly) and part of it is reflected. When it is reflected, it is reflected flipped (out of phase) Compare it to the phase it was it when it hit the boundary (on the far left)- you can see that its phase is now shifted. This reflected wave will head back to the surface. It won’t change phase when it leave the thin film (because the air is a less dense boundary. If the thin film is the right thickness, this reflected wave (the one on the far right) will actually interfere will the original reflected wave (in the middle), canceling out the reflection all together. Crazy stuff, right?

The Math of Thin FilmsOkay-pay close attention to this section because it can be a bit confusing. If you understand this, however, these can be straight plug n’ chug problems.

Whether the two reflecting waves constructively or destructively interfere has a lot to do with two things:

1. How many phase shifts the light undergoes as enters and leaves the thin film. 2. How thick the thin film is.

Let’s talk about that first one about phase shifting. This is just counting up the boundaries where the light would reverse its phase- in other words, how many times did it bounce off a more dense material? The example we talked about above involved two phase changes- the first one on the surface of the material. The next one from inside the thin film.

Interactive Animation: This applet let’s you play with the different index of refractions and the thickness of the material. It can really help you visualize how all this comes together. Play with it a little.

How about this guy? How many phase shifts here?

You should have said 1- because it will only shift when it enters a more dense medium- which in this case is just when it enters the soap. It won’t shift when it reflect on the bottom because it is entering a less dense medium.

The thickness will also come into play. In order for the two reflections to constructively interfere, the thin film will need to be a certain multiple of the wavelength.

Alright- here’s what you’ve been waiting for- the equations.

Number of Phase Shifts For Constructive Interference For Destructive Interference1 Phase Shift 2nt=(m+ 1

2) λ 2nt=mλ

0 or 2 Phase Shifts 2nt=mλ 2nt=(m+ 12) λ

..Where t is the thickness of the material, λ is the wavelength of the light, and m, though it looks scary, is just a way of saying that multiple thicknesses would create this effect. You will almost always be asked to find the “minimum thickness” in which case, m=0. If they asked you for the next thickest material that would cause interference, you could just enter 1 and so on.

Okay, let’s take this for a spin in an example problem.

What about the whole colorful soap bubble thing? Well, that’ is actually pretty easy. Notice that the whether there is constructive or destructive interference depends in large part on how thick the thin film is.. and it also depend on the wavelength of the material. So, if sunlight (which is made of lots of different wavelengths of light) hits a soap bubble (which seems smooth but has very thicknesses throughout) some of the colors will undergo constructive interference

(which will make the color appear very bright) and others will undergo destructive interference with their own reflection, canceling them out. This will create a rainbow or colors, where the color you see in each region is varied depending on the thickness of the film.

This technique is also used in solar panels, which try to reduce the amount of light reflected off of them, and in antiglare glass (such as you see in some eyeglasses, and in museums….)

AP Practice Problem

1990 Snells and Thin Film, water pool oil

1990B6. A beam of light from a light source on the bottom of a swimming pool 3.0 meters deep strikes the surfaceof the water 2.0 meters to the left of the light source, as shown above. The index of refraction of the water in thepool is 1.33.a. What angle does the reflected ray make with the normal to the surface?b. What angle does the emerging ray make with the normal to the surface?c. What is the minimum depth of water for which the light that strikes the surface of the water 2.0 meters to theleft of the light source will be refracted into the air?

In one section of the pool, there is a thin film of oil on the surface of the water. The thickness of the film is1.0 x l0–7

d. At which of the interfaces (air-oil and oil-water), if either, does the light undergo a 180° phase change uponreflection?meter and the index of refraction of the oil is 1.5. The light source is now held in the air and illuminatesthe film at normal incidence, as shown above.e. For what wavelengths in the visible spectrum will the intensity be a maximum in the reflected beam?

Solution

PolarizationTwu has got polarization covered. Watch the videos listed below. In class, I’m going to pull out some polarizers and show you examples of this. Also, remind me to tell you about one of the dumbest things I’ve ever done (it relates, trust me).

Video Examples and Demos: Twu Light 9, 10, and 11 Relax- they are all pretty short and full of fun visuals.