Chapter 26 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 26 with a basic discussion of refraction of light and Snell’s Law.

The index of refraction (n) is a property of a substance, like density or color or composition. “n” is typically between 1-2 for most substances. It can never be smaller than one.

“n” appears in a few different equations in this chapter. First, it tells you by what factor light slows down when it enters a substance:

Second, it tells you something about the geometry of how a ray of light will enter or exit a material, crossing a boundary between one substance and another.

As you can see, when a ray crosses a boundary between substances with different “n”, it will turn a bit from its original direction.

If it crosses into a higher “n”, it will bend towards the normal (top diagram).

If it crosses into a lower “n”, it will bend away from the normal (bottom).

Much of the time, when light hits a boundary like this, some of it will be reflected and some transmitted. We are usually only interested in the transmitted light, unless no light is transmitted at all.

That’s called total internal reflection.

Section 26.3 talks about how the total internal reflection case differs from our usual Snell’s Law case. Take a look at figure 26.8 at the beginning of that section, and watch animated figure 26.8, linked just below the figure. This will really help you remember what’s going on when we talk about the critical angle and total internal reflection.

When light passes from one substance (say, water) into another substance with a lower “n” (air), the light ray will bend away form the normal, as we saw. And you can solve for the angles by using Snell’s Law:

Take a look at the solved examples 1 and 2 in section 26.2 so you can get a feel for this equation and how to solve for the unknowns. Sometimes, instead of giving you an angle, the problem will give you two sides of a triangle, from which you have to deduce the angle (like in example 2). One you have looked at those examples, try solving worksheet 26.1.

For the rest of section 26.2, please ignore the part about displacement of light by a transparent slab and DEFINITELY ignore the derivation of Snell’s Law. Hard pass.

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 26.1, please post to the “Chapter 26 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 26 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.

So, did you have any trouble with that worksheet? One little problem a lot of students tend to have with this material is this convention we have of measuring angles with respect to the normal. Notice that we always measure angles in optics equations (like Snell’s Law) with respect to the normal (with only one exception I’ll get to later). That’s because we are often dealing with curved surfaces where measuring the angle between a ray and the surface can be problematic, but there is no mistaking the normal.

So now you know how to do basic Snell’s Law problems. They can get a little hairy when you have a light ray bouncing around inside a prism, and we’ll get to that.

For now, let’s go back to internal reflection. I assume you’ve watched animated figure 26.8 so you can see where the behavior of light transitions at what we call the Critical Angle. This is a phenomenon that can only occur when light is passing from a higher “n” substance into a lower “n” substance.

- If light strikes such a boundary at an angle less than the Critical Angle, it will transmit through and bend away from the normal according to Snell’s Law.

- If light strikes such a boundary at the Critical Angle or some higher angle of incidence, then the light will experience total internal reflection (and we use the law of reflection from chapter 25, which is very easy).

Note the discussion right below figure 26.8 where the formula for Critical Angle is derived.

I need to make a quick point here, because I don’t use the same convention as the book does. If you follow the book’s examples and discussion, the book always says that “n1” is the index of the substance where the ray originates. And “n2” is the index of the substance that the ray passes into after crossing the boundary.

I don’t like that. It’s not intuitive, and in my experience, it leads to a lot of confusion because the paths of light rays are reversible (Snell’s Law works no matter which way the ray is going).

Therefore, this is my rule:

n1 = lower index of refractionn2 = higher index of refraction

That way, the formula for Critical Angle is always exactly the way it is described on your formula sheet:

This formula always works, and it doesn’t matter which way the light is going, because n1 is always smaller than n2 if we are using my conventions.

If you have a situation where a light ray is RIGHT AT EXACTLY THE CRITICAL ANGLE, then you can use the equation above. For example, if a prism is surrounded by air (n1 = 1.0000000) and you know the Critical Angle, then you can easily find n2.

Take advantage of this fact and use some of your geometry from the last chapter (with triangle angles adding up to 180, right angles adding up to 90, etc) to solve worksheet 26.2. 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.

θcrit=sin−1 n1

n2

⎛⎝⎜

⎞⎠⎟

If you have any questions about worksheet 26.2, please post to the appropriate “Chapter 26 worksheets” forum on d2l.tcu.edu, and I will respond there (or send me an email).

—Now we’re going to get into the more complicated problems that involve a bit of judgement and a bit more geometry. The “judgement” is that when a light ray hits a boundary, you are going to have to decide whether the light ray passes through the boundary or totally internally reflects.

There are two ways to make this decision easy for you. The first way is to just compare the angle of incidence with the Critical Angle. Remember the rule, I’ll quote it again:

- If light strikes such a boundary at an angle less than the Critical Angle, it will transmit through and bend away from the normal according to Snell’s Law.

- If light strikes such a boundary at the Critical Angle or some higher angle of incidence, then the light will experience total internal reflection (and we use the law of reflection from chapter 25, which is very easy).

—But even if you mess this up, there is a safety net! Let’s say, for example, that a light ray is leaving water and passing into the air, and the angle of incidence is 71°. n2 for water is 1.33, and of course n1 for air is 1.00.

Try to use Snell’s Law to figure out the angle of refraction for the light that emerges from the water and into the air. I’ll wait.

Still waiting…

You can’t do it, because when you rearrange the equation to solve for θ1, you get θ1 = sin-1(1.2575), which gives an error! Because the sin function cannot return a result larger than 1. So when your calculator won’t let you solve a problem, that should give you pause and maybe make you rethink your approach!

I recommend the Physics Demonstration Video “Blackboard Optics - Total Internal Reflection”, which is a brief video showing an example of light crossing the boundary between transmission and total internal reflection as the geometry changes. This is at the very end of section 26.3.

It may help also for you to play around with Concept Simulation 26.1, just below the discussion in which Critical Angle is introduced in section 26.3.

Worksheet 26.3 is going to take things up a notch in complexity, so please take it slowly.

Do you see how I labelled the angles in worksheet 26.2? Every angle that might come up in the problem, I assigned it a simple capital letter so that it would be easy to refer to and easy to reconstruct my work later. I recommend you do something similar for later study.

You will definitely want to do this for worksheet 26.3, because (spoiler alert) the light ray is going to enter the prism, bounce off the left side and exit the top right side of the prism. There are a lot of angles along that path that you will need to consider to get the right answer.

So please try worksheet 26.3 now. 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 26.3, please post to the appropriate “Chapter 26 worksheets” forum on d2l.tcu.edu, and I will respond there (or send me an email).

—I hope you were able to follow my logic. Once I found that angle D1 was 49.22°, I knew the light had to reflect off the left surface because the Critical Angle is 41.47°.

The rule is that if the angle of incidence is greater than the Critical Angle, we get total internal reflection. And as I said before, if you think the ray emerges from the left side and you try to calculate the angle of refraction, you will see that it is impossible anyway.

Up on the top right, when the light ray hits that face, the angle of incidence there (F) is 40.77°, which is JUST a bit less than the critical angle, so the light gets out, and we are done. Note that is important to keep a few extra sig figs, like in most problems. Too many rounding errors here, and you might have gotten a value for F that is too large and mistakenly thought there is another internal reflection!

At this point, it may be a good idea for you to try some more problems like this before you tackle the homework. There are some good ones in WileyPlus, of course, and I’ve recommended them in the resources list. There are also some old exam videos you may find useful as additional practice problems:

https://www.youtube.com/watch?v=xzWzcZx2xUAhttps://www.youtube.com/watch?v=LQvgBlF0bIo (dispersion)https://www.youtube.com/watch?v=zh70mbtUkKwhttps://www.youtube.com/watch?v=wvc_axQgfBkhttps://www.youtube.com/watch?v=w5dANX4bnvc

That last one is the most like worksheet 26.3 if you want another crack at a problem like this while it is still fresh.

Note the 2nd video is marked (dispersion). That’s an example where light shines through a prism which (like most substances) has a wavelength-dependent “n”. So “n” is a little higher for blue light, making it turn more upon entering and exiting. This results in “white” light getting dispersed into the colorful spectrum. In practice, it just means you do the same problem twice, once for red light and once for blue light, so you can see the difference.

Your book’s section 26.5 has a little more detail on dispersion, including this helpful diagram and a discussion of how dispersion creates rainbows.

A slightly trickier Critical Angle problem is in this video linked below, and you might want to watch this video before you try worksheet 26.4.

https://www.youtube.com/watch?v=R-lpiKYyoIo

Worksheet 26.4 is a problem related to the physics of fiber optic cables. Your book gets into this about halfway through section 26.3. Light enters one end of a plastic/glass cable and then can’t get out because of the Critical Angle issue. It just internally reflects over and over.

The cable doesn’t have to be surrounded by cladding for this effect to work. It just has to have a pretty high “n”.

Now that you’ve taken a look at the youtube problem video I recommend and read a bit about fiber optic cables in your book (don’t worry about example 7 there … way too long and needlessly complex), please try worksheet 26.4.

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 26.4, please post to the appropriate “Chapter 26 worksheets” forum on d2l.tcu.edu, and I will respond there (or send me an email).

—I know the wording in this is a little tricky. It would have been easier if I had just asked: “Assume light enters the cable and hits the top surface at the Critical Angle … what must be the angle of incidence A?”

I asked the same question but in a broader (for applications, a more useful and practical) way, and part of your practicing with these kinds of problems is that I hope to make you see that the two statements are logically the same thing.

—We are skipping section 26.4 on Brewster’s Angle and Polarization by Reflection and we are skipping the associated worksheet 26.5. You will not be held responsible for this content on the exam, nor is it covered in the Chapter 26 homework.

Brewster’s Angle is the one exception where we measure the angle with respect to the surface (not the normal) in an optics equation. That’s a good reason right there to skip it.

I will just say a quick qualitative word about the subject, though, because it is a pretty cool effect. When light reflects off a surface at JUST the right angle, it turns out that will polarize the light parallel to that surface. For example, when light reflects off of water at an angle of 37°, some of the light transmits into the water and some reflects.

Well, the light that reflects is horizontally polarized. So if you want to reduce the amount of “glare” from reflected light off of water or wet surfaces (like roads), then you wear sunglasses with a vertically polarized filter. That zeroes out any light trying to get through that is horizontally polarized. Very useful for selectively filtering out annoying light while letting other light in.

Anyway, on to the 2nd half of the chapter, which deals with lenses and images.

We start this half of the chapter with section 26.6. It would be a good idea before proceeding if you would read, study and practice 26.6-26.9 at least.

We are going to deal with two different kinds of lenses, converging and diverging. I think it would be very helpful if you would have your formula sheet page 7 handy for this discussion so that you can follow along with the vocabulary and sign conventions for lenses.

A converging lens has convex faces (we will assume for now that all lenses are symmetric). When parallel light rays are incident on this kind of lens, they will pass through the lens and then turn toward each other and come to a focus behind the lens.

Like mirrors, we read lens diagrams from left to right. Anything to the left of the lens is “in front of” the lens. Anything to the right of the lens is “behind” the lens.

A diverging lens (symmetric concave faces) takes parallel rays and makes them spread apart. If you follow the rays backwards (the dashed lines), they come to a focus in front of the lens. We would say for this lens that there is a virtual image at the focal point F.

But is there actually an image at the location of F? Not at all. But saying that for a lens, an image distance is -13 cm, that just is describing how light rays leave the lens. They are diverging at a rate such that if you follow them backwards, they would converge 13 cm in front of the lens.

In photography, some lenses are referred to as “fast” and “slow”. That’s a measure of curvature. “Fast” lenses cause parallel rays to curve a lot, coming to a focus in a very short distance (or “quickly”) behind (or in front of) the lens.

For a converging lens, there isn’t always an image at the focal point F either. What if, for example, I put a mirror in between the lens and F? The rays can no longer converge at F, right? But we still say the lens forms an image at, say, 35 cm behind the lens. That’s just a description of how light rays are leaving the lens. They are converging at a rate so that they will come together 35 cm behind the lens if nothing intervenes.

Like with mirrors, we aren’t going to do any really methodical ray tracing on homeworks or exams, but ray tracing can be helpful to understand what is happening with the light rays passing through lenses. Take a look at these two figures from section 26.7 so you can understand how magnification works for converging and diverging lenses, and read the associated section.

This shows image formation for a converging lens (a magnifying glass). Since the image is further from the lens than the object, the magnification (q/p) is greater than one, making the text appear larger.

And now we see a similar ray trace for a diverging lens, which makes things appear smaller.

It may be helpful for you to play around with Concept Simulations 26.3 (Converging Lenses) and 26.4 (Diverging Lenses) so you can see how the location of object and image changes for different positions compared to the focal length.

Notice for lenses that every equation we are going to use is the same equation we used for mirrors, including the optics equation, the definition of magnification, etc. The only difference is in our sign conventions. See the diagram on page 7 of your formula sheet to learn these.

Now we need to do some practice problems with single lenses. I recommend you work through examples 9 and 10 in section 26.8. Keep in mind, again, that I’m using p and q for object and image distance, rather than do and di.

Like with mirror problems, we have two equations (optics equation, and magnification equation) and two unknowns. The easiest variant for solving occurs when we are given p and q. But you can solve for all four unknowns if you know any two of them.

Now it’s time to try our first worksheet for this section, which is 26.6 (remember we skipped 26.5). 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 26.6, please post to the appropriate “Chapter 26 worksheets” forum on d2l.tcu.edu, and I will respond there (or send me an email).

—Notice how I set that diagram up, with the illuminated slide in front of the lens and the screen behind the lens. I did that not because I knew what the answers were going to be, but because I know how slide projectors work. Any time a problem refers to a screen, you know that’s where the image is going to be projected, and it will be behind the lens (only real images can be projected onto a surface).

Now try the next worksheet. In this one, I don’t tell you the magnification directly but instead I give you the height of the object and height of the image. From this (and the fact that the image is inverted), you should be able to deduce the magnification. Once you know that plus the focal length (given), the rest of the problem should be easy.

Try worksheet 26.7 now. 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 26.7, please post to the appropriate “Chapter 26 worksheets” forum on d2l.tcu.edu, and I will respond there (or send me an email).

—For the last single-lens worksheet, go back to section 26.7 and study figures 26.26 and 26.27. They should help you understand the vocabulary in this problem and why the same lens can do two very different tasks. You can also see this played out if you mess with the variables in Concept Simulation 26.3, if you haven’t already tried that.

Now please try worksheet 26.8. 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 26.8, please post to the appropriate “Chapter 26 worksheets” forum on d2l.tcu.edu, and I will respond there (or send me an email).

—Please pay close attention to the signs for p and q in this problem and which numbers I assigned to p and q for each case. The wording on problems like this can be confusing initially and will take some practice for you to get comfortable with.

If you want a little more practice with a single-lens system, you can also try this old exam video:

https://www.youtube.com/watch?v=ZoRVaY-kQds

—One last kind of problem that I want you to examine is kind of a combination of the single lens image-formation problem and simple geometry, and these are problems involving “Apparent Depth”. This is optional (not going to be on any homework problem or on the exam), but it is a neat thing to understand.

Please take a look at the last half of section 26.2 again and work through exam 2. Study figure 26.4 closely so you can see where is the object and where is the image (the flat surface is the “lens” in this problem, which has a focal length of infinity). In problems like this, we have to modify the optics equation slightly to get the solution, and I won’t get into that, but this is why water looks shallow when it is deep.

When you are looking down into a pool, the drain you may see seems closer than it is. If a drain looks like to 6 feet below the surface, for example, it is usually more like 8 feet below. When crossing a mountain stream, it may look like it is only a few inches deep, but when you step in, you find yourself up to your ankle with a soaked shoe! The physics described in this section helps explain why, and I think it is useful to know.

Now we move on to double-lens problems in section 26.9.

For these problems, we have to deal with two lenses separately. It’s simple enough, if you just deal with one lens at a time.

The easiest way for me to show you methodically how this works is to ask you to watch one of my problem-solving videos from my old exam, so please watch:

https://www.youtube.com/watch?v=-OrFC66Rjig

Notice that I solve for q1 for the first lens and then use the position of q1 as the position of p2 for the second lens. Once I have q2 for the second lens, I can find the magnification for each lens and multiply those two numbers together to get the total magnification of the system.

Notice that q1 is in front of the first lens, so it almost feels like light goes through the first lens, then turns around and converges back at q1 and then has to go through the first lens *again* to get to the second lens! Of course, that’s not really happening. Remember our discussion about image distance back on pages 14-15 (maybe go back and read that again).

When we get q1 = -4 cm, all that means is that the light rays upon leaving lens 1 are diverging. That’s the minus sign. The “4” tells us how fast the rays are diverging. The second lens now uses that as information for what is the trajectory of the incoming light rays so we can calculate the location of the final image.

A couple of other video problems you may want to try like worksheets can be found at:

https://www.youtube.com/watch?v=vFug2A9f6V4https://www.youtube.com/watch?v=vFug2A9f6V4

Now it is time to try worksheet 26.9, which is a double-lens problem you have to solve in reverse. Instead of being given the initial position and asked to find the final image, you are given the final image location and asked to locate the initial object position.

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 26.9, please post to the appropriate “Chapter 26 worksheets” forum on d2l.tcu.edu, and I will respond there (or send me an email).

—The last section of Chapter 26 that we will cover deals with human eyesight and corrective lenses. So we start with myopia (nearsightedness) and how to fix it.

Myopia is caused by the lens of the eye having too much curvature. In figure (a) above, light from a distant object strikes the lens pretty much as parallel rays, but the lens brings these rays to a focus too quickly, in front of the retina, so you can’t focus on the object.

To see such an object comfortably, you would need for the object to be nearby so that it’s light rays diverge a bit upon entering the eye. That way, they won’t come to a focus quite so fast.A diverging lens placed in front of the eye accomplishes this in (b). It takes those nearly parallel rays of the distant object and causes those rays to diverge a bit. That means the diverging lens forms a virtual image of the object that becomes the object the eye sees.

To repeat: The distant object (p1) for the diverging lens results in a virtual image (q1). That virtual image becomes the object (p2) for your eye, resulting in a real image (q2) that falls on the retina. This arrangement is seen in part (c) of the figure.

The “far point” of the eye is the furthest distance away an object can be from the eye and still be seen comfortably. For a person with normal eyesight, the far point is infinity. For me, personally, my far point is pretty bad, about 36 cm in front of my eye. Anything outside of 36 cm away from my eye is a blur.

Thus, I wear diverging lens glasses. These lenses will take an object at infinity (p1) and put a virtual image (q1) at a distance of 34 cm in front of my glasses. Since my glasses are already 2 cm in front of my eye, that virtual image 34 cm in front of my glasses creates an object (p2) 36 cm in front of my eye, and now my eye can comfortably make an in-focus image (q2) on my retina.

The focal length of my glasses would be easy enough to find with the optics equation. Try it! It is -34 cm.

If I were wearing contact lenses, there is no gap between my contacts and the eye, so the needed focal length would be -36 cm.

Work through example 12 in section 26.10 and try this for yourself.

The other common problem found in human eyesight is farsightedness (hyperopia), described on the next page.

The farsighted person’s eye has too little curvature. So objects that are nearby can’t be seen clearly. When an object is too close to the eye, the rays diverse so “fast” that the lens can’t bring them to a focus quickly enough. They form an image behind the retina and thus, out of focus.

A converging lens will solve the problem (b) by making these diverging rays a little “slower” (now they aren’t spreading apart quite so much) so that when your eye’s lens transmits the rays, they end up focussed on the retina.

So a converging lens takes a real object (p1) and creates a virtual image (q1) a bit further away so that the rays aren’t diverging so quickly. This virtual image becomes the object (p22) for your eye, resulting in an image (q2) on the retina.

Take a look at example 13 and work through it so you can see an example of a simple lens correcting hyperopia.

The “near point” for a person is the minimum distance an object can be from the eye and still seen comfortably. If you get an object closer to the eye than the near point, even a normal eye can’t bring the quickly diverging rays into focus. The normal near point for a person is 25 cm, and that’s typically where we want corrective lenses to put the virtual image for our eye.

My near point (uncorrected) is about 21 cm, and my far point (uncorrected) is about 36 cm. That’s a pretty narrow range of distances that I can see clearly! But as I said before, the diverging lenses of my glasses magically extend my far point all the way out to infinity. They take an object at infinity and put an image of it 36 cm in front of my eye.

What about my near point? These glasses will change my far point from 36 cm to infinity. How will my near point change?

We can answer that by asking what object distance with these glasses results in an image at my eye’s near point. An image at my eye’s near point (21 cm in front of my eye) would be 19 cm in front of my glasses. If we plug in q1 = -19 cm and f = -34 cm, we get 43 cm for p1. YIKES! That means my new near point is 44 cm, which means it may be difficult to read a book or a screen held in my hand unless I hold it way out at arm’s length.

So that would not be a good way to live my life, only being able to read things that are that far away from my eye. Goodbye, email! Goodbye, fun little phone games!

Ah, but there is a way around it called bifocals. The top half of my glasses are for seeing things that are far away and the top half has a curvature such that the focal length is -34 cm. But the BOTTOM HALF of my lenses has a different curvature for things I want to see up close.

If I just want to leave my near point alone, I can just have that part of the lenses with no curvature (focal length infinity) so that p = q and my near point is unchanged.

If I want a normal near point of 25 cm, then we can work through it:

My eye needs the object to be at my near point, 21 cm in front of my eye, which is 19 cm in front of my glasses. So q1 for my glasses is -19 cm.

I want the object to be at a normal person’s near point, which is 25 cm in front of my eye and 23 cm in front of my glasses. So p1 for my glasses is 23 cm.

Using that p1 and q1 gives me a focal length f = -109 cm. Remember, the larger the number, the less curvature the lenses have. Smaller value of f = “faster” lens.

My glasses have a curvature of f = -34 cm on the top half andf = -109 cm on the bottom half. Bifocals! People with bifocals when reading things will often tilt their heads back or adjust their glasses so they sit at their proper height if they have slid down the nose. You’ve seen that, right? That’s so they are looking at the nearby text through the bottom half of their glasses.

Don’t laugh, this may be you someday!

Note also the little section at the end that defines a new unit called the diopter. You may see this from time to time in optics problems. It’s just another way to state the focal length, as “optical power”. The trick is that we often use cm for our optics distances, but the diopter relies on distances to be given in meters, so it’s easy to trip up.

Now it is time to try our last worksheet, 26.10, which is similar to example 13 from your book (with the added complexity of glasses 2 cm in front of the eye instead of contact lenses). 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 26.10, please post to the appropriate “Chapter 26 worksheets” forum on d2l.tcu.edu, and I will respond there (or send me an email).

—I don’t have any human eyesight video problems for some reason, but there are several examples on old recent exams. I’ll list them here, and I encourage you to work through them for additional practice.

Summer 2019, Exam 3A, #4Spring 2019, Exam 3A, #4Spring 2019, Exam 3E, #4Summer 2018, Exam 3, #4Spring 2018, Exam 3B, #3

That’s just from the last two years, and I am sure you could find more if you keep going back over old exam 3’s or 4’s.

There are also several related problems listed from WileyPLUS in your Chapter 26 resources file.

—This concludes my lecture notes for Chapter 26. Now it is time for you to get started on the Chapter 26 homework.

As always, I hope you will solve or seriously attempt each problem before asking for help. I will be checking the “Chapter 26 homework” forum on our course shell occasionally and contributing helpful comments in response to any questions you may have about the homework.

According to our timeline, here are the upcoming dates of interest for the class:

Tue Apr 14 11:59pm - Ch 26 HW dueWed Apr 15 (morning) - Exam 3 sent outFri Apr 17 11:59pm - Exam 3 due