Reflection of Light - Video Tutorials & Practice Problems

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concept

Law of Reflection

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Welcome back, everyone in this video. We're gonna continue talking about light. We're gonna talk about light as a ray rather than a wave. We'll also talk about something that's called the law of reflection, which is a very straightforward and pretty intuitive equation. Let's get started. So this actually sort of brings us to an area of physics which we call geometric optics, which basically says that light travels in a straight line and travels in a straight line as a ray as a light ray. So you'll see these types of diagrams all the time with these arrows, those are basically just light rays. And we'll see that lights interacting with a bunch of different objects like mirrors and lenses and things like that. So light travels as a ray, even though it's made up of a bunch of waves, we said that light is an electromagnetic wave, but it always just, it always just travels in a particular direction here at the, at a fixed speed. And so we can just model light like that as an arrow um that's traveling at a fixed speed. All right. So let's actually talk about the first situation that we'll encounter, which is called reflection. And that's what happens when you have light that hits a flat or shiny surface, something like a mirror or polished metal. So when light rays come in and hit a mirror at an angle, so this is gonna be some angle theta here, then they reflect. And you've probably seen this all the time before. If you've looked at a mirror or if you've ever shined like a laser pointer at a mirror or something, you can see a sort of like this beam of light and it kind of just bounces off the mirror. And so it reflects that's called reflection. And what's really interesting about this angle or this reflected ray is that it always reflects at the exact same angle that it came in. So in other words, this data is exactly equal to this data. They always will be all right. Now these two the have names, by the way, the first one is called the angle of incidence. You can kind of remember this because it sounds like incoming incident is incoming light. So this is given as theta I and the second one is called the angle of reflection, not very creative, but it's the reflected ray. And so this angle here is called theta R. Now these two say these two angles are exactly the same. So we can actually just write this as an equation and this is just theta I is equal to theta R this is sometimes called the law of reflection. And it's pretty straightforward, pretty intuitive, whatever it comes in as is whatever it has to leave. As. All right. Now, what's really important that you might have noticed about these diagrams is that these angles have been drawn not relative to the horizontal like we normally do in physics, they've been drawn relative to this black line here. Uh And that's actually really, really important. The one thing that you have to remember about geometric optics is that your angles should always be measured relative to the normal. Now, we've seen that word before normal. Normal just means perpendicular to the surface. So in other words, just make a 90 degree angle, these lines will always be drawn for you. So this normal line over here and it's basically just telling you where 90 degrees is all right. So you have to make sure that your angles are measured relative to this normal. Otherwise you're gonna get the wrong answer. All right. So be really, really careful about that. All right. But that's really all there is to it for reflection. So let's just go ahead and take a look at our first problem here. All right. So we have a laser that we're gonna shine in a flat mirror that's on the ground. We want the laser beam to hit this point on the wall and this point is actually 4 m away from where the laser point hits the mirror. So this is gonna be 4 m and it's also 2 m above the floor. So in other words, this is just 2 m over here. All right. So we wanna actually calculate what's the angle of incidence? OK. So this angle of incidence remember is just the incoming angle theta I. All right. So then how do we do that? How do we calculate theta I? Well, the one thing that we've seen about theta, I is that theta, I is equal to theta R, but they're exactly the same. So if you don't know one of them, then you obviously don't know the other one, right? So what's tricky about these problems is not the, theta I equals theta R. It's what you have to do to get those angles. All right. So take a look here because uh I have no information about any of the distances involved on the left side of the diagram. But I do have some information about the distances here. I've got 4 m and 2 m. So if I'm trying to figure out one of these angles here, hopefully you figured out by now that you're gonna have to solve some kind of a triangle. That's what's tricky about these problems. It's not really the theta, I equals theta R. It's what you have to do to get those equations, which is usually some trig. All right. So we can see here that this kind of look sets up a little triangle like this. But what's really, really important about this? Uh You know what I have to be really careful about here is that when you use your equations in this format for sine cosine and tangent, be very careful because usually you're gonna be solving for this angle over here. We've already said that this angle is gonna be the bad one. So you don't want to solve for that. Instead, what we can do is we can actually sort of flip this triangle and we could sort of draw like an inverted one and the distances are still the same. This is gonna be four and this is gonna be two. Now, what's good about this is that now when you solve for your sine cosine tangent, it's gonna have this angle which you want. That's the good one. All right. So which, which um angle, which co which formula are we gonna use to solve for this theta here, we don't have the hypotenuse. Um So all we can use is, is actually just the tangent. So this is gonna be to tangent of theta R is equal to the opposite over the adjacent side. Uh This is your opposite and this is your adjacent side over here. So this is just gonna be 4/2. So if you take the inverse tangent, theta R is equal to the inverse tangent of 4/2 and what you'll get is 63 degrees. All right. So that is your angle of incidence and it's also your angle of reflection. So this is 63 degrees and this is also 63 degrees. All right. So that's all there is to it. All right. So let's move on now because I have actually one last conceptual points. By the way, problems will always involve specular reflections. There's two different types of reflection. This is only a conceptual thing. You'll never have to solve any problems like this. Um But problems will always involve specular reflection from smooth surfaces instead of diffuse reflection from rough surfaces. So in other words, you can kind of assume that your surfaces are going to be perfectly flat and that light rays will always perfectly bounce at the right angle that's called specular reflection um versus diffuse reflection, which kind of like, you know, assumes that there's, you know, rough points or whatever. And um the light will just scatter off in different directions. All right. So it's just a conceptual point, you might need to know anyway. So that's if this one, let me know if you have any questions.

2

Problem

Problem

A light ray strikes a horizontal surface at a 60° angle. The reflected ray then strikes a wall 2.5m above the reflective surface. How far is the wall from the point where the incident light ray strikes the surface?

A

4.33m

B

1.44m

C

2.5m

D

0.69m

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