Diffuse reflection involves light rays reflecting from an uneven surface, as shown in the diagram. The diagram shows four points, D, E, F, and G, that the three light rays A, B, and C might possibly pass through after being reflected. Which of the points would the light ray A pass through?
First things first, let’s take a look at this diagram, which shows us this uneven surface and three light rays labelled A, B, and C incident on it. We see that each one of these three rays hits the surface at a different location. We also notice these four points, D, E, F, and G, marked out as candidate locations for the reflected rays from these three rays to pass through. In other words, after ray A, for example, has bounced off this surface, it might pass through point D, point E, point F, or point G. And in fact, that’s our first question. Just which of the points would this light ray, light ray A, pass through? As we start to figure this out, let’s clear some space on screen.
To help us answer this question, let’s recall a law in optics, known as the law of reflection. This law says that the angle at which a ray is incident on a surface is equal to its angle of reflection. This means that if we had some surface, like this one here, and a ray incident on that surface, like this one. Then if we were to draw a line perpendicular to the surface at 90 degrees to it, at the location where the ray strikes the surface, then the angle from this normal line to the incident ray is called the incident angle. We can call it 𝜃 sub 𝑖. And that if we go the same angular distance on the other side of that normal line, then what we found is the direction in which the reflected ray will travel. And we can call that angle on the other side 𝜃 sub 𝑟 for the angle of reflection.
Another way of writing the law of reflection using symbols rather than words is to say that 𝜃 sub 𝑖, the angle of incidence, is equal to 𝜃 sub 𝑟, the angle of reflection. So let’s now use this law in our diagram to figure out where light ray A will reflect. Now, we see that light ray A strikes this uneven surface right there at that point. So our first order of business is to draw a normal line to the surface at that location. With that normal line, the line perpendicular to the surface at that point sketched in, we can now figure out the angle of incidence of ray A by going from that normal line to ray A. Then, by the law of reflection, if we go the same angular distance on the other side of that normal line, then we’ll have figured out the direction in which the reflected ray will travel. And when we trace out that reflected ray, we see it goes through the point marked out E. So that’s our answer to this first question. The light ray A, when it reflects from this uneven surface, passes through point E.
Now, let’s move on to answer the same question about ray B. Which of these four points would the light ray B pass through? Once again, we’ll start off by marking on our uneven surface where ray B is incident. And then, we’ll draw in a perpendicular or normal line from that point on the surface. Now, if we move from that normal line that we’ve sketched in to ray B and what we’ve defined is the angle of incidence of this ray. And based on the law of reflection, we know that if we move the same angular distance on the other side of this line, then we’ll have figured out the angle of reflection. But notice something interesting. When ray B first reflects off this uneven surface, the reflected ray then runs into the surface once more, right at this location.
Since the ray is reflecting again, we’ll repeat this process. We’ll draw in a normal line from the uneven surface at this point of incidence. And then, we’ll draw an arrow from that normal line to the incident ray, ray B. That will define our angle of incidence in this second case. Once more, following the law of reflection, we’ll then measure the same angle on the opposite side of this normal line. It’s in this direction that the second reflection of ray B will travel. As we trace that line out, we see that it goes through point F. So that’s our answer to this part of the question. After bouncing twice off the uneven surface, ray B passes through point F.
And then, last but not least, we’ll answer the same question, this time for light ray C. Which of the points would this light ray pass through after reflecting off the surface? As we trace light ray C along, we see it runs into the surface right here. Drawing in our normal line as usual, we see that the angle of incidence for ray C is this angle here. Which means that if we measure the same angular distance on the other side of this normal line, then we’ll have found the direction the reflected ray travels. And if we trace this reflected ray out, we see that the point it passes through is point G. So it’s point G that light ray C passes through after reflecting off of this uneven surface.