Question Video: Calculating the Path of Light Rays Undergoing Diffuse Reflection Physics

Video Transcript

Diffuse reflection involves light rays reflecting from an uneven surface, as shown in the diagram. The diagram shows three points 𝐷, 𝐸, and 𝐹 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?

So, this question is asking us to consider three light rays which are labeled 𝐴, 𝐵, and 𝐶. These three light rays are all going to reflect off this surface here, and we’ve being given three possible points labeled 𝐷, 𝐸, and 𝐹 that each of these three rays might pass through after reflecting off the surface. This first part of the question is asking us about the light ray 𝐴. So that’s this top ray in the diagram. We need to work out how light ray 𝐴 reflects off this surface and then which of these three points the reflected ray passes through.

Let’s begin by recalling what happens when a light ray reflects off a surface. So, let’s suppose that this here is our surface and this is a light ray which is incident on that surface. At the point where the light ray meets the surface, we can draw a dashed line that is perpendicular to that surface; that is, it meets the surface at an angle of 90 degrees. This line that is perpendicular to the surface is known as the normal to the surface. The angle that an incoming light ray makes to this normal is known as the angle of incidence of that light ray. This is commonly labeled as 𝜃 𝑖. When the light ray reflects, the reflected ray makes an angle to the normal known as the angle of reflection. This is commonly labeled as 𝜃 𝑟.

It turns out that if we know the angle of incidence of a particular light ray, then we can find the angle of reflection by using our law known as the law of reflection. The law of reflection states that the angle of incidence is equal to the angle of reflection. For an angle of incidence 𝜃 𝑖 and an angle of reflection 𝜃 𝑟, the law of reflection says that 𝜃 𝑖 is equal to 𝜃 𝑟. It’s worth making it clear that the angles 𝜃 𝑖 and 𝜃 𝑟 will always be on opposite sides of the normal. That is, the reflected ray will always be on the opposite side of the normal to the incident ray.

In this diagram, we drew our surface as being flat or even. However, in the question, we’re told that we’re considering the case of diffuse reflection, which involves light rays reflecting from an uneven surface. If we look at the diagram given to us in the question, we see that this surface is not a single straight line. Instead, it consists of different sections with different directions. We can still define a normal at any point on the surface. The catch here is just that that normal will be in different directions at different positions.

For example, let’s consider this point here. The normal at this point is the line that is perpendicular to the direction of the surface at that point. Now let’s see what the normal looks like at this point here. We see that this normal is in a different direction to the one we considered earlier. Just as with the first normal, this second normal makes an angle of 90 degrees to the surface at the point at which the normal meets the surface. However, since this bit of surface and this bit of surface are at different angles to each other, then the two normals must also be at different angles to each other. Okay, let’s get back to the question and consider what happens to the light ray labeled 𝐴.

We’re going to start by extending light ray 𝐴 up until it meets the surface. Remember that in the absence of anything in its way, a light ray travels in a straight line. So, extending light ray 𝐴 in a straight line, we see that it meets the surface at this point here. We want to determine the angle of incidence of this light ray. So let’s draw in the normal to the surface at the point where the light ray meets it. Now that we have the normal at this point, we can measure the angle of incidence of light ray 𝐴. So, that’s the angle between the incoming ray and the normal. When we do this, we find that this angle of incidence is approximately equal to 43 degrees.

Then, the law of reflection tells us that the reflected ray will have this same angle, approximately 43 degrees, on the opposite side of the normal. Measuring out this angle of reflection, we can draw in the reflected ray, and extending this ray, we find that it passes through the point labeled 𝐸. So our answer to this first part of the question is that the light ray 𝐴 will pass through point 𝐸. Now let’s look at the second part of the question.

Which of the points would the light ray 𝐵 pass through?

This time we’re considering the light ray labeled 𝐵, so that’s the middle of the three rays in the diagram. The process we’re going to use is exactly the same as we did for light ray 𝐴. So, we’ll start by extending the ray 𝐵 up until the point where it meets the surface. We see that ray 𝐵 meets the surface at this point here. So let’s draw in the normal to the surface. At this point, this normal is in a different direction to the one that we drew when answering the first part of the question, as we expect, based on the fact that the surface is uneven.

Now that we’ve drawn the normal, we can measure the angle of incidence of light ray 𝐵. We find that this is approximately equal to 22 degrees. Once again using the law of reflection, we know that 𝜃 𝑟 should be the same as 𝜃 𝑖. Measuring out an angle of reflection of 22 degrees, we can draw in the reflected ray. Extending this reflected ray, we see that it passes through the point marked 𝐷. So our answer to the second part of the question is that light ray 𝐵 will pass through point 𝐷. Now let’s look at the third and final part of the question.

Which of the points would light ray 𝐶 pass through?

Okay, same process again but this time for light ray 𝐶, so that’s the lowest of the three lines in the diagram. We’ll start by extending light ray 𝐶 until it meets the surface. We find that the light ray hits the surface at this point here. Let’s add in the normal to the surface at this point. Then we can measure the angle of incidence between this incoming ray and the normal. We find that this angle is approximately 59 degrees, so the angle of reflection is also 59 degrees. Measuring out this angle of reflection, we can then draw in the reflected ray. Extending this ray, we find that it passes through the point labeled 𝐹. And so our answer to this final part of the question is that the light ray 𝐶 will pass through point 𝐹.