Lesson Explainer: Reflection of Light Physics • 9th Grade

In this explainer, we will learn how to describe the paths of light reflected from specular and diffuse surfaces, applying the law of reflection.

We can recall that light rays travel in straight lines. If a particular ray of light were to never run into anything, it would go on and on, forever. However, in reality, there is always eventually going to be some object in its path.

To get a sense of what happens when a light ray meets an object, it may be helpful to think of light as a particle traveling along the direction of that ray. Then, when we are thinking about light, we are picturing a solid object colliding with another object or boundary.

This is the same idea as throwing a ball at a wall. From experience, we know that, in this case, the ball will bounce off the wall. Likewise, the ray of light will bounce off the object that it hits. This process is known as reflection.

This process of reflection happens when light is traveling in a straight line through the air and encounters a solid object in its path. However, this is not the only situation in which it occurs. More generally, it can happen whenever there is a boundary between any two media. Recall that a medium is any material that light can travel through. So, we could be talking about a boundary between air and water, or between glass and plastic, and so on.

In fact, it is only because things reflect light that we can actually see them. Otherwise, the only things we would be able to see would be objects that emit light. We call these objects sources of light. These sources of light, such as the Sun or a light bulb, are relatively few and far between. The majority of objects around us do not emit light of their own but rather reflect light from an external source. And so, it is only because of this reflection that we can see them.

It turns out that when light reflects, it does so according to a law known as the law of reflection.

To see how this law works, we will begin by considering the following diagram, in which a ray of light reflects off a flat surface:

We have drawn the incident ray of light and the reflected ray of light on this diagram, with arrows representing the direction of travel of the light. We have also labeled two angles on the diagram. These angles are labeled 𝜃, which is known as the angle of incidence, and 𝜃, which is known as the angle of reflection.

Notice that these two angles are defined relative to the dashed line, which is perpendicular to the surface. This line is known as the “normal” to the surface.

The angle of incidence, 𝜃, is the angle of the incident or incoming light ray relative to the normal to the surface. Similarly, the angle of reflection, 𝜃, is the angle of the reflected light ray relative to this same normal.

The law of reflection lets us work out which direction the reflected ray is going to travel in. In other words, it tells us what the angle of reflection, 𝜃, will be for a given value of the angle of incidence, 𝜃. This law is as follows.

Definition: The Law of Reflection

The angle at which a ray of light is incident on a surface is equal to the angle at which it is reflected, and on the opposite side of the normal.

Mathematically, if the angle of incidence is 𝜃 and the angle of reflection is 𝜃, then 𝜃=𝜃.

It is worth pointing out that we have drawn a two-dimensional picture, while the real world is three dimensional. In fact, we can always consider a two-dimensional cross section like in our diagram. The reason for this is that the reflected ray always lies in the plane defined by the incident ray and the normal to the surface.

Let’s now look at an example problem in which we are asked to calculate the angle of reflection of a light ray.

Example 1: Calculating the Angle of Reflection of a Light Ray

A light ray falls on a reflective surface as shown in the figure. What will the angle of reflection be?

Answer

This question presents us with a diagram showing a ray of light incident on a flat, reflective surface and asks us to calculate the angle of reflection.

We are given an angle of 130 on the diagram, but we should notice that this is not, in fact, the angle of incidence. To find the angle of incidence, we need to add the normal to the surface to the diagram:

Then, the angle that we have marked 𝜃, which is the angle of the incident ray relative to this normal, is the angle of incidence.

Since we know that the normal to the surface is, by definition, perpendicular to the surface, we know that the angle between the normal and the surface itself must be 90.

We can see from the diagram that the angle of 130 is equal to this 90 angle between the surface and the normal plus the angle marked 𝜃.

This means we know that 130=90+𝜃.

We can rearrange this to make 𝜃 the subject by subtracting 90 from both sides: 𝜃=13090=40.

Now that we know that the angle of incidence of the light ray is 𝜃=40, we can use the law of reflection to work out the angle of reflection.

Recall that the law of reflection states that the angle of incidence is equal to the angle of reflection.

If we call this angle of reflection 𝜃, then the law of reflection tells us that 𝜃=𝜃=40.

So, our answer to the question is that the angle of reflection is 40. We point out that this reflected ray will be in the same plane as the incident ray but at an angle of 40 on the opposite side of the normal.

When we look at an object in a mirror, we know from experience that we do not see that object where it actually is. Instead, we see an image of that object that appears to us to be placed behind the mirror. This image is known as a virtual image, because the image is not “real”; it is simply where the light rays look like they are coming from to us.

Let’s see how this works by considering the light rays from an object when they get reflected in a mirror.

We will consider two differently angled rays coming from the object. For each ray, we know from the law of reflection that when it gets reflected off the mirror, the angle of incidence is equal to the angle of reflection. This is illustrated in the diagram below:

An observer looking at the mirror sees the light rays reflected from the mirror. To this observer, it appears as if the light ray is coming through the mirror. If we trace the reflected rays back through the far side of the mirror, we see that they meet at a point behind the mirror. This point is where the virtual image of the object is formed. This is shown in the diagram below:

We can see from the diagram that the virtual image is at the same distance from the mirror as the object. In other words, this virtual image is as far behind the mirror as the object is in front of it.

Let’s look at an example problem about virtual image formation.

Example 2: Reflection and Image Formation

The reflection of an object is seen in a mirror by an observer whose eye is shown in the diagram. At which of the points A, B, C, D, and E is the object’s image seen?

Answer

To answer this question, we need to remember that the observer is looking at the mirror. This means that they only see the reflected ray of light.

To the observer, this reflected ray appears to be coming through the mirror surface from behind the mirror.

To find the position of the image seen by the observer, we need to extend this reflected ray back behind the mirror.

Doing this gives us the following:

We have extended the reflected ray using a red dashed line.

We can see that this extension of the reflected ray passes through the point marked A. And so, we know that the image seen by the observer is formed at point A.

The law of reflection can also be used to determine the path that a ray will follow after reflecting off a surface.

For perfectly flat surfaces, this process is straightforward. We have already seen how to find the angle of reflection of such a ray. We draw the incident ray up to the point where it hits the surface. We then draw the normal to the surface at the point at which the ray is incident. Finally, the law of reflection tells us that the reflected ray is on the opposite side of the normal to the incident ray, with the same angle relative to the normal. We may then extend this reflected ray as far as we want, knowing that it will travel in a straight line.

The important thing to note is that, for a flat surface, the normal to the surface will be in the same direction at all points on that surface. This means that all rays coming in toward the surface at the same angle will reflect and leave at the same angle as each other.

When we have light reflecting off a flat surface in this way, it is known as specular reflection.

Let’s now look at an example problem featuring specular reflection.

Example 3: Calculating the Path of Light Rays Undergoing Specular Reflection

Specular reflection involves light rays reflecting from an even 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.

  1. Which of the points would the light ray A pass through?
  2. Which of the points would the light ray B pass through?
  3. Which of the points would the light ray C pass through?

Answer

The question is asking us to work out which of the three possible points D, E, and F each of the three rays A, B, and C pass through after reflecting off the even surface shown in the diagram.

To answer this question, we first need to extend each of the three rays until it hits the surface.

Then, we need to draw the normal to the surface at that point and measure the angle that this incident ray makes to the normal—this is the angle of incidence.

Finally, we appeal to the law of reflection to draw an outgoing ray with an angle relative to this normal (the angle of reflection) and equal to the angle of incidence.

We will work through this step by step for part 1, then apply the same process to answer the two following parts more concisely.

Part 1

In this part of the question, we are asked about the ray marked A. So, let’s extend that ray until it meets the surface and add the normal to the surface at the point where the ray meets it.

The next step is to measure the angle of incidence.

As shown in the diagram, this angle is approximately 45.

Then, the law of reflection tells us that the angle of reflection is the same as the angle of incidence but on the opposite side of the normal.

This means that we need to measure an angle of reflection of 45 on the right-hand side of the normal and draw the path our reflected ray will follow at this angle.

Extending this ray shows that it passes through the point marked F.

Part 2

This part of the question asks us to do the same thing for ray B.

As in part 1, we extend the ray to the surface and draw in the normal to the surface at the point where the ray hits it. Then, we note that ray B is parallel to ray A, so the angle of incidence will be the same value of 45.

Then, the law of reflection tells us that the angle of reflection is the same as the angle of incidence but on the opposite side of the normal, so we can draw in the path of the reflected ray.

Extending this ray shows that it passes through the point marked E.

Part 3

The final part of the question asks us to do the same thing again for ray C.

As in parts 1 and 2, we extend the ray to the surface and draw in the normal to the surface at the point where the ray hits it. Ray C is parallel to the rays A and B, so the angle of incidence will again be the same value of 45.

Again appealing to the law of reflection, we can draw in the path of the reflected ray.

Extending this ray shows that it passes through the point marked D.

We have seen that, when the surface is flat, we get specular reflection. In this case, all light rays coming in at a given angle leave at that same angle. It is important to note that this means that they leave at the same angle relative to each other.

The result in this case is that the reflected image closely resembles the object. An example of this is looking at the reflection of objects seen on the surface of water. In photos such as the one below, we see a glassy lake with a seemingly perfect image of the bridge behind it.

Bridge reflection in water

However, if the surface of the water is rough and bumpy, we no longer see such a neat reflection. Instead, we see a somewhat blurry mess of an image, as in the photo below:

Diffuse reflection in rough water

This second case, in which the surface is not flat, is known as diffuse reflection.

It is important to realize that the law of reflection does still apply. However, because the surface is rough and bumpy, the normal to the surface points in different directions at different points on the surface.

Imagine that we have several light rays incident on a bumpy surface. Let’s say that these rays are all at the same angle relative to each other but that they will hit the surface at different positions. The incident and reflected rays in this case are shown in the diagram below:

If we look at each incident ray individually, we see that it is reflected according to the law of reflection at the point at which it is incident. That is, we can draw the normal to the surface at each point and measure the incident angle of the ray relative to this normal. We then know that the angle of reflection relative to this normal is equal to this angle of incidence.

Looking at the reflected rays, we see that they are all pointing in different directions relative to each other. The incident rays were all at the same angle relative to each other. The law of reflection applied for each incident ray. However, in spite of these facts, the result in the case of diffuse reflection is all the reflected rays pointing in different directions. The reason is that each individual incident ray hits a surface with a different orientation and hence a different normal.

It is the fact that the reflected rays point in different directions from each other that causes the reflected image to look blurry in the case of diffuse reflection.

Let’s finish up by looking at an example problem involving diffuse reflection.

Example 4: Calculating the Path of Light Rays Undergoing Diffuse Reflection

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.

  1. Which of the points would the light ray A pass through?
  2. Which of the points would the light ray B pass through?
  3. Which of the points would the light ray C pass through?

Answer

The question is asking us to work out which of the three possible points D, E, and F each of the three rays A, B, and C pass through after reflecting off the uneven surface shown in the diagram.

To answer this question, we first need to extend each of the three rays until it hits the surface.

Then, we need to draw the normal to the surface at that point and measure the angle that this incident ray makes to the normal—this is the angle of incidence.

Finally, we appeal to the law of reflection to draw an outgoing ray with an angle relative to this normal (the angle of reflection) and equal to the angle of incidence.

The catch here, as compared to the case of specular reflection from an even surface, is that the normal to the surface will be in different directions at different points on the surface.

We will work through the process step by step for part 1, then apply the same approach to answer the two following parts more concisely.

Part 1

In this first part of the question, we are asked about the ray marked A. So, let’s extend that ray until it meets the surface and add the normal to the surface at the point where the ray meets it.

The next step is to measure the angle of incidence, which is the angle between the incident ray A and this normal line.

As shown in the diagram, this angle is approximately 43.

Then, the law of reflection tells us that the angle of reflection is the same as the angle of incidence but on the opposite side of the normal.

So, we need to measure an angle of reflection of 43 on the right-hand side of the normal and draw the path our reflected ray will follow at this angle.

Extending this ray shows that it passes through the point marked E.

Part 2

This second part of the question asks us to do the same thing for ray B.

As with ray A, we extend the ray until it meets the surface and add the normal line at this point. Notice that since the surface is uneven, the normal line points in a different direction where B meets the surface from the direction that it pointed in where A met the surface. Measuring the angle of incidence of ray B gives a result of 22.

The law of reflection tells us that the reflected ray makes the same angle of 22 on the opposite side of the normal. Extending this reflected ray shows that it passes through the point marked D.

Part 3

This final part of the question asks us to do the same thing again, but this time for ray C.

As with rays A and B, we extend the ray until it meets the surface and add the normal line at this point. Again, the normal line points in a different direction here as a result of the uneven surface. Measuring the angle of incidence of ray C gives a result of 59.

From the law of reflection, we know that the reflected ray makes the same angle of 59 on the opposite side of the normal. Extending this reflected ray shows that it passes through the point marked F.

Key Points

  • When a ray of light encounters a boundary between two media, some or all of the light may be reflected.
  • Whenever light is reflected, it obeys the law of reflection. This law states that the angle of incidence, 𝜃, is equal to the angle of reflection, 𝜃. Mathematically, this may be written as 𝜃=𝜃. The angles 𝜃 and 𝜃 are measured relative to the normal line, which is the line perpendicular to the surface.
  • When an observer sees the reflection of an object in a mirror, it appears to them as if the object is placed behind the mirror. This is because the observer sees a virtual image of the object. This virtual image is the same distance behind the mirror as the object is in front of it.
  • When light reflects off an even surface, we get specular reflection. In this case, the normal to the surface is the same at all points on the surface. A reflected image closely resembles the object.
  • When light reflects off an uneven surface, we get diffuse reflection. The normal to the surface is different at different positions on the surface, and so, reflected rays get scattered in different directions. A reflected image is blurry.

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