# 9.2.2: Reflection

- Page ID
- 2205

### What Is Reflection?

When a wave reaches the interface between two different media, typically some of the wave will bounce back into the original medium. This process is known as **reflection**. A familiar example of reflection is optical reflection in mirrors, where light waves reflect off a smooth surface. Another familiar example of reflection comes from water waves; as the waves travel they reflect off objects that are floating in the water, and also reflect off the walls of the container holding the water. Most of us are familiar with the concept of echoes, which are the reflections of sound waves. Any kind of wave can undergo reflection.

## Specular Reflection

###### The Normal to a Plane

Let us start by describing how light bounces off a mirror. If you have ever shone a flashlight or laser pointer at a mirror in a dark room, you may have noticed that the light reflects in a well-defined and regular direction. If not, you will get some experience with this in one of your physics labs.

One of the questions we would like to answer is “which direction does the light bounce off the mirror?” The answer to this question depends on the angle the light hit the mirror, and in order at answer it precisely we need to introduce the concept of a normal. For a flat plane, the **normal** to that plane is a line or vector that points directly into or out of the plane. This means that the normal will always be at 90° to the plane.

###### The Law of Reflection

If we shine a ray of light onto a mirror, the angle between the ray and the normal is denoted \(\theta_{inc}\), where “inc” is short for "incident ray." The ray of light that bounces off the mirror is on the other side of the normal, but at the same angle! The picture below may make this more clear:

The angle that the light reflects at is called \(\theta_{ref}\) where "ref" is short for "reflected ray." The **law of reflection** tells us that

\[\theta_{inc} = \theta_{ref}\]

We can apply this principle to smooth surfaces that are curved, like funhouse mirrors. To do this, consider only an incredibly small part of the mirror, so you can approximate that the mirror is flat. We can then find the normal and use the law of reflection to find where the light ray reflects from that part of the mirror. By repeating this for many light rays we can find out anything we want to know about a curved mirror. A quick sketch of how to do this is presented below.

Instead of drawing the magnified image, we can also consider drawing a* tangent* to the surface at a particular point. As you might expect, the normal is* perpendicular *to the tangent line*.* You should spend some time considering the definition of a tangent line to figure out why this “magnification procedure” and “tangent procedure” are both valid ways of locating the normal.

### Diffuse Reflection

When we first discussed geometric optics, we made a point of stating that, on most surfaces, light reflects in all directions. We called this phenomenon** diffuse reflection**. This behavior seems very different from the ideas discussed above, where reflected light waves have very well defined directions.

Above, we don't acknowledge that most materials are actually rough, not smooth. If you zoom in on a rough surface, you'll see that it's composed of very small smooth surfaces, all of which are misaligned. Light incident on rough surfaces reflects in mulitple, misaligned directions, and it becomes scattered (as shown below). This is why light shining on the tree in How We See Things, for example, reflects in all directions.

Example 2.2.1

A light ray traveling right hits a mirror as shown below. The mirror is slightly tilted. Where does the reflected ray go?

Solution

In this problem we have not been directly given the angle of incidence \(\theta_{inc}\) directly. We will need to do some geometry to figure out what \(\theta_{inc}\) is. One way of doing this is to construct a right-angled triangle using a vertical line as shown:

The angle in the upper right hand corner of the triangle must be 60°, because the sum of the interior angles in a triangle is always 180°. Because the original light ray is coming in horizontally and the dotted line is vertical, we can deduce that the angle between the mirror and the light ray is 30°.

Now we introduce the normal as a dotted line. Because the angle between the mirror and the normal must be 90° we can deduce that \(\theta_{inc} = 90°-30°=60°\). From the law of reflection, we must also have \(\theta_{ref} = 60°\). The final part of the solution is sketched below.

The difficult part of this problem is not the law of reflection itself, but rather the geometry of finding the incident ray. There are many less complicated ways to solve this problem, but we chose to do it this way because, when we introduce the more quantitatively complicated *refraction*, it will be important to make sure you are using the correct angles to do your calculations