Lesson Explainer: Deviation due to a Prism Physics • Second Year of Secondary School

In this explainer, we will learn how to determine the effect of the shape and refractive index of a prism on the path taken through it by light rays.

When a ray of light passes through a prism, it is refracted twice: once as it enters the prism, and again as it exits. These refraction events cause it to deviate from its original path. Through some careful geometric analysis, we will see how to derive an equation that enables us to calculate the angle by which a ray of light deviates as it passes through a prism. We will see what this angle depends on, and we will also see how we can make it as small as possible.

Let’s start by giving this angle a name. We will call it .

In this diagram, the lines of the incident ray and the emergent ray have been extended with dashed lines so that we can mark the angle between them.

Let’s take a closer look at what happens when a light ray passes through a glass prism. We will start by defining the apex angle at the top of the prism.

We will also say that the prism is surrounded by air and that it has a refractive index greater than that of air. Recall that , so we are saying that . This is generally the case for glass prisms.

When a ray of light enters the prism, the change in refractive index means that, rather than passing straight through, the light will be bent, or refracted. We can describe the refraction of the light ray in terms of a line normal to (i.e., perpendicular to) the surface of the prism.

Specifically, because the refractive index increases as a ray of light enters the prism, it will be refracted toward the normal line.

However, as the light exits the prism, the refractive index decreases again. This means that it refracts away from the normal at the point where it exits.

We can see that as a ray of light passes through the prism, its total deviation is due to two refraction events: once as it enters, and once as it leaves. The total angle of deviation, , is the angle between the path the ray was following before it entered the prism and the path the ray follows as it exits the prism. Our aim in the first part of this explainer is to derive an expression for in terms of the apex angle and a few other angles involved as the ray moves through the prism.

To start with, there are four angles we need to define:

• : the angle of incidence as the ray enters the prism,
• : the angle of refraction as the ray enters the prism,
• : the angle of incidence as the ray exits the prism,
• : the angle of refraction as the ray exits the prism.

According to our convention, refers to angles of incidence and refers to angles of refraction. Subscript 1 refers to the point where the light enters the prism, and subscript 2 refers to the point where the light exits the prism.

Remember that angles of incidence and angles of refraction are both measured relative to the normal to a surface.

There is a lot going on here, but we can see that the lines we have drawn have created a triangle in the middle of the prism: this consists of a solid pink line and two dashed pink lines. We are going to focus on this triangle for a while. Let’s take a closer look.

Consideration of the interior angles of the triangle formed here gives us a way of expressing the angle alpha in terms of the other angles we have defined. To start with, we can notice that the angle and the interior angle at the top of the triangle form a straight line. This means that the top interior angle of the triangle must be equal to .

Now, let’s look at the interior angle on the left of this triangle. We can come up with an expression for this by reintroducing the angles of incidence and refraction at the point where the light ray enters the prism ( and ).

is the angle between the incident ray and the normal. We can label this angle on the inside of the prism as well.

Doing this, we can see that the internal angle on the left of the triangle is equal to the difference between and (i.e., it is equal to ). Now that we have found an expression for this angle, we can turn our attention to the last internal angle of this triangle, located on the right. We will reintroduce the angles of incidence and refraction at the point where the ray exits the prism ( and ).

is the angle between the refracted ray and the normal at the point where the ray leaves the prism. Again, we can mark this angle inside the prism.

Now, we can see that the remaining interior angle of the triangle is equal to .

Now that we have come up with expressions for these angles, we are able to link them together by noting that the internal angles of a triangle always add up to . This means that

We can simplify this equation by subtracting from each side and then rearranging to make the subject:

This is an important result, as it relates to the other angles formed as the light ray travels through the prism. However, recall that we want to express in terms of the geometry of the prism itself—specifically, the apex angle . To do this, we need to consider another shape.

Let’s direct our attention to the quadrilateral (4-sided shape) formed by the normal lines and parts of the prism’s faces.

Once again, we are going to consider the interior angles of this shape. The top angle is the apex angle of the prism, . We also know that the interior angles on the left and right are both , since these angles are formed by the faces of the prism and the normal lines to these surfaces. This leaves us with one unknown angle: the interior angle at the bottom of the quadrilateral. Let’s call this angle .

We can establish a relationship between these angles by recalling that the interior angles of a quadrilateral always add up to . So,

Subtracting from each side, we have

Remember that our goal here is to express the angle of deviation in terms of the apex angle . In order to do this, we need to be able to express in terms of some of the same angles of incidence and refraction (i.e., , , , and/or ) that we have expressed in terms of previously.

We can achieve this by expressing in terms of the angles of incidence and refraction. To do this, we can look at how this quadrilateral is split into two triangles when we reintroduce the ray of light inside the prism.

If we consider the interior angles in the bottom triangle of the two, we can see that we have already given names to the two interior angles other than . On the left, we have the angle of refraction as the light ray enters the prism, , and on the right, we have the angle of incidence as the ray exits the prism, .

Again, we can use the fact that the interior angles of a triangle must add up to to write an equation:

Making the subject, we obtain an expression for that we can substitute into the equation that we found previously.

Making this substitution,

At this point, we have nearly reached our goal. We just need to substitute this expression for the apex angle into our original expression for the angle of deviation :

To make this substitution, we first regroup the terms like this:

Then, we take out a factor of from the last two terms:

We can now see that the terms inside the parentheses match our expression for the apex angle . This means that we can substitute in their place:

We have now accomplished our goal of expressing the angle of deviation in terms of the apex angle , along with the angle of incidence at which the ray enters the prism () and the angle of deviation at which the light ray leaves the prism ()!

Example 1: Identifying Angles Formed When a Ray of Light Passes through a Glass Prism

The diagram shows the path of a light ray through a triangular prism. The apex angle of the prism . What is the angle ? Answer to the nearest degree.

Answer

To answer this question, we will use some of the same geometric reasoning that we used to derive our equation for the angle of deviation of a prism. In this question, the apex angle of the prism is called , and we also have another angle . This angle is formed by the intersection of the normal lines with the prism’s surface at the points where the light ray enters and exits the prism. These two normal lines are shown as dashed lines in the diagram.

To answer this question, we can first recognize that these dashed lines, being normal lines, are perpendicular to the prism’s surfaces. That means we can mark some angles in the diagram.

In our diagram, we can see a quadrilateral formed by the normal lines and the top of the prism.

We can answer the question by recalling that the internal angles of any quadrilateral sum to . This enables us to write down an equation involving the internal angles of the quadrilateral shown in yellow:

We can simplify and rearrange the equation to make the subject:

All we need to do now is substitute , which was given to us in the question:

Next, let’s look at how we can derive an expression for the minimum angle of deviation () caused by a prism. The expression we derived in the previous section shows us that depends on , , and :

The apex angle of a given prism is fixed. This means that is a constant in this equation. Additionally, for a given prism with refractive index , the final angle of refraction depends entirely on the initial angle of incidence . This means that, for a given prism, the only variable on which depends is the initial angle of incidence .

In other words, we can see the full range of possible values of just by varying the angle of incidence at which the ray strikes the prism. We could imagine varying this angle of incidence and plotting the resulting angle of deviation on a graph. Doing so results in a curve like this.

We can see that the angle of deviation reaches a minimum value at some specific value of . We can call this minimum value of the angle of deviation , and we can say that the angle of incidence that causes this minimum deviation is called .

One characteristic of a light ray undergoing the minimum amount of deviation as it passes through the prism is that the ray will be parallel to the base of the prism as it passes through the prism.

Let’s now take a look at the initial angle of incidence and the final angle of refraction. Previously, we called these angles and , but here we will call them and to signify that they correspond to the minimum deviation of the light ray.

Since the prism is symmetrical and the ray inside the prism is parallel to the base of the prism, it turns out that and are equal:

We can now write down a special version of the equation we found before:

This version applies specifically to the situation where the light ray undergoes minimum deviation.

For this specific case, we will substitute , , and into the above equation. This gives us

Recognizing that and are equal, we can replace with :

Rearranging to make the subject, we have

Example 2: Identifying Angles of Incidence and Refraction in the Case of Minimum Deviation of a Light Ray through a Prism

The graph shows how the angle of deviation of the triangular prism shown in the diagram varies with the angle of incidence of light rays on it.

1. Where , the angle of deviation, has its minimum value, which of the angles in the diagram is equal to the angle ?
2. Where , the angle of deviation, has its minimum value, which of the angles in the diagram is equal to the angle ?

Answer

Part 1

In this question, we are asked to consider the specific case of minimum deviation (i.e., where has its minimum value). In order to answer it, we first need to recall that when a light ray undergoes minimum deviation through a prism, it will be parallel to the base of the prism as it passes through.

In fact, this means that the path of the ray is symmetrical, with a line of symmetry down the middle of the prism. Once we have recognized this, answering the question becomes straightforward.

For this first part of the question, we want to find the angle equal to , which we can see is the angle of incidence when the ray enters the prism. By recognizing the symmetry of our diagram for the case of minimum deviation, we can see that this angle must be the same as the angle of refraction where the ray leaves the prism, .

Part 2

Once again, we can answer this question by recognizing that the prism and light ray are symmetrical for the case of minimum deviation. Noticing this, we can see that must be equal to the corresponding angle on the right of the triangle. This is the angle of refraction at the point where the light ray exits the prism, labeled .

Another useful equation expresses the refractive index of the prism, , in terms of the minimum angle of deviation and the apex angle . The refractive index describes exactly how much a ray is diffracted when it enters or leaves a medium, so it is possible to derive this equation using Snell’s law, but we will not worry about this derivation here. Here is the equation.

In some cases, we can use a simplified version of this equation. Specifically, if the apex angle is small (i.e., if the prism is thin), we can use the following equation.

Let’s look at some examples of problems that can be solved using these two equations.