After working through this chapter, you should be able to:

  • Describe what a prism is

  • Explain what the minimum angle of deviation is

  • Define a ‘critical angle’

  • Discuss how light deviates through a prism (relative to the base and apex)

  • Calculate prismatic power


In the previous chapter, we introduced the topic of prisms and discussed that they are capable of dispersing white light. In this chapter, we will discuss prisms in more detail and learn all about their power, minimum angle of deviation, total internal reflection and critical angle.

What is a prism?

When I read the word ‘prism’, I think of a glass, triangular-shaped object with five sides like the one in Fig. 9.1 A. Technically my instinct is correct, and this is a prism, but in optics the true definition of the word ‘prism’ encompasses any object that possesses two flat refracting surfaces inclined at an angle towards each other. This means that all of the glass objects illustrated in Fig. 9.1 B are also technically prisms, but this chapter is going to focus on the triangularly shaped prisms.

• Fig. 9.1

Drawing of ‘classic’ triangular prisms (A) and examples of other types of prism (B).

Deviation of light

Now, if we take a cross section of a triangular-shaped prism, we can imagine that it would be made of a material other than air – suggesting it will have a different refractive index (n p ) relative to the surrounding air (n s ) ( Fig. 9.2 ). Importantly, prisms also possess an apical angle (a), which is defined as the angle that exists between the first and second refracting surface (where the light enters and leaves the prism). This is demonstrated in Fig. 9.2 with a monochromatic (single wavelength) beam of light (as this prevents dispersion, see chapter 8 for revision on this if necessary).

• Fig. 9.2

Illustration showing refractive index outside the prism (n s ) will differ to the refractive index inside the prism (n p ). If a monochromatic (single wavelength) beam of light is incident upon the surface, the light will refract at both the incident first surface and the emergent second surface, producing two angles of deviation (d 1 and d 2 ). The overall angle of deviation (d) is the difference between the original path of the light relative to the deviated path of light.

To start with, let’s consider the deviation of the light path. If you look at the solid lines in Fig. 9.2 , these correspond to the path that the light takes as it travels through the prism. An important feature of prismatic refraction (refraction through a prism) is that light will always deviate towards the base , shown by the emergent light ray being angled downwards in this example. For clarity, the ‘base’ of the prism is the side opposite to the apical angle, so in Fig. 9.2 it is the horizontal line along the bottom of the prism. Now, if you look instead at the dashed lines in Fig. 9.2 , you can see that they correspond to the original path the light would have taken if it hadn’t been deviated by this pesky prism being in the way. To this end, the angle between the original path of light and the true, deviated path of light is called the angle of deviation (d) . It is also crucial to note that this angle of deviation (d) is the sum of the angles of deviation at each surface of the prism (d 1 and d 2 ) – see Equation 9.1 for an example of this.


dev.=dev.at1stsurface+dev.at2nd surface

But, how do we calculate these angles of deviation in the first place? The answer is maths (as always) and cyclic quadrilaterals . Now, if you haven’t heard of cyclic quadrilaterals before, don’t worry, you can utilise the information in Box 9.1 to help you learn what they are.

• BOX 9.1

Opposite Angles, Cyclic Quadrilaterals and Prisms

If we think back to chapter 2 , we remember that angles of incidence (the angle at which rays of light approach a refractive surface) are measured relative to the ‘normal’ of the surface – a hypothetical line that exists at a perpendicular angle to the surface itself (at the point where light intersects the surface). We also learnt all about opposite angles (see Box 2.4 for revision on this), and with understanding deviation in prisms, opposite angles hold the key. For example, in Fig. B9.1 , we can see that we can apply the knowledge that vertically opposite angles are equal in order to calculate the deviation at the first surface (d 1 ), showing that:

  • d 1 = i 1 −i 1

• Fig. B9.1

Illustration showing how principle of opposite angles applies to the first surface of the prism. Image highlights that i 1 = d 1 + i 1 ′ and therefore d 1 = i 1 – i 1 ′.

The same is true for the second angle of deviation (d 2 ) at the second face of the prism ( Fig. B9.2 ) showing that:

  • d 2 = i 2 ′−i 2

• Fig. B9.2

Illustration showing how principle of opposite angles applies to the second surface of the prism. Image highlights that i 2 ′ = d 2 + i 2 and therefore d 2 = i 2 ′ – i 2 .

And as Equation 9.1 taught us, we know that total deviation is equal to d 1 + d 2 , so that means we can also say that:

  • d = (i 1 −i 1 ′) + (i 2 ′−i 2 )

Now, let’s take a moment to think about cyclic quadrilaterals . The definition of a cyclic quadrilateral is that it must have four sides, and all four vertices (corners) must connect to a hypothetical circle (see Fig. B9.3 B). Cyclic quadrilaterals also require all internal angles to add up to 360°, and opposite angles within the quadrilateral must add up to 180°. Now, you’re likely thinking ‘ Why is Sam telling us all this? ’, and the answer is because the ‘normal’ lines of each surface (at the point where light enters and leaves the prism) form a cyclic quadrilateral ( Fig. B9.3 A). This means that using our newfound knowledge of cyclic quadrilaterals, we know that the apical angle (a) and the angle formed by the two normal lines (b) must add up to 180°. However we also know that angles along a straight line equal 180°, which means that this hypothetical angle outside of the cyclic quadrilateral in Fig. B9.3 B would also be equal to that of the apical angle.

• Fig. B9.3

(A) Diagram showing how the two ‘normal’ lines (orange dashed lines) of the prism create a cyclic quadrilateral (blue). (B) Diagram showing properties of cyclic quadrilaterals to prove that the external angle (right) will be equal to the apical angle (a) of the prism. (C) Diagram showing that the apical angle (a) will be equal to the sum of the two internal angles formed by the light ray (i 1 ′ and i 2 ).

Another important mathematical fact about triangles is that external angles (along a straight line) will be equal to the sum of the opposite two interior angles (see Fig. B9.3 C). This is because the internal angles of a triangle all add up to 180°, and angles along a straight line add up to 180°, so if we consider Fig. B9.3 C for a moment, we know that:

  • 180° = b + i 1 ′ + i 2


  • 180° = b + a

Thereby proving that:

Equation B9.1:

a = i 1 ′ + i 2

Thus highlighting that we can rearrange our earlier equation to turn:

  • d = (i 1 − i 1 ′) + (i 2 ′ − i 2 )


The information provided in Box 9.1 highlights that we can edit Equation 9.1 to look like Equation 9.2 (depending on the information we have at the time):


dev.=(−apical ang.

This shows us that the angle of deviation (d) is directly related to the angle that the light enters the prism (i 1 ), the angle light leaves the prism (i 2 ′), and the apical angle (a) of the prism itself – neat.


A ray of light is incident upon a prism at an angle of 35° and leaves the prism at an angle of 67°. If the prism has an apical angle of 50°, what is the angle of deviation of the light?

Step 1: Determine what we need to calculate

  • angle of deviation, d

Step 2: Define variables

  • i 1 = 35° (angle of incidence at first surface)

  • i 2 ′ = 67° (angle of refraction at second surface)

  • a = 50° (apical angle)

Step 3: Determine necessary equation

Step 4: Calculate

  • d = (i 1 + i 2 ′) – a

  • d = (35 + 67) – 50

  • d = 52°

(don’t forget the units!)

Practice questions:

  • 9.1.1 A ray of light is incident upon a prism at an angle of 20° and leaves the prism at an angle of 40°. If the prism has an apical angle of 55°, what is the angle of deviation of the light?

  • 9.1.2 A ray of light is incident upon a prism at an angle of 21.2° and leaves the prism at an angle of 58.5°. If the prism has an apical angle of 30°, what is the angle of deviation of the light?

Minimum angle of deviation

By this point in the chapter, we’ve learned that prisms deviate light towards the base, and we know that the deviation of light partly relates to the angle of incidence (i 1 ) of the incoming light. This means that as the angle of incidence changes, the angle of deviation (d) will also change – but how are the two variables related? Well, thanks to Equation 9.2 , we can determine that if we know the apical angle of a specific prism, we can calculate the angle of deviation for lots of different angles of incidence. Don’t worry, though; I’ve done the maths for you, so all you need to do is enjoy looking at the data in Table 9.1 .


Data Showing That as Angle of Incidence at the First Face of a Prism (i 1 ) Increases, Angle of Refraction at the Second Face (i 2 ′) Decreases, but Total Angle of Deviation (d) Can’t Go Lower than 38.9° (See bold text)

i 1 (deg) i 2 ′ (deg) d (deg)
30 83.3 53.3
40 60.6 40.6
50 48.9 38.9
60 40.5 40.5
70 34.4 44.4
80 30.7 50.7

This table shows us that as angle of incidence at the first face of a prism (i 1 ) increases, angle of refraction at the second face (i 2 ′) decreases. However, it also shows us that the total angle of deviation will reach a point where it can’t go any lower. This point is called the minimum angle of deviation (d min ) of the prism, and as you might expect by now, it’s related to the apical angle (a) and the refractive indices of the prism (n p ) and the surroundings (n s ). We can calculate the minimum angle of deviation using Equation 9.3 :


Only gold members can continue reading. Log In or Register to continue

Stay updated, free articles. Join our Telegram channel

Feb 6, 2023 | Posted by in OPHTHALMOLOGY | Comments Off on Prisms

Full access? Get Clinical Tree

Get Clinical Tree app for offline access