## Objectives

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

Explain what a ray diagram is

Draw an accurate ray diagram for a convex lens

Draw an accurate diagram for a concave lens

Draw an accurate ray diagram for a convex mirror

Draw an accurate ray diagram for a concave mirror

## Introduction

By this point in the textbook you might have become a bona fide maths wizard, but there’s also a chance that you might be sick of looking at equations, so let’s take a short break to learn how to understand image formation using line drawings.

Yes, you heard me correctly, *line drawings* .

In the optics business, we refer to this as **ray tracing** because we can use scale line drawings to trace the path of the rays from the object to the image, and my best advice is to read this chapter with a pad of paper, a pencil and a ruler so that you can learn to apply the principles as you go.

## What is a ray diagram?

A **ray diagram** is a carefully measured line drawing that traces the path light takes from an object to its image formed by an optical system. These diagrams can be produced successfully for both thin lenses and spherical mirrors, but in all cases, we are assuming that we are using **paraxial rays** (rays that lie close to the optical axis).

## Ray tracing – single thin lens

Remember that with a thin lens, the effect of the refractive index of the material is considered so negligible that it is ignored completely. This means that with a thin lens in air, we can assume that the primary refractive index (n) and the secondary refractive index (n′) are identical (both = 1.00). If we think back to the equations from chapter 3 , we know that the focal length of a thin lens is determined by the power (F) of the lens and the refractive index in which it resides. To that end, given that the refractive index is identical on either side of the lens, we can assume that the focal length would be the same distance away from the lens in either direction. Now, for our equations we always assume that light is travelling from left to right, meaning that for a positively powered lens that *converges* light, the **secondary focal length** (f′) will exist to the right of the lens to form the **secondary focal point** (F′). However, for ray tracing, we can start to consider what happens if the light were to change direction and travel from right to left. Theoretically, because we’re assuming a thin lens, the light should behave identically in either direction, so if made to travel backwards through the system, it will produce a focus at the **primary focal point** (F). The distance between the lens and the primary focal point is called **the primary focal length** (f) and should be equivalent to the negative of the secondary focal length ( Equation 7.1 and Fig. 7.1 ). For example, a +5.00 D lens will have a secondary focal length of +0.2 m (to the right of the lens) and a primary focal length of −0.2 m (to the left of the lens).

For a negatively powered lens that *diverges* light, light originating from infinity will diverge after refracting through the lens, which means we have to draw the rays back to see where they appear to originate from. As Fig. 7.2 A shows, this produces a negative **secondary focal length** (f′) to the left of the lens to form the **secondary focal point** (F′), which is the exact opposite to that of a positive lens. If the light then travelled backwards through the system ( Fig. 7.2 B), it would diverge on the opposite side of the lens, meaning the rays would appear to originate from the **primary focal point** (F) on the right of the lens. The distance between the lens and the primary focal point is called **the primary focal length** (f) and, again, should be equivalent to the negative of the secondary focal length. For example, a −5.00 D lens will have a secondary focal length of −0.2 m (to the left of the lens) and a primary focal length of +0.2 m (to the right of the lens).

In these diagrams, you’ll also probably notice that we’ve added points referred to as 2f and 2f′. These are important points for ray diagrams and correspond to twice the secondary focal length (2f′) and twice the primary focal length (2f). However, as a word of warning, when we’re learning this content it can be very easy to mix up 2f with f′ as the description “twice the focal length” for 2f sounds like a similar idea to the “secondary focal length” (f′) as they both relate to the number 2. My best advice here is to make sure you understand what they each relate to, which may involve going back over chapter 3 for some revision on secondary focal lengths if necessary.

## Drawing lenses

One of the key aspects of a ray diagram is that the person who views it should know whether it’s a positively powered or negatively powered lens. This can be achieved through looking at the behaviour of the rays, but the easiest way to do this is to draw the lens in line with the set specification. The guidelines state that all lenses are drawn as straight, vertical lines, but positively powered lenses are drawn with outward facing arrows (as in Fig. 7.1 ) whilst negatively powered lenses are drawn with inward facing arrows (as in Fig. 7.2 ).

## Drawing objects and images

In ray diagrams, we will learn to draw our objects as upward-pointing arrows, the base of which starts at the optical axis – the idea is that it helps to make the diagram easy to interpret, but I do agree that it’s not very imaginative. Then, once we have our arrow-shaped object, we need to think about how we’ll draw the light rays. Technically, an object is an extended source with light rays emanating from every part of it ( Fig. 7.3 A), but it is simpler (and quicker) to limit our drawings to the light emanating from the **tip of the object** ( Fig. 7.3 B). This helps us to use ray tracing to determine the location of the **tip of the image** after refraction or reflection, which allows us to assume the distance of the tip of the image relative to the optical axis (which gives us the image height).

When we come to draw our image, this will also be depicted as an arrow with the base starting at the optical axis, but the image will be upright or inverted, depending on the type of lens and where the object is positioned.

## Rules for a positively powered lens

Now we’re ready to learn how to draw the rays in our ray diagrams. For positive- and negative-powered lenses there are three rays that we need to learn, but they’re drawn slightly differently for each lens type, so I’ll start with positively powered lenses in this section and then move onto negatively powered lenses in the next section.

The rays are outlined below and demonstrated in Fig. 7.4 :

**P-ray**– drawn from the tip of the object, parallel to the optical axis; refracts through F′

**C-ray**– drawn from the tip of the object, undeviated through the optical centre (intersection of lens and optical axis)

**F-ray**– drawn from the tip of the object, through F; refracts parallel to the optical axis

In the example in Fig. 7.4 you can see that the rays all come to a neat little intersection on the right side of the lens, which is the point corresponding to the tip of the image! Importantly, light traveling along the normal (optical axis) through the optical centre will just travel in a straight line, so we know the base of the object and the base of the image both form along the optical axis. We can draw our image by connecting the intersection of the rays to the optical axis (with a very straight, vertical line). In this example, the diagram shows that with this lens and object distance, the image forms slightly to the right of 2F′, and it’s inverted (upside-down) and slightly magnified.

Now, if drawn to scale (with the correct focal length and object distance, and even object height when specified), this diagram can tell you the *exact* image distance and *exact* image height for the height of the object. (You can try this for yourself at home with a ruler and a pencil.)

We can also do a quick mathematical confidence check to prove that our ray tracing makes sense. For this, we can just use sensible estimates; for example, if we assume the lens in Fig. 7.4 is +5.00D (a power that for this example has been chosen at random), then our secondary focal length (f′) would be +0.20 m and our primary focal length (f) would be −0.20 m (see chapter 3 for revision on this if needed). This means our 2F point on the left corresponds to −0.40 m (twice the primary focal length). The object itself looks to be slightly closer to the lens than 2F, so let’s estimate its distance (l) as −0.375 m. Using our vergence equations from chapter 2 , we can then determine that the object vergence (L) for our estimated object distance would be −2.67, which would make our image vergence (L′) +2.33. This means mathematically, the image (l′) *should* form +0.429 m right of the lens, which would correspond to slightly right of 2F′. . . Which looks pretty spot-on in Fig. 7.4 to me!

For positive lenses, the position of the image relative to the object can be determined by learning the contents of Table 7.1 , but the highlights are that:

- •

An object at infinity will form an image at F′

- •

An object at F will produce an image at infinity

- •

An object at 2F will produce an inverted image the same size at 2F′

- •

An object between F and the lens will produce an upright image left of the lens

Object Distance (l) | Image Location | Image Distance (l′) | Inverted/Upright | Image Size |
---|---|---|---|---|

>2F | Right of lens | >F′ & <2F′ | I | <O |

2F | Right of lens | 2F′ | I | = O |

>F & <2F | Right of lens | >2F′ | I | >O |

infinity | Right of lens | F′ | I | |

F | ∼ | Infinity | ∼ | |

<F | Left of lens | >l | U | >O |