## Objectives

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

Define ‘thin lens’ and apply the vergence equations from chapter 2 to thin lenses

Determine linear magnification of an image

Understand the difference between back vertex power, front vertex power, and equivalent power

Confidently apply step-along vergence equations and Newton’s Formulae to determine image formation with multiple lens systems

## Introduction

In chapter 2 we covered the basics of vergence and refraction at plane and spherical curved surfaces, but within clinical optics we also need to be able to understand how light interacts with lenses.

This chapter will seek to cover the fundamentals of thin lenses for calculating image location, power, vergence and focal lengths of the system.

## What is a ‘thin lens’?

Lenses, by definition, are refractive devices that possess two surfaces (one on the front and one at the back) ( Fig. 3.1 A) and are available in several **forms** ( Fig. 3.1 B). In the last chapter we discussed at length that the curvature of a surface and the refractive index of a material can have an impact on the refractive power of a material. However, if the thickness is slight enough (relative to the radius of curvature of each surface), then it’s assumed that the refractive index of the lens material has a negligible effect on the power, so it can be completely ignored. These types of lenses, in which we ignore the refractive index, are called **thin lenses** .

In optical diagrams, thin lenses are drawn as vertical lines with arrowheads, where positive (converging/ convex) lenses are simplified to be outward pointing arrows, and negative (diverging/ concave) lenses are simplified to be inward pointing arrows ( Fig. 3.2 ). We will cover this in more detail in chapter 7 , but it is good to be aware of this now as lenses will commonly be depicted as arrows throughout this chapter.

## Power of a thin lens

The refractive **power** (F) of a thin lens can be determined by either knowing the power or curvature of each surface, or by knowing the focal length (f′) of the lens. Remember, power is measured in **dioptres** .

## Power of a single thin lens

With a single thin lens, the approximate overall power can be calculated by adding the front (F _{1} ) and back surface powers (F _{2} ) together ( Equation 3.1 ), for example, in Fig. 3.3 , the front and back surface powers add together to suggest the lens depicted has an approximate power of −3.50D.

F=F1+F2

power=front surface+back surface

A biconvex thin lens has a front surface power of +5.00D and a back surface power of +4.25D. What is the overall power of this lens?

Step 1: Determine what we need to calculate

power, F

Step 2: Define variables

F

_{1 }= +5.00

F

_{2 }= +4.25

Step 3: Determine necessary equation

F = F

_{1 }+ F_{2 }*(**Equation 3.1**)*

Step 4: Calculate

F = F

_{1 }+ F_{2 }

F = +5.00 + +4.25

F = +9.25D

(don’t forget the ± sign and the units!)

## Practice questions:

**3.1.1**A biconcave thin lens has a front surface power of −4.00D and a back surface power of −8.00D. What is the overall power of this lens?

**3.1.2**A plus meniscus thin lens has a front surface power of +6.50D and a back surface power of −2.00D. What is the overall power of this lens?

## Focal length

In chapter 2 we learned the basics of a surface’s **focal length** by discussing that the **secondary focal length (f′)** describes the distance between the surface and the **secondary focal point (F′)** . We also learned that convex spherical surfaces will have a positive focal length whilst concave surfaces will have a negative focal length. For thin lenses, this principle is the same, with positively powered (convex) lenses converging light towards the optical axis, resulting in a secondary focal point on the right-hand side of the lens, and therefore a positive secondary focal length. In contrast, negatively powered (concave) lenses diverge light away from the optical axis and therefore possess a secondary focal point on the left of the lens, leading to a negative secondary focal length. However, we now also need to start to consider what happens if light travels backwards through the system, from right to left.

With a thin lens, we assume that the lens will do the same to light travelling from either direction, so the focal length will be identical on either side. The only difference lies in the nomenclature, as when light travels backwards through the system it produces a **primary focal point (F)** at a distance corresponding to the **primary focal length (f)** . This distinction will be very important for chapter 7 , but for now, please see Fig. 3.4 for an illustration of this in action.

We can use the secondary focal length (f′) and the **secondary refractive index** (of the surrounding material) (n′) to calculate the power of the lens using Equation 3.2 . Assuming that refractive indices of materials will always be ≥1, then if you’re a fan of maths, this equation should demonstrate that positively powered lenses will have a positive focal length (and vice versa), and negatively powered lenses will have a negative focal length (and vice versa).

F=n’f’

power=surrounding refr.indexsecondary focal length

A biconvex thin lens has a focal length of 16 cm. What is its power?

Step 1: Determine what we need to calculate

power, F

Step 2: Define variables

n′ = 1.00

*(nothing is specifically mentioned so we assume the surrounding medium is air)*

f′ = +0.16

*(the question stated the lens is biconvex, making it positively powered – this means the focal length will be to the right of the lens, making it positive. We also need to convert to metres)*

Step 3: Determine necessary equation

F = n′ / f′

*(**Equation 3.2**)*

Step 4: Calculate

F = n′ / f′

F = 1.00 / +0.16

F = +6.25D

(don’t forget the ± sign and the units!)

## Practice questions:

**3.2.1**A biconvex thin lens has a focal length of 30 cm. What is its power?

**3.2.2**A biconcave thin lens has a focal length of 11 cm. What is its power?

**3.2.3**A biconvex thin lens has a focal length of 25 cm. What is its power in water (refractive index 1.333)?

**3.2.4**A biconcave thin lens has a power of −4.00D. What is its focal length?

## Vergence relating to a thin lens

For thin lenses in a homogenous material – e.g. the entire lens is in air – you can use the same vergence equations we learned in chapter 2 to calculate where the image will form (for revision, see Equations 2.1 , 2.5 , and 2.6 ), but we can also use equations to determine how big (or small) the image will be relative to the object. This is called **linear magnification** (m).

## Linear magnification

The term ‘linear magnification’ is a measure of how big (or small) the image is relative to the *original* object. It’s important to note here that because we are using **paraxial rays** (which are, by definition, extremely close to the optical axis), the angle of incidence (i) and the angle of refraction (i′) are extremely small and hardly different from one another. Ultimately this helps us to use substitute tan into Snell’s Law and use trigonometry to determine the linear magnification of the image, but we don’t need to worry about how that works providing we can remember how to apply the equation to calculate linear magnification (m) ( Equation 3.3 ). It is also important here to understand that although magnification is related to power, it is not the same. For example, an object 10cm in front of a +5.00D lens will produce an image with different magnification to an object 12cm in front of the same lens. In that way, magnification is related to both the power of the lens and the distance of the object. A top tip here is to make a note to yourself to ensure you understand the difference between power of a lens and magnification of an image.

m=h’h=l’l=LL’

mag.=image heightobject height=image dist.object dist.=object verg.image verg.

The best feature of a magnification calculation is that the answer can tell you an awful lot about the characteristics of the image. To be specific, the numerical value indicates the size of the image relative to the object, and the sign (+/−) will tell you if the image is upright or inverted. For example, if the magnification is below zero (negative), the image is inverted (classified as a **real** image), whereas if it is above zero (positive), it is upright (classified as a **virtual** image). Similarly, if the magnification is between −1 and +1 (excluding zero), then it is **minified** (smaller than the object), and if it is below −1 or above +1, then it is **magnified** (bigger than the object). For example, a magnification of −0.45 would be inverted and minified, whilst a magnification of +2.30 would be upright and magnified. Importantly, if the magnification equates to a value of +1 or -1, then it is the same size as the object, and the sign will tell you if it is inverted or upright. See Fig. 3.5 for a handy way to remember this. For a real world example of magnification in action, if you know someone who wears glasses then you may have noticed that their eyes appear to be larger (magnified) or smaller (minified) when seen through their lenses. This is an example of the light from the object (in this example, the person’s eyes) being refracted to produce an image a different size (the image is the view of the person’s eyes that you see when you look at them). Hopefully though their eyes will always still be upright, indicating a magnification value above 0…

## Image distance

Utilising equations from chapter 2 and from the previous linear magnification section, we can calculate where an image forms after the light from the object refracts through a thin lens. Remember, with thin lenses we don’t consider the refractive index of the lens at all, and so the refractive index will usually be identical on each side (n≡n′), for example if the lens is “in air” then the refractive index would be 1.00 on both sides ( Fig. 3.6 ).

An object is placed 20 cm in front of a biconvex thin lens of power +4.50D.

- (a)

Where does the image form?

- (b)

What is the magnification of the image?

Step 1: Determine what we need to calculate

image distance, l′

magnification, m

Step 2: Define variables

l = −0.20m

*(negative distance and we need to convert to metres)*

F = +4.50D

n = 1.00

*(nothing is specifically mentioned, so we assume the primary medium is air)*

n′ = 1.00

*(nothing is specifically mentioned, so we assume the secondary medium is air)*

Step 3: Determine necessary equation

L = n / l

*(**Equation 2.1**)*

L′ = L + F

*(**Equation 2.5**)*

L′ = n′ / l′

*(**Equation 2.6**)*

m = (h′ / h) = (l′ / l) = (L / L′)

*(**Equation 3.3**)*

Step 4: Calculate

L = n / l

L = 1.00 / −0.20

L = −5.00

L′ = L + F

L′ = −5.00 + +4.50

L′ = −0.50

l′ = n′ / L′

*(rearranged)*

l′ = 1.00 / −0.50

l′ = −2.00m

- (a)

The image forms 200.00 cm left of the lens.

m = (h′ / h) = (l′ / l) = (L / L′)

m = L / L′

*(choose which part we want to use)*

m = −5 / −0.50

- (b)

m = +10.00X

(don’t forget the ± sign!)

## Practice questions:

**3.3.1**An object is placed 50 cm in front of a biconvex thin lens with a power of +10.00D. Where does the image form?

**3.3.2**An object is placed 35 cm in front of a biconcave thin lens with a power of −12.00D. What is the magnification of the image?

**3.3.3**An object is placed 15 cm in front of a biconcave thin lens with a focal length of 20 cm. Where does the image form?

## Multiple thin lenses

Up to this point we have considered thin lenses in isolation, but sometimes optical systems have more than one lens present, for example, in telescopes. This means we also need to think about systems with more than one thin lens, and how to calculate power, vergence and magnification for these systems.

## Multiple thin lenses in contact

The simplest form of system containing more than one thin lens is when all the lenses are **in contact with one another** . When positioned like this to have negligible distance between each lens, just like how we can ignore the refractive index, we can pretend the distance doesn’t exist! To that end, we can simply add up the powers of each of the lenses in contact in order to work out the overall power of the system ( Fig. 3.7 ), and then we can calculate vergence using the same equations as in the section titled “Vergence Relating to a Thin Lens”. This time, however, we will refer to the calculated power as the **equivalent power** (F _{e} ) as we are working out the equivalent *single* power of the multiple lenses, as shown in Equation 3.4 .

Fe=F1+F2

equivalent power=lens1power+lens2power

Two thin lenses of powers +3.00D and +2.00D are in contact with each other. If an object is placed 40 cm in front of the first lens, where does the image form?

Step 1: Determine what we need to calculate

image distance, l′

Step 2: Define variables

l = −0.40m

*(negative distance and we need to convert to metres)*

F

_{1 }= +3.00D

F

_{2 }= +2.00D

n = 1.00

*(nothing is specifically mentioned, so we assume the primary medium is air)*

n′ = 1.00

*(nothing is specifically mentioned, so we assume the secondary medium is air)*

Step 3: Determine necessary equation

F

_{e }= F_{1 }+ F_{2 }*(**Equation 3.4**)*

L = n / l

*(**Equation 2.1**)*

L′ = L + F

*(**Equation 2.5**)*

L′ = n′ / l′

*(**Equation 2.6**)*

Step 4: Calculate

F

_{e }= F_{1 }+ F_{2 }

F

_{e }= +3.00 + +2.00

F

_{e }= +5.00D

L = n / l

L = 1.00 / −0.40

L = −2.50

L′ = L + F

L′ = −2.50 + +5.00

L′ = +2.50

l′ = n′/ L′

*(rearranged)*

l′ = 1.00 / +2.50

l′ = +0.40m

The image forms 40.00 cm right of the second lens.

(don’t forget the ± sign (or the direction) and the units!)

## Practice questions:

**3.4.1**Two thin lenses of powers +1.00D and +5.50D are in contact with each other. If an object is placed 10 cm in front of the first lens, where does the image form?

**3.4.2**Two thin lenses of powers +6.50D and −4.25D are in contact with each other. If an object is placed 40 cm in front of the first lens, what is the linear magnification?

**3.4.3**Three thin lenses of powers +1.00D, −3.50D and +6.75D are in contact with each other. If an object is placed 25 cm in front of the first lens, where does the image form?

## Multiple thin lenses separated by a distance

If the thin lenses are separated by a distance, then we need to take that distance into consideration when performing our calculations. This is because vergence depends on the distance from the surface, so increasing the distance (d) of the second lens will have an impact on the sampled vergence of the light as it reaches the second lens ( Fig. 3.8 ).

Lens systems of this nature still have secondary and primary focal points, but they will no longer be equidistant on either side, as the powers of the individual lenses will likely be different, so light will travel differently in each direction. Instead, we need to consider what’s called the back and front vertex power.

## Vertex power

Vertex power refers to the vergence of light as it leaves the system, assuming that the incident light is parallel (zero vergence), leading to light focusing at one of the focal points. If the light travels forwards through the system from left to right, the vergence of light leaving the second lens will be called the **back vertex power** (F _{v} ′), sometimes referred to as emergent vergence at F _{2} . However, if the light travels backwards, from right to left, then the vergence of light leaving the first lens will be called the **front vertex power** (F _{v} ), or sometimes referred to as emergent vergence at F _{1} . Now, because the incident (approaching) light is parallel, the light will form a focus at the secondary (F′) or primary focal point (F), depending on the direction. The distance between the second lens (F _{2} ) and the secondary focal point (F′) is called the **back vertex focal length** (f _{v} ′), and the distance between the first lens (F _{1} ) and the primary focal point (F) is called the **front vertex focal length** (f _{v} ). Fig. 3.9 depicts these distances in a multiple lens system.

To calculate vertex power, we need to take the distance between the lenses (d) into consideration. See Equations 3.5 and 3.6 for back and front vertex calculations, but be careful because these equations are extremely similar and easy to confuse for one another. My top tip for successful application of these equations is to try to understand what the calculations are doing. For example, if we’re determining the emergent vergence at F _{2} after light has entered the system through F _{1} (back vertex power; F _{v} ′ ), then we need to know how the vergence from F _{1} will have changed across the distance (d) between the lenses. As the incident light is assumed to be parallel, the power of the lens at F _{1} will be identical to the vergence leaving it, and so the equation for back vertex power takes into account the power of F _{1} in the denominator (and the opposite is true for front vertex power; F _{v} ).

Fv’=F1+F2−dF1F21−dF1