## Objectives

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

Define what a ‘thick lens’ is, and calculate how to measure thickness

Understand the relationship between thickness, curvature of the surfaces, refractive index and how these contribute to the power of the lens

Calculate image position using the virtual object equations or the step-along vergence equations

Define what a ‘Fresnel lens’ is

## Introduction

Up to this point, we have considered lenses with such negligible thickness that it is possible to ignore the refractive index of the material and still reach reasonable estimates of image location and magnification, etc.; these are called thin lenses. However, it is more accurate to take both the thickness and the refractive index into account when performing these calculations. If the thickness is considered, the lens is referred to as a **thick lens** , and in these circumstances the individual lens surfaces need to be considered.

This chapter will cover some of the fundamentals of thick lens theory and calculations.

## Lens thickness

To measure the thickness of a lens, usually corresponding to the central point of the lens ( **centre thickness** ), it is first important to be able to determine the **sagitta (sag)** and the **edge thickness** . The sag is defined as the height of a segment of a circle (or sphere) from arc to base, whereas the edge thickness is the physical thickness at the edges of the lens. Fig. 4.1 shows a diagram to explain these concepts visually. It is important to know the thickness of the lens because this, in combination with the curvature and refractive index, helps to understand the power of the lens.

To start calculating the thickness, we need to calculate the sag (s) of a spherical lens, but in order to do this, we need to know the radius of curvature (r) of the surface and the value that equates to half the lens diameter (y). Fig. 4.2 shows these variables labelled on an example lens.

Sag can be calculated using Equation 4.1 .

s=r−(r2−y2)

sag=radius−(radius2−halflensdiameter2)

A 5-cm wide planoconvex lens has a radius of curvature of 15 cm. What is the sag of the lens?

Step 1: Determine what we need to calculate

sag, s

Step 2: Define variables

y = 0.025 m

*(good practice to convert to metres)*

r = 0.15

*(refractive index of lens)*

Step 3: Determine necessary equation

s = r – √(r

^{2 }– y^{2 })*(**Equation 4.1**)*

Step 4: Calculate

s = r – √(r

^{2 }– y^{2 })

s = 0.15 – √(0.15

^{2 }– 0.025^{2 })

s = 0.002098 m

s = 0.21 cm

(don’t forget the units!)

## Practice questions:

**4.1.1**A 7-cm wide planoconvex lens has a radius of curvature of 12 cm. What is the sag of the lens?

**4.1.2**A 4-cm wide planoconvex lens has a radius of curvature of 50 cm. What is the sag of the lens?

**4.1.3**A 5-cm wide planoconvex lens has a radius of curvature of 10 cm. What is the sag of the lens?

The centre thickness (t _{c} ) can only be calculated if you know the sag (s) and the edge thickness (t _{e} ), and the equation for this will change depending on the type of lens. As shown in Fig. 4.3 , the relationship between the front and back surface of the lens, and whether those surfaces are convex or concave, will affect the equation for centre thickness. Another important point to note here is that the sag is determined by the curvature of the surface, which helps to contribute to the power of the lens. That means that for high powered lenses, the thickness can increase so much that the lenses become heavy. In an optical imaging system this would be ok, but if you’re a person who needs glasses, heavy lenses are not ideal! Luckily, the thickness can be reduced (whilst maintaining the same power) by changing the material of the lens to something of a higher refractive index. Remember that higher refractive indices indicate that light is slowing more, leading to higher amounts of refraction. This ultimately means that higher index lenses won’t need to be as curved (leading to a smaller sag and smaller thickness) to produce the same power.

## Thick lens power

In general terms, any lens in which the thickness exceeds an amount that is acceptable for thin-lens assumptions is classed as a thick lens. However, in some cases, lenses that are physically very thin are classed as thick lenses if their front vertex and back vertex powers are substantially different, for example, contact lenses (lenses that sit on the frontsurface of the eye).

The key thing to note for thick lenses is that you need to take into account the refractive index of the material, so it is a little like having two thin lenses separated by a distance, with a refractive index change in the middle. Importantly, with thick lenses, we can’t assume that incident light rays will only refract once as they pass through; instead, we think of them as refracting at each individual surface of the lens. Fig. 4.4 depicts an example thick lens to show that the light will refract twice as it passes through the lens; once at the front surface, and again at the back surface.

To determine the overall power of a thick lens, we need to calculate the power of the front (F _{1} ) and back (F _{2} ) surfaces individually, and we need to know how thick (t) the lens is. To start with then, to calculate the power of each surface, we can use a familiar calculation from chapter 2 which has been updated in Equation 4.2 . This will involve using what we know about the radius of curvature of each surface (r _{1} and r _{2} ) and the refractive index of the lens and surrounding material (n) in order to calculate the respective powers. However, we need to be careful to remember that we always assume light travels from left to right, meaning that at the first surface (F _{1} ), the primary refractive index (n _{1} ) will be air, and the secondary refractive index (n _{1} ′) will be the lens; however at the back surface (F _{2} ) this will be swapped as the light will travel from the lens (n _{2} ) to the air (n _{2} ′). Fig. 4.5 explains the relationship between the radius of curvature values and the refractive index values.

F1=n1’−n1r1F2=n2’−n2r2

power=secondary refr.index−primary refr.indexradius of curvature

It’s key here to make sure we know whether the radius of curvature will be positive or negative – remember that measurements always start *from* the relevant surface *to* the point of measurement (in this case the centre of curvature, C). As we learned in chapter 2 , if the radius distance is measured from left to right (in the same direction as light) then it will be positive; otherwise, it will be negative. The trick is to remember that if the front surface is convex, it will have a positive radius of curvature, whereas if it’s concave, it will have a negative radius of curvature. Similarly, if the back surface is convex, it will have a negative radius of curvature, and if the back surface is concave, it will have a positive radius of curvature ( Fig. 4.6 ).

In order to determine how the thickness of the lens and the refractive index work together to refract the light, we need to calculate the reduced thickness ( <SPAN role=presentation tabIndex=0 id=MathJax-Element-5-Frame class=MathJax style="POSITION: relative" data-mathml='t¯’>t⎯t¯

t ¯

; pronounced tee-reduced or tee-bar). We do this by dividing the measured thickness (t) by the refractive index of the material (n _{g} ), as shown in Equation 4.3 .

t¯=tng

reducedthickness¯=thicknessrefractiveindex of lens

Once we have all the information needed, we can utilise and modify a familiar equation from chapter 3 to calculate the equivalent power of the lens ( Equation 4.4 ).

Fe=F1+F2−t¯F1F2

equivalentpower=lens1+lens2−(reducedthickness×lens1×lens2)

A 3-cm thick biconvex lens (refractive index 1.523) has a front surface with a radius of curvature of 15 cm and a back surface with a radius of curvature of 20 cm. What is the power of the lens?

Step 1: Determine what we need to calculate

equivalent power, F

_{e }

Step 2: Define variables

t = 0.03 m

*(we need to convert to metres)*

n

_{g }= 1.523*(refractive index of lens)*

r

_{1 }= +0.15 m*(because the front surface is convex the radius is positive and in metres)*

r

_{2 }= −0.20 m*(because the back surface is convex the radius is negative and in metres)*

n = 1.00

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

Step 3: Determine necessary equation

F

_{1 }= (n_{1 }′ – n_{1 }) / r_{1 }*(**Equation 4.2**)*

F

_{2 }= (n_{2 }′ – n_{2 }) / r_{2 }*(**Equation 4.2**)*

<SPAN role=presentation tabIndex=0 id=MathJax-Element-10-Frame class=MathJax style="POSITION: relative" data-mathml='t¯’>t⎯t¯

t ¯

= t / n_{g }*(**Equation 4.3**)*

F

_{e }= F_{1 }+ F_{2 }– <SPAN role=presentation tabIndex=0 id=MathJax-Element-11-Frame class=MathJax style="POSITION: relative" data-mathml='t¯’>t⎯t¯

t ¯

F_{1 }F_{2 }*(**Equation 4.4**)*

Step 4: Calculate

F

_{1 }= (n_{1 }′ – n_{1 }) / r_{1 }

F

_{1 }= (1.523 – 1.00) / 0.15

F

_{1 }= +3.49…D*(remember to keep the number long in the calculator)*

F

_{2 }= (n_{2 }′ – n_{2 }) / r_{2 }

F

_{2 }= (1.00 – 1.523) / −0.20

F

_{2 }= +2.615D

<SPAN role=presentation tabIndex=0 id=MathJax-Element-12-Frame class=MathJax style="POSITION: relative" data-mathml='t¯’>t⎯t¯

t ¯

= t / n_{g }

<SPAN role=presentation tabIndex=0 id=MathJax-Element-13-Frame class=MathJax style="POSITION: relative" data-mathml='t¯’>t⎯t¯

t ¯

= 0.03 / 1.523

<SPAN role=presentation tabIndex=0 id=MathJax-Element-14-Frame class=MathJax style="POSITION: relative" data-mathml='t¯’>t⎯t¯

t ¯

= 0.0197…*(remember to keep the number long in the calculator)*

F

_{e }= F_{1 }+ F_{2 }– <SPAN role=presentation tabIndex=0 id=MathJax-Element-15-Frame class=MathJax style="POSITION: relative" data-mathml='t¯’>t⎯t¯

t ¯

F_{1 }F_{2 }

F

_{e }= 3.49… + 2.615 – (0.0197… × 3.49 × 2.615)

F

_{e }= +5.92D

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

## Practice questions:

**4.2.1**A thick lens (refractive index 1.523) has a convex front surface with a radius of curvature of 25 cm. What is the power of the front surface?

**4.2.2**A thick lens (refractive index 1.523) has a concave back surface with a radius of curvature of 25 cm. What is the power of the back surface?

**4.2.3**A 5-cm thick lens (refractive index 1.523) has a front surface power of −2.00D and a back surface power of +4.00D. What is the power of the lens?

## Thick lens focal length

If it’s possible to calculate the equivalent power of a thick lens, then it is also possible to calculate the **secondary (f** _{e }**′) and primary (f** _{e }**) equivalent focal lengths** ( Equations 4.5 and 4.6 ). Remember that the equation for the primary equivalent focal length requires you to include a minus sign.

fe’=nFe