PHYS 6.3: Optical elements: prisms, lenses and spherical mirrors	PPLATO @
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1.1 Module introduction

You will probably be familiar with the action of a prism in splitting white light into the colours of the rainbow, or of a simple magnifying lens, or of the curved reflectors used in vanity mirrors, shaving mirrors and driving mirrors. This module describes these optical elements and introduces you to the equations which govern their operation and which can be used in their design. The operation of all these optical elements depends on the known behaviour of light rays when reflected from mirrors or refracted at the boundary between two transparent optical media. The magnifying glass is just one application of a thin lens and the mirrors listed above are examples of spherical mirrors. An understanding of the action of these simple optical elements opens the way to an understanding of more complex instruments, such as telescopes, microscopes, camera lenses and projection systems, which are discussed in other FLAP modules.

Section 2 describes refraction by a prism, using Snell’s law, and shows how prisms can be used to produce dispersion and total internal reflection. Section 3 describes refraction at a single spherical surface using a Cartesian sign convention and introduces the conjugate equations for refraction at this surface and for refraction by a thin lens. The focal length for a thin lens is introduced and the lens maker’s equation and the thin lens equation are derived. Convex and concave lens behaviour is then discussed using this formula, ray diagrams and principal rays. The transverse magnification and optical power of a lens are described, as is the process of image formation in a two–lens system.

In Section 4 we extend these ideas to convex and concave spherical mirrors using the same Cartesian sign convention. The spherical mirror equation is derived, and this is used with ray diagrams to discuss the operation and transverse magnification of spherical mirrors.

It is interesting to note that although the fundamental principles of reflection and refraction were discovered nearly three hundred years ago, the subject has seen something of a renaissance recently with the advent of fast computer systems; complex lens systems can now be designed quickly. This has led to a resurgence of interest in optics in general and in particular to the use of optical components in advanced technology and physics research. However, this creates plenty of scope for gettings things wrong on an impressive scale, as was shown in the case of the Hubble space telescope. This was launched in 1990 and had a wrongly configured main mirror. The problem has now been corrected with the help of an additional optical system, installed in orbit by NASA astronauts.

Study comment Having read the introduction you may feel that you are already familiar with the material covered by this module and that you do not need to study it. If so, try the following Fast track questions. If not, proceed directly to the Subsection 1.3Ready to study? Subsection.

1.2 Fast track questions

Study comment Can you answer the following Fast track questions? If you answer the questions successfully you need only glance through the module before looking at the Subsection 5.1Module summary and the Subsection 5.2Achievements. If you are sure that you can meet each of these achievements, try the Subsection 5.3Exit test. If you have difficulty with only one or two of the questions you should follow the guidance given in the answers and read the relevant parts of the module. However, if you have difficulty with more than two of the Exit questions you are strongly advised to study the whole module.

1.3 Ready to study?

2 Refraction by a prism

2.1 A parallel–sided glass block

We will start our discussion of how light is refracted by a glass prism by recalling what happens to light as it travels through a block of glass as shown in Figure 1.

When a ray of light of a single wavelength (i.e. a monochromatic ray) enters the glass (an optically dense medium) from the air (an optically rare medium), it is refracted towards the normal to the boundary according to Snell’s law of refraction: i

μ₁sinθ₁ = μ₂sinθ₂(1)

where μ₁ and μ₂ are the refractive indices of air and glass, and θ₁ and θ₂ are the angles of incidence and refraction at the boundary. The refractive index of air is about 1.0003, which we approximate to 1, so that Equation 1 becomes:

sinθ₁ = μ₂sinθ₂(2)

After the ray has crossed the block, its angle of incidence on the second boundary is θ₃, and it emerges, refracted away from the normal, at θ₄. However, since the block is parallel-sided, θ₃ = θ₂ so that:

μ₂sinθ₂ = μ₂sinθ₃ = sinθ₄(3)

Comparing Equations 2 and 3 shows us that θ₁ = θ₄, so the ray emerges in the same direction as it entered, but has suffered a lateral shift.

2.2 Refraction by a prism

When the second face of the block is not parallel to the first, the path of a monochromatic ray will in general be like that through the prism of Figure 2, and will no longer have the symmetry of Figure 1. The ray leaving the prism will not be parallel to the ray entering the prism, as it will have been deviated by it, having turned through the angle of deviation D.

However, there is a situation in which a different symmetry is present; this occurs when the ray in the glass makes equal angles with the two prism faces, as in Figure 3. This geometry may be shown to have the important consequence that it makes the angle of deviation a minimum for any given wavelength. A relationship between this angle of minimum deviation and the prism angle A, which is the angle between the refracting faces of the prism, can be obtained as follows. i

For triangle PXY we have D_min = 2(θ₁ − θ₂), and in triangle AXY, A + 2(90° − θ₂) = 180°, so that A = 2θ₂. Combining these equations gives the minimum deviation D_min = 2θ₁ − A. Then, using Snell’s law,

$\mu = \dfrac{\sin\theta_1}{\sin\theta_2} = \dfrac{\sin\left[\frac12 (D_{\rm min}+A)\right]}{\sin\left(\frac12 A\right)}$(4)

This equation can be rearranged to give an expression for D_min:

D_min = 2arcsin[μsin(A/2)] − A(5)

2.3 Prisms in optical instruments

There are two principal applications of prisms in optical instruments; they are dealt with in detail elsewhere in FLAP and so we will discuss them here only very briefly. They involve dispersion and total internal reflection – topics which are themselves introduced more fully elsewhere in FLAP.

Dispersion is the splitting up of a beam of light of mixed wavelengths into its constituent wavelengths. The reason why a prism can perform this task is that the refractive index μ varies with the wavelength of the light being refracted, as is shown for a common type of borosilicate glass in Figure 4. Now referring to Equation 2,

sinθ₁ = μ₂sinθ₂(2)

we see that for incidence at a given angle θ₁, the angle of refraction θ₂ depends on the refractive index μ₂, which in turn depends on the wavelength. Tables which summarize the physical properties of materials often quote a single value of μ for a particular type of glass; this single value is then usually for sodium light, with a mean wavelength of 589.3 nm. i

If we have a beam of light which contains a range of different wavelengths (e.g. white light) incident at angle θ₁ then there will be a range of refracted angles θ₂, with shorter wavelengths (e.g. violet) being refracted or deviated through larger angles than the longer wavelengths (e.g. red). If the surface where this refraction occurs is the first surface of a prism, then further dispersion of the different constituent wavelengths takes place at the second surface, as shown in Figure 5 (in which the angles of dispersion are exaggerated). A prism being used for dispersion is usually orientated so that it is in a position of minimum deviation for the middle of the range of wavelengths present in the beam. This arrangement produces the minimum dispersion, but it also causes the graph of angle of deviation against wavelength to be nearly linear, i.e. it corresponds to linear dispersion.

The second important application of prisms is their use as reflectors. If we restrict our discussion to a single wavelength, then we see that as the angle of incidence of a light ray on the second face of a prism is increased, an angle will be reached when the refracted ray travels along the surface of the prism (Figure 6a) corresponding to an angle of refraction θ₂ = 90°. The angle of incidence within the glass at which this happens is the critical angle θ_c and, if this angle is exceeded, then total internal reflection occurs (Figure 6b).

The type of prism usually employed for its ability to produce such internal reflections is the 45°/90°/45° prism, which may be used in the two orientations shown in Figure 7.

3 Refraction at a single spherical boundary

Our objective in this section of the module is to derive and apply equations governing the passage of rays of light through lenses, which are usually discs of glass or other transparent material, with one or both surfaces curved symmetrically about the optical axis of the lens. The optical axis is the line drawn normal to the disc and through its centre. We restrict our discussion to lenses having surfaces of a constant radius of curvature and therefore described as spherical lenses, and also to thin lenses, where the maximum distance between the faces is very small compared with the radius of curvature of either face. We include the possibility of one of the surfaces being flat, i.e. of having an infinite radius of curvature.

The first stage of our discussion is to establish the conjugate equation which describes the refraction of light at a single spherical surface, by which we mean the spherical boundary between regions of uniform refractive index μ and μ′. This is a system which has only limited practical application, although as we shall see, it does have important image–forming properties. However, when we have obtained an equation which describes what happens to light at a single spherical surface, then we can apply it again at a second spherical surface: two such surfaces in close succession, with appropriate changes of index, are what we mean by a lens.

3.1 The conjugate equation for a single spherical surface

We start by considering the refraction of a monochromatic light ray OA which leaves a point source of light O in Figure 8, situated on the optical axis at a distance l to the left of the vertex_of_a_lensvertex or pole_of_a_lenspole C of the curved boundary (which is the point where the optical axis intersects that boundary).

We will assume that the refractive index μ′ of the medium to the right of the boundary is greater than the index μ to the left. The ray is therefore refracted towards the normal to the boundary, and crosses the optical axis at O′, a distance l′ from C. Snell’s law of refraction when applied at A gives:

μsinθ = μ′ sinθ′(6)

where θ and θ′ are the angles of incidence and refraction measured with respect to the normal AR to the boundary at A.

We now introduce a most important restriction to our discussion and assume that:

In any system subject to this approximation, we may write the sine or tangent of the angles made by any ray to the axis or to any other ray or to any radius of curvature of the boundary, as equal to the angles themselves, measured in radians.

If we now apply the paraxial approximation to Equation 6 we obtain:

μθ = μ′θ′

From the triangles in Figure 8 we have:

θ = α + β   and   β = θ′ + γ

so that μ(α + β) = μ′ (β − γ)

We now replace the angles in this equation by their tangents, writing

α = tanα = h/l,   β = tanβ = h/r,   γ = tanγ = h/l′

where h is the distance of A from the optical axis and we have made the approximation that C is directly below A.

Then,$\mu\left(\dfrac hl + \dfrac hr \right) = \mu'\left(\dfrac hr - \dfrac{h}{l'}\right)$

Dividing by h and rearranging then gives:

 $\dfrac{\mu}{l} + \dfrac{\mu'}{l'} = \dfrac{\mu'-\mu}{r}$(7)

Equation 7 allows us to calculate the distance l′ from the boundary, of the point where the refracted ray AO′ crosses the optical axis. As explained below, points O and O′ are examples of an object and an image in an optical system; any light ray originating from O arrives at O′. An important property of Equation 7 is that it does not contain h. Thus all rays from O are refracted so as to pass through O′ (as in Figure 8), no matter how far from the optical axis they cross the boundary, provided they may be regarded as paraxial rays. O′ is then said to be the point image of the point object O. Moreover, if it were possible to place a screen at O′, then the image would be displayed on the screen and such a ‘captured’ image, where the rays converge and pass through a point, is called a real image. Note that the ray which coincides with the optical axis remains undeviated because it is at normal incidence to the boundary. i

A point object placed to the left of a curved boundary, as in Figure 8, does not always result in a real point image being formed to the right of the boundary. If the object is brought closer to the same boundary as in Figure 9, then again provided that μ′ > μ, rays from the object to the boundary will be refracted towards the normal but will not converge to a point on the optical axis – instead they appear to diverge from a point to the left of the object position. Such a point is called a virtual image. It cannot be captured on a screen as rays do not actually pass through it but only appear to have come from that point when viewed from the right of the boundary. You will notice that a convention has been introduced in Figure 9 whereby the paths followed by rays of light are shown by thick lines; similar analysis to that applied to Figure 8 leads to an equation containing identical terms to those in Equation 7, but with some differences in the signs of the terms. We will examine those differences shortly.

The boundary surfaces which we have been examining bulge towards the object position and, when seen from that side, are called convex surfaces. We can also examine the refracting properties of spherical boundaries which are hollow when seen from the left; these are called concave surfaces.

Another situation which could be analysed has the object closer to the concave surface than in Question T2. If we chose, we could also analyse four sets of object and image positions when the second medium had a lower refractive index than the first. However, we would find that the equations resulting from all eight possibilities differed only in the signs of the terms. We can surmise from this that if the different systems were specified more completely, then they could all be described by the same equation. This is in fact the case, and the additional information required is the location of object, image and centre of curvature positions with respect to the boundary vertex.

What we have used so far are the distances of O, O′ and R from C, when we should have used their displacements. This is easily rectified by introducing a sign convention which will enable us to attach a sign to these distances and which, in effect, converts them into displacements (although we will continue to call them distances because this is the normal practice in optics). There are several possible conventions and we will adopt the one most commonly used, which is the Cartesian sign convention. i Although it is being introduced here for refraction at a single curved surface, it will be applied to lenses in Subsection 3.2 and to mirrors in Section 4.

It is defined as follows:

We can now see what effect this convention has on Equation 7,

$\dfrac{\mu}{l} + \dfrac{\mu'}{l'} = \dfrac{\mu'-\mu}{r}$(Eqn 7)

In Figure 8, the object O is to the left of C so l becomes −l; the image O′ is to the right, so l′ retains its positive sign, as does r, since R is also to the right of C.

Now see for yourself what changes this convention makes to the result obtained in Question T2.

You should have obtained exactly the same result as the modified Equation 7 (see Answer T2) and it would also be obtained for the other six configurations. This new form of the equation, shown below, is called the conjugate equation for a single spherical surface; it is an equation which links together the object and image distances and the radius of curvature of the spherical surface at which refraction occurs.

To interpret this in the general case (convex or concave), with light moving from left to right, remember that μ is the refractive index to the left of the interface, μ′ is the refractive index to the right of the interface, l is the object distance, l′ is the image distance and r is the radius of curvature. The signs of l, l′ and r are determined by the Cartesian sign convention.

Even though we have incorporated the sign convention into Equation 8, when it is used for calculation, it is still necessary to allot signs to distances according to the sign convention before they are entered into the equation. Object and image points linked by the conjugate equation are called conjugate points, and planes perpendicular to the optical axis containing conjugate points are conjugate planes.

We can extend our discussion of point objects to include objects of finite size by regarding a finite object as a continuous succession of point objects. For example, consider the image of the line object AA′ which is perpendicular to the optical axis in Figure 11. We will assume that the end A is imaged at B by the spherical surface with centre of curvature at R. The line joining the end A′ to R may be regarded as another optical axis and, within the paraxial approximation, an image of A′ will be formed at B′. If AA′ and BB′ are small, they may be assumed to be perpendicular to the original optical axis, and BB′ is then the extended image of the extended object AA′.

3.2 Refraction by a thin convex lens

When confronted with a compound system (i.e. more than one optical component), we can apply Equation 8,

$\dfrac{\mu'}{l'} - \dfrac{\mu}{l} = \dfrac{\mu'-\mu}{r}$(Eqn 8)

to each component in turn, treating the image produced by one lens as an object for the next one along. If we follow this procedure, using a consistent convention of signs throughout, then the image for the final component will be the image for the entire system.

With this in mind, we can now consider the refraction produced by two spherical surfaces with radii of r₁ and r₂, respectively, with the second surface a distance t from the first, along the common optical axis. These surfaces form a spherical lens, which has a refractive index μ and is surrounded by air with a refractive index of 1. Figure 12 shows such a lens, in this case a biconvex lens, since it has both faces convex as seen from the outside. Often a biconvex lens is simply called a convex lens.

An axial point object O, is at a distance l from the vertex C₁ of the first surface, which, by itself would form an image O′ at a distance l′ from C₁. Equation 8 now relates object and image distances for this first (left-hand) surface:

$\dfrac{\mu}{l'} - \dfrac{1}{l} = \dfrac{\mu-1}{r_1}$ i

The image O′ now acts as an object for the second surface, being at a distance (l′ − t) from it, where t is the lens thickness. i The second surface produces an image O′′ of O′ at a distance l′′ from the vertex C₂. Applying Equation 8 a second time gives:

$\dfrac{1}{l''} - \dfrac{\mu}{l'-t} = \dfrac{1-\mu}{r_2}$ i

We see from these two equations that there is no simple relationship between object and image distances for a finite value of t. However, simplification follows if the thickness t is much less than the object and image distances and each of the radii of curvature and can therefore be neglected; this is our thin lens approximation. Putting t = 0, and adding the last two equations gives: i

$\dfrac{1}{l''} - \dfrac{1}{l} = (\mu-1)\left(\dfrac{1}{r_1}-\dfrac{1}{r_2}\right)$(9)

We now make a change in the symbols used for object and image distances to accord with common practice and write the object distance l = u and the final image distance l′′ = υ. We then obtain the conjugate equation for a thin lens, relating object and image distances for a thin lens to the radii of curvature of its surfaces.

Remember that according to the Cartesian sign convention any of the quantities u, υ, r₁ and r₂ may be negative in this formula.

3.3 Focal points of a thin convex lens

Now let us consider what happens when the object is moved further and further away from the lens, so that u is large compared to r₁ and r₂. We can take the limit as |u| → ∞ and 1/|u| → 0. i In this limit, all rays from the object which are incident on the lens will be parallel to the optical axis and will be refracted by the lens through a single image point on the axis, as shown in Figure 13a. This image point F₂ is called the image focus or second focus or second focal point of the lens and it is situated at a distance f from the lens. The distance from the centre of the lens to F₂ is called the focal length f of the lens and is the parameter by which the focusing property of the lens is quantified.

The focal length of a convex lens is the distance from the centre of the lens at which parallel rays from an object at infinity are brought to a focus.

This formal definition of focal length provides the basis of a simple method by which the focal length of a convex lens may be found. If the image of a distant object, such as a far window, tree, or house is focused on to a screen, then the distance from the lens centre to the screen will be slightly greater than the focal length. The further the object is from the lens, the closer the image distance will be to the focal length.

Equation 11 shows an expression for f obtained by putting 1/u = 0 in Equation 10,

$\dfrac{1}{\upsilon} - \dfrac{1}{u} = (\mu-1)\left(\dfrac{1}{r_1}-\dfrac{1}{r_2}\right)$(Eqn 10)

This is called the lens maker’s equation because it relates the surface radii required to produce a lens of a given focal length from a particular type of glass.

Within the Cartesian sign convention, with light travelling from left to right, a positive value for f indicates a converging lens that will form a real image of an infinitely distant object. A negative value of f indicates a diverging lens that can only produce a virtual image of such an object. Converging and diverging lens are also referred to as positive and negative lenses respectively.

The combination of Equations 10 and 11 (for a given lens) gives the thin lens equation (Equation 12) which relates object distance, image distance and the focal length of the lens.

This is probably the most important equation in elementary geometrical optics for it shows that if the focal length of a thin lens is known (and it has just been noted how easily that may be measured), then the image position resulting from any object position may readily be calculated.

An object position may be found from which all paraxial rays are refracted by the lens to become parallel to the optical axis, as in Figure 13b. The corresponding image position will be where these parallel rays meet (!) i.e. at infinity, so that 1/υ = 0. The object will then be at the object focus F₁, or first focus or first focal point of the lens. The distance of this point from the centre of the lens, obtained by putting 1/υ = 0 in Equation 12, is shown to be equal to −f. Thus the first and second focal points are symmetrically placed on the opposite sides of the lens.

The converging lenses of Figures 12 and 13 are biconvex and may have radii of curvature which are the same for each face or which may differ. It is however not essential for each face to be convex in order for a lens to be converging: the essential requirement is that it should be thicker at the centre than at the edges. This also occurs with two other forms of lens: the plano–convex lens and the convex meniscus lens – see Figure 14.

3.4 Concave lenses

There are three forms of concave lens, namely biconcave_lensbiconcave, plano_concave_lensplano–concave and concave_meniscus_lensconcave meniscus (see Figure 15). Equations 10, 11 and 12, which were derived for convex lenses,

 $\dfrac{1}{\upsilon} - \dfrac{1}{u} = (\mu-1)\left(\dfrac{1}{r_1}-\dfrac{1}{r_2}\right)$(Eqn 10)

The lens maker’s equation$\dfrac{1}{f} = (\mu-1)\left(\dfrac{1}{r_1}-\dfrac{1}{r_2}\right)$(Eqn 11)

The thin lens equation$\dfrac{1}{\upsilon} - \dfrac{1}{u} = \dfrac{1}{f}$(Eqn 12)

also apply to concave lenses provided that the correct (negative) sign for focal length is used. Once again the focal length is defined as the distance from the centre of the lens to the second focal point which is generated when incident rays are parallel to the optical axis.

However, a concave lens produces a very different ray pattern to that of a convex lens. As shown in Figure 16a, a concave lens causes rays to diverge away from the optical axis and the second focal point F₂ is now the point from which the rays appear to diverge.

For a convex lens the first focus is the location at which a point source will generate rays which are refracted by the lens so as to emerge parallel to the optical axis. For a concave lens, the first focus F₁ is that point towards which rays must be directed in order that, after refraction at the lens, they follow paths parallel to the optical axis – see Figure 16b. This is another (compare Subsection 3.2) example of a virtual_objectvirtual (point) virtual_objectobject – the rays do not actually meet at F₁ – only their projections. i

3.5 Constructing ray diagrams for lenses

We have seen how rays which are parallel to the optical axis pass through, or appear to pass through the first or second focal points of both convex and concave lenses. These rays, together with a third ray through the centre of the lens, are usually called the principal rays and are employed in drawing ray diagrams which enable the relative positions of object, lens and image to be found. If ray diagrams are drawn to scale then they may be used instead of Equation 12 to find the precise position of the image of a particular object. Alternatively, an approximately scaled free–hand sketch provides a quick and easy source of information about the image, as we shall see shortly. Principal rays may be used for both convex and concave lens ray diagrams and, in a modified form, for convex and concave mirrors – as is shown in Subsection 4.4.

The third principal ray is drawn passing through the centre of a lens without deviation or lateral displacement. What actually happens for a biconvex lens, with equal radii of curvature, is shown in Figure 17. In order to show the effect more clearly the thickness of the lens is exaggerated. Tangents XX′ and YY′ have been drawn at the points on the surface of the lens where the ray enters and leaves. Since the ray PQ must pass through the centre of the lens C and the lens is symmetric, the tangents are parallel and the refraction of the ray is the same as if it were passing through a parallel–sided block of glass, i.e. it experiences a small lateral shift but the emergent ray is parallel to the incident ray. For paraxial rays and thin lenses (which may even be asymmetric), this shift is negligible and a ray through the centre of a lens can be represented by a straight line.

Examples of principal rays for converging and diverging lenses are shown in Figure 18.

We are now equipped to find the position of images produced by thin convex and concave lenses for any object position. We have two methods for doing this: we can use either the thin lens equation (Equation 12),

The thin lens equation$\dfrac{1}{\upsilon} - \dfrac{1}{u} = \dfrac{1}{f}$(Eqn 12)

or a ray diagram. If we use the latter, then we will also obtain information about image size and orientation. Practice is required in order to use both methods with confidence, so we start with some examples and follow with some questions.

In the ray diagram of Figure 19, the usual practice has been adopted of showing the object as an upright arrow with its base on the optical axis, and it is then described as being erect_imageerect. For clarity and accuracy of drawing, the vertical scale has been much exaggerated – in reality all rays would be paraxial, making angles of not more than about 10° to the optical axis.

Three principal rays are drawn from the point B at the top of the object: BX, a ray parallel to the optical axis is refracted through the second focal point F₂; BC, through the centre C of the lens, continues undeviated; BY, an incident ray through the first focal point F₁, is refracted to become parallel to the optical axis. All three rays meet at M, which is the image of the point B.

There is no need for further construction to find the position of the remainder of the image as it will be perpendicular to the optical axis, for a perpendicular object.

There are three attributes which may be used to describe an image. These are

(i) 

its accessibility, i.e. whether it is real or virtual,

(ii) 

its orientation, i.e. whether it is erect or inverted,

(iii) 

its magnification, i.e. whether it is enlarged or diminished ( having a dimension perpendicular to the optical axis which is larger or smaller than that of the object).

In Figure 19 we see that IM represents a real image – it could be captured on a screen because the rays we have drawn all come together and pass through M. Secondly, it is upside down so it is inverted, and finally, it is bigger (by a factor of two) than the object, so it is enlarged. We return to this latter point in the next subsection.

As all three principal rays from a point on the object intersect at the corresponding image point, it is only necessary to draw two of them in order to define the image position. It is however instructive to construct all three rays, as this will show how precisely they must be drawn to locate an image position with precision.

All three principal rays are shown in Figure 20. BX is refracted so as to appear to come from F₂ and BY is refracted parallel to the optical axis; the back projections of these two rays intersect at M through which BC also passes. The image IM is erect and diminished. It can be seen by looking through the lens but cannot be obtained on a screen and so is virtual.

Now see if you can use these methods to answer the following question. i

3.6 Transverse magnification by a lens

In the ray diagrams of the previous subsection we saw how the image and object sizes could be very different. Clearly such changes of scale can be of practical importance – the action of many optical instruments such as cinema projectors and microscopes is to create enlarged images; that of the camera is usually to create diminished images.

The transverse dimensions of an image created by a single thin lens are simply related to the transverse object dimensions by the image and object distances.

This may be seen in the simple ray diagram of Figure 19 where the heights of object and image are shown as h and h′, respectively. From the similar triangles OBC and IMC we have h/u = h′/υ. Defining the lens transverse magnification m as the ratio of image height to object height we obtain

This expression is applicable to concave and convex lenses, and for both real and imaginary objects and images. If the signs of υ and u are included when calculating m, then it conveys more information than the numerical value of the transverse magnification. Conventionally the object is taken as erect and so if m is positive the image is erect; if m is negative the image is inverted. i

3.7 Optical power

The function of a simple spherical lens is to cause light to converge or diverge, and the shorter the focal length of a lens, the greater its ability to cause this convergence or divergence. Consequently we speak of strong or weak lenses, depending on whether they have short or long focal lengths. The optical power of a lens is defined as the reciprocal of its focal length, and so gives us a quantitative measure which increases with greater converging or diverging ability.

If the focal length is measured in metres, then the unit of optical power is the dioptre. Thus a converging lens with a focal length of (+)20 cm has an optical power of +5 dioptres: a diverging lens with a focal length of (−)2.5 m has a power of −0.4 dioptres. The importance of specifying a lens by its optical power will become apparent in the next subsection, when we deal with two lenses in combination.

3.8 Image formation in a two–lens system

Whenever we see an object through some optical system, the final image is formed on the retina of the eye. This means that even for something as simple as a magnifying glass, at least two lenses are involved – the magnifier and your eye lens. To examine the basic concepts involved in analysing multi–lens systems we will consider the simple case of two converging lenses, with the second one placed a distance d along the common optical axis from the first. If the same method of analysis were applied to other lens combinations, identical general expressions would be obtained. The procedure can be extended to systems of more than two lenses.

A double lens system is illustrated in Figure 22. The focal length of the first lens is f₁ and that of the second lens is f₂; the lens separation is d. i As usual, we place an extended object on the left of the first lens at a distance u₁ and the light rays go from left to right. The first image is formed at a distance υ₁ from the first lens. This image is then taken to be the object for the second lens for which we use a second Cartesian coordinate system with its origin at the centre of the second lens. The object distance for the second lens is u₂ = −(d − υ₁). Figure 22 shows that the overall linear magnification of this two–lens system is the product of the magnifications produced by each of the two lenses,

Since$\text{overall magnification} = \dfrac{\text{final image size}}{\text{initial object size}} = \dfrac{\rm I_2M_2}{\rm OB} = \dfrac{\rm I_2M_2}{\rm I_1M_1} \times \dfrac{\rm I_1M_1}{\rm OB} = \left(\dfrac{\upsilon_2}{u_2}\right)\left(\dfrac{\upsilon_1}{u_1}\right)$

Applying the thin lens equation (Equation 12),

$\dfrac{1}{\upsilon} - \dfrac{1}{u} = \dfrac{1}{f}$(Eqn 12)

to each lens, we obtain:   $\dfrac{1}{\upsilon_1} = \dfrac{1}{f_1} + \dfrac{1}{u_1}\quad\text{and}\quad\dfrac{1}{\upsilon_2} = \dfrac{1}{f_2} + \dfrac{1}{u_2}$

We can eliminate υ₁ and u₂ by combining these equations, using υ₁ = u₂ + d to obtain:

$\left(\dfrac{1}{f_1}+\dfrac{1}{u_1}\right)^{-1} - \left(\dfrac{1}{d}\right)^{-1} = \left(\dfrac{1}{\upsilon_2}-\dfrac{1}{f_2}\right)^{-1}$(15)

This is a clumsy equation to use and generally it is better to solve two–lens problems by calculating the position of the first image and then using that as the object position for the second lens calculation, which has been done in the example shown below. The reason that Equation 15 has been quoted here is because, by letting d tend to 0 and rearranging, we obtain a relationship between the initial object and final image distances and the focal lengths of two thin lenses when they are in contact:

$\dfrac{1}{\upsilon_2} - \dfrac{1}{u_1} = \dfrac{1}{f_1} + \dfrac{1}{f_2}$(16)

This equation shows that they behave like a single thin lens which has a focal length f given by:

$\dfrac{1}{f} = \dfrac{1}{f_1} + \dfrac{1}{f_2}$(16)

If we now represent the lenses by their optical powers rather than by their focal lengths, then Equation 17 takes an even simpler form to become:

P_resultant = P₁ + P₂(18)

which states that the resultant optical power of two thin lenses in contact is the sum of their individual powers.

This equation can be extended to include several lenses. i

Notice in Figure 22 that the principal rays (labelled 1 and 2) used in locating the image produced by the first lens are not useful for locating the position of the final image. Two further principal rays (3 and 4) are used for this purpose. These rays can be retraced back through the first lens to the tip of the object, as the figure shows.

The following is an example of a two stage calculation for a two–lens system, similar to that of Figure 22. Treat it as a question if you feel able to do so.

4 Reflection at spherical mirrors

The positions of the real or virtual images of extended objects formed by mirrors with surfaces of spherical form (i.e. spherical mirrors) may be studied using methods similar to those developed for thin lenses, keeping to the paraxial (small angle) approximation. However, it should be noted that it is relatively easy to make large mirrors with a parabolic shape (i.e. non-spherical) which make sharp images of point objects on the axis even when large angle rays are included. This is why very large parabolic mirrors are used in optical telescopes where a large light collecting area is desirable. The trigonometry of parabolic mirrors is more complicated than that for spherical mirrors so we will stay with the latter to learn the basic physics of curved mirrors.

The behaviour of light rays at curved mirrors can be summarized as:

4.1 A spherical convex mirror

Figure 23 shows a spherical convex mirror. The convex surface bulges out at the centre towards the object, which is conventionally placed to the left of the mirror. As usual, we use the paraxial approximation, which means that all rays considered are either parallel to or make small angles to the optical axis.

This in turn implies that the outward bulge of the mirror is much less than its diameter so that rays at small angles to the optical axis still make small angles to the optical axis after reflection. The radius of curvature of the mirror is r and the centre of curvature is at R.

The position of an image point is fully determined by the intersection of any two rays from an object point. Consider ray BP which is parallel to the axis from the tip B of the arrow which forms the extended object. This ray strikes the mirror at P where RP is the normal to the surface at the point of incidence. It is reflected away at angle β on the other side of the normal, as though from a point F on the axis. From the triangles PCF and PCR, putting angles equal to their tangents we then have:

2β = tan2β = h/CF

andβ = tanβ = h/CR

where h = CP

Putting CF = f and CR = r, then gives f = r/2.

This is an important result, for it shows that the ray parallel to the axis and reflected at P appears to come from a point halfway between the vertex of the mirror and its centre of curvature. In addition, the expression for f does not involve h, so in the paraxial approximation any ray parallel to the axis is reflected so as to appear to originate at F. We therefore define f as the focal length of the mirror and the point F as the mirror’s focal point. Ray BC, which is directed towards the vertex C of the mirror at an angle α to the axis, is reflected at the same angle below the optical axis, since the axis is normal to the mirror surface at C. Rays BP and BC appear to diverge from the virtual image point at M. Similar ray diagrams and analysis would apply for any object point between B and O and therefore an extended, erect virtual image IM of the extended object OB is formed.

From the trigonometry of triangles OBC and IMC we can obtain the image distance υ in terms of the object distance u, the distance CF = f, and the height of the object OB. Note that in doing so we will, for the moment, treat u, υ and f as positive quantities since we have not yet introduced the appropriate sign convention for mirrors.

We will eventually reintroduce the Cartesian sign convention, but not until it can be properly justified, in Subsection 4.3. For now, note that $\alpha = \dfrac{\rm OB}{u} = \dfrac{\rm IM}{\upsilon}$, so that ${\rm IM} = \dfrac{{\rm OB} \times\upsilon}{u}$.

Also, from above we have $2\beta = \dfrac{\rm CP}{\rm CF}= \dfrac{\rm OB}{f}$ and, from triangle IMF, $2\beta = \dfrac{\rm IM}{f-\upsilon}$

Combining these two expressions for IM, $\dfrac{{\rm OB}\times\upsilon}{u} = \dfrac{{\rm OB}\times( f-\upsilon)}{f}$

which, on dividing by OB × υ, then gives:

$\dfrac{1}{u} - \dfrac{1}{\upsilon} = -\dfrac1f$(19) i

4.2 A spherical concave mirror

A spherical concave mirror is shown in Figure 24 with its reflecting surface curving inwards at the centre away from the extended object OB. We can analyse such a system in much the same way as we did for the convex mirror. The two principal paraxial rays BP and BC are shown originating from the tip of the extended object at B. Ray BP starts parallel to the optical axis, is reflected at P towards the axis and passes through the focal point F. Ray BC strikes the mirror at C, making an angle α to the optical axis, and is then reflected at α below the axis. The point of intersection of the two rays defines M as the tip of the image, so that an extended, real, inverted image IM is formed at distance υ from the mirror.

Referring to Figure 24 we can find the image distance υ = IC in terms of the object distance u = OC, f = FC and h = OB.

Thus in triangles OBC and IMC (treating h, u, υ, f and IM as positive):

$\alpha = \dfrac hu = \dfrac{\rm IM}{\upsilon-f}$

so that${\rm IM} = \dfrac{h\upsilon}{u}$

In triangles PCF and IMF:

$2\beta = \dfrac hf = \dfrac{\rm IM}{\upsilon-f}$

giving${\rm IM} = \dfrac{h(\upsilon-f)}{f}$

Equating expressions for IM:

$\dfrac{h\upsilon}{u} = \dfrac{h(\upsilon-f)}{f}$

Then dividing by hυ gives us:

 $\dfrac{1}{\upsilon} + \dfrac{1}{u} = \dfrac1f$(20) i

Again we can use the equations β = h/r and 2β = h/f to show that the length f is half the radius of curvature. For paraxial rays, because f is independent of h, it is defined as the focal length and F as the focal point of the concave mirror.

4.3 The sign convention and mirror calculations

For lenses, it was possible to develop a single formula relating object and image distances for both convex and concave lenses by using a particular sign convention. The same is true for mirrors and we will adopt the same convention as before, which is to use Cartesian coordinates with origin at the mirror’s vertex. Distance of objects, images, focal points or centres of curvature to the right of the vertex are then taken as positive and those to the left as negative.

Equations 19 and 20,

$\dfrac{1}{\upsilon} - \dfrac{1}{u} = \dfrac1f$(Eqn 19)

$\dfrac{1}{\upsilon} + \dfrac{1}{u} = \dfrac1f$(Eqn 20)

may now be written in accordance with this convention.

Thus for the convex mirror of Figure 23, u is negative whereas both υ and f are positive. This modifies Equation 19, so that it becomes:

$\dfrac{1}{\upsilon} + \dfrac{1}{u} = \dfrac1f$(Eqn 21a)

For a concave mirror, with object and image positions as in Figure 24, u, υ and f are all negative so that Equation 20 is unchanged and is identical to Equation 21a.

There are other possible conjugate object and image positions for a concave mirror (e.g. with a virtual object or image) which have not been discussed here. However, analysis of all the possible arrangements will yield equations which, when modified by the Cartesian sign convention, are identical to Equation 21a.

This becomes the spherical mirror equation, the general equation for convex or concave spherical mirrors for paraxial object/image locations.

We need to note carefully that this mirror equation is not identical to Equation 12,

the thin lens equation$\dfrac{1}{\upsilon} - \dfrac{1}{u} = \dfrac{1}{f}$(Eqn 12)

because of a difference in signs.

4.4 Constructing ray diagrams for mirrors

As with lenses, there are three principal rays which can be used in the construction of ray diagrams for spherical mirrors, and these are summarized below. Only two of the three are needed to relate object, mirror and image positions in a ray diagram.

4.5 Transverse magnification by a mirror

The transverse magnification m produced in the paraxial approximation was defined for lens systems as the ratio of image height to object height. With this same definition and referring to the mirror diagrams of Figure 23 or 24 we see that:

$\dfrac{\rm IM}{\rm OB} = \dfrac{\upsilon}{u}$ so that:

If the signs of u and υ are included, then M is negative for erect images and positive for inverted images; note that this is opposite to that for lenses.

4.6 Examples of image formation by spherical mirrors

Two examples will illustrate some practical applications of spherical mirrors. The first is the concave ‘vanity’ mirror, frequently used for putting on make–up or for shaving. This has a typical radius of curvature of about 1 m and the object (i.e. your face) is placed between the focal point and the mirror, as shown in the ray diagram of Figure 25. i

Inserting the values r = −1 m, u = −30 cm and focal length f = r/2 = −50 cm into the spherical mirror equation,

$\dfrac{1}{\upsilon} + \dfrac{1}{u} = \dfrac1f = \dfrac2r$(Eqn 21b)

we obtain the image distance:

$\dfrac{1}{\upsilon} = \dfrac1f - \dfrac1u = \rm \dfrac{1}{-50\,cm}-\dfrac{1}{-30\,cm}$

$\phantom{\dfrac{1}{\upsilon} }= \rm \dfrac{-3+5}{150\,cm} = \dfrac{2}{150\,cm}$    so υ = 75 cm

From Equation 22,

$M = \dfrac{\upsilon}{u}$(Eqn 22)

the transverse magnification is:

$M = \dfrac{\upsilon}{u} = \rm \dfrac{+75\,cm}{-30\,cm} = -2.5$

The virtual image is located 75 cm behind the mirror, it is erect and magnified by a factor of 2.5. i

Figure 25 also shows how the virtual image can be seen by your eye. Two additional rays from the tip of the object are reflected into the eye and form a real image on the retina. The virtual image in the mirror acts as a real object for the eye lens. In the real world the vanity mirror is often quite large and our strict requirements for paraxial rays are violated. This is not too serious but results in a rather distorted image being produced, particularly at the edges. i

 

Our second example is the familiar convex ‘rear view’ mirror sometimes fitted to cars. Figure 23 shows the detail of the image position construction, the only difference in our example being that the object is much further away so that angle α is smaller and the image much reduced in size and very near to the focus of the mirror.

The mirror has a typical radius of curvature of +2 m and, in this example, we take as the object a car 100 m away. With r = +2 m and u = −100 m, Equation 21b,

$\dfrac{1}{\upsilon} + \dfrac{1}{u} = \dfrac1f = \dfrac2r$(Eqn 21b)

gives the focal length as f = r/2 = +1 m

The image distance υ can then be obtained:

$\dfrac{1}{\upsilon} = \dfrac1f - \dfrac1u = \rm \dfrac{1}{1\,m}-\dfrac{1}{-100\,m} \approx \dfrac{1}{1\,m}$

   so υ = 1 m

Equation 22,

$M = \dfrac{\upsilon}{u}$(Eqn 22)

gives the magnification:

$M = \dfrac{\upsilon}{u} = \rm \dfrac{+1\,m}{-100\,m} = -10^{-2}$

Assuming that a car is 2 m in width, the tiny 2 cm wide virtual erect image is located 1 m behind the mirror. You should now think carefully why this is a good idea: we will return to this topic in the Exit test.

5 Closing items

5.1 Module summary

1

A monochromatic ray of light passing obliquely through a parallel–sided glass block is Figure 1deviated by equal but opposite amounts at the two faces and so the entering and emerging rays are parallel. The deviations of a ray at the faces of a prism are usually both in the same sense so that quite large overall deviations may be produced. If a ray passes symmetrically through a prism, so that its path in the prism makes equal angles with the two prism faces, then the deviation of the ray is a minimum. Relationships between this angle of minimum deviation, the prism angle and its refractive index are given in Equations 4 and 5. The dependence of the deviation on the refractive index together with the variation of refractive index of glass with wavelength, give rise to the dispersion of white light passing through the prism.

2

Light rays leaving a point source and crossing a spherical boundary between two transparent media are refracted, and either converge towards an image point or appear to diverge from a point, provided that all rays are paraxial. In the paraxial approximation of geometric optics it is assumed that all rays are parallel to, or make small angles with, the optical axis.

3

The combination of equations describing refraction at two spherical boundaries which are close to each other, leads to a conjugate equation (Equation 10) relating object and image distances to the radii of curvature of the surfaces of a thin lens.

4

The image distance for an object at infinity, defines the focal length of a lens, and this, together with the adoption of a Cartesian coordinate system, enables the lens maker’s equation to be derived. This equation gives the focal length of a thin lens in terms of the refractive index of the material of the lens and the radii of curvature of its surfaces:

$\dfrac{1}{f} = (\mu-1)\left(\dfrac{1}{r_1}-\dfrac{1}{r_2}\right)$(Eqn 11)

5

Rays parallel to the optical axis are deviated by a convex lens towards the axis and converge to a real image point. Parallel rays passing through a concave lens are deviated away from the axis and appear to diverge from a virtual image point. These image points are Subsection 3.3focal points for their respective lenses.

6

Analysis of ray paths leads to the thin lens equation, which relates object and image distances and the focal length of a thin lens, and which is valid for both convex and concave lenses:

$\dfrac{1}{\upsilon} - \dfrac{1}{u} = \dfrac{1}{f}$(Eqn 12)

7

The lens transverse magnification is given by:

$m = \dfrac{h'}{h} = \dfrac{\upsilon}{u}$(Eqn 13)

and is positive for erect images. A quantitative description of the strength of a lens is its optical power P = 1/f, expressed in dioptres.

8

As an alternative to calculation, ray diagrams, which are scale drawings using three principal rays, provide a method of finding image positions and system magnifications. Images are described by three attributes: orientation (erect or inverted), magnification (enlarged or diminished), and accessibility (real or virtual).

9

Systems consisting of two thin lenses on a common optical axis are analysed by regarding the image of the object produced by the first lens, as an object for the second lens.

10

The paraxial approximation applied to analyse reflection from a convex mirror and a concave mirror produces relationships between object and image distances and focal lengths which are similar, but not identical, to those for lenses. The Cartesian sign convention leads to a paraxial spherical mirror equation

$\dfrac{1}{\upsilon} + \dfrac{1}{u} = \dfrac1f = \dfrac2r$(Eqn 21b)

The focal lengths of concave and convex mirrors are equal to half their radii of curvature.

11

Ray diagrams, using three principal rays, provide an alternative method of finding image positions and system magnifications for spherical mirrors.

12

The mirror transverse magnification is given by:

$M = \dfrac{\upsilon}{u}$(Eqn 22)

and is negative for erect images.

5.2 Achievements

Having completed this module, you should be able to do the following:

A1

Define the terms that are emboldened and flagged in the margins of the module.

A2

Describe and explain the behaviour of a light ray passing through a prism and show how this depends on the wavelength of the light. Calculate the angle of minimum deviation for known values of prism angle and refractive index.

A3

Describe: (i) the behaviour of paraxial light rays passing through spherical boundaries and through converging and diverging lenses and (ii) the formation of real and virtual images.

A4

Recall and use the thin lens equation for any object and image position with any kind of thin lens. Determine the position and size of extended images of extended objects using ray diagrams, and determine the magnification of a lens system.

A5

Recall the lens maker’s equation and use it to calculate the focal length of a lens.

A6

Determine the location of the final image in a two–lens system with a common axis by calculation and by ray diagram. Find the optical power of such a system when the lenses are in contact.

A7

Describe the behaviour of paraxial light rays reflected from convex and concave spherical surface mirrors and the formation of images.

A8

Recall and use the spherical mirror equation for any object and image position for convex and concave spherical mirrors. Determine the position and size of extended images of extended objects using ray diagrams, and determine the magnification in any single mirror system.

A9

Describe how convex and concave spherical mirrors are used in some common practical situations.

Study comment You may now wish to take the following Exit test for this module which tests these Achievements. If you prefer to study the module further before taking this test then return to the topModule contents to review some of the topics.

5.3 Exit test

Study comment Having completed this module, you should be able to answer the following questions each of which tests one or more of the Achievements.

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