We've already touched on the most important thing about spectacles: what power to use to correct myopia or hyperopia. In this chapter, we will be looking at the "side-effects" of spectacles: magnification and prism effects.
Spectacle correction introduces magnification as a side-effect. A myope wearing spectacles see the world slightly smaller than they would without spectacles, and a hyperope sees the world slightly larger.
As before, we define magnification as the tangent of the visual angle of an object observed while wearing spectacles, divided by the tangent of the visual angle of the object observed without spectacles:
Figure 1 shows a myope wearing spectacles at a distance of \( d \) metres. They are looking at a distant object. We pick two rays from the distant object, one coming from the top of the object and one from the bottom. These two rays are picked out because they are refracted by the spectacles so they both hit the optic centre of the patient’s eye. The size of the object is measured by the tangent of the visual angle \( \alpha \) between the rays when they hit the eye. This is
where \( h \) is the separation between the rays when they hit the spectacles.
If the two rays were not diverged by the spectacles, they would have converged a distance \( e \) from the spectacles. The size of the object without the spectacles is the tangent of the angle \( \beta \), shown in the bottom image of Figure 1c , which is given by
So the magnification produced is
Since \( e \) is less than \( d \) , the overall magnification must be less than 1, so myopes experience minification (things appear smaller) when they look through spectacles.
We can’t work out this magnification numerically unless we know the value of \( e \). However, \( e \) can be calculated using the paraxial equation. The two rays hitting the spectacles are converging to a point which is \( e \) metres from the spectacles ( Figure 1c). So their vergence is \( V_{in}=1/e \). The two rays leave the spectacles and converge to a point which is \( d \) metres from the spectacles Figure 1a, so the convergence of these rays leaving the spectacles is \( V_{out}=1/d \). Using the thin lens equation \( V_{in}+F=V_{out} \), we have
Solving for \( e \) gives \( e=d/(1-d F) \). Substituting this into the formula \( M=e/d \) gives us
For example, a myopic patient has spectacles with a power of \( -7.5 \text{D} \) worn at a distance of \( 15 \) mm. The magnification they experience is thus \( M=1/(1-d F)= \) \( 1/(1-0.015\times(-7.5))= \)\( 0.899 \). That is, things appear only about 90% of the size they would without spectacles. However, if the same patient wore contact lenses, then the spectacle distance \( d \) would be zero, and so the magnification would be 1. Thus contact lenses produce fewer magnification problems than spectacles.
There are a couple of other formulas for spectacle magnification. If we let \( K \) be the patient’s ocular refraction, then the magnification is also
or
These formulas come by combining the formula for magnification \( M=1/(1-d F) \) with the formula \( 1/F=1/K+d \) , which is the relationship between spectacle focal length \( 1/F \), spectacle distance \( d \), and far point distance \( 1/K \).
Exactly the same formula for magnification applies to hyperopia. However, in hyperopia, the spectacle power \( F \) is positive, so this yields magnification. For example, a hyperopic patient has spectacles with a power of \( +7.5 \text{D} \) worn at a distance of \( 15 \) mm. The magnification they experience is thus \( M=1/(1-d F)=1/(1-0.015\times7.5)=1.12 \). Thus, objects appear about 12% bigger to the hyperope wearing spectacles than without.
Spectacle magnification in astigmatism can be analyzed simply by calculating the magnification along the principal meridians. Because the spectacle lens powers are different along the different meridians, a patient with astigmatism wearing spectacles will experience different amounts of magnification in different directions. That is, their retinal image will be distorted by being stretched or squashed.
Spectacle magnification can lead to transient difficulties when a patient receives a new prescription. The new prescription may cause a different amount of magnification compared to the previous one, and it can take a while to get used to it.
The other serious consequence of magnification is when it leads to different image sizes in the left and right retina (called aniseikonia), perhaps because the refractive error is quite different in the two eyes. This leads to perceptual distortions, such as the world appearing to be slightly tilted, and may result in nausea, headaches, and giddiness. Sensitivity to aniseikonia and rate of adaption to it vary greatly between patients.
One side effect of spectacle magnification occurs when you look at someone wearing spectacles. If they are a myope, then you will not see their eyes through the spectacles, but instead a virtual image of their eyes. Since a myope is corrected with a negative lens, the virtual image of their eyes will be smaller and closer than their real eyes. For small degrees of myopia, this is not an issue, but for larger amounts of myopia, the effect may be unwanted.
The opposite effect occurs with hyperopes. They are corrected with positive lenses, so these act as magnifiers of their eyes, and their eyes then appear larger and further away.
A prism is a block of refracting material (e.g. glass) with a triangular cross-section and flat sides (see Figure 2). The triangular cross section has a base and an apex (top). The apical angle is the angle between two of the surfaces at the apex of the prism. Prisms change the direction of light.
Figure 2 shows a few rays passing through a prism. A ray is bent once when it enters the prism, then bent again (by about the same amount) when it leaves. The divergent rays leaving the prism can be traced backwards to the point where they seem to originate from. This point, a virtual image, is shifted towards the apex (peak) of the triangle.
The total angle that the light is bent by as it passes through a prism is called the deviating angle, which I'll write as \( \delta \). Here we will work out the relationship between the deviating angle and the apical angle of the prism.
Figure 3(a) shows a ray of light entering a symmetrical prism with apical angle \( a \). For convenience, the ray has been selected so that the refracted ray inside the prism is horizontal. The angle of incidence \( \theta_{in} \) and the angle of refraction \( \theta_{out} \) are related by Snell's Law, \( n_{in}\sin{(\theta_{in})}=n_{out}\sin{(\theta_{out})} \). For fairly small angles (less than about \( 5^o \) ) this can be approximated by
Figure 3(b) shows another angle \( x \) which is one of the base angles of the prism. Because of symmetry, the other base angle is also \( x \), and so \( x+x+a=180 \) , from which \( x=(180-a)/2 \). In addition, \( x+\theta_{out}=90 \), because those two angles make up the angle between the surface and the surface normal. Taking the second equation, substituting in \( x=(180-a)/2 \), and solving for \( \theta_{out} \) gives
Finally, Figure 3(c) introduces the angle \( \delta/2 \) which is the amount the ray has been bent by when travelling through the first surface. From Figure 4(c) , we have
The ray will be bent again by \( \delta/2 \) when it passes through the second surface, so the overall deviation will be \( \delta \). We can now work out the total deviation \( \delta \) as follows:
| Step | Reason | Statement |
|---|---|---|
| 1 | Equation (1) | \( n_{in}\theta_{in} \approx n_{out}\theta_{out} \) |
| 2 | From Eqn (2), \( \theta_{out}=a/2 \) | \( n_{in}\theta_{in} \approx n_{out} a/2 \) |
| 3 | From Eqn (3), \( \theta_{in}=\theta_{out}+\delta/2 \) | \( n_{in}(\theta_{out}+\delta/2) \approx n_{out}a/2 \) |
| 3 | From Eqn (2), \( \theta_{out}=a/2\tag{2} \) | \( n_{in}(a/2+\delta/2) \approx n_{out}a/2 \) |
| 4 | Expand | \( n_{in}a/2+n_{in}\delta/2 \approx n_{out}a/2 \) |
| 5 | Multiply both sides by 2 | \( n_{in}a+n_{in}\delta \approx n_{out}a \) |
| 6 | Subtract \( n_{in}a \) from both sides | \( n_{in}\delta \approx n_{out}a-n_{in}a \) |
| 7 | Gather right hand side | \( n_{in}\delta \approx (n_{out}-n_{in})a \) |
| 8 | Divide by \( n_{in} \) | \( \delta \approx (n_{out}-n_{in})a/n_{in} \) |
Thus the deviating angle \( \delta \) is
That is, the deviation \( \delta \) produced by a prism is simply a multiple of the apical angle \( a \). For glass prisms in air, this simplifies to \( \delta\approx a/2 \) because \( n_{in}=1 \) and \( n_{out}\approx 1.5 \).
The deviation created by a prism causes objects, seen through a prism, to be displaced from their true position. This is because the prism creates a virtual image of the object which is shifted towards the prism's apex, and when we look through the prism, we only see the virtual image (Figure2 ). The apparent position of the object is found by tracing the light rays that leave the prism backwards.
In optometry, another measure of prism deviation is sometimes useful, called the prism dioptre, whose symbol is \( ^{\Delta} \). One prism dioptre (written \( 1^{\Delta} \)) of displacement means that an object which is 1 metre from the prism appears displaced by 1 cm towards the prism's apex. One prism dioptre is also a displacement of 2cm for an object which is 2 metres from the prism, and in general
The reason why this deviation is called dioptres comes from Prentice’s Rule. Figure 4 shows an object placed at the focal point of a thin lens. One ray of light enters the lens at a distance \( h \) in metres above the optic axis. Since this ray of light comes from the focal point, it is bent to emerge parallel to the optic axis. By tracing the ray backwards, this ray of light then seems to come from a point which is \( h \) metres above the true object location. Expressed in prism dioptres, this object displacement is
where \( f \) is the focal length and \( F \) the focal power (The 100 is because we have to convert the height \( h \), which is in metres, to centimetres). If \( h \) is 1cm and \( F \) is 1D, then this is a displacement equal to one prism dioptre. Note that Prentice’s Rule is usually expressed with the distance \( h \) on the lens in millimetres, giving
Any lens is, very roughly, the same as a pair of prisms. A positive lens is roughly like two prisms stuck base-to-base, while a negative lens is roughly like two prisms stuck apex-to-apex ( Figure 5). Because lenses are like prisms, they deviate light as it passes through them. Some of that deviation is useful - it’s what lenses do when they change the vergence of light - but some is not.
When a patient views an object through the centre of their spectacles, the objects is not displaced. But when they view an object through points other than the centre, the object appears to be displaced because the lens is acting rather like a prism. Prentice’s Rule tells us how much the object appears displaced. For example, suppose a patient wears spectacles with power \( -6D \). They view an object through a point 5mm above the lens centre. By Prentice’s rule, the object position is deviated by \( 5\times -6/10 = -3 \) prism dioptres.
The minus sign in the displacement is usually ignored. The displacement is always towards the apex of the prism, so instead of fiddling about with signs we just say which way around the prism is. The convention is to say whether the base of the prism is up or down. In the above example, the prism is Base-Up (BU). As another example, the prismatic effect shown on the right hand side of Figure 6 would be described as Base Down (BD).
When the pupil distance (pd) of the prescribed lenses is not correct, this induces an unwanted prism effect. Consider a myopic patient with a pd of 68mm, who is given spectacles with a pd of 71mm. In this case, when they look straight ahead, they are not looking through the centres of the lenses, but 1.5mm from the centres (assuming the 3mm error in the pd is equally distributed between the two lenses). If the prescription in the left eye is \( -5.5D \), then by Prentice’s rule, their left eye gets a prismatic effect of \( 0.825 \) prism dioptres, written \( 0.825^{\Delta} \). This is called Base In (BI) since the part of the lens they look through acts like a prism with the base pointing inwards (i.e. nasally). If the prescription in the right eye is \( -4D \), then the prismatic effect is \( 0.675^{\Delta} \). Again this is Base In, but the base is pointing in the opposite direction. These prismatic effects are pointing in opposite directions, so the total prismatic effect (which is the difference in displacement between the two eyes) is the sum of these two, or \( 1.5^{\Delta} \). This would be unacceptable in spectacles; for prescriptions up to \( 10D \), no more than \( 0.33^{\Delta} \) is allowed.
In Chapters 9 and 10 we assumed that spectacles were thin lenses, and it was then fine to use the thin lens equation to work out the relationship between spectacle power, spectacle distance, and the patient's far point. Often, however, it isn't reasonable to think of spectacles as thin. What happens then?
Luckily enough, there is only a small change to the relationship we derived for thin lenses. Instead of using the focal length of the spectacles (as if they were thin lenses) we use the back vertex focal length ( \( =1/BVP \)) instead (Chapter 4). Instead of using the spectacle distance, we use the back vertex distance, which is the distance from the back vertex of the spectacles to the eye. That is,
The reason we make this change is that the relationship between the spectacle power and spectacle distance in Chapters 9 and 10 depended only on the vergence of light leaving the spectacles when parallel light entered them. For a thick lens, the vergence leaving a lens when parallel light entered it is just the back vertex power.
The magnification produced by a thick lens is not, however, the magnification produced by a thin lens. Thick lens magnification is the product of two factors, called the shape factor and the power factor. The power factor is just the magnification formula \( 1/(1-dF) \) derived earlier with \( F \) replaced by the back vertex power. The shape factor is \( 1/(1-(t/n)F_1) \), where \( t \) is the lens thickness, \( n \) the refractive index, and \( F_1 \) the front surface power.