After working through this chapter, you should be able to:
Outline all the important features of a reduced eye
Outline the difference between a spherical refractive error and a cylindrical refractive error
Explain what the far point of the eye is and be able to calculate it
Understand how power of a lens will need to be altered as its distance from the eye increases
Interpret a power cross
Now that we know the basics of thin and thick lenses, we need to spend a little bit of time focusing on clinical examples of corrective lenses. However, in order to do that effectively, we need to first consider the power of the human eye.
The human eye
As you will no doubt be aware, the human eye is a particularly complex anatomical structure, and one of its many functions is to focus light onto the back of the eye (the retina). Interestingly for us, the human eye is able to focus light by adding convergence to the incoming light. This means it must be positively powered, which (if we think back to chapter 2 ) suggests that its front surface must be curved and, in particular, convex. Now, given that it’s likely that you have at least one eye, this probably comes as no surprise to you because we know that the front surface (the cornea) is domed and pokes outwards in a convex way from the front surface of the eye ( Fig. 5.1 ).
Importantly, we can imagine that if light travelled from one section of air into another section of air that was curved but had the same refractive index, it probably wouldn’t do very much to the path of the light (we can even confirm this mathematically using Equation 2.4 from chapter 2 (F = (n′ – n) / r) – here we can see that if the primary (n) and secondary refractive indices (n′) are identical then no matter the curvature, the power will always be zero). This means that a key feature of the human eye focusing light is that the cornea has a different refractive index to the air, and this has been measured to be roughly 1.376. This means that the human eye can focus light because (1) it is curved, and (2) it is a different refractive index to air (please see chapter 22 for a do-it-yourself demonstration of this). This also explains why your vision is blurry if you try to open your eyes underwater – the refractive index difference between water (1.333) and the cornea (1.376) is much smaller than that of air (1.00) and the cornea (1.376), so the light doesn’t focus properly.
As one final fact about the human eye before we get into the depths of chapter 5: light can enter the eye through the convex cornea because the cornea is completely transparent (please see biology textbooks for a cool explanation of how this is possible), which means the light will then travel through the pupil to be focused on the retina . The retina is the physiological structure at the back of the inside of the eye that processes the light signal and transduces it into a signal the brain can interpret. The central part of the retina is called the fovea , and this is important because this part of the eye provides us with our central vision.
A reduced eye
The previous section has briefly covered the complexities of human vision to explain that the human eye has a positive power and that the goal is to focus the incoming light onto the retina at the back of the eye. Now, as we’ll remember from chapter 2 (vergence), the amount of power of a surface will greatly affect where the image forms. In the human eye, we need the image to form on the back of the eye, which is approximately 22.22 mm behind the front surface of the eye, and this means there’s little room for error. However, this distance (from the front of the eye to the back) will vary depending on individual factors such as height and age, etc.
Providing the incoming light is focused with the right power for the distance of the retina from the cornea, then the eye is appropriately proportioned and the image will be in focus ( emmetropic eye ). If, however, the light focuses too far in front of the retina ( myopic ) or too far behind the retina ( hyperopic ), then this is called a refractive error , meaning the eye has refracted the light inappropriately. Fig. 5.2 shows examples of this in a simple diagram of an eye.
The important thing to remember here is that because the emmetropic eye (in general terms) is powered to focus light from infinity (zero vergence) at the retina, this means the distance between the cornea and the retina is the secondary focal length of the eye, and the fovea corresponds to the secondary focal point (F′) .
However, if we wanted to calculate how light travels through the eye mathematically, there would be a lot to consider – the power of the cornea, the refractive index changes (cornea, aqueous, lens and vitreous), the power of the lens and the relative distances of all the front and back surfaces of the structures. Luckily for us, we can use a simplified model of an eye in order to make this easier; this simplified model is referred to as a reduced eye .
The reduced eye assumes a dioptric corneal power of +60.00D, meaning the convex corneal radius of curvature is +5.55 mm. This also then assumes an axial length (length of the eye from cornea to retina) of +22.22 mm. Importantly, the eye also possesses a different refractive index (n′ = 1.333) relative to the air outside. Fig. 5.3 shows how the reduced eye can be drawn and Box 5.1 shows how this can be calculated.
As discussed in the main text, the reduced eye assumes a refractive index of 1.333 and a corneal curvature of 5.55 mm. As our model cornea would be a convex spherical surface relative to the light entering the eye, we know the centre of curvature needs to be positioned to the right of the cornea (within the eye) and that it will therefore have a positive distance.
F = n′ – n / r
F = 1.333 – 1.00 / +0.00555
F = +60.00D
We can then use the power of the cornea to calculate the emmetropically appropriate distance of the retina, whilst being careful to remember that the refractive index of the eye is not the same as air:
f = n′ / F
f = 1.3333 / +60.00
f = +0.022216 m
f = +22.22 mm (converted to mm)
Now, we know that not all objects in the world will be an ‘infinite’ distance away from us, but we are able to account for changes in distance by using our lens (accommodation). This book will not discuss accommodation, but it is good to be aware that we are only discussing distant objects as examples.
Spherical refractive error
As discussed previously, in nature not all eyes are appropriately powered for their length, which can lead to myopia or hyperopia, both of which are examples of a spherical refractive error (measured in dioptres; DS or D ). A spherical error is defined by a refractive error that is the same in all meridians, so for example, if an eye had a spherical error of −2.00D, the error would be −2.00D along all possible axes, as shown in Fig. 5.4 .
Myopia (shortsightedness) is characterised by the eye either being too powerful or the axial length of the eye being too long – both of which result in an image that forms in front of the retina ( Fig. 5.5 ). The result is that people with myopia will see blurry images of objects at far-away distances. However, they will still be able to see objects closer to them (although, how close will depend on how large the refractive error is). For example, I am myopic, and I can only see clearly (without my lenses) to about 14.81 cm away from my eye before things start to get a little blurry, meaning that 14.81 cm corresponds to the far point of my eye. However, as I can still see objects clearly at 14.81 cm, the vergence of an object at that location will be focused on my retina, and we can use that information to calculate my refractive error.
L=n / lL=1.00 /−0.1481L=−6.75
Equation 2.1 (from chapter 2 ) helps us to determine that an object 14.81 cm in front of my eye will have a far point vergence of −6.75 at my cornea, which is then focused neatly onto the retina of my eye. This means that my eye needs −6.75 of vergence removed from the visual signal in order to focus correctly, so we can therefore refer to the refractive error of my eye as −6.75D. If we assume that my cornea is appropriately powered at +60.00D, we can calculate the image vergence and then the image distance (which will equate to the axial length of the model eye).
L’=L+FL’=−6.75+60.00L’=+53.25l’=n’/ L’l’=1.333 /+53.25l’=+0.02503ml’=+25.03mm (converted to mm)