Lens Corrections

It turns out glasses correct our vision by "enhancing" the natural performance of our own eyes. Even if an eye shows either near or far-sightedness, there is a still a range in which that person can see clearly. The closest point at which a person can see an object in perfect focus is called the "near point".  Similarly, there is a distance called a "far point" which represents the farthest distance that a person can see a clear, focused image. For a person with "perfect" vision, this range of clear vision is around 25 cm for the near point, all the way out to "infinity". While an eye cannot see all the way out to infinity, in terms of the size of our eyes, anything over five meters represents infinity.

For someone who is nearsighted, they are likely to have a near point that is even closer than 25 cm. This is due to the fact that as an image moves closer to one's eye, it focuses further and further back. The natural elongation of the back of the eye in a near-sighted person allows it to focus even when these objects are close. However, the far point for a near sighted person is likely to be quite short; around 17 to 25 cm is average. For a far sighted person, they can see off into "infinity", but can only focus on objects that are a distance away from them (usually a meter or more away).

The way that lens helps us see is by adjusting light rays so that it gives the "impression" that it came from that distance. At this point, it would be handy to introduce some technical aspects.

Lens Equation
This commonly used formula is used to describe the relationship between three objects. An object located dO units away from a lens with a focal length f, will create an image with dI units away. They are related as such:
1/(f) = 1/(dO) + (1/dI)

where:
f is the focal length of the lens
dO is the distance from the lens to the object
dI is the distance from the lens to the object
There is a technical term given the left hand side; 1/f , when f is measured in meters is called the power of lens. The units have a special name called a diopter (D), and it is defined as inverse meters or m-1. There is also a handed convention when dealing with the two distance measurements. If an object and its image are on the same side of a lens, then the image distance, dI, is considered negative. If an object and its image are on opposite sides of a lens, then both are considered positive. It turns out this strange convention allows the lens equation to work under almost all basic circumstances.

Let's take a look at the near sighted case again:


In this case, the lens that will be used is called a diverging lens. What it is going to do is bend the rays so that light from infinity "appears" as though it is coninciding with a person's far point. In doing so, it allows the person to see a clear image, without altering their depth perception. Here's a look at what the lens does:


The lens actually bends the light outwards so that it looks like that it coming from a source that is quite close. The following a quick sample calculation that uses the lens equation.

EXAMPLE: Let's assume that there is a near-sighted person which has a far point of 17 cm away from their eye. How strong does the lens have to be so that the rays coming in from "infinity" look like they are coming in from 17 cm? For this example, let's say that the lens is about 2 cm away from the eye.

We know we would like the object distance to be "infinity". We want the image to appear 17 cm in front of the person's eye; this happens to be 15 cm (=17 cm - 2 cm) in front of the lens. Since the object and image distance are on the same side, we have to place a negative sign with the image distance, so we use -15 cm instead of +15 cm. We can figure out the power of the lens:
1/f = -1/0.15 m + 1/∞

While it may not be terribly obvious, we will treat 1/∞ as being zero for this calculation. If one were to put in a real number, it would only make a small contribution to the overall result. So we end up getting:
1/f = -1/0.15 m + 0
1/f = -1/0.15 m = 6.667 m-1 = 6.667 D

As such, we need to use a lens with a focal length of 0.15 meters, or a power of 6.667 D. Now that we've let the person see images from far away, can they still see images that are up close? Now that we have a lens focal power, we can use information about the person's near point to determine how it has moved.

EXAMPLE: Let's say that the same person has a near point of 12 cm from their eye. With the lens on, how far does an object 12 cm away appear to be?

As with before, the lens and  the resultant image are still on the same side, so  we assign a  negative sign for the image distance. Since the near point is 12 cm with respect to the eye, it happens to be 10 cm (= 12 cm - 2cm) away from the front of the lens. We can plug in what we know into the lens equation.

1/(f) = 1/(dO) + (1/dI)

1/(0.15 m) = 1/(0.10 m) - (1/dI)

1/(0.15 m) - 1/(0.10 m) = - (1/dI)

- (1/dI) = 2/(0.30 m) - 3/(0.60 m)

(1/dI) =0.30 m
The net result is that the person sees the image as being about 30 cm away. This is fairly close to how a person with "perfect" vision sees. What (a proper) lens seems to do is take the furthest and nearest point at which we can see, and adjusts it so that the world falls within the limitations of our eye. Everything in the real world remains its proper distance; all that happens is that the lens adjusts it so that the light enters our eye that way.

Let's take a look at far-sightedness now:


With a corrective lens, the following thing happens:


In this case, the reverse happens from the near-sighted case. It takes all the rays coming in from a source that is very close, and "straightens" them out so that they look like they are coming in from infinity. The lens can be treated in much the same way as the near sighted case.

EXAMPLE: For a particular far-sighted person, they have a near point of 102 cm. How powerful does the lens need to be if they person wants to see an object that 27 cm away from their eye?

As with the first case, the image and the actual object are on the same side, so the image is considered negative. The real object is 25 cm (= 27cm - 2cm) away from the front of the lens, and we want to produce an image that appears to be 100 cm away (= 102 cm - 2 cm) from the lens. Plug everything into the equation:
1/(f) = 1/(dO) + (1/dI)

1/(f) = 1/(0.25 m) - (1/1m)

1/(f) = 4/(1 m) - (1/1m) = 3/(1m) = 3 m-1 = 3.00 D
The resulting lens has a positive power, and is known as a converging lens. It "pulls in" rays that are spreading out. Now that we have "fixed" this person's vision at closer distances, what happens to objects at infinity? Since all that the lens does is "straighten" out light, rays that are already going in a straight line will only be effected slightly. This, in turn, keeps the far point at infinity.

Astigmatism
There is one last type of eye defect that we have not addressed. Astimagtism is an eye defect in which people cannot  properly on lines because their corneas are out of shape.  It  turns out that  people with astimatism have corneas that are not spherical. As a result, light that comes from a single point does not focus at a single point when it goes through their eyes. It actually turns into a line.



In this image, it  the green lens is not a sphere; it is actually part of a cylinder. This deviation from spherical shapes causes the light to continue in a straight path when hitting a lens, instead of focusing together. Unfortunately, there was insufficient time for me to fully address this section, so I will have to leave it at that!

Introduction
Colour Vision
Colour Math
Approximations
Focal Lengths and Distances
GRIN Systems
Human Vision
Vision Problems
Corrections