It turns out glassescorrect our vision by "enhancing" the natural performance of our owneyes. Even if an eye shows either near or far-sightedness, there is astill a range in which that person can see clearly. The closest point atwhich a person can see an object in perfect focus is called the "nearpoint". Similarly, there is a distance called a "far point" whichrepresents the farthest distance that a person can see a clear, focusedimage. For a person with "perfect" vision, this range of clear vision isaround 25 cm for the near point, all the way out to "infinity". While aneye cannot see all the way out to infinity, in terms of the size of oureyes, anything over five meters represents infinity.
For someone who isnearsighted, they are likely to have a near point that is even closerthan 25 cm. This is due to the fact that as an image moves closer toone's eye, it focuses further and further back. The natural elongationof the back of the eye in a near-sighted person allows it to focus evenwhen these objects are close. However, the far point for a near sightedperson is likely to be quite short; around 17 to 25 cm is average. For afar sighted person, they can see off into "infinity", but can only focuson objects that are a distance away from them (usually a meter or moreaway).
The way thatlens helps us see is by adjusting light rays so that it gives the"impression" that it came from that distance. At this point, itwould be handy to introduce some technical aspects.
LensEquation
This commonly usedformula is used to describe the relationship between three objects. Anobject located dO units away from a lens with a focal lengthf, will create an image with dI units away. They are relatedas such:
1/(f)= 1/(dO) + (1/dI)
where:
f is the focal length of the lens
dO isthe distance from the lens to the object
dI isthe distance from the lens to the object
There is a technicalterm given the left hand side; 1/f , when f is measured in meters iscalled the power of lens. The units have a special name called a diopter (D), and it isdefined as inverse meters or m-1. There is also a handedconvention when dealing with the two distance measurements. If an objectand 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 oppositesides of a lens, then both are considered positive. It turns out thisstrange convention allows the lens equation to work under almost allbasic circ*mstances.
Let's take alook at the near sighted case again:
In this case, thelens that will be used is called a diverginglens. What it is going to do is bend the rays so that light frominfinity "appears" as though it is coninciding with a person's farpoint. In doing so, it allows the person to see a clear image, withoutaltering their depth perception. Here's a look at what the lens does:
The lens actuallybends the light outwards so that it looks like that it coming from asource that is quite close. The following a quick sample calculationthat uses the lens equation.
EXAMPLE: Let's assume that there is anear-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 wouldlike the object distance to be "infinity". We want the image to appear17 cm in front of the person's eye; this happens to be 15 cm (=17 cm - 2cm) in front of the lens. Since the object and image distance are on thesame side, we have to place a negative sign with the image distance, sowe 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 beterribly obvious, we will treat 1/∞ as beingzero for this calculation. If one were to put in a real number, it wouldonly make a small contribution to the overall result. So we end upgetting:
1/f = -1/0.15 m + 0
1/f = -1/0.15 m = 6.667 m-1 =6.667 D
EXAMPLE: Let's say that the sameperson has a near point of 12 cm from their eye. With the lens on, howfar does an object 12 cm away appear to be?
Assuch, we need to use a lens with a focal length of 0.15 meters, or apower of 6.667 D. Now that we've let the person see images from faraway, can they still see images that are up close? Now that we have alens focal power, we can use information about the person's near pointto determine how it has moved.
As with before, the lens and the resultant image are still on thesame side, so we assign a negative sign for the imagedistance. Since the near point is 12 cm with respect to the eye, ithappens to be 10 cm (= 12 cm - 2cm) away from the front of the lens. Wecan plug in what we know into the lens equation.
1/(f)= 1/(dO) + (1/dI)
1/(0.15m) = 1/(0.10 m) - (1/dI)
1/(0.15m) - 1/(0.10m) = - (1/dI)
- (1/dI)= 2/(0.30m) - 3/(0.60m)
(1/dI)=0.30 m
The net result isthat the person sees the image as being about 30 cm away. This is fairlyclose to how a person with "perfect" vision sees. What (a proper) lensseems 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 oureye. Everything in the real world remains its proper distance; all thathappens is that the lens adjusts it so that the light enters our eyethat way.
Let's take a look atfar-sightedness now:
With a corrective lens, the following thing happens:
In this case, thereverse happens from the near-sighted case. It takes all the rays comingin from a source that is very close, and "straightens" them out so thatthey look like they are coming in from infinity. The lens can be treatedin much the same way as the near sighted case.
EXAMPLE: For a particular far-sightedperson, they have a near point of 102 cm. How powerful does the lensneed to be if they person wants to see an object that 27 cm away fromtheir eye?
As with the firstcase, the image and the actual object are on the same side, so the imageis considered negative. The real object is 25 cm (= 27cm - 2cm) awayfrom the front of the lens, and we want to produce an image that appearsto be 100 cm away (= 102 cm - 2 cm) from the lens. Plug everything intothe 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 lenshas 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'svision at closer distances, what happens to objects at infinity? Sinceall that the lens does is "straighten" out light, rays that are alreadygoing in a straight line will only be effected slightly. This, in turn,keeps the far point at infinity.
Astigmatism
There is one lasttype of eye defect that we have not addressed. Astimagtism is an eyedefect in which people cannot properly on lines because theircorneas are out of shape. It turns out that peoplewith astimatism have corneas that are not spherical. As a result, lightthat comes from a single point does not focus at a single point when itgoes 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 acylinder. This deviation from spherical shapes causes the light tocontinue in a straight path when hitting a lens, instead of focusingtogether. Unfortunately, there was insufficient time for me to fullyaddress this section, so I will have to leave it at that!
Introduction
Colour Vision
Colour Math
Approximations
Focal Lengths andDistances
GRIN Systems
HumanVision
Vision Problems
Corrections
