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19.2 Vision correction  (Page 29/12)

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  • Identify and discuss common vision defects.
  • Explain nearsightedness and farsightedness corrections.
  • Explain laser vision correction.

The need for some type of vision correction is very common. Common vision defects are easy to understand, and some are simple to correct. [link] illustrates two common vision defects. Nearsightedness , or myopia    , is the inability to see distant objects clearly while close objects are clear. The eye overconverges the nearly parallel rays from a distant object, and the rays cross in front of the retina. More divergent rays from a close object are converged on the retina for a clear image. The distance to the farthest object that can be seen clearly is called the far point    of the eye (normally infinity). Farsightedness , or hyperopia    , is the inability to see close objects clearly while distant objects may be clear. A farsighted eye does not converge sufficient rays from a close object to make the rays meet on the retina. Less diverging rays from a distant object can be converged for a clear image. The distance to the closest object that can be seen clearly is called the near point    of the eye (normally 25 cm).

Part a shows two figures of cross-sectional area of eye depicting myopia. In both the figures, parallel rays coming from an object placed at infinity are converging in front of the retina. Figure on the left shows the lens of the eye too strong and figure on the right illustrates the shape of the eye too long. Part b shows two figures of cross-sectional area of eye depicting hyperopia. In both the figures, rays coming from a close object are shown which are converging at the back of the retina. Figure on the left shows the lens of the eye too weak and figure on the right illustrates the shape of the eye too short.
(a) The nearsighted (myopic) eye converges rays from a distant object in front of the retina; thus, they are diverging when they strike the retina, producing a blurry image. This can be caused by the lens of the eye being too powerful or the length of the eye being too great. (b) The farsighted (hyperopic) eye is unable to converge the rays from a close object by the time they strike the retina, producing blurry close vision. This can be caused by insufficient power in the lens or by the eye being too short.

Since the nearsighted eye over converges light rays, the correction for nearsightedness is to place a diverging spectacle lens in front of the eye. This reduces the power of an eye that is too powerful. Another way of thinking about this is that a diverging spectacle lens produces a case 3 image, which is closer to the eye than the object (see [link] ). To determine the spectacle power needed for correction, you must know the person’s far point—that is, you must know the greatest distance at which the person can see clearly. Then the image produced by a spectacle lens must be at this distance or closer for the nearsighted person to be able to see it clearly. It is worth noting that wearing glasses does not change the eye in any way. The eyeglass lens is simply used to create an image of the object at a distance where the nearsighted person can see it clearly. Whereas someone not wearing glasses can see clearly objects that fall between their near point and their far point, someone wearing glasses can see images that fall between their near point and their far point.

Two illustrations of cross-sectional view of an eye are shown. In the first figure, a diverging spectacle lens is placed in front of the eye structure. A ray diagram for the diverging lens is also shown. Parallel rays from a distant object, taken as tree, are striking the lens and then diverging. A smaller image of the tree is shown in front of the lens. In the second figure, a ray diagram with respect to the diverging lens within the eye structure is shown. Parallel rays from a distant object are striking the diverging lens, entering the lens of the eye, and converging at retina. This explains the correction of nearsightedness using a diverging lens.
Correction of nearsightedness requires a diverging lens that compensates for the overconvergence by the eye. The diverging lens produces an image closer to the eye than the object, so that the nearsighted person can see it clearly.

Correcting nearsightedness

What power of spectacle lens is needed to correct the vision of a nearsighted person whose far point is 30.0 cm? Assume the spectacle (corrective) lens is held 1.50 cm away from the eye by eyeglass frames.

Strategy

You want this nearsighted person to be able to see very distant objects clearly. That means the spectacle lens must produce an image 30.0 cm from the eye for an object very far away. An image 30.0 cm from the eye will be 28.5 cm to the left of the spectacle lens (see [link] ). Therefore, we must get d i = − 28.5 cm size 12{d rSub { size 8{i} } = - "28" "." 5"cm"} {} when d o ≈ ∞ size 12{d rSub { size 8{o} } approx infinity } {} . The image distance is negative, because it is on the same side of the spectacle as the object.

Solution

Since d i size 12{d rSub { size 8{i} } } {} and d o size 12{d rSub { size 8{o} } } {} are known, the power of the spectacle lens can be found using P = 1 d o + 1 d i size 12{P= { {1} over {d rSub { size 8{o} } } } + { {1} over {d rSub { size 8{i} } } } } {} as written earlier:

P = 1 d o + 1 d i = 1 ∞ + 1 − 0 . 285 m . size 12{P= { {1} over {d rSub { size 8{o} } } } + { {1} over {d rSub { size 8{i} } } } = { {1} over { infinity } } + { {1} over { - 0 "." "285 m"} } } {}

Since 1/ ∞ = 0 size 12{"1/" infinity " = 0"} {} , we obtain:

P = 0 − 3 . 51 / m = − 3 . 51 D . size 12{P=0 - 3 "." "51"/m= - 3 "." "51 D"} {}

Discussion

The negative power indicates a diverging (or concave) lens, as expected. The spectacle produces a case 3 image closer to the eye, where the person can see it. If you examine eyeglasses for nearsighted people, you will find the lenses are thinnest in the center. Additionally, if you examine a prescription for eyeglasses for nearsighted people, you will find that the prescribed power is negative and given in units of diopters.

Questions & Answers

A golfer on a fairway is 70 m away from the green, which sits below the level of the fairway by 20 m. If the golfer hits the ball at an angle of 40° with an initial speed of 20 m/s, how close to the green does she come?
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can someone explain to me, an ignorant high school student, why the trend of the graph doesn't follow the fact that the higher frequency a sound wave is, the more power it is, hence, making me think the phons output would follow this general trend?
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Nevermind i just realied that the graph is the phons output for a person with normal hearing and not just the phons output of the sound waves power, I should read the entire thing next time
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Follow up question, does anyone know where I can find a graph that accuretly depicts the actual relative "power" output of sound over its frequency instead of just humans hearing
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Source:  OpenStax, Physics 101. OpenStax CNX. Jan 07, 2013 Download for free at http://legacy.cnx.org/content/col11479/1.1
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