The Human Eye and the Colourful World • Topic 2 of 3

Defects of Vision & their Correction

A normal eye sees clearly from $25\,\text{cm}$ (near point) all the way to infinity (far point). When the eye loses part of this range, we say it has a defect of vision. Most defects happen because the image forms slightly in front of or behind the retina. The good news: simple spectacle lenses of the correct power shift the image back onto the retina.

1. Myopia (near-sightedness): A myopic person sees nearby objects clearly but distant objects appear blurred. The image of a distant object forms in front of the retina. Causes: either the eyeball is too long, or the eye lens is too curved (focal length too short). The far point shifts from infinity to a closer point. Correction: a concave (diverging) lens of suitable negative power. It diverges the rays slightly before they enter the eye so the image moves back onto the retina.

2. Hypermetropia (far-sightedness): A hypermetropic person sees distant objects clearly but nearby objects appear blurred. The image of a near object forms behind the retina. Causes: either the eyeball is too short, or the eye lens is too flat (focal length too long). The near point shifts beyond $25\,\text{cm}$. Correction: a convex (converging) lens of suitable positive power.

3. Presbyopia: With age, the ciliary muscles weaken and the eye lens loses flexibility, so the power of accommodation falls. The near point recedes and reading becomes difficult. A person may have both myopia and presbyopia, needing bi-focal lenses (concave upper part for distance, convex lower part for reading).

4. Cataract: The eye lens becomes cloudy/milky due to a membrane growing over it, causing foggy vision or even blindness. It cannot be cured by glasses; it is treated by surgery in which the cloudy lens is replaced by an artificial transparent lens.

Lens power: The power $P$ of a lens is the reciprocal of its focal length $f$ in metres: $P=\frac{1}{f}$. The unit is the dioptre (D). A concave lens has negative power; a convex lens has positive power. So myopia correction needs a lens of negative power and hypermetropia correction needs a lens of positive power.

Correction of myopia using a concave lens so the image forms on the retinaRetinaConcave lensEye lensDiverged rays now converge on the retina (image corrected)
Solved Examples
1
Worked Example
A person can see distant objects clearly but cannot read a book held at the normal reading distance. Name the defect and the lens used to correct it.
Solution
  1. Distant vision is fine but near vision is blurred, so the image of a near object forms behind the retina.
  2. This defect is hypermetropia (far-sightedness).
  3. It is corrected with a convex (converging) lens of suitable positive power.

Answer: Hypermetropia; corrected by a convex lens of positive power.

2
Worked Example
A myopic person can clearly see objects only up to $0.5\,\text{m}$. Find the nature and power of the corrective lens.
Solution
  1. For myopia, the corrective concave lens must bring the image of an object at infinity to the eye's far point.
  2. So the lens must form a virtual image at the far point: object distance $u=-\infty$, image distance $v=-0.5\,\text{m}$.
  3. Using $\frac{1}{f}=\frac{1}{v}-\frac{1}{u}=\frac{1}{-0.5}-\frac{1}{-\infty}=-2$, so $f=-0.5\,\text{m}$.
  4. Power $P=\frac{1}{f}=\frac{1}{-0.5}=-2\,\text{D}$.

Answer: A concave lens of power $P=-2\,\text{D}$.

3
Worked Example
Find the power of a corrective lens whose focal length is $+25\,\text{cm}$, and state the defect it corrects.
Solution
  1. Convert focal length to metres: $f=+25\,\text{cm}=+0.25\,\text{m}$.
  2. Apply $P=\frac{1}{f}=\frac{1}{0.25}=+4\,\text{D}$.
  3. The power is positive, so the lens is convex, used to correct hypermetropia.

Answer: $P=+4\,\text{D}$ (convex lens), corrects hypermetropia.

4
Worked Example
The far point of a myopic eye is $80\,\text{cm}$ in front of the eye. What power of lens is required to correct this defect?
Solution
  1. The concave lens must form a virtual image of a distant object at the far point: $v=-80\,\text{cm}=-0.8\,\text{m}$, $u=-\infty$.
  2. $\frac{1}{f}=\frac{1}{v}-\frac{1}{u}=\frac{1}{-0.8}-0=-1.25$, so $f=-0.8\,\text{m}$.
  3. $P=\frac{1}{f}=\frac{1}{-0.8}=-1.25\,\text{D}$.

Answer: A concave lens of power $P=-1.25\,\text{D}$.

5
Worked Example
A hypermetropic person has a near point of $1\,\text{m}$ and wishes to read at $25\,\text{cm}$. Find the power of the spectacle lens needed.
Solution
  1. The convex lens must form a virtual image at the person's near point of an object placed at $25\,\text{cm}$.
  2. So $u=-25\,\text{cm}=-0.25\,\text{m}$ and $v=-100\,\text{cm}=-1\,\text{m}$.
  3. $\frac{1}{f}=\frac{1}{v}-\frac{1}{u}=\frac{1}{-1}-\frac{1}{-0.25}=-1+4=3$, so $f=\frac{1}{3}\,\text{m}$.
  4. $P=\frac{1}{f}=+3\,\text{D}$.

Answer: A convex lens of power $P=+3\,\text{D}$.

6
Worked Example
Why does an elderly person often need bi-focal lenses, and what type of lens is each part?
Solution
  1. With age the eye may develop presbyopia along with an existing defect such as myopia.
  2. Distant vision needs one type of lens while near (reading) vision needs another, so a single lens cannot serve both.
  3. Bi-focal lenses combine two parts: a concave upper portion for distant vision and a convex lower portion for reading.

Answer: Bi-focal lenses are used; upper part concave (distance), lower part convex (reading).

Key Points

  • Myopia: distant objects blurred, image forms in front of retina; corrected by a concave lens (negative power).
  • Hypermetropia: near objects blurred, image forms behind retina; corrected by a convex lens (positive power).
  • Presbyopia: age-related loss of accommodation; often needs bi-focal lenses.
  • Cataract clouds the lens and is corrected only by surgery, not by spectacles.
  • Lens power $P=\frac{1}{f}$ in dioptre (D), with $f$ in metres; concave $\Rightarrow$ negative, convex $\Rightarrow$ positive.
Quick Check — 4 MCQs
Tap an option to check your answer0 / 4
Q1.Myopia is corrected by using a
Explanation: A concave (diverging) lens moves the image back onto the retina.
Q2.The power of a concave lens of focal length 0.5 m is
Explanation: Concave lens: $f=-0.5\,\text{m}$, $P=1/f=-2\,\text{D}$.
Q3.In hypermetropia, the image of a near object forms
Explanation: Far-sighted eyes converge near rays too late, behind the retina.
Q4.Cataract is corrected by
Explanation: The clouded lens is surgically replaced; glasses cannot help.
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