Defects of vision and their correction

Spectacles

Many people wear spectacles. The lens of a person who cannot see the blackboard is not the same as the lens of a person who cannot read a book. The two people have two different defects of vision. Both can be corrected with a lens of the right power.

Short sight: myopia

A person with myopia sees near objects clearly, but far objects are blurred. The far point is nearer than infinity, for example $120\ \text{cm}$.

Cause: either the eyeball is too long, or the lens is too curved (the eye lens has too short a focal length). The rays from a far object meet at a point in front of the retina, and they spread out again before they reach it.

Correction: a concave lens of the right power. It makes the rays spread a little before they enter the eye, so they meet on the retina.

The lens must form a virtual image of a far object at the far point of the eye. So its focal length is the far point, with a negative sign: $f = -(\text{far point})$.

Example. A short-sighted student has a far point of $120\ \text{cm}$. The lens needs $f = -120\ \text{cm}$, so $P = \dfrac{1}{-1.2\ \text{m}} = -0.83\ \text{D}$.

Long sight: hypermetropia

A person with hypermetropia sees far objects clearly, but near objects are blurred. The near point is farther than $25\ \text{cm}$, for example $40\ \text{cm}$.

Cause: either the eyeball is too short, or the lens has too long a focal length. The rays from a near object would meet at a point behind the retina, so they have not come together on the retina.

Correction: a convex lens of the right power. It makes the rays meet a little earlier, so they meet on the retina.

The lens must take an object at $25\ \text{cm}$ and form a virtual image at the near point of the eye. So $u = -25\ \text{cm}$ and $v = -(\text{near point})$.

Example. A long-sighted person has a near point of $40\ \text{cm}$. We want $f$, so $\dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u} = -\dfrac{1}{40\ \text{cm}} + \dfrac{1}{25\ \text{cm}} = \dfrac{-5 + 8}{200\ \text{cm}} = \dfrac{3}{200\ \text{cm}}$. So $f = +66.7\ \text{cm}$, and $P = +1.5\ \text{D}$.

You should find that a long eyeball focuses in front of the retina and needs a negative (concave) lens. A short eyeball focuses behind it and needs a positive (convex) one. A difference of a fraction of a millimetre is enough to blur the view.

Presbyopia

With age, the ciliary muscles become weaker and the lens becomes less flexible. The near point moves away, as in long sight. This is presbyopia. A person can have it together with myopia, and then needs two kinds of lens in the same spectacles. The upper part is concave, for far objects, and the lower part is convex, for reading. This is a bifocal lens.

Cataract

In cataract, the lens of the eye becomes cloudy, and less light passes through. Vision is blurred, and may be lost. It is cured by an operation, in which the cloudy lens is taken out and replaced by a clear artificial lens.

Two defects and their lenses

Practise with new numbers

Try it in Python

The program finds the power of the lens for a short-sighted eye and for a long-sighted eye.

Remember these

• Myopia: the image forms in front of the retina. A concave lens corrects it.
• Hypermetropia: the image forms behind the retina. A convex lens corrects it.
• Presbyopia comes with age, and a bifocal lens helps. Cataract is a cloudy lens, and it is cured by an operation.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. A man reads a far road sign easily, but cannot read his phone at $30\ \text{cm}$. What is the defect, and what lens helps?

2. What is the power of the lens for a far point of $2\ \text{m}$?

3. Which spectacles have a negative power: those for short sight or for long sight?

4. A woman's near point is $80\ \text{cm}$. What power of lens lets her read at $25\ \text{cm}$?