The human eye and its defects

Reading a bus number, then your phone

A bus is far down the road. You read its number. Then you look at the phone in your hand, and the text is sharp too. The same eye saw both. Yet a camera has to move its lens to change from far to near. How does the eye do it? And why do some people need spectacles?

The parts of the eye

Each part has a job in the optics.

• Cornea. The clear front window. Light bends most here. It gives about two thirds of the focusing power of the whole eye.
• Aqueous humour. A watery fluid between the cornea and the lens.
• Iris and pupil. The iris is a coloured ring of muscle. The hole in its middle is the pupil. The iris makes the pupil smaller in bright light and larger in dim light.
• Lens. A soft, clear lens that can change its shape.
• Ciliary muscles. Muscles round the lens. They change its shape.
• Vitreous humour. A clear jelly that fills the large space behind the lens.
• Retina. The screen at the back. It holds light-sensitive cells, the rods and the cones. The fovea is the small spot where the image is sharpest.
• Optic nerve. It carries the signals to the brain.

The image on the retina

The cornea and the lens together act as one converging lens. For an object that is far away, the eye forms a real, inverted and smaller image on the retina. The brain has learned to turn this picture the right way up, so we do not see the world upside down.

In the simple model of NCERT, the distance from the lens to the retina is about $2.5\ \text{cm}$. This distance cannot change. So the focal length of the eye must change instead, if the object distance changes.

Why this model is a simplification. The real eye has a power of about $60\ \text{D}$ when relaxed, and the cornea alone gives about $43\ \text{D}$. The fluids inside the eye have a refractive index of about $1.34$, not $1.00$. In the model of a thin lens in air, with $2.5\ \text{cm}$ from lens to retina, the power is $40\ \text{D}$. The model keeps the sums simple, and the changes in power that we find from it are correct.

Accommodation

When you look at a far object, the ciliary muscles relax. The lens is then thin and has its longest focal length. When you look at a near object, the muscles squeeze the lens. It becomes thicker and its focal length becomes shorter. This change of the focal length of the eye lens is called accommodation.

Let us check the numbers in the model. A relaxed eye has $f = 2.5\ \text{cm}$. Now look at a book at $25\ \text{cm}$. The image must stay on the retina at $v = 2.5\ \text{cm}$, so

and $f = 2.27\ \text{cm}$. The focal length fell from $2.50\ \text{cm}$ to $2.27\ \text{cm}$. The power rose by $\dfrac{1}{0.25\ \text{m}} = 4.0\ \text{D}$. This is the power of accommodation that a normal eye needs for a book at $25\ \text{cm}$.

The eye cannot keep squeezing. There is a nearest point that it can still bring into sharp focus.

• The near point is the nearest point that the eye can see clearly. By convention it is $25\ \text{cm}$ for a normal eye. It is also called the least distance of distinct vision, and we write it $D$. A young person may do better, and an older person worse.
• The far point is the farthest point that the eye can see clearly. For a normal eye it is at infinity.

So a normal eye sees clearly everything from $25\ \text{cm}$ to infinity. This is its range of vision.

The power of a lens

Opticians do not quote a focal length. They quote the power $P = \dfrac{1}{f}$, with $f$ in metre. The unit is the dioptre, symbol D. A converging (convex) lens has a positive power. A diverging (concave) lens has a negative power. A lens with $f = 0.50\ \text{m}$ has $P = +2.0\ \text{D}$. Lenses close together add their powers.

A prescription such as $-2.50\ \text{D}$ means a concave lens with $f = -40\ \text{cm}$. In the lessons here the spectacle lens is thin and is treated as sitting right at the eye, unless a problem says otherwise.

Sign conventions. We use the New Cartesian convention of NCERT. A-Level and IB use "real is positive", $\dfrac{1}{u} + \dfrac{1}{v} = \dfrac{1}{f}$. For the eye both conventions give the same powers. A virtual image at the far point has $v = -d_f$ in both of them.

Short sight, or myopia

A short-sighted person sees near things clearly but distant things as blurred. The far point is not at infinity. It is at a finite distance $d_f$. Either the eyeball is too long or the lens is too strong, so the rays from a far object meet in front of the retina.

The cure is a concave lens. It must take a very distant object, at $u = -\infty$, and make a virtual image at the far point, $v = -d_f$. Then the eye can see that image. We want $f$, so

Example. Amir cannot see clearly beyond $40\ \text{cm}$. Then $f = -40\ \text{cm}$ and $P = \dfrac{1}{-0.40\ \text{m}} = -2.5\ \text{D}$.

If the lens is not at the eye. Spectacles sit about $1\ \text{cm}$ to $2\ \text{cm}$ in front of the eye. Then the image must be at the far point of the eye, which is $(d_f - s)$ from the lens, so $f = -(d_f - s)$. For Amir with $s = 2\ \text{cm}$, $f = -38\ \text{cm}$ and $P = \dfrac{1}{-0.38\ \text{m}} = -2.63\ \text{D}$. The difference is small for weak lenses and larger for strong ones.

Start with the defaults. What is the power for a far point of $100\ \text{cm}$? Now make the far point $200\ \text{cm}$. Does the lens become stronger or weaker? Then move the lens $4\ \text{cm}$ from the eye. Which way does the power change?

Long sight, or hypermetropia

A long-sighted person sees far things clearly but near things as blurred. The near point is farther than $25\ \text{cm}$, at a distance $d_n$. Either the eyeball is too short or the lens is too weak, so the rays from a near object would meet behind the retina.

The cure is a convex lens. It must take a book at $25\ \text{cm}$, so $u = -25\ \text{cm}$, and make a virtual image at the near point, $v = -d_n$. Then the eye can see that image:

Example. A woman has a near point of $75\ \text{cm}$. Then $P = 4.0\ \text{D} - \dfrac{1}{0.75\ \text{m}} = 4.0\ \text{D} - 1.33\ \text{D} = +2.67\ \text{D}$. The focal length is $\dfrac{1}{2.67\ \text{D}} = 0.375\ \text{m} = 37.5\ \text{cm}$.

Start with the defaults. A near point of $100\ \text{cm}$ gives $3.00\ \text{D}$. Now drag the near point to $300\ \text{cm}$. The curve flattens. Why can the power never reach $4\ \text{D}$?

Practise with new numbers

Presbyopia, bifocals and astigmatism

Presbyopia comes with age. The lens becomes stiff and the ciliary muscles are weaker, so the near point moves away. A person with normal distance vision then needs convex reading glasses, as in long sight. A person who is also short-sighted needs two lenses: a concave one for distance and a convex one for reading. A bifocal lens holds both. Its upper part is concave and its lower part is convex.

Example. A man has a far point of $100\ \text{cm}$ and a near point of $60\ \text{cm}$. The upper part of his bifocal is $P = -\dfrac{1}{1.0\ \text{m}} = -1.0\ \text{D}$. The lower part must make a book at $25\ \text{cm}$ look as if it were at $60\ \text{cm}$, so $P = 4.0\ \text{D} - \dfrac{1}{0.60\ \text{m}} = +2.33\ \text{D}$.

Astigmatism is a different kind of defect. The cornea, or the lens, is not equally curved in every direction. It is more curved in one plane than in another, a little like a spoon. Lines in one direction look sharp and lines in the other direction look blurred. A cylindrical lens, curved in one plane only, corrects it. It can be combined with a spherical lens if the eye is also short-sighted or long-sighted.

Other eye problems, such as cataract, a cloudy lens, are medical and are not cured by a lens of the right power.

The eye in biology too

The eye is a topic of biology as well. There you learn about the rods and the cones, the blind spot and the way a signal travels to the brain. Here we need only the optics: a real image on the retina, accommodation, and the lenses that repair the defects.

Try it in Python

The program finds the lens for four short-sighted eyes and four long-sighted eyes. The lens is thin and sits at the eye.

Common mistakes

• Using the wrong lens. Short sight needs a concave lens with a negative power. Long sight needs a convex lens with a positive power.
• Mixing up the two points. The far point matters for short sight. The near point matters for long sight.
• Using the near point itself as the power. The lens makes a book at $25\ \text{cm}$ look as if it were at the near point. So $P = \dfrac{1}{0.25\ \text{m}} - \dfrac{1}{d_n}$, not $\dfrac{1}{d_n}$.
• Forgetting to change cm to m. The dioptre needs the focal length in metre.
• Thinking that the retina moves. It does not. The focal length of the lens changes.
• Saying that a short-sighted eye cannot see anything far away. It can see up to its far point. Beyond it, things are blurred.

Remember these

• The eye forms a real, inverted, smaller image on the retina. The cornea gives most of the bending, and the lens fine-tunes it.
• Accommodation changes the focal length of the lens. A normal eye sees from $25\ \text{cm}$ (near point) to infinity (far point).
• Power $P = \dfrac{1}{f}$ in dioptre, with $f$ in metre. Convex is positive and concave is negative.
• Short sight: far point at $d_f$. Concave lens, $f = -d_f$, $P = -\dfrac{1}{d_f}$.
• Long sight: near point at $d_n$. Convex lens, $P = \dfrac{1}{0.25\ \text{m}} - \dfrac{1}{d_n}$.
• Presbyopia is helped by bifocals. Astigmatism is helped by a cylindrical lens.

Check yourself

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

1. In the model with $2.5\ \text{cm}$ from the lens to the retina, what is the focal length of the eye lens when it looks at a page $40\ \text{cm}$ away?

2. A girl cannot see clearly beyond $150\ \text{cm}$. Which lens does she need, and what is its power?

3. A long-sighted man sees distant objects clearly but cannot read a newspaper held at $25\ \text{cm}$. Explain why, in terms of the focus on the retina.

4. The near point of a woman is $80\ \text{cm}$. What is the power of the reading glasses that she needs?

5. A person sees vertical lines clearly but horizontal lines look blurred. Which defect is this, and which lens is used?

References

• NCERT Physics Part II (Class 12), Chapter 9, the section on the human eye and the optical instruments.
• OpenStax College Physics 2e, Chapter 26, the sections on vision correction.