Optical Instruments

Lessons

Senior 4 · Physics

Optical Instruments

4 lessons

Lesson 4 of 4

The Human Eye and Defects of Vision

25 min

The instrument you look through

A pupil at the back of the class squints at the board but reads her notebook easily; her grandfather reads the newspaper at arm's length, or not at all without his glasses. Both are using an optical instrument with one variable lens, the eye, and both have a fault that a single extra lens can fix. In this lesson you model the eye as a single lens and screen, find out how it focuses, and calculate the lenses that correct its common defects.

The eye as a single-lens system

Figure 4.1: The main parts of the human eye.

Part	Job	Camera equivalent
cornea	transparent curved front; does most of the refracting (about two-thirds of the eye's power)	front element of the lens
iris and pupil	the iris muscles change the size of the pupil to control the light entering	diaphragm (aperture)
lens and ciliary muscles	fine focusing: the muscles change the lens's shape and so its focal length	focusing mechanism
retina	light-sensitive layer of rods (dim light) and cones (colour); densest at the fovea	sensor or film
optic nerve	carries signals to the brain; where it leaves there are no receptors (the blind spot)	cable to the processor

For calculations we use a reduced eye: one thin converging lens, standing for the cornea and lens together, a fixed distance of about 2.0 cm from the retina. The image distance is therefore fixed at $v = 2.0\,\mathrm{cm}$ , unlike in a camera. To focus objects at different distances the eye changes its power. This is accommodation: to view near objects the ciliary muscles contract, the lens becomes fatter and its power increases; for distant objects they relax and the lens becomes thinner.

The near point is the closest point the eye can focus on (for a normal young adult, about 25 cm; for a child it can be under 10 cm). The far point is the farthest (for a normal eye, infinity). The range of accommodation is the change in power between them.

Definition

Accommodation is the ability of the eye to change the power of its lens so that objects at different distances are focused on the retina. The near point is the nearest point at which an object can be seen clearly; the far point is the farthest. The amplitude of accommodation is the difference between the eye's powers when focused at its near and far points. Example: a normal eye (near point 25 cm, far point infinity) has an amplitude of 4.0 D.

Worked example 1: the power of a normal eye

Model the eye as a lens 2.00 cm from the retina. Find its power when viewing (a) a distant object, (b) an object at the near point, 25.0 cm away.

\text{(a)}\;\; P = \frac{1}{u} + \frac{1}{v} = 0 + \frac{1}{0.0200\,\mathrm{m}} = 50.0\,\mathrm{D} \qquad \text{(b)}\;\; P = \frac{1}{0.250} + \frac{1}{0.0200} = 4.0 + 50.0 = 54.0\,\mathrm{D}

The eye's power is large (its focal length is only 2 cm), but accommodation changes it by only 4.0 D, about 8%. That small change is all that separates reading a book from looking at the hills.

Defects of vision and their correction

Figure 4.2: Myopia and hypermetropia, and their correction with spectacle lenses (the lenses are drawn close to the eye).

Myopia (short sight)

The eyeball is too long or the eye too powerful, so parallel light from distant objects focuses in front of the retina (Figure 4.2a). The far point is closer than infinity; near objects are seen clearly. Correction: a diverging lens that takes light from a very distant object and makes it appear to come from the eye's far point. For a lens in contact with the eye (a contact lens), with far point at distance $x$ :

u = \infty,\; v = -x \;\Rightarrow\; f = -x, \qquad P = -\frac{1}{x}

Hypermetropia (long sight)

The eyeball is too short or the eye too weak, so light from near objects would focus behind the retina (Figure 4.2c). The near point is farther than 25 cm; distant objects can usually be seen. Correction: a converging lens that takes an object at 25 cm and forms a virtual image at the eye's near point $N$ :

u = 0.25\,\mathrm{m},\; v = -N \;\Rightarrow\; P = \frac{1}{0.25} - \frac{1}{N}

Presbyopia and astigmatism

Presbyopia is the loss of accommodation with age: the lens stiffens and the near point moves out (to a metre or more in some older people), so many people need reading glasses (converging) from middle age, and some need bifocals with a distance part above and a reading part below. Astigmatism occurs when the cornea is curved more in one direction than another, so vertical and horizontal lines cannot be in focus together; it is corrected with a cylindrical lens, curved in one direction only.

Defect	Symptom	Cause	Correcting lens
myopia	distant objects blurred	eyeball too long or eye too powerful	diverging (negative power)
hypermetropia	near objects blurred	eyeball too short or eye too weak	converging (positive power)
presbyopia	near objects blurred, with age	lens loses elasticity	converging reading glasses or bifocals
astigmatism	lines in one direction blurred	cornea not spherical	cylindrical lens

Worked example 2: correcting myopia

A student's far point is 50.0 cm. Find the power of (a) a contact lens, (b) spectacles worn 1.5 cm in front of the eye, that let her see distant objects. (c) Her near point is 12.0 cm; where is it with the contact lens in?

\text{(a)}\;\; P = -\frac{1}{0.500} = -2.00\,\mathrm{D} \qquad \text{(b)}\;\; f = -(50.0 - 1.5) = -48.5\,\mathrm{cm},\quad P = \frac{1}{-0.485} = -2.06\,\mathrm{D}

\frac{1}{u} = P - \frac{1}{v} = -2.00 + \frac{1}{0.120} = 6.33\,\mathrm{m^{-1}} \;\Rightarrow\; u = 15.8\,\mathrm{cm}

The correction moves her near point out a little, from 12.0 cm to 15.8 cm, which is still well within normal reading distance.

Worked example 3: correcting hypermetropia

A man's near point is 1.00 m. Find the power of (a) a contact lens, (b) reading glasses worn 2.0 cm in front of the eye, that let him read at 25 cm.

\text{(a)}\;\; P = \frac{1}{0.25} - \frac{1}{1.00} = 4.00 - 1.00 = +3.00\,\mathrm{D}

(b) Measured from the spectacle lens, the book is at $u = 0.25 - 0.02 = 0.23\,\mathrm{m}$ and the image must be at the near point, $v = -(1.00 - 0.02) = -0.98\,\mathrm{m}$ :

P = \frac{1}{0.23} - \frac{1}{0.98} = 4.35 - 1.02 = +3.33\,\mathrm{D}

Opticians quote power in dioptres for exactly this reason: the power needed depends on where the lens sits.

Experiment 4.1: finding your near point and your blind spot

Near point. Cover one eye. Bring a page of small print slowly towards the open eye until the letters just begin to blur. Ask a partner to measure the distance from the page to your eye. Repeat three times and average. Compare results across the class and with older adults.
Blind spot. On a card draw a cross and, 8 cm to its right, a dot. Close your left eye and stare at the cross with your right eye from about 40 cm. Move the card slowly closer: at one distance the dot disappears, because its image falls on the blind spot where the optic nerve leaves the eye.

Never test vision with lasers or bright lights; this experiment uses only printed paper.

Warning

Common mistakes. (1) Correcting myopia with a converging lens because the person "cannot see far": the eye is too strong, so a diverging lens is needed. (2) Treating the eye like a camera that moves its lens: the eye changes its focal length. (3) Forgetting the minus sign for a virtual image at the near or far point. (4) Ignoring the gap between spectacles and eye when a question gives it. Health: anyone who has trouble seeing the board or reading should have their eyes tested by a trained eye-care worker; wearing someone else's glasses is not a safe substitute.

Tip

Myopia: far point too near, fix the far point with $P = -\frac{1}{\text{far point}}$ . Hypermetropia: near point too far, fix the near point with $P = \frac{1}{0.25} - \frac{1}{\text{near point}}$ .

Vision correction (PhysicsOMG). Watch how the near and far points describe an eye, why a myopic eye needs a diverging lens and a hyperopic eye a converging one, and how the lens power is calculated.

✦

Practice

Name the part of the eye that (a) does most of the refracting, (b) controls the amount of light entering, (c) changes the eye's focal length.
Explain how the eye focuses on near and on distant objects. How is this different from a camera?
Using the reduced eye (lens 2.0 cm from the retina), find the power of an eye focused on an object 50 cm away.
A person's far point is 80 cm. What contact lens does she need to see distant objects clearly?
A person's near point is 75 cm. Find the power of a contact lens that lets him read at 25 cm.
The student in question 4 has a near point of 15 cm. Where is her near point when she wears the correcting lens?
A 50-year-old can see distant objects clearly but needs reading glasses of +2.0 D. A friend says "your eyes are long-sighted, the same as hypermetropia". Evaluate this statement.

Show answersHide answers

(a) cornea; (b) iris (by changing the pupil size); (c) the lens, through the ciliary muscles.
For near objects the ciliary muscles contract, the lens becomes fatter and its power increases; for distant objects they relax and it becomes thinner. The lens-to-retina distance is fixed. A camera keeps its focal length and moves the lens to change $v$ .
$P = \frac{1}{0.50} + \frac{1}{0.020} = 2.0 + 50 = 52\,\mathrm{D}$ .
$P = -\frac{1}{0.80} = -1.25\,\mathrm{D}$ (diverging).
$P = \frac{1}{0.25} - \frac{1}{0.75} = 4.00 - 1.33 = +2.67\,\mathrm{D}$ (converging).
$\frac{1}{u} = -1.25 + \frac{1}{0.15} = 5.42\,\mathrm{m^{-1}}$ , so $u = 18\,\mathrm{cm}$ .
Partly right. Both cause near objects to blur and both are corrected with converging lenses. But the cause differs: hypermetropia is an eye that is too short or too weak, present at any age, and may also need correction for distance; presbyopia is the loss of accommodation as the lens stiffens with age. This person's distance vision is normal, so the condition is presbyopia.

Quick check

How does the eye focus on a nearer object?

Key points

The eye is a variable-power lens with a fixed image distance (about 2.0 cm); accommodation changes its power by about 4 D in a normal adult.
Normal eye: near point 25 cm, far point infinity.
Myopia: far point too near, corrected by a diverging lens, $P = -\frac{1}{\text{far point}}$ .
Hypermetropia: near point too far, corrected by a converging lens, $P = \frac{1}{0.25} - \frac{1}{\text{near point}}$ ; presbyopia is age-related loss of accommodation; astigmatism needs a cylindrical lens.

1The instrument you look through

2The eye as a single-lens system

3Defects of vision and their correction