Watch a grandparent read the morning newspaper without spectacles. The paper goes out to arm's length, and the head tilts back. Hand the same paper to a school student. The student reads it comfortably at a hand's width from the face. The print, the light and the paper are the same. Something inside the two pairs of eyes is different.
A camera made of living tissue
Light enters the eye through a clear, curved window called the cornea. Behind it lies a watery liquid and then the coloured iris. The hole in the middle of the iris is the pupil. In bright sunlight the iris narrows the pupil, and in a dark room it widens it. Behind the pupil hangs the eye lens, a soft, clear body held by a ring of ciliary muscles. At the back of the eye lies the retina, a thin layer of light-sensitive cells.
The cornea and the eye lens together act as a converging lens. They form a real, inverted and much smaller image on the retina, as Fig 1.1 shows. The cells of the retina turn the light into signals, and the optic nerve carries them to the brain. The brain reads the inverted image as an upright world.
Most of the bending happens at the cornea, not at the eye lens. Light bends where the refractive index changes, and the largest change is from air, \(1.00\), to the cornea, about \(1.38\). Inside the eye the lens sits between liquids of nearly its own index, so it bends light less. Its job is the fine adjustment, as the next topic shows.
The eye as one lens
The eye works like a camera. The cornea and the lens take the place of the camera's lens, the iris is its diaphragm, and the retina is its sensor. There is one big difference. A camera focuses by moving its lens towards or away from the sensor. The retina cannot move, so the eye focuses by changing the shape, and so the power, of its lens.
In this chapter the eye is modelled as a single thin lens with the retina \(2.5\,\text{cm}\) behind it. For a distant object the image must form at the focus, so \(f = 2.5\,\text{cm}\). The power of the relaxed eye is then \(P = 1/0.025\,\text{m} = 40\,\text{D}\).
In the real eye the light ends in a clear jelly of refractive index about \(1.34\), not in air. So the real eye has a power of about \(60\,\text{D}\), of which the cornea supplies about \(40\,\text{D}\). The simple model gives the wrong total power but the right changes of power. The spectacle calculations of the next lesson never use the eye's power at all, only its near and far points.
Which part of the eye bends the incoming light the most?
What kind of image does the eye form on the retina of a tree 20 m away?
Focusing near and far
The retina stays \(2.5\,\text{cm}\) behind the lens, so the image distance is fixed at \(v = +2.5\,\text{cm}\). Put an object at a distance \(d\) in front, so \(u = -d\). The lens formula says what power the eye needs. \[ P = \dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{0.025\,\text{m}} + \dfrac{1}{d} = 40\,\text{D} + \dfrac{1}{d} \] With \(d\) in metres, a nearer object needs more power. An object \(1\,\text{m}\) away needs \(41\,\text{D}\). One at \(50\,\text{cm}\) needs \(42\,\text{D}\).
The eye gets this extra power from the ciliary muscles. When they relax, the lens is pulled thin and flat, and the eye is set for distant objects. When they contract, the lens is free to bulge, and its surfaces curve more strongly. A more curved lens has a shorter focal length and a larger power. This change of power is called accommodation.
Slide the object in from \(3\,\text{m}\). The eye lens thickens little by little, and the image stays on the retina. At \(25\,\text{cm}\) the readout reaches \(44\,\text{D}\), and the lens can bulge no more. Bring the object nearer still, to \(20\,\text{cm}\) or \(15\,\text{cm}\). The rays now meet behind the retina, and the image on it is a blur.
The near point is the nearest point that the eye can see clearly with its greatest accommodation. For a normal young eye it is about \(25\,\text{cm}\) away, and this distance is called the least distance of distinct vision, \(D = 25\,\text{cm}\). The far point is the farthest point that the relaxed eye can see clearly. For a normal eye it is at infinity.
The range of accommodation of an eye is the power it can add between its far point and its near point. For a normal eye this is \(44 - 40 = 4\,\text{D}\). In symbols, with both distances in metres, \[ A = \dfrac{1}{\text{near point}} - \dfrac{1}{\text{far point}}. \] A normal eye has \(A = 1/0.25 - 0 = 4\,\text{D}\). This book takes \(D = 25\,\text{cm}\) throughout, as the NCERT does.
A student looks up from a book at \(25\,\text{cm}\) to a hill far away, and back. Using the one-lens model of the eye, find the power and the focal length of the eye at each distance.
The image distance is fixed: \(v = +2.5\,\text{cm} = +0.025\,\text{m}\). For the hill, \(u \to -\infty\); for the book, \(u = -0.25\,\text{m}\).
\(P = \dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{0.025} - \dfrac{1}{u}\).
For the hill, \(P = 40\,\text{D}\) and \(f = 2.5\,\text{cm}\). For the book, \(P = 40 + 4 = 44\,\text{D}\) and \(f = 1/44\,\text{m} = 2.27\,\text{cm}\).
Looking from the hill to the book, the eye adds \(4\,\text{D}\) and shortens its focal length by only \(2.3\,\text{mm}\). The ciliary muscles make this change in a fraction of a second.
Take the one-lens model, with the retina \(2.5\,\text{cm}\) behind the lens. What power must a normal eye have to see an object \(50\,\text{cm}\) away, in dioptres?
A normal eye looks at a small ant only \(10\,\text{cm}\) away. What happens?
Now the grandparent's newspaper makes sense. With age the eye lens slowly hardens, and the ciliary muscles weaken. The range of accommodation falls from about \(4\,\text{D}\) in a student to about \(1.5\,\text{D}\) at sixty. The near point then moves out to \(1/1.5 = 0.67\,\text{m}\), which is arm's length. Print nearer than that forms its image behind the retina. The next lesson shows what spectacle lens brings it back.
Apparent size is an angle
How big does an object look? The answer is not its height. A \(2\,\text{m}\) tall man across a playground looks smaller than your thumb held at arm's length. What matters is the size of the image on the retina. A ray from the top of the object through the centre of the eye lens goes straight on. So the height of the retinal image is \(2.5\,\text{cm}\) times the angle that the object makes at the eye.
This angle is the visual angle. For an object of height \(h\) at a distance \(d\), the angle in radians is \(\theta = \tan^{-1}(h/d)\). For small objects the tangent equals the angle, so \(\theta \approx h/d\). A letter \(5\,\text{mm}\) tall at \(25\,\text{cm}\) makes \(\theta = 0.5/25 = 0.02\,\text{rad}\). Its image on the retina is \(2.5\,\text{cm} \times 0.02 = 0.5\,\text{mm}\) tall.
Watch the readout in Fig 1.2. From \(1\,\text{m}\) to \(25\,\text{cm}\) the angle grows from \(0.29°\) to \(1.15°\), four times, as \(1/d\) grows four times. Bringing the letter closer is the simplest way to see it larger. But the near point stops this. At \(4.2\,\text{cm}\) the angle is \(6.84°\), and the letter is a blur. The best the unaided eye can do is the angle at \(25\,\text{cm}\).
The end of Fig 1.2 shows the way round the limit. A convex lens at the eye lets the letter sit at \(4.2\,\text{cm}\) and still be seen sharp. Its virtual image stands at \(25\,\text{cm}\), where the eye can focus, and makes the same large angle, \(6.84°\). That is about six times the best angle without the lens. Every optical instrument in this chapter works this way: it enlarges the angle.
The magnifying power \(M\) of an optical instrument is the ratio of two angles. \(\beta\) is the angle that the final image makes at the eye. \(\alpha\) is the angle that the object makes at the unaided eye: at \(D = 25\,\text{cm}\) for a microscope, and where it stands for a telescope. \[ M = \dfrac{\beta}{\alpha} \]
Why a ratio of angles, and not of heights? The image seen in an instrument may be at \(25\,\text{cm}\) or at infinity. An image at infinity has no finite height, yet it looks large, because it makes a large angle. The image on the retina, and so the apparent size, depends only on the angle.
A coin \(2.0\,\text{cm}\) across is held \(40\,\text{cm}\) from the eye. Find the angle it makes at the eye, in radians.
Why is the magnifying power of an instrument defined by angles rather than by the heights of image and object?
The full Moon makes an angle of about \(0.5°\) at the eye. How far away must a coin \(2\,\text{cm}\) across be held to just cover the Moon, in metres?
For two thousand years scholars argued about how the eye sees. Some thought it sent out rays of its own. In 1604 the German astronomer Johannes Kepler traced the rays through the eye as a lens problem. He showed that the eye forms a real, inverted image on the retina, like the picture in a dark room with a pinhole. The brain, he said, had the separate task of reading it. A few years later Christoph Scheiner cut away the back of an ox's eye and saw the small inverted picture himself.
- Thinking the eye focuses by moving its lens, like a camera.The retina and the lens stay in place. The eye changes the shape, and so the power, of its lens.
- Thinking the eye lens does most of the bending.The cornea does most of it, because the jump in refractive index from air to cornea is the largest. The lens fine-tunes.
- Judging apparent size by the height of an object.Apparent size is the angle at the eye, \(\theta \approx h/d\). A small near object can look larger than a large far one.