Short sight
A myopic eye sees near objects clearly but distant objects as a blur. Its relaxed lens is too strong for the length of the eyeball. Either the lens curves too much or the eyeball is too long. Parallel rays from a distant object then meet in front of the retina and spread again before they reach it. This defect is myopia, or short sight.
A myopic eye still has a far point, but it is near. Only objects nearer than the far point send rays that spread enough for the eye to focus. Suppose the far point is \(50\,\text{cm}\) away. In the one-lens model the relaxed eye then has a power of \(40 + 1/0.5 = 42\,\text{D}\), \(2\,\text{D}\) too many for a distant object.
Choose the chip "myopic" and slide the object from \(3\,\text{m}\) to \(15\,\text{cm}\). At \(3\,\text{m}\) and \(1\,\text{m}\) the image lies in front of the retina. From \(50\,\text{cm}\) inwards the image is sharp, down to the near point of this eye, about \(17\,\text{cm}\). Now choose the chip with the \(-2\,\text{D}\) lens. Everything from \(3\,\text{m}\) down to \(25\,\text{cm}\) is sharp: the eye behaves like a normal eye.
The correcting lens has one job. It must take light from a distant object and send it on as if it came from the far point. A diverging lens does exactly this with a parallel beam: the beam leaves as if from its focus. So the focus must be at the far point. The lens needs \(f = -x_f\), where \(x_f\) is the distance of the far point.
The far point of a myopic eye is \(2\,\text{m}\) away. Where do parallel rays from a distant object meet? Why is the object blurred? Find the focal length and power of the correcting lens.
The relaxed eye is too strong, so parallel rays meet in front of the retina and spread again before reaching it. Only objects within \(2\,\text{m}\) form sharp images.
A distant object must appear at \(2\,\text{m}\): \(u \to -\infty\), \(v = -2\,\text{m}\). Then \(\dfrac{1}{f} = \dfrac{1}{-2} - 0\), so \(f = -2\,\text{m}\) and \(P = -0.5\,\text{D}\), a weak concave lens.
A myopic eye usually has a near point nearer than \(25\,\text{cm}\) too. The spectacle lens moves that near point outwards. Take an eye with its far point at \(50\,\text{cm}\) and its near point at \(12\,\text{cm}\). The lens for distant vision is \(f = -50\,\text{cm}\), or \(-2\,\text{D}\). With it on, the nearest clear object is the one whose image falls at \(12\,\text{cm}\): \(\dfrac{1}{u} = \dfrac{1}{v} - \dfrac{1}{f} = -\dfrac{1}{12} + \dfrac{1}{50}\), so \(u = -15.8\,\text{cm}\).
A myopic student cannot see clearly beyond \(80\,\text{cm}\). Find the power of the spectacle lens she needs, in dioptres, with its sign.
Long sight
A hypermetropic eye has the opposite fault. Its lens is too weak for the length of the eyeball, or the eyeball is too short. Even with full accommodation it cannot focus a near object: the rays would meet behind the retina. Its near point is farther than \(25\,\text{cm}\), so reading is hard. This defect is hypermetropia, or long sight.
A hypermetropic eye can often see distant objects clearly, by using part of its accommodation. The eye in the figure above, on the chip "hypermetropic", relaxes to \(38\,\text{D}\). It adds \(2\,\text{D}\) for a distant hill and \(4\,\text{D}\) at most. So its near point is where \(40 + 1/x_n = 42\), at \(x_n = 50\,\text{cm}\).
The correcting lens again does one job. It must take light from a page at \(25\,\text{cm}\) and send it on as if from the near point \(N\). So an object at \(u = -25\,\text{cm}\) must give a virtual image at \(v = -N\). The lens formula then fixes the focal length.
\[ u = -25\,\text{cm}, \qquad v = -N \]
Why. the page is held at the least distance of distinct vision. Its virtual image must be at the eye's near point, on the same side.
\[ \dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u} = -\dfrac{1}{N} + \dfrac{1}{25} \]
Why. the lens formula, with the two signed distances put in.
\[ \dfrac{1}{f} = \dfrac{N - 25}{25N} \]
Why. bring the two fractions over the common denominator \(25N\).
\[ f = \dfrac{25N}{N - 25} \]
Read it. take the reciprocal. For \(N > 25\,\text{cm}\), \(f\) is positive: the lens is convex.
For a near point of \(50\,\text{cm}\), \(f = 25 \times 50/25 = +50\,\text{cm}\), a convex lens of \(+2\,\text{D}\). On the chip "hypermetropic, +2 D lens", the figure shows the eye seeing clearly from far away to \(25\,\text{cm}\). For a near point of \(1\,\text{m}\), \(f = 2500/75 = 33.3\,\text{cm}\), or \(+3\,\text{D}\).
A boy cannot read print nearer than \(75\,\text{cm}\). What power of lens lets him read at \(25\,\text{cm}\), in dioptres?
A person's spectacles make a page at \(25\,\text{cm}\) appear \(50\,\text{cm}\) away. What are the lens and the defect?
The ageing eye
The last lesson's grandparent has neither an eye that is too strong nor one that is too weak. The relaxed eye still focuses distant objects. What has gone is accommodation. The eye lens stiffens with age and the ciliary muscles weaken, so the lens can no longer bulge. This loss of accommodation is presbyopia. It begins in most people in their forties.
Take a range of accommodation of \(1.5\,\text{D}\), so the near point is at \(67\,\text{cm}\). The reading lens must make a page at \(25\,\text{cm}\) appear at \(67\,\text{cm}\). Its power is \(P = 1/0.25 - 1/0.667 = 4 - 1.5 = +2.5\,\text{D}\). On the chip "ageing, +2.5 D lens", the figure shows the newspaper sharp at \(25\,\text{cm}\).
The same chip shows a new problem. Slide the object back to \(50\,\text{cm}\) or \(3\,\text{m}\), and it blurs. The relaxed eye already focuses a distant object, and the extra \(2.5\,\text{D}\) is now too much. With the reading lens on, the far point is only \(1/2.5 = 40\,\text{cm}\) away. So older readers look over the top of their reading glasses to see across a room.
The answer is the bifocal lens. Its upper part carries the power for distance, and a small lower part carries the reading power. The wearer looks straight ahead through the top and down through the bottom to read. An eye that is also a little myopic might need \(-1.00\,\text{D}\) above and \(+1.50\,\text{D}\) below. Modern progressive lenses change power smoothly from top to bottom instead.
A sixty-year-old sees distant objects clearly but has a near point of \(67\,\text{cm}\). Which statements are true?
Astigmatism
A perfect cornea is part of a sphere, curved equally in every direction. Many corneas are shaped more like the back of a spoon, curved more steeply in one plane than in the plane at right angles. Rays in the steeper plane then focus nearer than rays in the flatter plane. Vertical lines may be sharp while horizontal lines are blurred. This defect is astigmatism.
A spherical lens cannot fix this, because it adds the same power in every plane. The cure is a cylindrical lens, cut from a cylinder of glass. It bends light only in the plane across its axis and not at all along it. Chosen with the right power and turned to the right angle, it evens out the two planes.
This is why an optician's slip has three numbers for each eye. A slip may read "SPH −1.50, CYL −0.75, AXIS 90". The first is the spherical power, for myopia or hypermetropia. The second is the extra power of the cylinder, and the third is the angle, in degrees, at which its axis is set.
A student sees the vertical bars of a window grille sharply but its horizontal bars blurred. What lens corrects this?
The young English astronomer George Airy found that his left eye saw a point of light as a short line. He studied his own eye as he would a telescope. It focused lines in one direction at a different distance from lines at right angles. In 1825 he had a Cambridge optician grind a lens with a cylindrical face, of the power his own measurements asked for. It worked, and he wore such spectacles for the rest of his life. He later became the Astronomer Royal.
An optician in any Indian town first measures the eye's far and near points with charts and trial lenses in a frame. The trial lenses are marked in dioptres, and powers of lenses held together simply add. A slip of \(-2.00\,\text{D}\) means a concave lens of \(f = -50\,\text{cm}\), for a far point of \(50\,\text{cm}\). The lenses are ground to that power and set about \(1.5\,\text{cm}\) in front of the eye; for small powers the gap barely matters.
- Giving a myopic eye a convex lens because it "cannot see far".A myopic eye is too strong. It needs a concave lens, of negative power, to weaken it.
- Using \(f = N\), the near point, as the focal length for long sight.Use the lens formula with \(u = -25\,\text{cm}\) and \(v = -N\). For \(N = 50\,\text{cm}\) this gives \(f = +50\,\text{cm}\), but for \(N = 100\,\text{cm}\) it gives \(33.3\,\text{cm}\).
- Expecting reading glasses to help with distant objects.A reading lens adds power that a distant object does not need. It pulls the far point in, to \(40\,\text{cm}\) for \(+2.5\,\text{D}\).