Prescribing from Limited Spectacle Stock at an Eye Camp
At a free eye-screening camp in a block town, two cases are referred to the optician. P, a 42-year-old bus driver, can see clearly only up to 1.5 m; beyond that everything is blurred, and he cannot read the number plate of a truck 10 m away. Q, a 55-year-old tailor, can read a newspaper only when it is held at least 50 cm from her eyes; at the normal reading distance of 25 cm the print blurs. The camp has ready-made lenses of power +0.5, +1.0, +1.5, +2.0, +2.5, −0.5, −1.0 and −1.5 dioptre. The optician must decide what to prescribe from this stock for each person and record whether the prescription is exact or merely the nearest available. P asks whether the −0.5 D lens from stock will let him drive safely, since it is the closest available to his requirement.
1. Find P's required lens power and decide what should be dispensed from the available stock.
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P is myopic: his far point is 1.5 m, so the lens must take an object at infinity and form its image at 1.5 m on the same side of the lens. The focal length is therefore −1.5 m.
P = 1/f = 1/(−1.5) = −0.67 D (concave, diverging).
From the stock, the neighbours are −0.5 D and −1.0 D; the exact value is not available. The usual rule is to under-correct rather than over-correct a myope, because an over-minus lens makes the eye accommodate even for distant objects and produces strain, headache and a receded near point. On that principle the weaker lens, −0.5 D, is the safer of the two. However, the calculation in the last part shows that −0.5 D leaves distant vision blurred, so for a bus driver the correct decision is to order a −0.67 D (or −0.75 D) lens rather than dispense the stock item.
2. Calculate Q's required power and state whether the stock can supply it exactly.
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Q is hypermetropic: her near point is 50 cm and she wants to read at 25 cm, so the lens must accept an object at 25 cm and produce a virtual image at 50 cm from the lens.
Using the lens formula with the sign convention that distances on the object side are negative: u = −25 cm, v = −50 cm.
1/f = 1/v − 1/u = (−1/50) − (−1/25) = −1/50 + 2/50 = +1/50.
f = +50 cm = +0.5 m, so P = +2.0 D, a converging lens.
The stock contains +2.0 D, so this prescription is exact. Cross-check: the eye must be given an extra 2 D of convergence to bring the near point from 50 cm to 25 cm, which agrees with the value obtained.
3. If the −0.5 D lens is dispensed to P, where does the image of a distant object form, and will he read a board at 4 m? Give a judgement on dispensing it.
Read the worked answer
With P = −0.5 D, f = 1/(−0.5) = −2 m. A distant object is therefore imaged 2 m in front of the lens, but P's eye can focus only up to 1.5 m, so the image falls outside his range and remains blurred - the lens is too weak.
For a board at 4 m: 1/v = 1/f + 1/u = −0.5 + (−0.25) = −0.75, so v = −1.33 m. That image is inside his 1.5 m far point, so he would see the board fairly clearly.
Judgement: the −0.5 D lens only helps for objects nearer than about 2 m and leaves true distance vision uncorrected. It under-corrects by 0.17 D - tolerable for reading a classroom board but not acceptable for a bus driver who must read number plates at 10 m and beyond. The camp should not dispense it to P; the −0.67 D lens must be ordered.