An optician tests a patient's eye and places two corrective lenses of power +1.5 D and +0.75 D in contact inside the trial frame. (a) What is the equivalent power of this combination? (b) What focal length single lens could replace the combination? (c) Why is it more practical for opticians to work with powers rather than focal lengths when combining lenses?
Generated by claude-sonnet-4-6 · 2026-06-26 01:16 · grounding rag
Model Answer
(a) Equivalent power of the combination:
$$P = P_1 + P_2 = +1.5 \text{ D} + 0.75 \text{ D} = +2.25 \text{ D}$$
(b) Focal length of the equivalent single lens:
$$f = \frac{1}{P} = \frac{1}{2.25} \approx 0.44 \text{ m}$$
(c) When lenses are placed in contact, their net power is simply the algebraic sum of individual powers. This makes calculation quick and easy. Using focal lengths instead would require a more complex formula, making it inconvenient during eye-testing.
Source: Chapter 9, Section 9.3.8 Power of a Lens
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Explanation
- Part (a): Use $P = P_1 + P_2$ — this is the key formula for lenses in contact.
- Part (b): $f = 1/P$; remember $f$ must be in metres when $P$ is in dioptres.
- Part (c): The examiner wants the word "algebraic sum" and the idea that addition is simpler than the focal-length equivalent. The textbook explicitly states this as the reason opticians prefer powers. Don't overthink it — one or two sentences suffice for a 1-mark sub-point.