The power of a lens is –0.25 D. Based on this information, find out
Generated by claude-sonnet-4-6 · 2026-06-15 06:53 · grounding rag
Model Answer
(a) Since power P = –0.25 D (negative), the lens is a concave (diverging) lens.
Focal length: $f = \dfrac{1}{P} = \dfrac{1}{-0.25} = \mathbf{-4 \ m}$
(b) A concave lens is used to correct myopia (near-sightedness). A myopic eye can see near objects clearly but cannot focus on distant objects.
(c) For a concave lens, the image is always virtual, erect, and diminished, regardless of the object's position. So when the object is placed between F and 2F, the image formed is virtual, erect, and smaller than the object.
Source: Chapter 9, Sections 9.3.6, 9.3.8; Chapter 10
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Explanation
- Part (a): Negative power → concave lens. Use $f = 1/P$ directly; keep the negative sign.
- Part (b): Concave lenses diverge light, helping myopic eyes focus distant objects on the retina.
- Part (c): A concave lens always forms a virtual, erect, diminished image on the same side as the object, no matter where the object is placed — this is a key fact examiners test.
- Write the focal length with its unit (metres) and include the negative sign — both are award-worthy details.