A lens has a power of –2.5 D. (a) Calculate its focal length. (b) What type of lens is it? (c) Where should an object be placed in front of this lens so that the image is formed at the focus?
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Model Answer
(a) $f = \frac{1}{P} = \frac{1}{-2.5} = -0.4\ \text{m}$
(b) It is a concave (diverging) lens (negative focal length).
(c) The object should be placed at infinity so that the image forms at the focus.
Source: Chapter 9, Section 9.3.7 (Power of lens) and Table 9.5
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Explanation
- Power formula: $P = 1/f$ (f in metres). Negative power → negative focal length → concave lens.
- For a concave lens, when the object is at infinity, the image forms at focus F₁ (highly diminished, virtual and erect) — directly from Table 9.5.
- Examiners expect all three parts answered; missing any part loses the mark. Keep calculations brief and state the sign convention result clearly.