(b) Define power of a lens. The focal length of a lens is −10 cm. Write the nature of the lens and find its power. If an object is placed at a distance of 20 cm from the optical centre of this lens, according to the New Cartesian Sign Convention, what will be the sign of magnification in this case ?
Generated by claude-sonnet-4-6 · 2026-06-15 06:49 · grounding rag
Model Answer
Power of a lens: The power of a lens is defined as the reciprocal of its focal length. $P = \dfrac{1}{f}$. Its SI unit is dioptre (D).
Nature of lens: Since $f = -10$ cm $= -0.1$ m, the lens is a concave (diverging) lens.
Power:
$$P = \frac{1}{f} = \frac{1}{-0.1 \text{ m}} = -10 \text{ D}$$
Sign of magnification:
Object distance $u = -20$ cm (by sign convention). Using lens formula with $f = -10$ cm:
$$\frac{1}{v} = \frac{1}{f} + \frac{1}{u} = \frac{1}{-10} + \frac{1}{-20} = \frac{-3}{20}$$
$$v = -\frac{20}{3} \text{ cm}$$
$$m = \frac{v}{u} = \frac{-20/3}{-20} = +\frac{1}{3}$$
The sign of magnification is positive, indicating the image is virtual and erect.
Source: Chapter 9, Sections 9.3.7 and 9.3.8
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Explanation
- Always define power clearly and state its SI unit — examiners award 1 mark for this.
- A negative focal length always means a concave lens — state this explicitly.
- For sign convention: object is placed to the left, so $u$ is negative. Use the lens formula to find $v$, then compute $m = v/u$. A concave lens always gives a virtual, erect image, so $m$ is always positive for a real object.
- Show the calculation step-by-step for full marks; don't skip to the answer.