A convex lens forms an 80 cm long image of a 20 cm long object which is kept at a distance of 60 cm from the optical centre of the lens. If the object and the image are on the same side of the lens, find (i) the nature of the image, (ii) the position of the image, and (iii) the focal length of the lens.
Generated by claude-sonnet-4-6 · 2026-06-15 06:49 · grounding rag
Model Answer
Given: Object size (h) = 20 cm, Image size (h') = 80 cm, Object distance (u) = –60 cm
Magnification: $m = \frac{h'}{h} = \frac{80}{20} = +4$
(i) Nature of the image: Since the image and object are on the same side of the lens, the image is virtual and erect (magnification is positive).
(ii) Position of the image:
$$m = \frac{v}{u} \Rightarrow +4 = \frac{v}{-60} \Rightarrow v = -240 \text{ cm}$$
The image is formed 240 cm from the optical centre, on the same side as the object.
(iii) Focal length: Using lens formula $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$:
$$\frac{1}{f} = \frac{1}{-240} - \frac{1}{-60} = \frac{-1+4}{240} = \frac{3}{240}$$
$$f = +80 \text{ cm}$$
Source: Chapter 9, Section 9.3.6
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Explanation
- When object and image are on the same side of a convex lens, the image is virtual and erect — this matches Table 9.4 (object between F₁ and O).
- Apply sign convention: u is negative (object on left); virtual image also on the same (left) side, so v is negative too.
- Magnification $m = v/u = h'/h$; since the image is virtual and erect, m is positive, confirming both v and u have the same sign.
- Examiners award one mark each for (i) nature, (ii) position, and (iii) focal length. Show formula and substitution clearly.