(B) An object of height 10 cm is placed 25 cm away from the optical centre of a converging lens of focal length 15 cm. Calculate the image-distance and height of the image formed.
Generated by claude-sonnet-4-6 · 2026-06-15 06:50 · grounding rag
Model Answer
Given: h = +10 cm, u = –25 cm, f = +15 cm (converging lens)
Using Lens Formula:
$$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$$
$$\frac{1}{v} = \frac{1}{15} + \frac{1}{(-25)} = \frac{1}{15} - \frac{1}{25} = \frac{5-3}{75} = \frac{2}{75}$$
$$v = +37.5 \text{ cm}$$
The image is formed 37.5 cm on the other side of the lens. It is real and inverted.
Height of image:
$$h' = h \times \frac{v}{u} = 10 \times \frac{+37.5}{-25} = -15 \text{ cm}$$
The image height is 15 cm (negative sign indicates it is inverted).
Source: Chapter 9, Section 9.3.7 – Lens Formula and Magnification
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Explanation
- Always apply sign convention: object on left → u is negative; converging lens → f is positive.
- The formula for lenses is $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$ (not the mirror formula).
- A positive v confirms real image on the opposite side; negative h′ confirms inverted image.
- Examiners award marks for: correct substitution with signs (1 mark), correct v (1 mark), correct h′ (1 mark).