A 5 cm tall object is placed 25 cm from a converging lens of focal length 10 cm. What are the image distance and the nature of the image?

1. Sign convention: u = −25 cm, f = +10 cm. 2. $\frac{1}{v} = \frac{1}{u} + \frac{1}{f} = \frac{1}{-25} + \frac{1}{10} = \frac{-2 + 5}{50} = \frac{3}{50}$. 3. So v = 50/3 ≈ +16.7 cm. 4. The positive image distance means the image is real and inverted; a negative v would be needed for a virtual image. _Source: Balbharati (Maharashtra Board) Class 10 Science & Technology, Ch 7 "Lenses", p.101_