An object is placed 10 cm from a converging lens of focal length 15 cm. Find the position, nature and magnification of the image.
- 1/v = 1/f − 1/u = 1/15 − 1/10 = (2 − 3)/30 = −1/30.
- v = −30 cm: the minus sign means a virtual image on the same side as the object.
- Magnification = 30 ÷ 10 = 3, and a virtual image from a single lens is upright.
Answer: A virtual, upright image 30 cm from the lens, magnified 3 times.