Question

A convex lens of focal length \( f = 15\,cm \) forms a real image of a 6 cm tall object placed at 30 cm. Find the position, nature, and height of the image.

Hint:

For lenses, always follow sign conventions strictly. Real images are inverted and have negative height; virtual images are upright and positive.

Answer 1

Step 1: Use the lens formula: \[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \Rightarrow \frac{1}{v} = \frac{1}{f} + \frac{1}{u} \] \[ f = 15\,cm, \quad u = -30\,cm \quad (\text{object on left}) \] \[ \frac{1}{v} = \frac{1}{15} - \frac{1}{30} = \frac{2 - 1}{30} = \frac{1}{30} \Rightarrow v = 30\,cm \] Step 2: Find magnification: \[ m = \frac{v}{u} = \frac{30}{-30} = -1 \Rightarrow h' = mh = -1 \cdot 6 = -6\,cm \] Negative sign implies real, inverted image.
Final Answer: \[ \boxed{\text{Image at } 30\,cm \text{ on opposite side, real, inverted, height } = 6\,cm} \]