Question

An object of 3 cm height is placed at a distance of 8 cm from a concave mirror which produces a virtual image of 4.5 cm height. Find the position of the image and focal length of the mirror.

Hint:

For a concave mirror, a virtual image is formed when the object is placed between the pole and the focus.

Answer 1

Using the magnification formula: \[ m = \frac{-v}{u} = \frac{h'}{h} \] Given \( h' = 4.5 \) cm, \( h = 3 \) cm, and \( u = -8 \) cm: \[ \frac{-v}{-8} = \frac{4.5}{3} \Rightarrow v = 12 \text{ cm} \] Using the mirror formula: \[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \] \[ \frac{1}{f} = \frac{1}{12} - \frac{1}{-8} = \frac{1}{12} + \frac{1}{8} \] \[ \frac{1}{f} = \frac{2 + 3}{24} = \frac{5}{24} \] \[ f = -24 \text{ cm} \]