Example

Question:

An object is placed at (i) 10 cm, (ii) 5 cm in front of a concave mirror of radius of curvature 15 cm. Find the position, nature, and magnification of the image in each case.

Solution:

The focal length is \(f = -\frac{15}{2} = -7.5\,\text{cm}\).
(i) Object distance \(u = -10\,\text{cm}\): \[ \frac{1}{v} + \frac{1}{u} = \frac{1}{f} \implies \frac{1}{v} + \frac{1}{-10} = \frac{1}{-7.5} \] \[ v = \frac{10 \times 7.5}{-2.5} = -30\,\text{cm} \] The image is 30 cm from the mirror on the same side as the object.
Magnification: \[ m = -\frac{v}{u} = -\frac{-30}{-10} = -3 \] The image is magnified, real, and inverted.

(ii) Object distance \(u = -5\,\text{cm}\): \[ \frac{1}{v} + \frac{1}{-5} = \frac{1}{-7.5} \] \[ v = \frac{5 \times 7.5}{7.5 - 5} = \frac{37.5}{2.5} = 15\,\text{cm} \] The image is formed at 15 cm behind the mirror (virtual image).
Magnification: \[ m = -\frac{v}{u} = -\frac{15}{-5} = 3 \] The image is magnified, virtual, and erect.