A needle of length 4.5 cm is placed vertically 12 cm away from a convex mirror of focal length 15 cm. The magnification of the image produced is

5/9

The correct answer is Option (2) → 5/9

Given: $ f = +15 \text{ cm}, \ u = -12 \text{ cm} $

From mirror formula: $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$

$ \frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{1}{15} - \left( -\frac{1}{12} \right) = \frac{1}{15} + \frac{1}{12} = \frac{4 + 5}{60} = \frac{9}{60} $

$ v = \frac{60}{9} = 6.67 \text{ cm} $

Magnification, $ m = \frac{v}{u} = \frac{6.67}{-12} = -0.56 $

Hence, the magnification = 0.56 (image is virtual and erect).