Using the mirror formula: \[ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \] Given: \[ f = -15 \text{ cm}, \quad u = -20 \text{ cm} \] Substitute: \[ \frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{1}{-15} - \frac{1}{-20} = -\frac{1}{15} + \frac{1}{20} = -\frac{4}{60} + \frac{3}{60} = -\frac{1}{60} \] So, \[ v = -60 \text{ cm} \] Magnification $(m)$ is given by: \[ m = -\frac{v}{u} = -\left( \frac{-60}{-20} \right) = -3 \] So, the magnification is $-3$ (image is real, inverted and magnified).