The end of a long glass rod (n = 1.50) is ground into a convex surface of radius 3.0 cm…

Step 1. The image distance. $\dfrac{n_2}{v} = \dfrac{n_2 - n_1}{R} + \dfrac{n_1}{u} = \dfrac{0.50}{3.0\ \text{cm}} - \dfrac{1.00}{12\ \text{cm}} = 0.167\ \text{cm}^{-1} - 0.0833\ \text{cm}^{-1} = 0.0833\ \text{cm}^{-1}$. We want $v$, so $v = \dfrac{1.50}{0.0833\ \text{cm}^{-1}} = +18\ \text{cm}$.

Step 2. The magnification. $m = \dfrac{n_1 v}{n_2 u} = \dfrac{1.00 \times 18\ \text{cm}}{1.50 \times (-12\ \text{cm})} = -1.0$.