To determine the refractive index ($\mu$) of a thin convex lens, we use the lens maker's formula:

$ \frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) $

Given:

Focal length ($f$) = 12 cm

Radii of curvature ($R_1$ and $R_2$) = 10 cm and -15 cm, respectively

Substituting these values into the formula, we have:

$ \frac{1}{12} = (\mu - 1) \left( \frac{1}{10} - \frac{1}{-15} \right) $

Simplifying the equation:

$ \frac{1}{12} = (\mu - 1) \left( \frac{3 + 2}{30} \right) $

$ \frac{1}{12} = (\mu - 1) \times \frac{5}{30} $

$ \frac{1}{12} = (\mu - 1) \times \frac{1}{6} $

Solving for $\mu$:

$ \mu - 1 = \frac{1}{12} \times 6 $

$ \mu - 1 = \frac{1}{2} $

$ \mu = \frac{3}{2} $

Thus, the refractive index of the lens material is $\mu = 1.5$.