## Fundamentals of Physics Extended (10th Edition)

$+10cm$
First we consider equation (34-8): $\frac{n_1}{p}+\frac{n_2}{i}=\frac{n_2-n_1}{r}$ Solving for the object distance, $p$, we obtain: $p=n_1(\frac{n_2-n_1}{r}-\frac{n_2}{i})^{-1}$ Using the given values in the table, we have: $p=1.5(\frac{1.0-1.5}{-30cm}-\frac{1.0}{-7.5cm})^{-1}$ $p=+10cm$