## Fundamentals of Physics Extended (10th Edition)

$1.0$
We can use the values given on table 34-5 in equation 34-8 to find the second index of refraction, $n_2$. $\frac{n_1}{p}+\frac{n_2}{i}=\frac{n_2-n_1}{r}$ Solving for $n_2$, we obtain: $n_2=-\frac{in_1(p+r)}{p(r-i)}$ $n_2=-\frac{600cm*1.5(100cm+(-30cm))}{100cm(-30cm-600cm)}$ $n_2=1.0$