Answer
$2.00$
Work Step by Step
First we consider equation (34-8):
$\frac{n_1}{p}+\frac{n_2}{i}=\frac{n_2-n_1}{r}$
We note that the object distance is infinite ($p=\infty$), and obtain:
$\frac{n_1}{\infty}+\frac{n_2}{i}=\frac{n_2-n_1}{r}$
$\frac{n_2}{i}=\frac{n_2-n_1}{r}$
Solving for the index of refraction of the sphere, $n_2$, we obtain:
$n_2=n_1(1-\frac{r}{i})^{-1}$
We know the first index of refraction is that of air, so $n_1=1.00$ and the image distance is twice the radius of the sphere ($i=2r$)
Using this information with the equation, we obtain:
$n_2=1.00(1-\frac{r}{2r})^{-1}$
$n_2=1.00(1-\frac{1}{2})^{-1}$
$n_2=2.00$
