$$
begin{align*}
n_1 = n_{air} &= 1 \
n_2 &= 2.42 \
theta_1 &= 60 text{textdegree}
end{align*}
$$

The index of refraction for the diamond from Table 17.1 is $n_2 = 2.42$. Therefore the angle of refraction when the light beam enters the diamond at $60 text{textdegree}$ can be calculated by applying Snell’s Law :

$$
begin{align*}
n_1 sin theta_1 &= n_2 sin theta_2 \
theta_2 &= {sin}^{-1} left( dfrac{n_1 sin theta_1}{n_2} right) \
&= {sin}^{-1} left( dfrac{1 cdot sin 60 text{textdegree}}{2.42} right)
end{align*}
$$

$$
{boxed{theta_2 = 20.97 text{textdegree}}}
$$

$$
begin{align*}
n_1 = n_{air} &= 1 \
n_2 = n_{ice} &= 1.31 \
theta_2 &= 31 text{textdegree}
end{align*}
$$

The index of refraction for the ice cube from Table 17.1 is $n_2 = 1.31$. Therefore the angle of incidence of a ray of light traveling through the cube at $31 text{textdegree}$ can be calculated by applying Snell’s Law :

$$
begin{align*}
n_1 sin theta_1 &= n_2 sin theta_2 \
theta_1 &= {sin}^{-1} left( dfrac{n_2 sin theta_2}{n_1} right) \
&= {sin}^{-1} left( dfrac{1.31 cdot sin 31 text{textdegree}}{1} right)
end{align*}
$$

$$
{boxed{theta_1 = 42.43 text{textdegree}}}
$$