$n_{air}=1$

$n_{eth}=1.36$

$theta_{air}=37text{ $^circ$}$

$textbf{Approach}$

In this problem we are going to use Snell’s law.

$textbf{Solution}$

The definition of Snell’s law of refraction is

$$
begin{align}
{n_1cdotsin theta_1}={n_2cdot sin theta_2}
end{align}
$$

where indexes $1$ and $2$ represent two different mediums. So we write

$$
begin{align}
&{n_{air}cdotsintheta_{air}}={n_{eth}cdotsintheta_{eth}} \
&{theta_{eth}}=arcsin{left({frac{n_{air}}{n_{eth}}cdot sin theta_{air}}right)} \
&{theta_{eth}}=arcsin{left({frac{1}{1.36}cdot sin37^circ}right)} \
&boxed{{theta_{eth}}=26{^circ}16′}
end{align}
$$

$n_{air}=1$

$n_{cg}=1.52$

$theta_{air}=45text{ $^circ$}$

$textbf{Approach}$

In this problem we are going to use Snell’s law.

$textbf{Solution}$

The definition of Snell’s law of refraction is

$$
begin{align}
{n_1cdotsin theta_1}={n_2cdot sin theta_2}
end{align}
$$

where indexes $1$ and $2$ represent two different mediums. So we write

$$
begin{align}
&{n_{air}cdotsintheta_{air}}={n_{cg}cdotsintheta_{cg}} \
&{theta_{cg}}=arcsin{left({frac{n_{air}}{n_{cg}}cdot sin theta_{air}}right)} \
&{theta_{cg}}=arcsin{left({frac{1}{1.52}cdot sin 45^circ}right)} \
&boxed{{theta_{cg}}=27{^circ}43′}
end{align}
$$

$n_{a}=1$

$n_{w}=1.33$

$theta_{a}=30text{ $^circ$}$

$textbf{Approach}$

In this problem we are going to use Snell’s law.

$textbf{Solution}$

The definition of Snell’s law of refraction is

$$
begin{align}
{n_1cdotsin theta_1}={n_2cdot sin theta_2}
end{align}
$$

where indexes $1$ and $2$ represent two different mediums. So we write

$$
begin{align}
&{n_{a}cdotsintheta_{a}}={n_{w}cdotsintheta_{w}} \
&{theta_{w}}=arcsin{left({frac{n_{a}}{n_{w}}cdot sin theta_{a}}right)} \
&{theta_{w}}=arcsin{left({frac{1}{1.33}cdot sin 30^circ}right)} \
&boxed{{theta_{w}}=22{^circ}5′}
end{align}
$$

$n_{a}=1$

$n_{d}=2.42$

$theta_{a}=45text{ $^circ$}$

$textbf{Approach}$

In this problem we are going to use Snell’s law.

$textbf{Solution}$

The definition of Snell’s law of refraction is

$$
begin{align}
{n_1cdotsin theta_1}={n_2cdot sin theta_2}
end{align}
$$

where indexes $1$ and $2$ represent two different mediums. So we write

$$
begin{align}
&{n_{a}cdotsintheta_{a}}={n_{d}cdotsintheta_{d}} \
&{theta_{d}}=arcsin{left({frac{n_{a}}{n_{d}}cdot sin theta_{a}}right)} \
&{theta_{d}}=arcsin{left({frac{1}{2.42}cdot sin 45^circ}right)} \
&boxed{{theta_{d}}=16{^circ}59′}
end{align}
$$

$n_{w}=1.33$

$theta_w=31text{ $^circ$}$

$theta_b=27text{ $^circ$}$

$textbf{Approach}$

In this problem we are going to use Snell’s law.

$textbf{Solution}$

The definition of Snell’s law of refraction is

$$
begin{align}
{n_1cdotsin theta_1}={n_2cdot sin theta_2}
end{align}
$$

where indexes $1$ and $2$ represent two different mediums. Since the motion is in the water we write

$$
begin{align}
&{n_{w}cdotsintheta_{w}}={n_{b}cdotsintheta_{b}} \
&{n_b}={n_{w}cdot frac{sintheta_{w}}{ sin theta_{b}} } \
&{n_b}={1.33cdot frac{sin31^circ}{ sin27^circ} } \
&boxed{{n_b}=1.5}
end{align}
$$