$n_{w}=1.33$

$n_{cg}=1.52$

$textbf{Approach}$

In this problem we will discuss Snell’s law.

$textbf{Solution}$

A light ray in both situations enters liquid in a same angle, but the angle of refraction is different. In the first situation a light ray enters a liquid from the water and it is bent toward the normal which means that the angle of refraction $theta_l$ is lower that the angle of incidence $theta_w$. If we put it in a Snell’s law we have:

$$
begin{align}
{n_{w}sintheta_{w}}&={n_{l}sintheta_{l}} \
frac{n_{w}}{n_{l}}&=frac{sintheta_{l}}{sintheta_{w}}<1 \
n_{w}&1 \
n_{cg}&>{n_{l}}
end{align}
$$

So, our solution is

$$
begin{align}
&{n_w}<{n_l}<{n_{cg}} \
&boxed{{1.33}<{n_l}<{1.52}}
end{align}
$$

$n_{a}=1$

$theta_a=30text{$^circ$}$

$theta_m=20text{$^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 air we write

$$
begin{align}
&{n_{a}cdotsintheta_{a}}={n_{m}cdotsintheta_{m}} \
&{n_m}={n_{a}cdot frac{sintheta_{a}}{ sin theta_{m}} } \
&{n_m}={1cdot frac{sin30^circ}{ sin20^circ} } \
&boxed{{n_b}=1.46}
end{align}
$$

$$
begin{align}
{n}=frac{c}{v}
end{align}
$$

An index $n$ would be less than 1 if $v$ is greater than $c$. Since $c$ is the speed of light in vacuum and since there is no higher speed than $c$, $n$ can not be less than 1.

$n=1.51$

$c=3times10^8 frac{text{ m}}{text{$ s$}}$

$textbf{Approach}$

In this problem we will use a definition of index of refraction.

$textbf{Solution}$

The definition of index of refraction is

$$
begin{align}
{n}=frac{c}{v}
end{align}
$$

where index $c$ is the speed of light and $v$ is the speed in the medium.

$$
begin{align}
&{v}=frac{c}{n} \
&{v}=frac{3times10^8 frac{ m}{ s}}{1.21} \
&boxed{{v}=1.99 frac{ m}{ s}}
end{align}
$$

The indexes of refraction for the crown glass and quartz are $n_{cg}=1.52$ and $n_q=1.54$. The index of quartz is greater so, by Snell’s law, the angle of incidence for quartz is smaller. For the cladding layer, we need a material with a greater angle of incidence because it has to be greater than the critical angle which means we will choose $textbf{a crown glass}$.

$n_{w}=1.33$

$n_{p}=1.5$

$theta_{w}=57.5text{ $^circ$}$

$textbf{Approach}$

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

$$
begin{align}
&{n_{w}cdotsintheta_{w}}={n_{p}cdotsintheta_{p}} \
&{theta_{p}}=arcsin{left({frac{n_{w}}{n_{p}}cdot sin theta_{w}}right)} \
&{theta_{p}}=arcsin{left({frac{1.33}{1.5}cdot sin 57.5^circ}right)} \
&boxed{{theta_{p}}=48{^circ} 24′}
end{align}
$$

$n_{w}=1.33$

$n_{g}=1.52$

$textbf{Solution}$

The critical angle is defined as

$$
begin{align}
{sintheta_c}=frac{n_2}{n_1}
end{align}
$$

Since $sintheta_c$ is equal or less than 1, angle exists only for ${n_2}<{n_1}$. $n_1$ represents first medium and $n_2$ the second one so the critical angle exists for ${n_1}={n_g}$ respectively ${n_2}={n_w}$.