$n_{a}=1$

$n_{w}=1.33$

$theta_{a}=46text{ $^circ$}$

$textbf{Approach}$

In this problem, we will use Snell’s law.

$textbf{Solution}$

The definition of Snell’s law of refraction is

$$
begin{align}
{n_1cdot sin theta_1}={n_2cdot sin theta_2}
end{align}
$$

so we write

$$
begin{align}
{n_{a}cdotsintheta_{a}}&={n_{w}cdotsintheta_{w}} \
{theta_{w}}&=arcsin{left( frac{n_a}{n_{w}} cdot sintheta_a right)} \
{theta_{w}}&=arcsin{left( frac{1}{1.33} cdot sin 46^circ right)} \
&boxed{{theta_{w}}=33^circ}
end{align}
$$

$v=1.24times10^8frac{text{ m}}{text{ s}}$

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

$textbf{Approach}$

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

$textbf{Solution}$

The definition of index of refraction is

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

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

$$
begin{align}
&{n}=frac{3times10^8frac{text{ m}}{text{ s}}}{1.24times10^8frac{text{ m}}{text{ s}}} \
&boxed{{n}=2.42}
end{align}
$$

$h_{o}=2.25times10^{-2}text{ m}$

$d_{o}=1.86text{ m}$

$f=4.7times10^{-2}text{ m}$

$textbf{Approach}$

We will find the solutions by using the thin lens equation.

$textbf{Solution}$

Thin lens equation is

$$
begin{align}
frac{1}{f}=frac{1}{d_i}+frac{1}{d_o}
end{align}
$$

so for image position $d_i$ we have

$$
begin{align}
{d_i}&=frac{d_ocdot f}{d_o -f} \
{d_i}&=frac{1.86mcdot 4.7times10^{-2}m}{1.86m- 4.7times10^{-2}m}{} \
&boxed{{d_i}=4.82times10^{-2}m}
end{align}
$$

$d_{o}=4.15text{ m}$

$d_{i}=5times10^{-2}text{ m}$

$textbf{Approach}$

We will find the solutions by using the definition of the magnification.

$textbf{Solution}$

The magnification is defined as

$$
begin{align}
m&=frac{-d_i}{d_o} \
end{align}
$$

So we write

$$
begin{align}
{m}&=frac{5times10^{-2}text{ m}}{4.15text{ m}} \
&boxed{{m}=-0.012}
end{align}
$$

$d_{o}=-3text{ m}$

$f=-2text{ m}$

$textbf{Approach}$

We will find the solutions by using the thin lens equation and equality.

$textbf{Solution}$

Thin lens equation is defined as

$$
begin{align}
frac{1}{f}=frac{1}{d_i}+frac{1}{d_o}
end{align}
$$

so for image position $d_i$ we have

$$
begin{align}
{d_i}&=frac{d_ocdot f}{d_o -f} \
{d_i}&=frac{-3mcdot (-2)m}{-3m- (-2)m}{} \
&boxed{{d_i}=-6m}
end{align}
$$

$n_{w}=1.33$

$n_{g}=1.52$

$textbf{Approach}$

In this problem, we are going to use the law of total refraction.

$textbf{Solution}$

The definition of total refraction is

$$
begin{align}
{sin theta_c}=frac{n_2}{n_1}
end{align}
$$

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

$$
begin{align}
&{sintheta_{c}}=frac{n_w}{n_g} \
&{theta_{c}}=arcsin{left(frac{n_{w}}{n_{g}}right)} \
&boxed{{theta_{c}}=61{^circ}}
end{align}
$$

$n_{a}=1$

$theta_{c}=24.4text{ $^circ$}$

$theta_{d}=20text{ $^circ$}$

$textbf{Approach}$

We will find the angle of refraction in the air by using the definition of total refraction and Snell’s law.

$textbf{Solution}$

From the definition of total refraction, we will find the index of refraction of a diamond.

$$
begin{align}
{sin theta_c}&=frac{n_2}{n_1} \
{sintheta_{c}}&=frac{n_a}{n_d} \
{n_d}&=frac{n_a}{sintheta_{c}} \
{n_d}&=frac{1}{24.4^circ} \
{n_d}&=2.42
end{align}
$$

Now, we can use Snell’s law and find the angle of refraction in the air.

$$
begin{align}
{n_1cdotsin theta_1}&={n_2cdot sin theta_2} \
{n_{d}cdotsintheta_{d}}&={n_{a}cdotsintheta_{a}} \
{theta_{a}}&=arcsin{left({frac{n_{d}}{n_{a}}cdot sin theta_{d}}right)} \
{theta_{a}}&=arcsin{left({frac{2.42}{1}cdot sin 20^circ}right)} \
&boxed{{theta_{w}}=55{^circ} 52′}
end{align}
$$

The type of lens is defined by the focal length.

$$
begin{align}
frac{1}{f}&=frac{1}{d_i}+frac{1}{d_o} \
{f}&=frac{d_ocdot d_i}{d_o +d_i} \
{f}&=frac{6.98times10^{-2}mcdot 2.95times10^{-2}m}{6.98times10^{-2}m+ (-2.95)times10^{-2}m}{} \
{f}&=-5.11times10^{-2}m
end{align}
$$

Since the focal length is negative, the lens is $textbf{concave}$.