From the graph, a 2 cm object distance $d_{text{o}}$ corresponds to an image distance $d_{text{i}}$ with a negative value. The sign convention for lenses stated in Section 17.3 says that when $d_{text{i}}$ is negative, the images are on the same side of the lens as the object and they are called virtual images.

Additionally, the magnification $m$ of an image is given by:

$$
begin{align*}
m=-frac{d_{text{i}}}{d_{text{o}}}
end{align*}
$$

wherein if $m$ is positive, the image is upright whereas if it is negative, the image is inverted.

Substituting the corresponding signs for $d_{text{i}}$ and $d_{text{o}}$ to the equation, $m$ would be positive and therefore upright. Thus, $textbf{the image formed is upright and on the same side of the lens as the object.}$

The graph shows that when the object has a distance of 4 cm in front of the lens, it corresponds to an image distance $d_{text{i}}$ with a positive value. From the sign convention for lenses, when $d_{text{i}}$ is positive, a real image is formed.

The approximated image distance can be solved through the thin-lens equation:

$$
begin{align*}
frac{1}{d_{text{o}}}+frac{1}{d_{text{i}}} = frac{1}{f}
end{align*}
$$

We then rearrange the equation such that the image distance $d_{text{i}}$ is isolated to one side. The following quantities are then plugged into the equation: $d_{text{o}} = 4;text{cm}$ and $f=3;text{cm}$.

$$
begin{align*}
frac{1}{d_{text{i}}} &= frac{1}{f} – frac{1}{d_{text{o}}} \ &=frac{1}{3;text{cm}} – frac{1}{4;text{cm}} \ &=frac{1}{12;text{cm}}\
{d_{text{i}}} &= 12;text{cm} approx 10;text{cm}
end{align*}
$$

In summary, $textbf{the image is real and located about 10 cm from the lens.}$

It is noted that the lens used in the problem is a concave lens; thus, we apply the rules for ray-tracing appropriate for this lens. For concave lenses, the parallel ray approaches the lens parallel to its axis. Afterwards, the ray extends back to a focal point for which in this problem, Ray A exhibits this behavior. Thus, $textbf{Ray D is not correct}$ in this diagram because the ray goes through the focal point instead of extending back.

The speed of light $v$ in a material is calculated through this equation:

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

where $c$ is the speed of light in vacuum and $n$ is the index of refraction of a material.

Air has an index of refraction approximately equal to 1. Thus, its speed is approximately equal to the speed of light in vacuum.

$$
begin{align*}
v_{text{air}}approx c
end{align*}
$$

For a material with an index of refraction value of 1.5, its corresponding speed of light is:

$$
begin{align*}
v_{text{new}} &=frac{c}{1.5}\
v_{text{new}} & approx frac{v_{text{air}}}{1.5}
end{align*}
$$

From the result, the $textbf{speed decreases by a factor of 1.5}$. Larger value of index of refraction results to smaller value of the speed of light.