$$
tt{In the graph we plotted the image $d_i$ as a function of the object distance $d_o$, to determine the focal length of the mirror we will use the mirror equation that relates all of theses three variables:}
$$

$$
frac{1}{f}=frac{1}{d_o}+frac{1}{d_i}Rightarrow frac{1}{d_i}=frac{1}{f}-frac{1}{d_o}=frac{d_o-f}{d_of}
$$

$$
Rightarrow d_i=frac{d_of}{d_o-f}
$$

We can now determine the special value of $d_i$ when we put the object at the focal point of the mirror; in other words $d_olongrightarrow f$

$$
lim_{d_olongrightarrow f}{d_i}=lim_{d_olongrightarrow f}{frac{d_of}{d_o-f}}=infty
$$

We can see on the graph that the value of $d_i$ approaches infinity when $d_o$ is approaching 2.

$$
Rightarrow f=2
$$

$$
tt{(B)}
$$

$$
tt{The object is placed at 4cm, thefore $d_o=4cm$, based on the graph the image distance $d_i$ should be around +4cm, The image is real because $d_i>0$}
$$

$$
tt{(A)}
$$

$tt{ for the values of the object distance $d_o=[0.5cm;1cm;1,5cm]$ the values of the image distance $d_i$ is negative therefore the image is virtual}$

for the values of $d_o=2.5cm$ the values of the image distance $d_i$ is positive therefore the image is real.

$$
tt{(D)}
$$

$$
tt{The magnification equation states:}
$$

$$
m=frac{h_i}{h_o}=-frac{d_i}{d_o}
$$

In order for the image to be upright, the magnification factor need to be positive ($m>0$), The object is always infront of the mirror therfore $d_o$ is always postive, Consequencially the magnification factor sign depends from the image distance $d_i$ sign:

if $d_i0$

if $d_i>0Rightarrow m<0$

Based on the graph for the values of $d_o$=[2.5cm,3.5cm,4.5cm] The corresponding values of $d_i$ are positive therefore m is negative, Consequentially the image is inverted.

for $d_o$=1.5cm The corresponding value of $d_i$ is negative therefore m is positive, Consequentially the image is upright.

$$
tt{(A)}
$$

tt{A surface appear to be dull when illuminated if the light reflected off this surface is scattered in various directions, $text{color{#4257b2}{Figure 16.5}}$ depicts a rough surface reflecting incoming light in various directions, this phenomenon is called diffuse reflection.

$$
tt{(B)}
$$

$tt{First we proceed by calculating the focal length $f$ of a convex mirror:}$

$$
f=-frac{R}{2}=-frac{20}{2}=-10cm
$$

To determine the type of image, we need to calculate the image distance to the mirror $d_i$:

$$
frac{1}{f}=frac{1}{d_o}+frac{1}{d_i}Rightarrow frac{1}{d_i}=frac{1}{f}-frac{1}{d_o}=frac{d_o-f}{d_of}
$$

$$
Rightarrow d_i=frac{d_of}{d_o-f}=frac{10*(-10)}{10+10}=-5cm
$$

$d_i0Rightarrow$ The image is upright.

$$
m=frac{h_i}{h_o}=frac{1}{2}Rightarrow h_i=frac{h_o}{2}
$$

$Rightarrow$ The image size is half the original size

$$
tt{(C)}
$$