1) Image is at -2.23 cm (virtual)
2) Height of the image: 0.058 m
Explanation:
1)
The location of the image can be found by using the mirror equation:
[tex]\frac{1}{f}=\frac{1}{p}+\frac{1}{q}[/tex]
where
f is the focal length of the mirror
p is the distance of the object from the mirror
q is the distance of the image from the mirror
In this problem, we have:
d = 9.3 cm is the diameter of the mirror, so its radius is
r = 4.65 cm
For a curved mirror, the focal length is half the radius, so
[tex]f=\frac{r}{2}=\frac{4.65}{2}=0.0233 m[/tex]
Moreover, the mirror is part of a sphere, so we can assumed it is curved outward; therefore it is a diverging mirror, so its focal length is negative:
[tex]f=-0.0233 m[/tex]
The distance of the boy from the mirror is
[tex]p=0.5 m[/tex]
So, we can find the distance of the image from the mirror:
[tex]\frac{1}{q}=\frac{1}{f}-\frac{1}{p}=\frac{1}{-0.0233}-\frac{1}{0.5}=-44.92 m^{-1}[/tex]
[tex]q=\frac{1}{-44.92}=-0.0223 m[/tex]
So, the image is at -2.23 cm, and it is a virtual image (due to the negative sign)
2)
We can find the height of the image by using the magnification equation:
[tex]M=\frac{y'}{y}=-\frac{q}{p}[/tex]
where here we have:
y' = size of the image
y = 1.3 m is the height of the boy
p = 0.5 m is the distance between the boy and the mirror
q = -0.0223 m is the distance between the image and the mirror
And solving for y', we find:
[tex]y'=-\frac{qy}{p}=-\frac{(-0.0223)(1.3)}{0.5}=0.058 m[/tex]
