Missing info in the text: the image is inverted
1) Magnification: -3
The magnification can be calculated with the equation
[tex]M=\frac{y'}{y}[/tex]
where
y' is the size of the image
y is the size of the object
In this problem, we have
y' = -60.0 mm (the sign is negative since the image is inverted)
y = 20.0 mm
Substituting,
[tex]M=\frac{-60}{20}=-3[/tex]
2) Focal length: 0.862 m (converging)
We can also rewrite the magnification as follows
[tex]M=-\frac{q}{p}[/tex] (1)
where
q is the distance of the image from the mirror
p is the distance of the image from the mirror
Here we know that the distance between the image and the object is 2.30 m, so
[tex]q-p=2.30[/tex]
which means
[tex]q=p+2.30[/tex]
Substituting into (1), we can find p:
[tex]M=-3=-\frac{p+2.30}{p}\\3p=p+2.30\\2p = 2.30\\p = 1.15 m[/tex]
And also q:
[tex]q=p+2.30=1.15+2.30 = 3.45 m[/tex]
So now we can finally find the focal length by using the lens equation:
[tex]\frac{1}{f}=\frac{1}{p}+\frac{1}{q}[/tex]
Where f is the focal length. Solving for f,
[tex]f=\frac{pq}{p+q}=\frac{(1.15)(3.45)}{1.15+3.45}=0.862 m[/tex]
And since the focal length is positive, it means that the mirror is converging.
