(a) -17 cm
Magnification equation:
[tex]\frac{h_i}{h_o}=-\frac{q}{p}[/tex]
where
[tex]h_i[/tex] is the size of the image
[tex]h_o[/tex] is the size of the object
[tex]q[/tex] is the distance of the image from the mirror
[tex]p[/tex] is the distance of the object from the mirror
In this problem, we have:
p = 51 cm (the object is 51 cm in front of the mirror)
[tex]\frac{h_i}{h_o}=\frac{1}{3}[/tex]
So, we can find q, the image distance:
[tex]q=-\frac{h_i}{h_o}p=-\frac{1}{3}(51 cm)=-17 cm[/tex]
and the negative sign means the image is virtual.
(b) -25.5 cm
Mirror equation:
[tex]\frac{1}{f}=\frac{1}{p}+\frac{1}{q}[/tex]
where
p = 51 cm
q = -17 cm
Substituting and re-arranging the equation, we find the focal length of the mirror:
[tex]\frac{1}{f}=\frac{1}{51 cm}-\frac{1}{17 cm}=-\frac{2}{51 cm}\\f=-\frac{51 cm}{2}=-25.5 cm[/tex]
and the negative sign is due to the fact it is a convex mirror.
