ANSWER:
The image distance is 4.11 cm
The image height is 1.8 cm
STEP-BY-STEP EXPLANATION:
The image distance is found by using the mirror-thin lens equation, just like this:
[tex]\frac{1}{s}+\frac{1}{s^{\prime}}=\frac{1}{f}[/tex]We solve and calculate for s':
[tex]\begin{gathered} \frac{1}{s^{\prime}}=\frac{1}{f}-\frac{1}{s} \\ \frac{1}{s^{\prime}}=\frac{s-f}{f\cdot s} \\ s^{\prime}=\frac{f\cdot s}{s-f} \\ \text{ replacing} \\ s=3.2 \\ f=1.8 \\ s^{\prime}=\frac{1.8\cdot3.2}{3.2-1.8} \\ s^{\prime}=4.11 \end{gathered}[/tex]The lateral magnification of the lens is founds in terms of the object and image distances, just like this:
[tex]\begin{gathered} M=-\frac{s^{\prime}}{s} \\ \text{ replacing} \\ M=-\frac{-4.11}{3.2}=1.29 \end{gathered}[/tex]The magnitude of the lateral magnification of the lens is found in terms of the objetc and image heights:
[tex]\begin{gathered} |M|=\frac{h^{\prime}}{h} \\ h^{\prime}=|M|\cdot h \\ \text{ replacing} \\ h^{\prime}=1.29\cdot1.4 \\ h^{\prime}=1.8 \end{gathered}[/tex]