Object height \(=3.00 cm\),
Object distance \(u=10 cm\),ocal length \(f=15 cm\), image distance \(v=\) ?
\(\begin{aligned} \frac{1}{f}=\frac{1}{u}+\frac{1}{v} \\ \frac{1}{15}-\frac{1}{10}=\frac{1}{v}, v=&-30 cm \\ \text { Magnification } m=\left|\frac{v}{u}\right|=\left|\frac{-30}{10}\right| \\=\frac{30}{10}=3 \end{aligned}\)
\(M=\frac{\text { image height }}{\text { object heihght }}\)
\(=\frac{\text { image distance }}{\text { object distance }}\)
\(3=\frac{\text { image height }}{3}\);
image height \(=3 \times 3=9 cm\)
since \(v\) is negative, the image is
virtual and \(9.0 cm\) tall