Given,
object distance = Given that, an object placed 45.0 cm from a convex lens having a focal length of 15.0 cm
and the height of the image is 5.0 cm
we know,
[tex]\begin{gathered} \frac{1}{f}=\frac{1}{v}-\frac{1}{u}\frac{}{} \\ \\ \frac{1}{15}=\frac{1}{v}-\frac{1}{-45} \\ \\ \frac{1}{v}=\frac{1}{15}-\frac{1}{45} \\ \\ \frac{1}{v}=\frac{2}{45} \\ \\ \therefore\text{ v=22.5cm} \end{gathered}[/tex]we know,
[tex]\begin{gathered} \frac{h_i}{h_o}=\frac{v}{u} \\ \\ \frac{\begin{equation*}h_i\end{equation*}}{5}=\frac{22.5}{-45} \\ \\ h_i=-2.5cm \end{gathered}[/tex]result,
image distance = 22.5cm towards right side of convex lens .
image height=2.5cm, {Enlarged and inverted}
