We know that the lens equation is given as:
[tex]\frac{1}{f}=\frac{1}{d_o}+\frac{1}{d_i}[/tex]where f is the focal length, do is the distance of the object and di is the distance of the image.
In this case we know:
The focal length is 0.14 m.
The distance of the object is 0.31 m.
Plugging these values in the lens equation we have:
[tex]\begin{gathered} \frac{1}{0.14}=\frac{1}{0.31}+\frac{1}{d_i} \\ \frac{1}{d_i}=\frac{1}{0.14}-\frac{1}{0.31} \\ \frac{1}{d_i}=\frac{0.31-0.14}{(0.14)(0.31)} \\ d_i=\frac{(0.14)(0.31)}{0.31-0.14} \\ d_i=0.255 \end{gathered}[/tex]Hence, the distance of the image is 0.255 m.
Now, the magnifying equation states that:
[tex]\frac{h_i}{h_o}=-\frac{d_i}{d_o}[/tex]We know that the height of the object is 0.33 m, plugging the values we have:
[tex]\begin{gathered} \frac{h_i}{0.33}=-\frac{0.255}{0.31} \\ h_i=-(0.33)(\frac{0.255}{0.31}) \\ h_i=-0.27 \end{gathered}[/tex]Therefore, the image will be -0.27 m tall. (The minus sign indicates that the image is inverted)
