We know that the magnification is given by:
[tex]M=\frac{h^{\prime}}{h}[/tex]
where h is the height of the object and h' is the height of the image, then in this case we have:
[tex]M=\frac{6.2}{4.1}[/tex]
On the other hand we also know that the magnification is given by:
[tex]M=-\frac{i}{o}[/tex]
where i is the distance of the image to lens and o is the distance of the object to the lens. From this we have:
[tex]\begin{gathered} \frac{6.2}{4.1}=-\frac{i}{10.3} \\ i=-\frac{10.3\cdot6.2}{4.1} \end{gathered}[/tex]
Once we have the distance of the image we can use the lens equation to find the focal point:
[tex]\begin{gathered} \frac{1}{10.3}+\frac{1}{-\frac{10.3\cdot6.2}{4.1}}=\frac{1}{f} \\ \frac{1}{10.3}-\frac{4.1}{10.3\cdot6.2}=\frac{1}{f} \\ f=(\frac{1}{10.3}-\frac{4.1}{10.3\cdot6.2})^{-1} \\ f=30.41 \end{gathered}[/tex]
Therefore the focal distance of the lens is 30.41 cm and this can be round to 30 cm, hence the answer is D
