Question

A plant that is 4.1 cm tall is 10.3 cm from a converging lens. You observe that the image of this plant is upright and 6.2 cm tall. What is the focal length of the lens?

Answer 1

We know that the magnification is given by:

$M=\frac{h^{\prime}}{h}$

where h is the height of the object and h' is the height of the image, then in this case we have:

$M=\frac{6.2}{4.1}$

On the other hand we also know that the magnification is given by:

$M=-\frac{i}{o}$

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:

$\begin{gathered} \frac{6.2}{4.1}=-\frac{i}{10.3} \\ i=-\frac{10.3\cdot6.2}{4.1} \end{gathered}$

Once we have the distance of the image we can use the lens equation to find the focal point:

$\begin{gathered} \frac{1}{10.3}+\frac{1}{-\frac{10.3\cdot6.2}{4.1}}=\frac{1}{f} \\ \frac{1}{10.3}-\frac{4.1}{10.3\cdot6.2}=\frac{1}{f} \\ f=(\frac{1}{10.3}-\frac{4.1}{10.3\cdot6.2})^{-1} \\ f=30.41 \end{gathered}$

Therefore the focal distance of the lens is 30.41 cm and this can be round to 30 cm, hence the answer is D