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=(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=(6.2)/(4.1)`

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

`M=-(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} (6.2)/(4.1)=-(i)/(10.3) \\ i=-(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} (1)/(10.3)+(1)/(-(10.3\cdot6.2)/(4.1))=(1)/(f) \\ (1)/(10.3)-(4.1)/(10.3\cdot6.2)=(1)/(f) \\ f=((1)/(10.3)-(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