Question

A converging lens has a focal length of 0.36 m. If an object is placed at 0.21 m from the lens, where will its image be located?

Answer 1

ANSWER

$0.504\text{ }m\text{ in front of the lens}$

EXPLANATION

We want to determine the position where the image will be formed.

To do this, we apply the formula:

$\frac{1}{f}=\frac{1}{u}+\frac{1}{v}$

where f = focal length

u = distance of the object from the lens

v = distance of the image from the lens

Solving for v, we have the position that the image is formed:

$\begin{gathered} \frac{1}{0.36}=\frac{1}{0.21}+\frac{1}{v} \\ \frac{1}{0.36}=\frac{v+0.21}{0.21v} \\ 0.21v=0.36(v+0.21) \\ 0.21v=0.36v+0.0756 \\ 0.21v-0.36v=0.0756 \\ -0.15v=0.0756 \\ v=\frac{0.0756}{-0.15} \\ v=-0.504\text{ }m \end{gathered}$

Since this distance is negative, we say that the image formed is a virtual image and it is formed 0.504 m in front of the lens.