Question

A cheese enchilada is$\mathbf{4}\mathbf{.}\mathbf{00}\mathbf{}\mathit{c}\mathit{m}$in front of a converging lens. The magnification of the enchilada is$\mathbf{-}\mathbf{2}\mathbf{.}\mathbf{00}$. What is the focal length of the lens?

Short answer

Focal length of the lens is$\mathrm{f}=2.67\mathrm{cm}$

Step-by-step solution

Listing the given quantities 

$\begin{array}{l}m=-2.0\\ p=4.0cm\end{array}$

Understanding the concepts of magnification

We will use the relation between object distance, image distance, and focal length for the converging lens.

Formula:

$\frac{\mathbf{1}}{\mathbf{p}}\mathbf{+}\frac{\mathbf{1}}{\mathbf{i}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{f}}$

Magnification

$\begin{array}{c}\mathbf{m}\mathbf{=}\mathbf{-}\frac{\mathbf{i}}{\mathbf{p}}\\ \mathbf{=}\frac{{\mathbf{H}}_{\mathbf{i}}}{{\mathbf{H}}_{\mathbf{p}}}\end{array}$

Calculation of the focal length of the lens

We have magnification as

$\begin{array}{l}m=-\frac{i}{p}\\ 2p=i\end{array}$

Distance of the object is$i=8.0cm$

Now, to find focal length, we will use the following relation:

$\begin{array}{c}\frac{1}{f}=\frac{1}{p}+\frac{1}{i}\\ f=\frac{1}{0.375}\\ =2.67cm\end{array}$

Focal length of the lens is$f=2.67cm$