Question

A converging mirror forms a real image of height 1 cm of an object of height 2 cm, placed 20 cm away from the mirror. What is the focal length of the mirror

Answer 1

Dear Student,
Answer to your question is given below:
$$\begin{gathered} Given,objectdis\tan ce,u=-20cm(\sin ceontheleftsideofthemirror) \\ heightoftheimage,h_i=-1cm(negativesignimpliesrealimage) \\ heightoftheobject,h_0=2cm \\ Nowweknowthat,magnification, \\ m=-\frac{v}{u} \\ \Rightarrow \frac{h_i}{h_0}=-\frac{v}{u} \\ \Rightarrow \frac{-1}{2}=-\frac{v}{\left(-20\right)} \\ \Rightarrow v=-\frac{20}{2} \\ \Rightarrow v=-10cm(ontheleftsideofthemirror) \\ Now,weknowthat, \\ \frac{1}{f}=\frac{1}{v}+\frac{1}{u} \\ \Rightarrow \frac{1}{f}=\frac{1}{\left(-10\right)}+\frac{1}{\left(-20\right)} \\ \Rightarrow \frac{1}{f}=-\frac{1}{10}-\frac{1}{20} \\ \Rightarrow \frac{1}{f}=\frac{-2-1}{20} \\ \Rightarrow f=-\frac{20}{3}cm \\ \Rightarrow f=-6.67cm \\ Hencethefocallenghtofthemirroris-6.67cm. \end{gathered}$$
Regards,