Question

A diverging mirror of radius of curvature 40 cm forms an image which is half the height of the object. Find the object and image positions.

Answer 1

Given,
The mirror is convex.

Magnification 'm' = $\frac{1}{2}$

Radius of curvature 'R' = 40 cm

Focal length of the convex mirror 'f' = $\frac{\mathrm{R}}{2}$ = 20 cm

We have to find the position of the image 'v' and distance of the object from the mirror 'u'.

Using the mirror formula, we get $$\begin{gathered} \frac{1}{f}=\frac{1}{u}+\frac{1}{v} \\ Given,m=\frac{1}{2} \\ Again,m=\frac{-v}{u} \\ Therefore,\mathrm{v}=-\frac{1}{2}\mathrm{u} \\ \mathrm{By}\mathrm{putting}\mathrm{the}\mathrm{value}\mathrm{of}\text{'}\mathrm{v}\text{'}\mathrm{in}\mathrm{the}\mathrm{mirror}\mathrm{formula},\mathrm{we}\mathrm{get} \\ \frac{1}{f}=\frac{1}{u}-\frac{2}{u}=\frac{-1}{\mathrm{u}} \\ \Rightarrow \frac{1}{20}=\frac{-1}{\mathrm{u}} \\ \Rightarrow u=-20cm \\ \Rightarrow v=\frac{-(-20)}{2} \\ \Rightarrow v=10\mathrm{cm} \\ \mathrm{If}\mathrm{an}\mathrm{object}\mathrm{is}\mathrm{placed}\mathrm{at}\mathrm{a}\mathrm{distance}\mathrm{of}20\mathrm{cm}\mathrm{in}\mathrm{front}\mathrm{of}\mathrm{th}\mathrm{e}\mathrm{mirror},\mathrm{the}\mathrm{image}\mathrm{will}\mathrm{be}\mathrm{formed}10\mathrm{cm}\mathrm{behind}\mathrm{the}\mathrm{mirror}. \end{gathered}$$