Question

A concave mirror produces three times enlarged virtual image of an object placed at 10 cm in front of it. Calculate the radius of curvature of the mirror.

Answer 1

Given,
Distance of the object 'u' = -10 cm
Magnification 'm' = 3
We have to find the focal length 'f' and the radius of curvature 'R'.
Using the magnification formula, we get
$$\begin{gathered} m=\frac{-v}{u} \\ 3=\frac{-\mathrm{v}}{-10} \\ v=30\mathrm{cm} \\ \mathrm{Thus},\mathrm{the}\mathrm{distance}\mathrm{of}\mathrm{the}\mathrm{image}\text{'}\mathrm{v}\text{'}\mathrm{is}30\mathrm{cm}\mathrm{behind}\mathrm{the}\mathrm{mirror}. \\ Now,u\sin gthe\mathrm{mirror}\mathrm{formula},weget \\ \frac{1}{f}=\frac{1}{v}+\frac{1}{u} \\ \frac{1}{f}=\frac{1}{30}+\frac{1}{-10} \\ \mathrm{or}\frac{1}{f}=\frac{1}{30}-\frac{3}{30}=\frac{-2}{30} \\ \mathrm{or}\frac{1}{\mathrm{f}}=-\frac{1}{15} \\ \mathrm{or}\mathrm{f}=-15\mathrm{cm} \\ \mathrm{Thus},\mathrm{the}\mathrm{focal}\mathrm{length}\mathrm{of}\mathrm{the}\mathrm{mirror}\text{'}\mathrm{f}\text{'}\mathrm{is}-15\mathrm{cm}. \\ \mathrm{Therefore},\mathrm{the}\mathrm{radius}\mathrm{of}\mathrm{curvature}\mathrm{of}\mathrm{the}\mathrm{mirror}\text{'}\mathrm{R}\text{'}\mathrm{will}\mathrm{be}2f=2\times -15 \\ =-30\mathrm{cm} \\ \mathrm{Radius}\mathrm{of}\mathrm{curvature}\mathrm{of}\mathrm{the}\mathrm{mirror}\mathrm{is}30\mathrm{cm}. \end{gathered}$$