Question

a concave mirror produces three times magnified real image of an object placed at 10cm in front of it. where is the image located and also calculate the radius of curvature of the mirror

Answer 1

Dear Student,

$$\begin{gathered} \mathrm{magnification}\mathrm{produced}\mathrm{by}\mathrm{spherical}\mathrm{mirror}, \\ \mathrm{m}=-\frac{\mathrm{v}}{\mathrm{u}} \\ \mathrm{magnification}\mathrm{of}\mathrm{concave}\mathrm{mirror},\mathrm{m}=-3 \\ \mathrm{u}=-10\mathrm{cm} \\ -3=\frac{-\mathrm{v}}{-10} \\ \mathrm{v}=-30\mathrm{cm} \\ \mathrm{here},\mathrm{negative}\mathrm{sign}\mathrm{indicates}\mathrm{that}\mathrm{image}\mathrm{is}\mathrm{formed}\mathrm{at}\mathrm{a}\mathrm{distance}\mathrm{of}30\mathrm{cm}\mathrm{in}\mathrm{front} \\ \mathrm{of}\mathrm{concave}\mathrm{mirror}. \\ \mathrm{According}\mathrm{to}\mathrm{mirror}\mathrm{formula}, \\ \frac{1}{\mathrm{f}}=\frac{1}{\mathrm{v}}+\frac{1}{\mathrm{u}} \\ \frac{1}{\mathrm{f}}=\frac{1}{-30}+\left(\frac{1}{-10}\right) \\ \frac{1}{\mathrm{f}}=\frac{-1-3}{30} \\ \frac{1}{\mathrm{f}}=\frac{-4}{30} \\ \mathrm{f}=-\frac{30}{4}\mathrm{cm} \\ \mathrm{here},\mathrm{negative}\mathrm{sign}\mathrm{shows}\mathrm{that}\mathrm{it}\mathrm{is}\mathrm{a}\mathrm{concave}\mathrm{mirror}. \\ \mathrm{Radius},\mathrm{R}=2\mathrm{f} \\ \mathrm{R}=2\times \frac{30}{4}=15\mathrm{cm} \end{gathered}$$