Question

Find the size, nature and position of image formed when an object of size 1cm is placed at a distance of 54cm from a concave mirror of focal length 36cm.

Answer 1

Hi,

$$\begin{gathered} \mathrm{Given}, \\ \mathrm{Focal}\mathrm{length}\mathrm{of}\mathrm{concave}\mathrm{mirror},f=-36\mathrm{cm} \\ \mathrm{Distance}\mathrm{of}\mathrm{object}\mathrm{from}\mathrm{the}\mathrm{concave}\mathrm{mirror},u=-54\mathrm{cm} \\ \mathrm{Size}\mathrm{of}\mathrm{object},h_1=1\mathrm{cm} \\ \mathrm{Using}\mathrm{mirror}\mathrm{formula} \\ \frac{1}{f}=\frac{1}{v}+\frac{1}{u} \\ \Rightarrow \frac{1}{-36}=\frac{1}{v}+\frac{1}{-54} \\ \Rightarrow \frac{1}{v}=\frac{1}{-36}+\frac{1}{54}=\frac{-3+2}{108} \\ \Rightarrow \frac{1}{v}=\frac{-1}{108} \\ \Rightarrow v=-108\mathrm{cm} \\ \mathrm{Thus},\mathrm{the}\mathrm{image}\mathrm{will}\mathrm{be}\mathrm{real}.\mathrm{Also},\mathrm{the}\mathrm{image}\mathrm{will}\mathrm{be}\mathrm{formed} \\ \mathrm{beyond}\mathrm{the}\mathrm{centre}\mathrm{of}\mathrm{curvature}\mathrm{of}\mathrm{the}\mathrm{mirror}. \\ \mathrm{Magnification}\left(m\right)\mathrm{is}\mathrm{given}\mathrm{by} \\ m=\frac{-v}{u}=\frac{\mathrm{height}\mathrm{of}\mathrm{image},h_2}{\mathrm{height}\mathrm{of}\mathrm{object},h_1} \\ \Rightarrow h_2=\frac{-v}{u}\times h_1 \\ \mathit{\Rightarrow }{h}_{\mathit{2}}=\frac{-\left(-108\right)}{-54}\times 1=-2\mathrm{cm} \\ \mathrm{Negative}\mathrm{sign}\mathrm{shows}\mathrm{that}\mathrm{the}\mathrm{image}\mathrm{will}\mathrm{be}\mathrm{inverted}.\mathrm{Also},\mathrm{the}\mathrm{height}\mathrm{of} \\ \mathrm{the}\mathrm{image}\mathrm{will}\mathrm{be}2\mathrm{cm}. \end{gathered}$$