Question

An object when placed in front of convex lens forms a real image of 0.5 magnification.If the distance of the image from the lens is 24cm, calculate the focal length of the lens.

Answer 1

Dear Student,

$$\begin{gathered} \mathrm{magnification},\mathrm{m}=-0.5\left(\mathrm{for}\mathrm{real}\mathrm{image}\right) \\ \mathrm{v}=24\mathrm{cm} \\ \mathrm{m}=\frac{\mathrm{v}}{\mathrm{u}} \\ -0.5=\frac{24}{\mathrm{u}} \\ \mathrm{u}=-\frac{24}{0.5} \\ \mathrm{u}=-48\mathrm{cm} \\ \frac{1}{\mathrm{v}}-\frac{1}{\mathrm{u}}=\frac{1}{\mathrm{f}} \\ \frac{1}{24}-\left(\frac{1}{-48}\right)=\frac{1}{\mathrm{f}} \\ \frac{1}{\mathrm{f}}=\frac{1}{24}+\frac{1}{48} \\ \frac{1}{\mathrm{f}}=\frac{2+1}{48} \\ \mathrm{f}=\frac{48}{3}=16\mathrm{cm} \end{gathered}$$