Question

A lens used as a magnifier give magnification of 6. what is its power and focal length.

Answer 1

Dear Student,

$$\begin{gathered} \mathrm{magnification},\mathrm{m}=\frac{\mathrm{D}}{\mathrm{f}} \\ \mathrm{m}=6 \\ \mathrm{where},\mathrm{D}=\mathrm{least}\mathrm{distance}=25\mathrm{cm} \\ \mathrm{f}=\mathrm{focal}\mathrm{length} \\ 6=\frac{25}{\mathrm{f}} \\ \mathrm{f}=\frac{25}{6}=4.17\mathrm{cm} \\ \mathrm{or},\mathrm{f}=4.17\times {10}^{-2}\mathrm{m} \\ \mathrm{Power},\mathrm{P}=\frac{1}{\mathrm{f}} \\ \mathrm{P}=\frac{1}{4.17\times {10}^{-2}}=23.98\mathrm{D} \end{gathered}$$