Question

A short sighted person can read distinctly a book placed at a distance of 20 cm. what is the nature and power of the lens in his spectacles to read the book 25 cm distance.

Answer 1

Hi,

$$\begin{gathered} \mathrm{As}\mathrm{the}\mathrm{defect}\mathrm{is}\mathrm{myopia}, \\ \mathrm{Here},v=-20\mathrm{cm} \\ \mathrm{and}u=-25\mathrm{cm} \\ \mathrm{Using}\mathrm{lens}\mathrm{formula} \\ \frac{1}{v}-\frac{1}{u}=\frac{1}{f} \\ \Rightarrow \frac{\mathit{1}}{\mathit{f}}=\frac{1}{-20}+\frac{1}{25} \\ \Rightarrow \frac{1}{f}=\frac{-5+4}{100}=\frac{-1}{100} \\ \Rightarrow f=-100\mathrm{cm} \\ \mathrm{Now}, \\ \mathrm{Power},\mathit{}P=\frac{100}{f\left(\mathrm{in}\mathrm{cm}\right)} \\ \Rightarrow P=\frac{100}{-100}=-1\mathrm{D} \\ \mathrm{The}\mathrm{lens}\mathrm{used}\mathrm{is}\mathrm{diverging}\mathrm{lens}\mathrm{whose}\mathrm{power}\mathrm{is}-1\mathrm{D}. \end{gathered}$$