Question

The power of a converging lens is 10D. If it forms a real image at a distance of 40 cm from the lens, calculate the position of the object.

Answer 1

\(f=\frac{1}{10}=0.1\,m=10\,cm\)

\(\frac{1}{40}-\frac{1}{u}=\frac{1}{10}\)

\(\frac{1}{40}-\frac{1}{10}=\frac{1}{u}\)

\(\frac{1-4}{40}=\frac{1}{u}\)

\(\frac{-3}{40}=\frac{1}{u}\)

\(u=\frac{-40}{3}\)

Comments

Thanks