An object is kept at a distance of 40 cm from a convex lens which forms a real image at a distance of 20 cm.

Question

Calculate the power and focal length of the given lens.

Answer 1

\(\frac{1}{f}=\frac{1}{-20}-\frac{1}{-40}\) \(=+\frac{1}{20}+\frac{1}{40}\) \(=\frac{+2+1}{40}\) \(=\frac{+3}{40}\)

\(f=\frac{+40}{3}\)

\(P=\cfrac{100}{\frac{40}{3}}=\frac{300}{40}=\frac{30}{4}=7.5\)

Comments

Thanks