Question

A concave lens has a focal length of 15cm. At what distance should the object from the lens be placed so that it forms an erect and virtual image at 10cm from the lens?

• A. 30cm
• B. 15cm
• C. 60cm
• D. 10cm

Answer 1

The correct option is A 30cm

From lens formula we have,

$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}\frac{1}{u}=\frac{1}{v}-\frac{1}{f}\frac{1}{u}=\left(\frac{-1}{10}\right)-\left(\frac{-1}{15}\right)\frac{1}{u}=\frac{-1}{30}\Rightarrow u=-30cm$