Question

A lens of focal length ‘f’ is positioned at 80 cm mark of a meter scale. If the lens forms an image, half the size of the object, on a screen placed at 100 cm mark on the same scale,
a) identify the nature of the lens
(b) the position of the object on the scale and
(c) focal length of the lens

Answer 1

$$\begin{gathered} \left(a\right)itsaconvexlens \\ \left(b\right)magnificationm=\frac{v}{u}=\frac{1}{2} \\ u=2v=2\times (100-80)=40cm \\ \left(c\right) \\ now \\ \frac{1}{v}-\frac{1}{u}=\frac{1}{f} \\ \frac{1}{20}-\frac{1}{40}=\frac{1}{f} \\ f=40cm \end{gathered}$$