Question

An object is 2 m from a lens which forms an erect image one-fourth (exactly) the size of the object. Determine the focal length of the lens. What type of lens is this ?

Answer 1

u = -2 m ....

m = $\frac{1}{4}$

$\frac{v}{u}=m$

$=>\frac{v}{-2}=\frac{1}{4}$

$=>v=\frac{-2}{4}=\frac{-1}{2}=-0.5m$

$\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$

$=>\frac{1}{f}=\frac{1}{-0.5}-\frac{1}{-2}$

$=>\frac{1}{f}=\frac{1}{-0.5}+\frac{1}{2}$

$=>\frac{1}{f}=-2+0.5=-1.5$

$=>f=\frac{1}{-1.5}=-0.67m$

The lens is concave lens