Question

a lens forms
a virtual image of
object at a
distance of 30cm to the left
the object is placed at a distance of 90 cm on the same
side. Determine the focal length and the nature of the
lens.
​

Answer 1

Given:

$\sf{v=-30}$

$\sf{u=-90cm}$

Solution:

From lens formula:

$\sf{{\dfrac{1}{f}}={\dfrac{1}{v}}-{\dfrac{1}{u}}}$

$\sf{{\dfrac{1}{f}}=-{\dfrac{1}{30}}+{\dfrac{1}{90}}}$

$\sf{{\dfrac{1}{f}}={\dfrac{-3+1}{90}}}$

$\sf{{\dfrac{1}{f}}=-{\dfrac{2}{90}}}$

$\sf{f=-45cm}$

Since the focal length of the lens is negative, the lens would be concave in nature.