Refraction Through a Lens — Question 15

As the magnification (m) of the lens is given as +0.5, which implies that the image is virtual and erect because m is positive for a virtual and erect image.

The image is smaller in size as the magnitude of magnification is smaller than 1.

As the image formed is virtual, erect and smaller in size so the lens used is concave in nature.

As we know, the relation between u and v is as shown below —

$m = \dfrac{v}{u} \\[0.5em]$

Given,

magnification by a lens is +0.5

Substituting the values in the formula

$0.5 = \dfrac{v}{u} \\[0.5em] \Rightarrow v = 0.5u \\[0.5em]$

The lens is concave in nature and the image distance v = 0.5u.