Question

An object is placed at a distance $24\text{cm}$ in front of a convex lens of focal length $8\text{cm}$.
$\left(\text{i}\right)$ What is the nature of the image so formed?
$\left(\text{ii}\right)$ Calculate the distance of the image from the lens.
$\left(\text{iii}\right)$ Calculate the magnification of the image.

Answer 1

Given:
focal length of lens, $f=8\text{cm}$
object distance, $u=24\text{cm}=3f$ (beyond $2f$)
$\left(\text{i}\right)$ For the given convex lens and the object distance from the lens, the image formed will be real, inverted, diminished and between $f$ and $2f$.

$\left(\text{ii}\right)$ We have focal length and object distance from the lens so, using lens formula,
$\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$ $⟹$ $\frac{1}{v}=\frac{1}{f}+\frac{1}{u}$
$⟹\frac{1}{v}=\frac{1}{8}+\frac{1}{-24}=\frac{3-1}{24}=\frac{1}{12}$
$⟹v=12\text{cm}$
So, the image is formed at $12\text{cm}$ on the other side of the lens between $f$ and $2f$.


$\left(\text{iii}\right)$ We know that, the magnification of a lens is given by, $m=\frac{v}{u}=\frac{12}{-24}=\frac{-1}{2}=-0.5$
So, the magnification is 0.5 and the negative sign indicates that the image formed is inverted and since the magnification is less than 1, the image will be diminished as we mentioned in the first part of the question.