Question

How far an object must be placed in front of a converging lens of focal length $10\,cm$ in order to produce an erect image of linear magnification $4?$
A. $15.8\,cm$
B. $2.5\,cm$
C. $12.5\,cm$
D. $7.5\,cm$

Answer 1

Hint: In order to solve this question, we will first use the lens equation formula and make a relation between object distance, image distance and focal length of a converging lens which is a convex lens and then using magnification formula we will solve for object distance be eliminating all other parameters value.

Formula used:
The lens equation formula is,
$\dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u}$
where,$f$ is focal length of lens, $v$ is image distance from lens and $u$ is object distance from lens.
Linear Magnification m is written as $m = \dfrac{v}{u}$

Complete step by step answer:
According to the question, we have given that $f = + 10\,cm$ focal length of convex lens is positive. Let $u, v$ be the object distance and image distance from the lens.Now, linear magnification is given to us as $m = \dfrac{v}{u} = 4$ so we can write,
$v = 4u$
Now, using lens equation formula
$\dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u}$
Put value of $f = + 10\,cm$ and $v = 4u$ ,and solve for $u$ we get,
$\dfrac{1}{{10}} = \dfrac{1}{{4u}} - \dfrac{1}{u}$
$\Rightarrow \dfrac{1}{{10}} = \dfrac{{ - 3u}}{{4{u^2}}}$
$\Rightarrow u = - \dfrac{{30}}{4}$
$ \therefore u = - 7.5\,cm$
Negative sign implies that the object is placed on the left side of the lens.

Hence, the correct option is D.

Note: It should be remembered that, a converging lens is convex lens whereas diverging lens is concave lens and focal length of convex lens is always taken positive and always remember all the proper sign conventions while putting values of u, v and f in lens equation formula.