Question

A convex lens of focal length 10cm forms a virtual image of an object at 30cm from the lens. The magnification produced will be
A. 2
B. 3
C. 3.5
D. 4

Answer 1

As a first step, one could note down all the given quantities from the question by using the general sign convention. Then you could recall the lens formula and then substitute these quantities into it to get the object distance. Then, you could recall the expression for magnification in terms of image distance and object distance and thus find the answer.
Formula used:
Lens formula,
$\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}$
Magnification,
$m=-\dfrac{v}{u}$

Complete answer:
In the question, we are given a convex lens having a focal length of 10cm. It is being said that a virtual image of an object is formed at 30cm from the lens. We are supposed to find the magnification produced by the lens.
Let us recall lens formula which is given by,
$\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}$
Where,
$f=10cm$
$v=30cm$
$\Rightarrow u=\dfrac{fu}{f-v}$
Substituting these we get,
$\Rightarrow u=\dfrac{10\times 30}{10-30}=\dfrac{300}{-20}=-15cm$
So, now we have found the object distance to be 15cm.
Now we have the expression for magnification which is given by,
$m=-\dfrac{v}{u}$
Now, by substituting the values we get,
$m=-\left( \dfrac{30}{-15} \right)=2$
So we found the magnification of the image formation to be 2 which would mean that the image formed will be larger in size when compared to the size of the object.

So, the correct answer is “Option A”.

Note: We have followed the general conventions used while solving the problems related to optics. We normally consider the distances measured to the left of the optic axis of the lens to be negative and the measurements to the right are taken positive. We got the object distance to be negative and which would mean that object is kept to the left.