Using the lens formula, show that an object is placed between the optical centre and the focus of a convex lens produces a virtual and enlarged image.

Hint: Before going to the question let us first know about the lens formula which we will use for the given query to answer it and we will also use the magnification formula. The lens formula is an equation that relates the object distance \[\left( u \right)\] , image distance \[\left( v \right)\], and focal length \[\left( f \right)\] of a lens and The linear magnification created by the lens is the ratio of the size of the image generated by refraction from the lens to the size of the object. The symbol \[m\] is used to represent it.

Complete step by step answer:
By using the Lens formula-
$\dfrac{1}{f} = \dfrac{1}{v} + \dfrac{1}{u}$ and $v > u$
Here, $f$ is the focal length of the given convex lens, $v$ is the image distance and $u$ is the object distance.
As we know that, for convex lens- their focal length is always positive, whereas image and object distance is negative.So,
$\dfrac{1}{f} = - \dfrac{1}{v} + ( - \dfrac{1}{u}) \\
\Rightarrow \dfrac{1}{f} = - \dfrac{1}{v} - \dfrac{1}{u} \\
\Rightarrow \dfrac{1}{v} = - \dfrac{1}{f} - \dfrac{1}{u} \\ $
$\Rightarrow \dfrac{1}{v} = - (\dfrac{{u + f}}{{uf}}) \\
\Rightarrow v = - \dfrac{{uf}}{{f + u}} \\ $
Therefore, the negative sign indicates that the image of a convex lens is virtual.
Now, the magnification of a convex lens:
$\because m = - \dfrac{v}{u} = \dfrac{f}{{f + u}}$
Thus, $m = \left( {\dfrac{f}{{f + u}}} \right)\, < 1$ , so the image formed is enlarged.

Hence, it is proved that an object is placed between the optical centre and the focus of a convex lens produces a virtual and enlarged image.

Note: We must use the necessary signs in relation to the object and the image when measuring this. The symbol, on the other hand, only indicates the location in the resultant value. By comparing the value of the object size to the value indicated with the sign, you can calculate how many times the object has been magnified/diminished.