The same size images are formed by a convex lens when the object is placed at $20$ cm or at $10$ cm from the lens. Find the focal length of the convex lens in cm?

Hint: It is clearly stated in the question that the images formed are of the same sizes. Hence, they have the same height or the same magnification. We will use the Lens formula to find the answer of this question. We have to derive the Lens formula in terms of magnification to find the desired result.

Formula used:
$\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}$
Where,
$v=$ distance of the image from the lens
$u=$ distance of the object from the lens
$f=$ focal length of the convex lens.

Complete answer:
Here, we are going to present the complete step by step process to the find the focal length of the given convex lens whose image sizes are same. Magnification in simple terms is defined as the height of the object to that of the image.

Here, we will use the Lens formula first. The Lens formula is given as,
$\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}$---(a)
Where, $v=$ distance of the image from the lens, $u=$ distance of the object from the lens and $f=$ focal length of the convex lens.

Magnification of a lens is formulated as,
$m=\dfrac{v}{u}$---(b)
where $m=$ magnification of the lens used.

From the equation (a) and (b) we can derive,
$\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}$

Taking $\dfrac{1}{u}$ to the right side of the equation and simplifying it, we get,
$\dfrac{1}{v}=\dfrac{f+u}{fu}$--(c)

Multiplying the whole equation by $u$ we get,
$\dfrac{u}{v}=\dfrac{f+u}{f}$

Inverting the equation we get,
$\dfrac{v}{u}=\dfrac{f}{f+u}$---(d)

Hence, we know, $m=\dfrac{v}{u}$. So, substituting the value in equation (d) we get,
$m=\dfrac{f}{f+u}$--(e)

Let the magnification of one be ${{m}_{1}}$ and the other be ${{m}_{2}}$. From the given question we know that,
${{m}_{1}}=-{{m}_{2}}$

From equation (e) we get,
$\dfrac{f}{f-20}=-\dfrac{f}{f-10}$
$\Rightarrow f-20=-f+10$

Simplifying the equation we get,
$\Rightarrow 2f=30$
$\Rightarrow f=15$

Hence, the focal length of the convex lens is $15$ cm when the object is placed at $20$ cm or at $10$ cm from the lens.

Note: It must be noted that we have to consider the sign convention when we use the mirror or the lens formula. Anything which is placed to the left of the lens is considered to be negative and anything placed to the right of the lens is considered to be positive. Also, when placed above the axis it is considered to be positive and below the axis as negative.