An object is placed perpendicular to the principal axis of a convex lens of focal length $8cm$. The distance of the object from the lens is 12cm. Find the position and nature of the image.

Hint: Here we have to apply the lens formula with appropriate signs. Then we will get the position of the image. Depending upon the sign and value of the image distance we will get an idea about the nature of the image. We will calculate magnification for that.We will follow this way to solve this question.
Formula used: $\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f},M=\dfrac{v}{u}$

Complete answer:
If $u$ is the object distance,$v$ is the image distance and$f$ is the focal length, then we have the formula
$\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}$. Here $u=-12cm,f=+8cm,v=?$
In case of a convex lens the focal length is positive. Putting these values we get
$\begin{align}
  & \dfrac{1}{v}-(-\dfrac{1}{12})=\dfrac{1}{8} \\
 & or\dfrac{1}{v}=\dfrac{1}{8}-\dfrac{1}{12} \\
 & or\dfrac{1}{v}=(\dfrac{3-2}{24}) \\
 & orv=24cm \\
\end{align}$
Now as the image distance is greater than the object distance the magnification $M$ is given by
$M=\dfrac{v}{u}=\dfrac{24}{-12}=-2$ . As its magnitude is $>1$ but its sign is negative, the image will be real, magnified and inverted.
Additional information: In case of a convex lens we get different types of image depending upon the object distance, whether the object is in infinity, between infinity and radius of curvature etc. But in the case of a concave lens we always get a virtual, diminished and erect image.

Note:
For this kind of problem we must put the values of different quantities with appropriate signs, otherwise we will not get the correct answer. For a convex lens the focal length is positive and for a concave lens the focal length is negative. Also one tends to mistake $\dfrac{1}{v}$ with $v$ in the calculation part. Also if the value of magnification is $<1$ then the image would be diminished and if it has positive value, it means the image is erect.