πFor better understanding I had added an image which shows,
- An infinitesimally thin grounded sheet AB (I know the diagram is quite different but just assume it)
- A charge $q$ present at right side at distance $l$ from the sheet
"Let me tell you that our purpose of grounding is just to develop zero potential over the sheet. There is nothing more use of it."

So the Image Method let you assume that in this situation (i.e. In which there is zero potential sheet and there is a charge $q$ present at distance $l$ from it) then you can assume that there will be another charge of same magnitude as of $q$ but will be of opposite sign (i.e. $-q$ here).
I had tried to show that totally 'imaginary' image in the figure boxed inside.
Now the main point,
You must be thinking that what is the use of it.
Well, from image method you can find the net force acting here which will be,
$$F=\frac{1}{4πε_0}.\frac{q.q}{(2l)^2}$$
which implies,
$$F=\frac{1}{4πε_0}.\frac{q^2}{4l^2}$$
Similarly you can find Electric field and other electrostatic terms.