How is loose coupling achieved using interfaces in Java when an implementation class is mandatory and bound to interface contract?

Question

How is loose coupling associated with interfaces when we are bound to create an implementation class regardless? The implementation class is forced to implement all those me

Answer 1

Might below explanation can answer this :

Into class A we need to have Object of class B. If directly exposure of B into A is there means there is Tight coupling. ex: one can add more methods into B or anything. Which means Behavior of A can be changed based on more exposure of B. But if B class is implementing any Interface and we are passing that ref of Interface into A. means whatever changes in B class further user A class is not bother because we use ref of Interface for accessing B and got only required access. Ex : class A { public void add( B b){ // Implementation } } class B{

    }
            this is Tight coupling. Because user can make any changes to class B which are directly exposed to class A and this defines Tight Coupling(Greater the exposure of depending Object , more Tight Coupling).  To resolve this one refer :
    Interface L{

    }
    class B implements L{}
    class A{
        public void add(L  b){
            // Implementation
        }
    }

        Since we are passing the ref of Interface L, adding changes into Implementation of B will not make any difference to class A.