Please note that you have to fully qualify the class name in the second parameter.
A use statement will not resolve namespace dependencies in that is_a() function.
<?php
namespace foo\bar;
class A {};
class B extends A {};
?>
<?php
namespace har\var;
use foo\bar\A;
$foo = new foo\bar\B();
is_a($foo, 'A'); // returns false;
is_a($foo, 'foo\bar\A'); // returns true;
?>
Just adding that note here because all examples are without namespaces.