WE ALL KNOW THAT ETHENE WITH IT'S CARBONS HAVING ALL DIFFERENT GROUPS ATTACHED TO THE IS OPTICALLY INACTIVE.BUT IS PROPENE WITH DIFFERNT GROUPS ATTACHED TO THE DOUBLE BONDED CARBONS ALSO OPTICALLY INACTIVE?

NO,BOTH THE CASES R DIFFERENT,in ehtene case the molecule can be cut into 2 by a plane.But in the propene case the molecule cannot be cut into 2 by a plane because the geometry of a carbon in propene in SP3 also?

In propene, the methyl gruop can turn round a single bond. As a result we will obtain population of conformations that will show the same plane of symmetry as double bond.
But each individual conformation can be opticalyy active if fixed.

As AWK alluded to, optical activity is a little more complicated than just having a chiral center.  For example, single molecules (even achiral ones) are optically active because the interaction with the polarized light is dependent on the orientation of the molecule relative to the plane of the light (such as parallel to the plane versus perpendicular to the plane).  When organic chemists measure optical activity they are dealing with many millions of molecules, so essentially every orientation has an equal population and they all cancel each other out.  When you have a chiral center, however, you can't populate some of the mirror image orientations needed to even everything out, so you can observe an optical rotation.

I hope that makes sense....

Don't confuse optical and geometry ISOMERISM!!!
 an essential condition for optical active molecule is presence rotation-reflecting axis higher then 2 order/. How could you find in any alkene this axis?
Alkens are planar, so alken molecule with 4 different functions will be agree with it's  mirror image. People why you do't like reading books?

I totally agrre with u dexangles.we r here to help each other and not to make fun.besides that noncommentes it is symmytry that is involved in optical activity,mirror image is just a mere way to express that,infact u haven't even understood my question!!!