[Ahmed Abdullah]

Planar molecules are optically inactive. So for being optically active this molecule must be non-planar.

[Ahmed Abdullah]

We know cyclobutane is planar, coz each of the carbon in the compound is equivalent so all the bond length and bond angle must be equal ; only planar square satisfy these conditions. But in this compound below all bonds length and bond angle aren't equal so its structure is not nessesarily planar. Though it can still be planar. Anyway  I think it should not be planar but not sure why  . 
*delete me*!

[sjb]

Cyclobutane is not planar - consider things like bond angles and Bayer strain ( http://en.wikipedia.org/w/index.php?title=Cyclobutane&oldid=210992629 )

There are other geometric shapes in 3-d space that have equivalent internal angles and side lengths.

I'd suggest that cyclobutene, however is, consider a isosceles trapezium, with the shorter of the parallel sides equivalent to the C=C bond.

As to the molecule you've drawn, this is optically active, why?

S

[ng]

The molecule you have drawn is NOT planar. The four-membered ring is planar but the methyl group is out of that plane.
The molecule is not superimposable on its mirror image so it is chiral.
You have drawn a single enantiomer (S) so yes, it is optically active.


Cyclobutane is not planar, the ring is puckered.