Diffraction
The same effect applies to water waves, radio waves and other electromagnetic radiation.
Consider a wave proceeding through a narrow aperture. Assume that the aperture is large compared to the wavelength and call the ratio between them R.
If the wave propagates at an angle of arcsin(1/R) to the left then the resulting wave form will interfere with itself, completely cancelling out. The part of the wave going through at the right edge will be exactly one cycle behind the part of the wave going through at the left edge.
[If you measure angles in radians then arcsin(1/R) ~= 1/R and you can simplify the formula accordingly]
For apertures that are small compared to the wavelength, there is no such interference regardless of angle and the wave simply spreads out.
And now I start waving my hands furiously...
Consider an obstacle around which a wave is propagating. The situation is exactly symmetric. The "shadow" of this object will have the same shape as the diffraction pattern from an aperture.
So if you have an obstacle that is 5 wavelengths in extent, you will have a shadow whose edges are at an angle of about 1/5 radians.
And if you have an obstacle that is less than one wavelength in extent, you will not have a noticible shadow at all.
For a fixed size object, this means that large wavelengths go around and for short wavelengths, the object casts shadows.
I'm no expert on this stuff and am working from first principles here, so forgive me if there are more apt descriptions or better formulas.