I think, no!

Of course it's depend on that how you are aiming the laser beam at wire? Perpenducular to the wire or parallel along it. In this case the top of wire is a disk but you will avoid the interference behind the disk and you can not get the patterns as they will be behind a disc. 

Dear  Rıza Demirbilek 

Thank you for your answer.Then what kind of pattern do you think this will be?

   In my opinión, if the laser is aimed perpendicularly to the wire, the phenomenon is one of scattering of  light by a cylinder...

The wire is e cylindrical object so your question, I think is diffraction of the coherent beam on a cylinder. If the incident beam is perpendicular to the axis of the cylinder ( your wire) on the transparent mode you will have diffraction ( interference) pattern of a long object, that's analogues with the diffraction on a 1-D slit. If the incident coherent light is along  the cylinder or perpendicular to the end surface of the cylinder) the obtained diffraction is circular, that is  corresponding diffraction pattern like on the circular opining or a circular hole.

Chi,

Jahja gave half the answer but its important to remember that obstructions can be modeled as negative apertures with respect to the field.  So for this case model the 1D slit and before you square the field to get the interference pattern take 1 minus the solution then square it to get the result.  

For example a 1D slit illuminated with a plane wave will give you a sinc field distribution and therefore a sinc^2 interference pattern.  So a wire illuminated by a plane wave will give you (1-sinc)^2 interference pattern.

I have attached the plot of diffraction of a plane wave from a wire for your reference. 

I hope this helps,

-Robert  

Of course it will not be a circular disk diffraction, since the obstacle isn't a circular disk: if the wire is a perfect cylindrical conductor, its cross section is an opaque rectangle, whose width is the diameter of the wire. Standard calculations from Fraunhofer diffraction lead to the answer. The point is that the excluded region is that occupied by the wire, so the amplitude is that of two semi-infinite regions, separated by the diamter of the wire. Standard exercise in wave optics, illustrating Babinet's principle. 

It will, of course, be the result of diffraction from an opaque circular disk, if the beam is coaxial with the wire.

Dear  Robert Chimenti， Stam Nicolis and Jahja Kokaj

Thanks for your answers.

If I didn"t misunderstand your meanings,all of you thought if the beam light is perpendicular to the end surface of the cylinder,it will get a pattern same with the disk diffraction.However ,do you think the length of the  will affect the result.?

The length of the wire will determine if you are in the Fresnel or the Fraunhofer diffraction region, assuming the detection plane is at the other side of the wire.  But the basic physics wont change.  

Dear Robert Chimenti

How the length affect the result?If i don"t misunderstand you ,you think in this question,

Fresnel or the Fraunhofer diffraction is not suitable.Am i right?

In Fraunhofer diffraction you assume you are far enough from the diffraction source that the pattern can be simplified to a Fourier Transform of the diffracting object.  If the wire is short enough and the detection plane is at the end of the wire, you may not be able to make this assumption so the math will be a bit more complicated.  But the procedure I laid out will still be the same.  

 Thank you