How to implement open filedialogbox in c#

C# Code:

String ImagePath=””;

OpenFileDialog OpenFileDialog1 = new OpenFileDialog();

OpenFileDialog1.InitialDirectory = Convert.ToString(Environment.SpecialFolder.MyDocuments);

OpenFileDialog1.Filter = "(*.png)|*.png|All Files (*.*)|*.*";

OpenFileDialog1.FilterIndex = 1;

if (OpenFileDialog1.ShowDialog() == DialogResult.OK)

{

    ImagePath= OpenFileDialog1.FileName;

}