In young's double slit experiment performed using a monochromatic light of wavelength $\lambda ,$ when a glass plate $\left(\mu =1.5\right)$ of thickness $x\lambda$ is introduced in the path of the one of the interfering beams, the intensity at the position where the central maximum occurred previously remains unchanged. The value of x will be:

Not really. The electric dipole moment vector directs or points from the negative charge to the positive charge. But the electric field lines that a dipole creates will point away from the positive and move to the negative charge.

Yes, the cube, which is a closed surface containing only one electric dipole will make electric flux zero. This follows Gauss's Law when the total charge inside it is zero. The field lines entering the surface will exit, and that would result in zero net flux. 

The magnitude of each charge and the distance that separates them. 

Gauss Law is only concerned with the total enclosed charge that finally tells us the total flux. The charges outside may change field patterns. They not affect the total flux. It's actually incorrect to assume the field due to the external charges should also affect the flux through the Gaussian sur

Gauss Law does not directly give the electric field in all cases. It can only be used in calculations for symmetrical surfaces: spherical, cylindrical, or planar.