Amith2006 said:
When I referred Griffith's book on Electromagnetism I came across the fact that the polarisation vector defines the plane of vibration. Doesn't this mean that the plane of vibration is parallel to the plane of polarization?
I'll explain with an example (since you're reading about EM waves anyway):
[tex]\vec{E} = E_{0}\sin(\vec{k}*\vec{r} - \omega t)\hat{n}[/tex]
Now, [itex]\vec{k}[/itex] is the
propagation vector whereas [itex]\hat{n}[/itex] defines the direction of vibration (the plane of polarization is the plane containing the propagation vector and n itself).
As EM waves are
transverse the direction of vibration is orthogonal to the direction of propagation. So, [itex]\vec{k}\dot\vec{r} = 0[/itex].
To cite a specific example, consider
[tex]\vec{E} = E_{0}\sin(kz - \omega t)\hat{x}[/tex]
The oscillations are in the xz plane with the positive z axis as the direction of propagation of the wave.
It also says that an electromagnetic wave propagating across the z-direction with its electric field vector along the x-direction and magnetic field vector along the y direction is said to be polarized in the x-direction.(Because by convention we use the direction of Electric field vector to specify the polarization of an electromagnetic wave).
The electric field component is written above. The magnetic field is given by
[tex]\vec{B} = \frac{\vec{k}X\vec{E}}{\omega} = \frac{E_{0}}{c}\sin(kz - \omega t)\hat{y}[/tex]
Does that convince you? As you said, by convention the direction of polarization of the EM wave is the direction of polarization of the electric field component.
Hope that helps...