# Polarisation of EM waves

My textbook reads 'The plane of polarisation of an electromagnetic wave is defined as the plane in which the electric field oscillates.'

What exactly does this mean? I understand that an EM wave consists of an electric field and magnetic field oscillating perpendicular to each other, does this mean the oscillations in the magnetic field gets polarised when passed through a polarising filter?

• – anna v May 11 '19 at 4:30

Yes. If the electric field after the polarizer is along $$\hat x$$ Only for a wave propagation along $$\hat z$$, the magnetic field will be only along $$\hat y$$. Since the polarization is defined in terms of the electric field, the wave would be (linearly) polarized along $$\hat x$$.

For an electromagnetic wave, you know that the electric field is perpendicular to the magnetic field, and both oscillate sinusoidally. If the amplitude of the electric field is $$E_0$$, and the amplitude of the magnetic field is $$B_0$$, then these two values are related by

$$B_0=\frac{E_0}{c}$$

where $$c$$ is the speed of light. From this the only logical result is that yes, the magnetic field is also polarized.

The plane in which the electric field oscillates is a plane in space in which the electric field can move. If your electromagnetic wave travels along the $$\hat{y}$$ direction, then it might be the case that $$\vec{E}\parallel\hat{z}$$. Thus the plane in which the electric field oscillates has a basis of $$\hat{y}$$ and $$\hat{z}$$: it is in the yz-plane. I have attached a diagram below to help illustrate this. The red curve is the electric field component of the wave, and you can see that the amplitude of this frozen snapshot of the wave at some point can be written as $$\langle0, y, \sin(y)\rangle$$. It should then be apparent that the wave exists in the yz-plane. • In the case of a vanilla plane wave the direction of propagation, the electric field, and the magnetic field are all perpendicular to one another. So if the wave propagates along the $\hat{x}$ direction the $E$-field will oscillate in the $yz$-plane. – Julius May 10 '19 at 23:22
• @Julius I've added a diagram to my answer to perhaps explain why what you have said is incorrect. – Kraig May 11 '19 at 4:34
• Thank you for your drawing, this helps me understand your reasoning better. While the surface that is swept out by the $E$-field indeed lies in the $yz$ plane, the field as you've drawn it just oscillates along $z$. When you change the polarization you rotate it along the $y$ axis, meaning the $E$-field vector has to lie in the $xz$ plane. – Julius May 11 '19 at 11:54

My textbook reads 'The plane of polarisation of an electromagnetic wave is defined as the plane in which the electric field oscillates.'

This isn't quite right. For a plane wave that's propagating along a clear direction $$\mathbf k$$, a better definition of the plane of polarization is the plane in which the electric field can oscillate. This matters because...

I understand that an EM wave consists of an electric field and magnetic field oscillating perpendicular to each other

... this isn't quite right either. Your description is correct for linearly polarized light, but other types of light are also possible: specifically, elliptical and circular polarizations, where both the electric and magnetic fields trace out ellipses (and, as a special case, circles). When this is the case, both the electric and magnetic fields are confined to a single plane, which is orthogonal to the propagation direction $$\mathbf k$$. This plane is what's known as the plane of polarization.