# Homework Help: Polarisation past paper question

1. Dec 30, 2013

### KiNGGeexD

Question

For (i)

I assumed that the light was plainly polarised as there is only a difference in amplitude

For (ii)

There is a negative sign infront of the amplitude which I ignored as it doesn't change the polarisation but one is cos and the other is sin, so there is a phase shift of pi/2 and hence the light is lefty circularly polarised?

For (iii)

There is a cos so a phase change of pi/2 but also another change of pi/2 so again the light would be plainly polarised??

Any help with this would be great as it is for my exam!:) I should also say I don't know how to sketch the tip of the resultant vector not sure what this means!!

Thanks a bunch!!!

2. Jan 1, 2014

### Simon Bridge

You are looking at Q5(b)?

The way to do these problems is to sketch the path traced out by the tip of the electric field vector ... oh you don't know what that means? Well I'll tell you:

You know what a vector is? It's just an arrow.
The electric field, in each case, is a vector of form $\vec E = E_y(t)\hat \jmath + E_z(t)\hat k$
When you plot Ey vs Ez on a graph you get a dot - this is the tip of the electric field vector. Over time, that dot moves about.

i.e. if $E_y=E\sin\omega t$ and $E_z=0$, then the graph starts out as $(E_y,E_z)=(0,0)$ - at the origin - then, as $t$ increases, the point travels up the $E_y$ axis until it reaches point $(E_y,E_z)=(E,0)$ at $t=\pi /2\omega$ then it starts heading back down again. Eventually it reaches point $(E_y,E_z)=(-E,0)$ where it starts moving up again to the origin. From there the motion repeats.
The result is a line that goes only up and down ... for linearly polarized light.