- #1
KiNGGeexD
- 317
- 1
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!
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!