2
$\begingroup$

...The electric field produced by the $i$-th laser is described by $$\vec{E_i}=E_{0,i}\vec{ε_i}\cos(\vec{k_i}\cdot\vec{r}-\omega t)$$ with the amplitude $E_{0,i}$, the wave vector $\vec{k_i}$ and the unit-length polarization vector $\varepsilon_i$

...

A typical setup is shown in Fig. 2a. Here, all three beams are polarized in the $z$ direction, propagate in the $𝑥𝑦$ plane and intersect at equal angles of $120°$.

This is from the 2021 IPhO problem 3. The main problem I am encountering is that I can't visualise where the polarisation vector is pointing and thus can't decide how the waves add up. Also what does the polarisation vector signify? Is it like the oscillation axis? It's not related to the hard part of the question anyway, but I just wanted to know where it points since we already have a wave vector showing propagation. The Wikipedia page also didn't help me much

Improve this question
$\endgroup$
0

1 Answer 1

Reset to default
3
$\begingroup$

The polarization vector here represents the direction of the electric field. In this problem, it is $\hat{\mathbf{z}}$ for all three lasers. It's important to consider here since field amplitudes will add; in the middle, the polarization vector allows us to see that the electric field contributed by each laser is the same, and will add accordingly.

Note that the magnetic field of each laser will not all point in the same direction, unlike the electric field.

Improve this answer
$\endgroup$
2
  • 1
    $\begingroup$ Can you please explain with a diagram what this vector is? $\endgroup$
    – TheSmartestNoob
    Commented Sep 16, 2021 at 18:59
  • 1
    $\begingroup$ Plane EM waves are comprised of a perpendicular electric field and magnetic field component, both of which are perpendicular to the direction of propagation. The polarization vector is in the plane perpendicular to the direction of propagation and points along the direction of electric field oscillation. $\endgroup$
    – DanDan面
    Commented Sep 17, 2021 at 0:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.