Answer to Question #188155 in Optics for TUHIN SUBHRA DAS

Question

The electric field vectors of two light waves propagating along the positive zdirection

are given as

E₁(z,t) = xˆ E₀₁cos(kz - (omega)t)

E₂(z,t) = yˆ E₀₂cos(kz-(omega)t + (si))

where xˆ and yˆ are unit vectors along x and y-axes, respectively. Show that when

these two waves superpose, we obtain elliptically polarised light. Also show that

the linear and circular polarisations are special cases of elliptical polarisation

Answer 1

Given,

$E_1(z,t) = \hat{x} E_{01}\cos(kz - (\omega)t)$

$\cos(kz - (\omega)t)=\frac{E_1(z,t)}{E_{01}}...(i)$

$E_2(z,t) =\hat{y} E_{02}\cos(kz-(\omega)t + (\psi))$

$=\hat{y}E_{02}(\cos((kz-(\omega)t)+(\psi)))$

$=\hat{y}E_{02}[\cos(kz-(\omega)t)\cos(\psi)-\sin(kz-(\omega)t)\sin(\psi)]$

$\Rightarrow$ $\sin(kz-(\omega)t)\sin(\psi)=\cos(kz-(\omega)t)\cos(\psi)-\frac{E_2(z,t)}{E_{02}}...(ii)$

From equation (i) and (ii)

$\sin(kz-(\omega)t)\sin(\psi)=\frac{E_1(z,t)}{E_{01}}\cos(\psi)-\frac{E_2(z,t)}{E_{02}}$

Now, squaring both side,

$\sin^2(kz-(\omega)t)\sin^2(\psi)=(\frac{E_1(z,t)}{E_{01}})^2\cos^2(\psi)+(\frac{E_2(z,t)}{E_{02}})^2-2\times \frac{E_1(z,t)}{E_{01}}\cos(\psi)\times\frac{E_2(z,t)}{E_{02}}$

Hence, from the above, we can conclude that it is a form of ellipse.

As here the value of $E_{01}$ and $E_{02}$ are not given, so we can not conclude about the polarization.