Answer to Question #182698 in Optics for Khushi

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

We have given electric field vectors of two light waves propagating along the positive z-direction.

$E_1(z,t) = xˆ E_{01}cos(kz - \omega t)-----(1)$

$E_2(z,t) = yˆ E_{02}cos(kz-\omega t + \phi)-----(2)$

Where, E_{01} and E_{02} are amplitudes, \omega is cyclic frequency ,

From equation (1),

$cos(kz- \omega t) = \dfrac{E_1}{E_{01}}-----(3)$

From equation (2),

$E_2 = E_{02}[cos(kz- \omega t)cos \phi - sin(kz- \omega t)sin \phi]-----(4)$

From equation (4),

$sin(kz-\omega t)sin \phi = cos(kz - \omega t)cos\phi - \dfrac{E_2}{E_{02}}-----(5)$

Putting (3) equation in (5), We get:

$sin(kz- \omega t)sin \phi = \dfrac{E_1}{E_{01}} cos\phi - \dfrac{E_2}{E_{02}}----(6)$

Squaring both sides,

$(sin(kz- \omega t)sin \phi)^2 = (\dfrac{E_1}{E_{01}} cos\phi - \dfrac{E_2}{E_{02}})^2---(7)$

$(sin(kz- \omega t)sin \phi)^2 = (\dfrac{E_1cos \phi}{E_{01}})^2+ (\dfrac{E_2}{E_{02}})^2 - 2\times \dfrac{E_1 cos \phi}{E_{01}} \times \dfrac{E_2}{E_{02}}---(8)$

From equation (3),

Multiplying both sides by $sin \phi$ we get,

$cos(kz- \omega t)sin\phi = \dfrac{E_1}{E_{01}} sin\phi---(9)$

Squaring both sides,

$(cos(kz- \omega t)sin\phi)^2 = (\dfrac{E_1}{E_{01}} sin\phi)^2---(10)$

Adding Equation (8) and (10)

We get,

$sin^2 \phi = (\dfrac{E_1}{E_{01}})^2 + (\dfrac{E_2}{E_{02}})^2 - 2\times \dfrac{E_1 cos \phi}{E_{01}} \times \dfrac{E_2}{E_{02}}$

which is the equation of Ellipse.

In general, projections of light vectors of polarized light on the axes X and Y

$E_x = E_0cos(\omega t- kz)$

$E_y = E_0cos(\omega t - kz+ \phi)$

they satisfy the equation

$(\dfrac{E_x}{E_0})^2 - 2\dfrac{E_xE_y}{E_0^2}cos\phi + (\dfrac{E_y}{E_0})^2 = sin^2 \phi$

This equation is an equation of an ellipse whose axes are oriented relative to the coordinate

axes X and Y arbitrarily. The orientation of the ellipse and magnitude of its semiaxes depends only on the angle $\phi$ (phase difference).

If $\phi = \pi$

$E_x + E_y = 0$

we have the equation of the line. That is, light is linearly or plane polarized.

If $\phi = 0$

$E_x^2+ E_y^2 = E_0^2$

We have the equation of the circle. That is, light is circularly polarized.